Arihant CBSE Class 11 Book

84 CBSE Term II Applied Mathematics XII

SOLUTIONS

Objective Questions Now, Z = x-m = x - np
s npq
1. (a) Clearly, to estimate the values of unknown population
parameters by using sample data is called estimation. = 12 - 10 = 2 = 0.89
2.24 2.24
2. (b) The process through which inference about the
population are drawn which is based on population \P( Z < 0.89) = F( 0.89) = 0.8133
parameter is called statistical inference.
16. (b) Let the sample mean be x and margin of error be E.
3. (b) A member of the population is called element.
Then, x - E = 152 …(i)
4. (d) All the given options are correct.
and x + E = 160 …(ii)
5. (d) Hypothesis is a statement made about a population, it is
tested and corresponding accepted, if true and rejected, if On solving Eqs. (i) and (ii), we get
false.
E=4
6. (a) If the null hypothesis is false, then alternative hypothesis
is accepted, it is also called as research hypothesis. Given, 1 - a = 95% = 0.95

7. (b) The values (measurable characteristics) obtained from the Þ a = 0.05
study of population such as the population mean (m),
population variance ( s2 ), population standard deviation ( s) \ Za/ 2 = Z0.025 = 1.96
and etc. are called parameters. s
Margin of error = Za/ 2 . n
8. (a) When selection of objects from the population is random,
so objects of the population has an equal. Þ 4 = 1.96 ´ 15
n
9. (d) A hypothesis which defines the population distribution is
called as simple hypothesis. It specifies all parameter values. Þ n = 1.96 ´ 15 = 7.35
4
10. (b) We know that, testing of hypothesis Type 1 error occurs
when we reject H0, if it is true. Þ n = ( 7.35)2 = 54.0225 Þ n = 54
Since, the probability of H0 is a.
\Error probability will be 1 - a. 17. (c) In two-tailed test the critical region is evenly distributed.
One region contains the area where null hypothesis is
accepted and another contains the area where it is rejected.

11. (b) Given, sample data is 3, 6, 8, 10, 12, 15 18. (c) Null hypothesis asserts that there is no true difference in
sample statistics and population parameter under
\The point estimate of population mean is sample mean. consideration.
\ x = Sxi = 3 + 6 + 8 + 10 + 12 + 15 = 54 = 9
19. (a) Given, n = 36, x = 44, S = 5.2 and m 0 = 45,
n 66 \ t = x - m 0 = 44 - 45
S / n 5.2 / 36
12. (b) In a statistical hypothesis test, when the hypothesis is = - 1 ´ 6 = - 1.15 < 0
true but our test rejects it, is called Type I error. 5.2

13. (a) Let us take the hypothesis that the die is not biased

\p = 1 , q = 1, n = 700 and np = 700 ´ 1 = 350 and degree of freedom = 36 - 1 = 35
22 2
\p-value = 2 (Area under the standard normal curve to the
then, Z = x - np left of Z)

npq = 2 ´ Area under the t-distribution curve

= 400 - 350 = 50 = 3.78 to the right of t.
700 ´ 1 ´ 1 175
22 From the t-distribution table, we find the t = 1.15 lies
between 1.306 and 1.690 for which area lies between 0.05
Since, the computed value of z is greater than the table and 0.10.

value (1.96 at 5% level of significance, the hypothesis is So, p-value lies between 2 ´ 0.05 and 2 ´ 0.10 i.e. between
0.10 and 0.20.
rejected).
So, 0.10 < p-value < 0.20
14. (a) We have, m 0 = 20, n = 40, x = 24.3 and s = 5
Since, p-value > 0.01
\Test statistics, = x- m0
Z s/ n So, we do not reject H0.
20. (a) We have, p0 = 0.84, n = 400 and p = 0.75
= 24.3 - 20 = 4.3 ´ 24.3 = 4.24
5 / 24.3 5 \ Z= p - p0 = 0.75 - 0.84
p0(1 - p0 ) 0.84 (1 - 0.84)
15. (c) Given, m = np = 20 ´ 0.5 = 10
n 400
s = npq = 20 ´ 0.5 ´ 0.5
= - 0.09 ´ 20 = - 4.86
= 5 = 2.24 0.37

CBSE Term II Applied Mathematics XII 85

21. Given, m 0 = 36, n = 60, x = 38.6, s = 7 and a = 0.01 6. A p-value or probability value, is a number describing how

(i) (b) We know that, likely it is that your data would have occurred by random
change. The level of statistical is often expressed as p-value
x = Sxi Þ Sxi = x×n between 0 and 1. The smaller the p-value, the stronger the
n evidence that you should reject the null hypothesis.

= 38.6 ´ 60 = 2316 7. The t-distribution is a probability distribution method where
in the hypothesis of the mean of a small sample is tested,
(ii) (a) \ Z = x- m0 = 38.6 - 36 which is drawn from the systematic population whose
s/ n 7 standard deviation is unknown. It is a statistical measure
used to compare the observed data with data expected to be
60 obtained from a specific hypothesis.

= 2.6 ´ 60 = 2.88 8. In statistics, the t-distribution is most often used to
7
(i) Find the critical values for a confidence interval when
(iii) (c) Since, Z > 2.88 > 0 the data is approximately normally distribute

So, p-value of 2.88 = Area under the standard normal (ii) Test of hypothesis about the population mean.
curve to the right of Z = 1 - 0.9980 = 0.002
(iv) (a) Since, p-value > 0.01 (iii) Test of hypothesis about coefficient of correlation.

So, we do not reject H0. 9. If two groups come from a single population, performed
(v) (b) We know that, Reject H0, if Z ³ Za paired t-test. The paired t-test is also known as the
dependent sample t-test.
Þ Z ³ Z 0.01
Þ Z ³ 2.326 10. Point estimation involves the use of sample data to calculate
Þ 2.88 ³ 2.33, which is true. a single value of an unknown population parameters. Single
value of sample data is known as point estimate of
So, we do not reject H0. corresponding population parameter.
Subjective Questions
11. The two-sample t-test (also known as the independent
1. The measurable characteristics of population is called sample t-test) is a method used to test whether the unknown
parameter and measurable characteristics of sample is called population means of two groups are equal or not.
statistics. Population parameter is a constant that is not
generally known and needed to be estimated from statistics. 12. The central limit theorem states that, if you have population
The sample value are subjected to change, as it depends with mean m and standard deviation s and take sufficiently
upon sample value which are chosen at random, hence large random samples from the population with
statistics are not constant. replacement. Then, the distribution of the sample means will
be approximately normally distributed. This will hold true
2. If S1, S2, S3, ....., Sn are values of statistics S (like mean, regardless of whether the source population in normal or
skewed, provided the sample size is sufficiently large.
variance etc.) obtained from n independent random samples.
13. Here, mean ( x) = Sxi
If a definite size selected from a given population, then n

S1, S2, S3, ...., Sn form a sampling distribution of statistics. = 6 + 7 + 9 + 12 + 16 + 10
6
åThen, = 1 n
Mean (S) n = 60 = 10
Si 6

i =1 Now, S( xi - x)2 = ( 6 - 10)2 + ( 7 - 10)2
+ ( 9 - 10)2 + (12 - 10)2
åand = 1 n - S)2
Var(S) n + (16 -10)2 + (10 - 10)2
(Si
= 16 + 9 + 1 + 4 + 36 = 66
i =1

3. Systematic sampling is relatively a simple technique and The point estimate of population standard deviation,
may be more efficient than simple random sampling,
provided the lists are arranged wholly as random. If \ S= S( xi - x )2 = 66
populations are sufficient large, systematic sampling can n -1 6-1
often be expected to yield result that are similar to those
obtained by proportional stratified sampling. = 66 = 13.2 = 3.63
5
4. When selection of objects from the population is random so
objects of the population has an equal chance of selection. 14. The point estimate of the proportion in the population who
Probability sampling is more complex, more time consuming respond ‘No’
and usually more costly than non-probability sampling. p = xNo
n
5. The significance level, also known as a, is a measure of the = 230 = 0.46
strength of the evidence that must be present in your sample 500
before you will reject the null hypothesis and conclude that
the effect is statistically significant, the statistian determines
the significance level before conducting the experiment. The
significance level is the probability of rejecting the null
hypothesis.

86 CBSE Term II Applied Mathematics XII

15. We have, m 0 = 295, x = 297.6, s = 12, a = 0.05 and n = 50 19. Let us take the null hypothesis that there is no significant
difference between x and m.
Now, Za / 2 = Z0.025 = 1.96 s \ t = x -m
n S/ n
\Confidence interval = x ± Za/ 2

= 297.6 ± 1.96 ´ 12 Given, x = 5.02, m = 5, S = 0.002 and n = 10
50
\ t = 5.02 - 5 ´ 10
= 297.6 ± 3.3 0.002

= ( 297.6 - 3.3, 297.6 + 3.3) = 0.02 ´ 3.162
0.002
= ( 294.3, 300.9)
= 31.62
Since, the hypothesized value of population m 0 = 295 lies in
the confidence interval (294.3, 300.9). Therefore, H0 cannot For degree of freedom = 9, t0.05 = 1.833
be rejected.
The calculated value is much higher than table value. The

16. We have, n = 28, x = 20, S = 5.68 and m 0 = 18 null hypothesis is rejected, therefore the adjustment in the
\ Test statistics, t = x - m 0
S/ n machine is not order.

20. We have, D = 0, n1 = 43, n2 = 48, x1 = 16.2,

= 20 - 18 = 2 ´ 28 = 1.86 x2 = 14.5, S1 = 6.4 and S2 = 9.6
5.68 / 28 5.68 \ Test statistics, t = ( x1 - x2 ) - D

\ t = 1.86 S12 + S22
and degree of freedom = 28 - 1 = 27 n1 n2

Q t = 1.86 = (16.2 - 14.5) - 0
Þ p-values of 1.86 = Area under the t-distribution curve ( 6.4)2 + ( 9.6)2
43 48
to the right of t.
= 1.7 = 1.7 = 1.06
From the distribution table, t = 1.86 lies between 1.703 and 0.95 + 1.92 2.57
2.052 for which area lies between 0.025 and 0.05, therefore
p-value lies between 0.025 and 0.05. \ t = 1.06
21. Given, X1 = 1000, X2 = 1050 ,
17. We have, m 0 = 22, n = 65, x = 20 and S = 6.4
\Test statistic = x - m 0 S1 = 10, S2 = 11, n1 = 100 and n2 = 120
S/ n
= 20 - 22 Let there is no significant difference in the mean life of the
6.4 / 65
two types of batteries i.e. consider the null hypothesis.

= -2 ´ 65 \ H0 : m1 = m 2 X1 - X2
6.4 Test statistics, Z = S12 + S22
n1 n2
= - 2 ´ 8.06
6.4

= - 2.52 = 1000 - 1050 = -50

18. Given, p0 = 0.55, n = 250 and p = 0.48 (10)2 + (11)2 100 + 121
100 120 100 120
\ Z= p - p0 = 0.48 - 0.55
p0(1 - p0 ) 0.55 (1 - 0.55)
= -50 = -50
n 250 1 + 1.0083 2.0083

= -0.07 = -50 = - 35.21
0.55 ´ 0.45 1.42

250 Since, the computed value of Z = 35.21 is greater than the
critical value of Z = 1.96 (5% level of significance).
= - 0.07 ´ 250
0.2475 So, reject our null hypothesis, therefore, we conclude that

= - 0.07 ´ 15.81 the two types of batteries are having different mean life.
0.5
22. Test statistics The test statistics is a number calculated from
= - 2.21 < 0 a statistical test of a hypothesis. It shows how closely your
\ p-value of - 2.21 = Area under the standard normal curve observed data match the distribution expected under the
null hypothesis of that statistical test. The test statistic is
to left of Z used to calculate the p-value of your results, helping to
decide whether to reject your null hypothesis.
= 0.136
\ p-value = 0.136

CBSE Term II Applied Mathematics XII 87

Now, we calculate x and S from given sample values. 6 2, 4 1/10 1490 1640 1565
7 2, 5 1/10 1490 1550 1520
x x - 65 = d d2 8 3, 4 1/10 1660 1640 1650
9 3, 5 1/10 1660 1550 1605
66 1 1 10 4, 5 1/10 1640 1550 1595

65(a) 0 0

69 4 16

70 5 25

69 4 16 Average 1580

71 6 36 Here, clearly the average of sample means (1580) is equal to

the population mean. This shows the unbiased nature of the

70 5 25 sample mean as an estimator of population mean.

63 - 2 4 24. Given, p0 = 0.20, p = 0.175 and n = 400
64 - 1 1
(i) Test statistics, Z = p - p0
p0(1 - p0 )
68 3 9
n
Sd = 25 Sd2 = 133
= 0.175 - 0.20
Here, x = a + Sd = 65 + 25 = 67.5 0.20(1 - 0.20)
n 10
400

Now, S= S( x - x)2 = Sx2 - (Sx)2 = - 0.025 = - 0.025 ´ 20
n -1 n -1 n( n - 1) 0.20( 0.8) 0.20 ´ 0.80

= Sd2 - (Sd)2 400
n -1 n( n -1) = - 0.025 ´ 20 = - 1.25 < 0

0.4

= 133 - ( 25)2 (ii) Since, value of Z is negative.
9 10 ´ 9 \ p-value of - 1.28 = Area under the standard normal

= 14.78 - 6.94 = 7.84 = 2.81 curve to the left of Z

\t-statistics = x - m = 0.1003
S/ n (iii) Given, a = 0.05

= 67.5 - 65 Since, p-value < 0.05
2.81 10
So, we reject H0.
= 2.5 = 2.81 25. Given, n = 30, x = 18, S = 5.14
0.89
and m 0 = 16
23. \Sample mean = 1560 + 1490 + 1660 + 1640 + 1550
5 and degree of freedom = 30 - 1 = 29
(i) Test statistics = x - m 0 = 18 - 16
= 7900 = 1580 S / n 5.14 / 30
5
= 2 ´ 30 = 2.13
All possible samples and their corresponding samples in this 5.14
are presented in following table.
(ii) Since, t = 2.13 > 0

Sample Units in Probability Sample Sample Therefore, p-value of 2.13
Number Sample Observations mean = Area under the t-distribution curve to the right of t.
x = x1 + x2
x1 x2 Since, t = 2.13 lies between 2.045 and 2.462 for which
2 area lies between 0.01 and 0.025.
1 1, 2 1/10 1560 1490 1525
2 1, 3 1/10 1560 1660 1610 \ 0.01 < p-value < 0.025
3 1, 4 1/10 1560 1640 1600
4 1, 5 1/10 1560 1550 1555 (iii) Sum of minimum value and maximum value of
5 2, 3 1/10 1490 1660 1575 p = 0.01 + 0.025 = 0.035

(iv) Given, a = 0.05

Since, p-value < 0.05

So, we reject H0

(v) Reject H0, if t ³ ta
Here, ta = t0.05

Chapter Test 9. What is a Hypothesis?

Multiple Choice Questions 10. A simple random samples of 50 items from a population
with s = 6 resulted in a sample mean of 32. Provided a
1. Sample value is called 90% confidence interval for the population mean.

(a) data (b) parameter (c) statistics (d) variable 11. What is degree of freedom?

2. A simple random sample of 500 individuals provides Long Answer Type Questions
150 yes responses. The point estimate of the
population proportion that would provides yes 12. The 9 items of a sample had the following values.
responses is
45, 47, 50, 52, 48, 47, 49, 53, 50
(a) 0.3 (b) 0.4 (c) 0.6 (d) 0.9
The mean is 49 and the sum of square of deviation
3. A population consists of four observations 1, 5, 10, 15, taken from mean is 52. Can this sample be regarded as
then variance is taken from the population having 47 as mean? Also
obtain 95% and 99% confidence limits of the
(a) 60 (b) 27.69 (c) 65 (d) 72 populations mean.

4. Consider a hypothesis H0, where m 0 = 25 against H1 13. Ten students are selected at random from a college and
where m < 25, the test is their heights are found to be 100, 104, 108, 110, 118, 120,
122, 124, 126 and 128 cm. In the light of these data,
(a) right tailed (b) center tailed discuss the suggestion that the mean height of the
(c) left tailed (d) None of these students of the college is 110 cm. (given, t9(0.05) = 2.262)

5. Consider the following hypothesis test 14. Consider the following hypothesis test
H0 : m = 8; Ha : m ¹ 8
H0 : m £ 22; Ha : m > 22
A sample of 120 provided a sample mean of 8.4. The
A sample of 32 provided a sample mean x = 27 and population standard deviation is 3.2
standard deviation S = 5.2. The value of the test
(i) Compute the value of the test statistics.
statistics is
(ii) What is the p-value?
(a) 3.52 (b) 4.32 (c) 5.44 (d) 5.36 (iii) At a = 0.05, what is your conclusion?

Case Based MCQs (iv) Compute 95% confidence internal for the population
mean. Does it support your conclusion?
6. A company is interested in knowing investor sentiment
survey every month to determine the number of 15. A soap manufacturing company was distributing a
investors, who are Bullish, Bearish or Neutral on the particular brand of a soap through a large number of
stock market for next few months. retail shops. Before a heavy advertisement campaign,
the mean sales per week per shop was 140 dozens.
The results for the month ending October 2021 are as After the campaign a sample of 26 shops was taken
follows: and mean sales was found to be 147 dozens with
standard deviation 16. Can you consider the
Bullish 208, Bearish 150, Neutral 142. advertisement effective? (given t25 (0.05) = 2.06)

Based on above information, answer the following 16. The average middle class family spends ` 9000 per
questions. month. A random sample of 25 families in a city,
showed a sample mean monthly expenditure of ` 8450
(i) The point estimate of the population proportion who with a standard deviation of ` 1450. Test H0 : m = ` 9000
and Ha ¹ ` 9000 with a = 0.05. Use two tailed test.
are Bullish, is
(i) What are the critical values of the test statistics and
(a) 0.519 (b) 0.416 (c) 0.369 (d) 0.345 what is the rejection region?

(ii) The point estimate of the population proportion who (ii) Compute the value of the test statistics.

are Bearish, is (iii) What is the your conclusion?

(a) 0.2 (b) 0.8 (c) 0.3 (d) 0.4 Answers

(iii) The point estimate of the population proportion who 1. (c) 2. (a) 3. (b) 4. (c) 5. (c)

are Neutral, is 6. (i) (b) (ii) (c) (iii) (b) (iv) (a) (v) (d)

(a) 0.193 (b) 0.284 (c) 0.43 (d) 0.431 10. 1.39 12. 47.04, 50.96 13. 110

(iv) A 95% confidence interval for the population 14. (i) 1.37 (ii) 0.0853 (iii) Do not reject H0 (iv) Do not reject H0
16. (i) ± 2.064 (ii) - 1.897
proportion, who are Bullish, is
For Detailed Solutions
(a) (0.373, 0.459) (b) (0.628, 0.741)
Scan the code
(c) (0.291, 0.429) (d) (0.282, 0.362)

(v) A 95% confidence interval for the population

proportion who are Bearish, is

(a) (0.41, 0.57) (b) (0.42, 0.56)

(c) (0.34, 0.43) (d) (0.26, 0.34)

Short Answer Type Questions
7. Explain cluster sampling?
8. What are sampling errors?

CBSE Term II Applied Mathematics XII 89

CHAPTER 06

Index Number and
Time Based Data

In this Chapter...

l Time Series
l Trend Analysis
l Equation of the Trend Line

Time Series (ii) Seasonal Variation

A time series is a sequence of observations which are ordered This type of variation depends on weather and other natural
in time (space). If observation are made on some forces or social conditions. Like sale of ice-creams increase in
phenomenon throughout time it is most sensible to display summer season.
the date in the order in which they arose, particularly, since
successive observations will properly be dependent. (iii) Cyclical Variation

Component (Variations) of Time Series Many industrial or financial sectors face ups and down
(booms or slumps) over regular period of time. It is very short
The various reasons or the forces that affect the values of an period variation and they do not affect the trend such
observation in a time series are called the components of a periodic variations are known as cyclical variation.
time series. There are four categories of components of a time
series as follows (iv) Irregular Variation

(i) Trend (ii) Seasonal l Irregular/random variation constitute one of four components
of a time series.
(iii) Cyclic (iv) Irregular
l Irregular variation do not follow a particular model and are not
(i) Trend Variation predictable.

A trend exists when there is a long-term increase or decrease Time Series Analysis
in the data. It does not have to be linear. Some time we will
refer to a trend as “change direction” when it might go from Time series analysis is a statistical technique that deals with
an increasing trend to a decreasing trend. time series data or trend analysis. Time series data means that
data is in a series of particular time periods or intervals.
e.g. A new apartment construction is going on, and someone The data is considered in three types
has opened a shop of hardware materials, so until all the
apartments get constructed, people will make purchases from (i) Time series data A set of observations on the values that
that store. This will result in the trend of the sales revenue a variable takes at different times.
increasing.

90 CBSE Term II Applied Mathematics XII

(ii) Cross-sectional data Data of one or more variables, Calculation of Moving Average
collected at the same point in time. Procedure for calculating the moving average in two cases
Case I
(iii) Pooled data A combination of time series data and (Period of moving average is an odd number i.e. 3, 5, 7,K).
cross-sectional data. For 3-yearly moving average

Model of Time Series Analysis

There are two models which are generally used for the Year Value 3-yearly moving average
decomposition of time series into its four components.

(i) Additive Model In this model, we represent a particular 1st x1
2nd x2
observation in a time series as, if O represents the 3rd x3 x1 + x2 + x3 = m1
4th x4 3
original data, T represents the trend, Srepresents the 5th x5
6th x6
seasonal variations, C represents the cyclical variations x2 + x3 + x4 = m2
3
and I represents the irregular variations.

Then, O =T +S+C + I x3 + x4 + x5 = m3
3
(ii) Multiplicative Model In this model, four components
have a multiplicative relationship. We represent a x4 + x5 + x6 = m4
particular observation in a time series as the product of 3
these four components.
x5 + x6 + x7 = m5
i.e. O = T ´ S´ C ´ I, 3

where O, T, S, C and I represents the terms as in additive 7th x7
model.

Trend Analysis Note m1, m2, K, m5 are moving average points (mean point).

A trend is a recurring pattern and trend analysis is the Case II
practice of collecting data in an attempt to spot that pattern.
When the user needs and behaviour are changing rapidly. (Period of moving average is an even number i.e. 4, 6, 8, K).
Trend analysis is a method that can act as window into the For 4-yearly moving average
future demands of users.
Year Value 4-yearly moving average 4-yearly moving
average centered

Methods of Measuring Trend 1st x1 x1 + x2 + x3 + x4 = m1
2nd x2 4
Trend is measured using by the following methods 3rd x3
(i) Graphical method 4th x4 x2 + x3 + x4 + x5 = m2 m1 + m2 = A
(ii) Semi-averages method 5th x5 4 2

(iii) Moving averages method x3 + x4 + x5 + x6 = m3 m2 + m3 =B
(iv) Method of least squares. 4 2
Here, we discuss (iii) moving averages methods and (iv)
method of least squares. x4 + x5 + x6 + x7 = m4 m3 + m4 = C
4 2
Purpose of Moving Averages
6th x6
The following points are main purpose of moving averages 7th x7
l It helps in technical traders like tract the trend of financial
Note A, B and C are centered points m1, m2 and m3 are moving average
assets by smoothing out the day-to-day price fluctuations. points.
l By identifying trend, moving averages allow traders, to make
Method of Least Square
these trends works in their favour and increases the number of
winning trades. It is helps in trading strategies. The least square method is the method of determining the
least-fitting curve or line for the processing data. This method
Period of Cycles provides a visual demonstration of the relation between the
data points which determines the trend line for the given
The time-interval covering the number of years is called the time series. This trend line is known as best fitting line.
period of the cycles.

Note Cycles always found in series.

CBSE Term II Applied Mathematics XII 91

Hence, the least square method is the process of finding the Eq. (i) can be written as …(iii)
least fitting line for data points through reduction of the sum xyt = ax + bx2
of squares of the offsets points from the curve.
and Sxy = aSx + bSx2
Let us assume that ( t1 , y1 ), ( t 2 , y2 ), ( t 3 , y3 ) … ( t n , yn )
denotes the time series, where t’s are the independent values Eqs. (ii) and (iii) are called normal equations from where we
and y’s are the dependent variable. get the value of a and b.

(i) The sum of deviations of y from their corresponding Assume the mid-point of time is taken as origin, then
trend value is zero.
Sx = 0
i.e. S( y - yt ) = 0
\From Eq. (ii), we get
(ii) The sum of the squares of deviations of y must be
minimal. Sy = na
Þ a = Sy
i.e. S( y - yt ) 2 is minimal.
n

Equation of the Trend Line Here, a represents the mean of y

The equation of the trend line can be written as and from Eq. (iii), we get
Sxy = bSx2

yt = a + bx …(i) Þ b = Sxy
Sx 2
where, yt is the trend value of y, x is the deviation of time t
from origin, a and b are constants. Here, b represents the slope of the trend line.

Now, Syt = Sa + bSx …(ii) Note If n is odd, then we take middle year as the origin and, if n is even
and Sy = na + bSx then there will be two middle years, then we take mean of the
two middle years as origin.
where, n be the number of years given in the data.

92 CBSE Term II Applied Mathematics XII

Solved Examples

Example 1. What is non-linear trend? Here, a = Sy = 439 = 87.8
n5
Sol. A time series may have faster or slower increase at early
stage and have a slower or faster increase at more rescent and b = Sxy = 21 = 2.1
time. In such case, better description of the time series is Sx2 10
given by a non-linear curve rather than straight line.
\ Trend line equation, yt = a + bx
Example 2. Write any two limitations of moving = 87.8 + 2.1x

average. Example 6. Coded monthly sales figures of a particular

Sol. (i) The effect of averaging is to give a smoother curve, brand of TV for 18 months commencing January
lessening the influence of the fluctuations. 2005 are as follows

(ii) The moving average method is most commonly applied Year Jan Feb Mar Apr May June July Aug Sept Oct Nov Dec
data, which are characterised by cyclical movements.
2005 18 16 23 27 28 19 31 29 35 27 28 24
Example 3. The average number, in lakhs of working 2006 24 28 29 30 29 22

days last in strikes during each year of the period Calculate 6-monthly moving averages and display
(1981-88) was an under. these and the original figures on the same graph
using the same axis for both.
1981 1982 1983 1984 1985 1986 1987 1988
Sol. According to the question,
1.5 1.8 1.9 2.2 2.6 3.7 2.2 6.4

Calculate the second 3-yearly moving averages. Months of 2005 Sales (Total)
Jan 18
Sol. Second 3-yearly moving average = 1.8 + 1.9 + 2.2 Feb 16
3 Mar 23
Apr 27
= 5.9 =1.97 May 28
3 Jun 19
Jul 31
Example 4. If Sy = 91, n = 7, Sx 2 = 28 and Sxy = 33. Aug 29
Sep 35
Then, find the equation of trend line. Oct 27
Sol. Here, a = Sy = 91 = 13 Nov 28
n7 Dec 24
24
and b = Sxy = 33 = 1.179 Jan 06 28
Sx2 28 Feb 06 29
Mar 06 30
\ Equation of trend line, yt = a + bx = 13 + 1.179x Apr 06 29
May 06 22
Example 5. Write a trend line equation to the following Jun 06

time series data.

Year 2008 2009 2010 2011 2012

Sale of sugar (in tonnes) 80 90 92 83 94

Sol.

Year Sales (y) x xy x2 Let the 6 months moving average is m,
2008 80 -2 - 160 4
2009 90 -1 - 90 1 So, m1 = 18 + 16 + 23 + 27 + 28 + 19 = 131 = 21.83
2010 92 0 0 6 6
2011 83 1 0 1
2012 94 2 83 4 m2 = 16 + 23 + 27 + 28 + 19 + 31 = 144 = 24.00
n =5 Sx = 0 188 å x2 = 10 6 6
Sy = 439 Sxy = 21
m3 = 23 + 27 + 28 + 19 + 31 + 29 = 157 = 26.2
6 6

CBSE Term II Applied Mathematics XII 93

m4 = 27 + 28 + 19 + 31 + 29 + 35 = 169 » 28.2 m6 + m7 = 28.2 + 29.00 = 28.6
6 6 22

m5 = 28 + 19 + 31 + 29 + 35 + 27 = 169 » 28.2 m7 + m8 = 29.00 + 27.83 = 28.415
6 6 22

m6 = 19 + 31 + 29 + 35 + 27 + 28 = 169 » 28.20 m8 + m9 = 27.83 + 27.66 = 27.745
6 6 22

m7 = 31 + 29 + 35 + 27 + 28 + 24 = 174 = 29.00 m9 + m10 = 27.66 + 26.66 = 27.16
6 6 22

m8 = 29 + 35 + 27 + 28 + 24 + 24 = 167 = 27.83 m10 + m11 = 26.66 + 27.16 = 26.91
6 6 22

m9 = 35 + 27 + 28 + 24 + 24 + 28 = 166 = 27.66 m11 + m12 = 27.16 + 27.33 = 27.245
6 6 22

m10 = 27 + 28 + 24 + 24 + 28 + 29 = 160 = 26.66 m12 + m13 = 27.33 + 27 = 27.165
6 6 22

m11 = 28 + 24 + 24 + 28 + 29 + 30 = 163 = 27.16 On the basis of above data we can draw the following graph
6 6

m12 = 24 + 24 + 28 + 29 + 30 + 29 = 164 = 27.33 35
6 6
m7 m9 m11
m13 = 24 + 28 + 29 + 30 + 29 + 22 = 162 = 27.00 30 m4 m5 m6 m8 m10 m13
6 6 m12

For centred moving average, 25 m3

m1 + m2 = 21.83 + 24.00 = 22.91 20 m1 m2
22
15
m2 + m3 = 24.00 + 26.2 = 25.1
22 10

m3 + m4 = 26.2 + 28.2 = 27.2 5
22
J FMAM J J A SOND J FMAM J
m4 + m5 = 28.2 + 28.2 = 28.2
22 2005 2006

m5 + m6 = 28.2 + 28.2 = 28.2
22

94 CBSE Term II Applied Mathematics XII

Chapter
Practice

PART 1 9. Seasonal variations are (b) sudden variation
Objective Questions (d) None of these
(a) long term variation
(c) short term variation

10. In the time series, shortage of certain consumer

l Multiple Choice Questions goods before the annual budget is due to

(a) irregular variation (b) cyclical variation

1. An orderly set of data arranged in accordance with (c) seasonal variation (d) secular trend

their time of occurrence is 11. Additive model for time series O is equal to

(a) time series (b) arithmetic series (a) T + S + C + I (b) T - S + C - I

(c) geometric series (d) harmonic series (c) T + S - C - I (d) - T + S - C + I

2. Which of the following is an example of time series 12. Multiplicative model for time series O is equal to

model? (a) T + S ´ C ´ I (b) T ´ S ´ C + I

(a) Moving average (b) Native approach (c) T ´ S ´ C ´ I (d) TS + S ´ I

(c) Exponential smoothing (d) All of these 13. If the demand is 150 during April 2017, 250 in May,

3. Which of the following can not be component for a 350 in June, 450 in July. Then, the 4 month simple

time series? moving average for August 2017 is

(a) Seasonality (b) Cyclical (a) 400 (b) 300 (c) 500 (d) 600

(c) Noise (d) None of these 14. If x1, x 2, x 3, …, x n is the given annual time series,

4. The time series analysis helps then 3-yearly moving average are

(a) to make predictions (a) x1 + x2 + x3 , x2 + x3 + x4 , x3 + x4 + x5 …, which are
(b) to compare the two or more series 333
(c) to know the behaviour of business
(d) All of the above placed against years 2, 3, 4, … respectively
(b) x1 + x2 + x3, x2 + x3 + x4 , x2 + x4 + x5 , which are
5. In a time series values of the variables y(say)
33 3
depend on time t represented as y = F(t), has against 1, 2, 4, … respectively.
(c) x1 + x3 + x5 , x2 + x4 + x5 , x1 + x2 + x4 , …which are
(a) four components (b) two components
333
(c) three components (d) five components against … respectively.

6. The component of a time series attached to (d) None of the above

long-term variations is termed as 15. The general pattern of increases or decreases in

(a) seasonal variations (b) irregular variations economics or social phenomena is know by

(c) secular trend variations (d) cyclic variations (a) irregular trend (b) secular trend

7. Prosperity, recession and depression in a business is (c) seasonal trend (d) cyclical trend

an example of 16. Method of moving average is used to find the

(a) seasonal trend (b) cyclical trend (a) cyclical variation (b) irregular variation

(c) irregular trend (d) secular trend (c) seasonal variation (d) secular trend

8. A fire in a factor delaying production for some 17. The best fitting trend is one which the sum of

weeks is squares of residuals is

(a) secular trend (b) irregular trend (a) maximum (b) zero

(c) cyclical trend (d) seasonal trend (c) minimal (d) None of these

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CBSE Term II Applied Mathematics XII 95

18. Moving average method is used for measurement of Based on the above information, answer the
following questions.
trend when

(a) trend is non linear (b) trend is linear (i) The value of constant ‘a’ in Eq. (i), is

(c) trend is curvilinear (d) trend is parabolic (a) 5.12 (b) 6.2
(c) 5.18 (d) 3.69
19. In the trend line equation y = a + bx, b is the

(a) slope (b) mean (ii) The value of constant ‘b’ in Eq. (i), is
(c) x-intercept (d) y-intercept
(a) 0.74 (b) 0.69
20. Value of b in the trend line y = a + bx is (c) 0.52 (d) 0.48

(a) may be positive or negative (iii) The equation of the trend line is
(b) always positive
(c) always negative (a) yt = 5.18 + 0.69x (b) yt = 3.12 + 0.54x
(d) always zero (c) yt = 1.45 + 0.52x (d) yt = 2.42 + 0.72x

(iv) The trend value for year 1962 is

21. In a trend line equation y = a + bx, where a is the (a) 0.52 (b) 0.39

(a) mean of x (b) mean of y (c) 2.14 (d) 1.73
(c) Slope (d) None of these
(v) The trend value for year 1970 is

22. The actual sales of a product in different months of (a) 7.25 (b) 4.2
a particular year are given below
(c) 3.92 (d) 5.1

Sep Oct Nov Dec Jan Feb

180 280 250 190 240 ? PART 2
Subjective Questions
Using the 4 month moving average method forecast

of the sale, for the month of February is

(a) 350 (b) 240 (c) 300 (d) 280

23. The demand of a product over 6 weeks is 90, 110, l Short Answer Type Questions

105, 130, 85 and 102. If 3 month moving averages is 1. What is a Time Series?
‘x’ and 5 month moving average is ‘y’, then (x + y) is 2. Write the components for time series.
3. What is the Trend?
(a) 202.70 (b) 105.67 4. Define linear and non-linear trend.
5. Describe the purpose of moving averages.
(c) 212.07 (d) 180.67 6. Describe the types of variation.
7. Explain the method of least squares.
l Case Based MCQs 8. Write the slope of the trend line.
9. Define irregular variations in the time series.
24. Fit a straight line trend by using the method of least 10. Write the equation of trend line.
square for the following data 11. Write any two uses of moving averages?
12. If (t1, y1 ), (t2, y2 ), …… , (tn , yn ) denote the given
Year Production (in tonnes)
1962 2 time series. Then, write the mathematical
1963 4 relationship between the trend value yt of the
1964 3 variable y.
1965 4 13. Consider the following series of observations.
1966 4
1967 2 Years 1995 1996 1997 1998 1999
1968 4
1969 9 Annual sales 3.6 4.3 4.3 3.4 4.4
1970 7 (in Thousand)
1971 10
1972 8

If the equation of the trend line can be written as Calculate the five year trend value.

yt = a + bx …(i)

96 CBSE Term II Applied Mathematics XII

14. The number of letters (in hundred) received in a Plot these moving averages on the same graph
town on each day of a fortnight is given below: paper and comment on the general trend.

Days of week Sun Mon Tue Wed Thu Fri Sat 18. The table given below shows the daily attendance
First week 35 70 36 59 62 60 71 in thousand at a certain exhibition over a period of
Second week 39 72 38 56 63 71 75 two weeks

Calculate the seven day moving average and display Week 1 52 48 64 68 52 70 72
it on a graph. Week 2 55 47 61 65 58 75 81

l Long Answer Type Questions Calculate seven day moving averages and illustrate
15. Explain model of time series analysis. these and original information on the same graph
16. Consider the following data using the same scales.

Year Production l Case Based Questions
1983 137 19. Consider the following data.
1984 140
1985 134 Years 2004 2005 2006 2007 2008 2009 2010
1986 137
1987 151 Profit (in Thousand) 114 130 126 144 138 156 164
1988 121
1989 124 and by using method of least square, answer the
1990 159 following questions.
1991 157
1992 169 (i) Find the equation of the trend line?
1993 172
1994 150 (ii) Determine the slope of the trend line?

Calculate a suitable moving average and show on a (iii) Find the trend value for the year 2007?
graph against the original data.
(iv) The sum of the trend value for year 2009 and
17. The table shows the number of students on the roll 2006?
of a school on the first day of each term
(v) The difference of trend value for the year 2010 and
2005?

20. The average number (in lakh) of working days lost
in strikes during each year of the period 1981-90
was

1981 1982 1983 1984 1985 1986 1987 1988 1989 1990
1.5 1.8 1.9 2.2 2.6 3.7 2.2 6.4 3.6 5.4

Term 1981 1982 1983 1984 1985 1986 Using 3-yearly moving averages answer the
Jan 252 261 288 299 312 312 following question
Apr 284 258 312 322 306 324
Sep 254 261 312 274 306 284 (i) Determine the trend value for the year 1982.

Draw the graph of these figures. Calculate a set of (ii) Determine the moving averages for the year 1985.
moving averages using the most number of
observations. Give reasons for your choice. (iii) The sum of the moving averages for the year 1984
and 1986.

(iv) Calculate the 3-yearly moving averages for the
year 1981.

(v) Draw the moving average graph.

CBSE Term II Applied Mathematics XII 97

SOLUTIONS

Objective Questions 18. (b) Moving average method is used for measurement of
trend when trend is linear.
1. (a) Time series is a set of statistical observations taken at
different point of time. 19. (a) ‘b’ represents the slope of trend line.

2. (d) The moving average model is a common approach for 20. (a) Value of slope may be positive or negative.
modeling univariate time series. 21. (b) In equation y = a + bx, ‘a’ represents mean of y.

Native approach estimating technique in which the last 22. (b) When n = 4, then 4 month moving average method, for
period’s actuals are used as the period’s forecast. the month of February = 280 + 250 + 190 + 240
4
In exponential smoothing older data is given = 960 = 240
progressively-less relative importance whereas never data is 4
given progressively-greater importance.
23. (c) We know that,
3. (d) A seasonal pattern exists when a series is influenced by 3 month moving average = Sum of latest 3 demand
seasonal factors. e.g. the month, the day or day of the week. 3

A cyclic pattern exists when data exhibit rises and falls that and similarly for 5 month moving.
are not of fixed period. Now, x = 130 + 85 + 102 = 317 = 105.67

In discrete time, white noise is a discrete signal whose 33
samples are regarded as a sequence of serially uncorrelated and y = 110 + 105 + 130 + 85 + 102
random variables with zero mean and finite variance.
5
4. (d) The time series analysis helps to make predictions, to = 532 = 106.4
compare the two or more series and to know the behaviour
of business. 5
\ x + y = 105.67 + 106.4 = 212.07
5. (a) There can be four types of components in the values of 24. Here, n = 11 (odd number)
the variables
Year Production x = ti - 1967 x2 xy yt = a + bx
(a) trend (b) cyclical (c) seasonal (d) irregular (t) (y) 25 - 10 1.73

6. (c) Secular trend variations are considered as long term (in tonnes)
variation, attributable to factor such as population change,
technological progress and large-scale shifts in consumer 1962 2 -5
tastes.
1963 4 - 4 16 - 16 2.42
7. (b) Time series show rise and fall of the values in a short
period time. The variations in a time series which operate 1964 3 - 3 9 - 9 3.11
themselves over a span of more than one year are called
cyclic variation. 1965 4 - 2 4 - 8 3.8

8. (b) Since, the irregular component of the time series is the 1966 4 - 1 1 - 4 4.99
factor which causes the variations that are random.
1967 2 0 0 0 5.18
9. (c) Seasonal variations represents the variability in the data
due to seasonal influence. 1968 4 1 1 4 5.87

10. (c) Shortage of certain consumer goods before the annual 1969 9 2 4 18 6.56
budget is due to seasonal variations.
1970 7 3 9 21 7.25
11. (a) In this model, we represents a particular observation in a
time series as the addition of these four components. 1971 10 4 16 40 7.94

12. (c) In this model, we represents a particular observations in 1972 8 5 25 40 8.63
a time series as the product of these four components. Sy = 57 Sx = 0 Sx2 = 110 Sxy = 74

13. (a) 4 month simple moving average for August 2017 Here, a = Sy = 57 = 5.18
= 150 + 250 + 350 + 450 n 11
4
= 1200 = 300 b = Sxy = 76 = 0.69
4 Sx2 110

14. (a) Clearly, 3-yearly moving average is represented by \The required equation of the trend line
option (a).
yt = 5.18 + 0.69x
15. (b) Secular trend movements are considered as long term
movements.

16. (d) Method of moving average is used to find the secular
trend (long term).

17. (c) The sum of the square of deviations of y must be minimal.

98 CBSE Term II Applied Mathematics XII

(i) (c) 5.18 8. We know that, the equation of the trend line

(ii) (b) 0.69 yt = a + bx,
(iii) (a) yt = 5.18 + 0.69x
where yt is the trend value of y, x is the deviation of time t
(iv) (d) For year 1962, x = - 5 from origin.
yt = 5.18 + 0.69 ´ ( - 5)
= 5.18 - 3.45 = 1.73 and a = Sy = represents the mean of y
n
(v) (a) For year 1970, x = 3
yt = 5.18 + 0.69 ´ 3 b = Sxy = represents the slope of the trend line.
= 5.18 + 2.07 = 7.25 Sx2

Subjective Questions 9. Irregular variations in the time series is the factor which
causes the variations that are randomly occurs.
1. Time series is simply a series of data points ordered in time.
In a time series, time is often the independent variable and e.g. due to floods, earthquakes, sudden ban on sale.
the goal is usually to make a forecast for the future.
10. The equation of trend line is
2. The four categories of the components of time series are
yt = a + bx,
(i) Trend Sy Sxy
(ii) Seasonal variations where a = n and b = Sx2
(iii) Cyclic variations
(iv) Random or Irregular variations 11. Moving average method is used to

3. The Trend shows the general tendency of the data to (i) reduce fluctuation (ii) obtain trend value
increase or decrease during a long period of time. A trend is
a smooth, general, long-term, average, tendency. 12. (i) The sum of deviations of y from their corresponding
trend value is least
4. If we plot the time series value on a graph in accordance i.e. S( y - yt )2 is zero.
with time t. The pattern of the data clustering shows the
type of trend. If the set of data cluster more or less round a (ii) The sum of squares of the deviations of the values of y
straight line, then the trend is linear otherwise it is from their corresponding trend values is least
non-linear. i.e. S( y - yt )2 is least.

5. The following points are main purpose of moving average 13. For given observations five year trend value

(i) It helps in technical traders like tract the trend of = 3.6 + 4.3 + 4.3 + 3.4 + 4.4
financial assets by smoothing out the day-to-day price 5
fluctuations.
= 20.0 = 4
(ii) By identifying trend, moving averages allow traders, to 5
make these trends works in their favour and increases
the number of wining trades. It is helps in trading 14. According to the question,
strategies.
Week 1
6. Variation can be divided into three types Days Sun Mon Tue Wed Thu Fri Sat

(i) Seasonal variation This type of variation depends on Number of
weather and other natural forces or social conditions.
Like sale of ice-creams increase in summer season. letters 35 70 36 59 62 60 71

(ii) Cyclical variation Many industrial or financial sectors (in hundred)
face ups and down (booms or slumps) over regular
period of time. It is very short period variation and they Week 2
do not affect the trend such periodic variations are Days Sun Mon Tue Wed Thu Fri Sat
known as cyclical variation.
Number of
(iii) Irregular variation Irregular/random variation
constitute one of four components of a time series. letters 39 72 38 56 63 71 75

Irregular variation do not follow a particular model and (in hundred)
are not predictable.
Let the moving average for seven days is m, then
7. The least squares method is the method of determining
the least-fitting curve or line for the processing data. m1 = 35 + 70 + 36 + 59 + 62 + 60 + 71 = 393 = 56.14
This method provides a visual demonstration of the relation 7 7
between the data points which determines the trend line
for the given time series. This trend line is known as best m2 = 70 + 36 + 59 + 62 + 60 + 71 + 39 = 397 = 56.71
fitting line. 7 7

m3 = 36 + 59 + 62 + 60 + 71 + 39 + 72 = 399 = 57.00
7 7

CBSE NEW Pattern Mathematics XII (Term-II) 99

m4 = 59 + 62 + 60 + 71 + 39 + 72 + 38 = 401 = 57.28 Year Production 3-Yearly Total
7 7 1985 134 411
1986 137 422
m5 = 62 + 60 + 71 + 39 + 72 + 38 + 56 = 398 = 56.85 1987 151 409
7 7 1988 121 396
1989 124 404
m6 = 60 + 71 + 39 + 72 + 38 + 56 + 63 = 399 = 57.00 1990 159 440
7 7 1991 157 485
1992 169 498
m7 = 71 + 39 + 72 + 38 + 56 + 63 + 71 = 410 = 58.57 1993 172 491
7 7 1994 150 -

m8 = 39 + 72 + 38 + 56 + 63 + 71 + 75 = 414 = 59.14
7 7

On the basis of above data we can draw the following graph
for moving average

7 day’s Actual
moving graph
70 average

65 Let M be 3-yearly moving average. So,

60 m8 m1 = 411 = 137.00 Þ m2 = 411 = 137.00
55 m1 m2 m3 m4 m5 m6 m7 3 3

m3 = 422 = 140.67 Þ m4 = 409 = 136.33
3 3
50
396 404
45 m5 = 3 = 132.00 Þ m6 = 3 = 134.67

40 m7 = 440 = 146.67 Þ m8 = 485 = 161.67
3 3
35
498 491
0 m9 = 3 = 166.00 Þ m10 = 3 = 163.67
S M T W Th F Sa S M T W Th F Sa
On the basis of above data we can draw the following
15. Model of Time Series Analysis There are two models which
are generally use for the decomposition of time series into its moving average graph
four components.
200Production
(i) Additive Model In this model, we represent a 180
particular observation in a time series as, if O 160
represents the original data, T represents the trend, S 140
represents the seasonal variations, C represents the 120
cyclical variations and I represents the irregular 100 Actual data
variations 80 Trend by 3 yr
Then, O = T + S + C + I
moving average
(ii) Multiplicative Model In this model four components
have a multiplicative relationship. We represent a 60
particular observation in a time series as the product of 40
these four components. 20
i.e. O = T ´ S ´ C ´ T, 0

where O, T, S, C and I represents the terms as in 83 84 85 86 87 88 89 90 91 92 93 94
additive model. Year

Trend Analysis A trend is a recurring pattern and trend 17. According to the question, we can arrange the given data in
analysis is the practice of collecting data in an attempt to following table
spot that pattern. When the user needs and behaviour are
changing rapidly, trend analysis is a method that can act as Years 1981 1982 1983
window into the future demands of users. Terms
Jan Apr Sep Jan Apr Sep Jan Apr Sep
16. According to the question, we can arrange the following
table Number of 252 284 254 261 258 261 288 312 312
students
1984 1985 1986
Years
Year Production 3-Yearly Total Jan Apr Sep Jan Apr Sep Jan Apr Sep
1983 137 - Terms
1984 140
411 Number of 299 322 274 312 306 306 312 324 284
students

100 CBSE NEW Pattern Mathematics XII(Term-II)

Let m be the moving average. 18. According to the question,

Then, Week 1
Days Sun Mon Tue Wed Thu Fri Sat
m1 = 252 + 284 + 254 = 790 = 263.3
3 3

Þ m2 = 284 + 254 + 261 = 799 = 266.3
3 3
Attendance 52 48 64 68 52 70 72
Þ m3 = 254 + 261 + 258 = 773 = 257.7 (in thousand)
3 3
Week 2
261 + 258 + 261 780
Þ m4 = 3 = 3 = 260
Days Sun Mon Tue Wed Thu Fri Sat

Þ m5 = 258 + 261 + 288 = 807 = 269 Attendance 55 47 61 65 58 75 81
33 (in thousand)

Þ m6 = 261 + 288 + 312 = 861 = 287
3 3
Let the 7 days moving average is m.

Þ m7 = 288 + 312 + 312 = 912 = 304 So, m1 = 52 + 48 + 64 + 68 + 52 + 70 + 72 = 60.85
3 3 7

Þ m8 = 312 + 312 + 299 = 923 = 307.7 Þ m2 = 48 + 64 + 68 + 52 + 70 + 72 + 55 = 61.28
3 3 7

Þ = 312 + 299 + 322 = 933 = Þ m3 = 64 + 68 + 52 + 70 + 72 + 55 + 47 = 61.14
3 3 7
m9 311

299 + 322 + 274 895 Þ m4 = 68 + 52 + 70 + 72 + 55 + 47 + 61 = 60.71
3 3 7
Þ m10 = = = 298.3

322 + 274 + 312 908 Þ m5 = 52 + 70 + 72 + 55 + 47 + 61 + 65 = 60.28
3 3 7
Þ m11 = = = 302.6

274 + 312 + 306 892 Þ m6 = 70 + 72 + 55 + 47 + 61 + 65 + 58 = 61.14
3 3 7
Þ m12 = = = 297.3

312 + 306 + 306 924 Þ m7 = 72 + 55 + 47 + 61 + 65 + 58 + 75 = 61.85
3 3 7
Þ m13 = = = 308

306 + 306 + 312 = 924 = 308 Þ m8 = 55 + 47 + 61 + 65 + 58 + 75 + 81 = 63.14
3 3 7
Þ m14 =

Þ = 306 + 312 + 324 = 942 = 314 On the basis of above data we can draw the following
3 3
m15 moving average graph

Þ m16 = 312 + 324 + 284 = 920 = 306.7
3 3
Number of students

Attendance in thousand
MoOvriingignaalveatrtaegnedagrnacpeh
On the basis of above data we can draw the following 88
moving average graph 84
80
310 76
300 Trend 72
68
290 64
60
280 56
52
270 Actual 48
44
260
250 1 2 3 4 5 6 7 8 9 1011121314

0X Week 1 Week 2
JAS JAS JAS J ASJASJ AS
1981 1982 1983 1984 1985 1986
Years

CBSE NEW Pattern Mathematics XII (Term-II) 101

19. Here, n = 7 odd 20. According to the question,

Year Profit x = ti - 2007 x2 xy Trend Years Average (in Lakh)
(ti ) (y) 9 - 342 values 1981 1.5
yt = a + bx 1982 1.8
2004 114 -3 1983 1.9
115.9 1984 2.2
1985 2.6
2005 130 -2 4 - 260 123.58 1986 3.7
1987 2.2
2006 126 -1 1 - 126 131.22 1988 6.4
1989 3.6
2007 144 0 0 0 138.86 1990 5.4

2008 138 1 1 138 146.50

2009 156 2 4 312 154.14

2010 164 3 9 492 161.78 Let the 3 yr moving average is m.

Sy = 972 Sx = 0 Sx2 = 28 Sxy = 214 So, m1 = 1.5 + 1.8 + 1.9 = 1.73
3
Here, a = Sy = 972 = 138.86
n7 Þ m2 = 1.8 + 1.9 + 2.2 = 1.96
3
Sxy 214
and b = Sx2 = 28 = 7.64 Þ m3 = 1.9 + 2.2 + 2.6 = 2.23
3

(i) The equation of the trend line Þ m4 = 2.2 + 2.6 + 3.7 = 2.83
yt = 138.86 + 7.64x 3

(ii) Slope of trend line, b = 7.64 Þ m5 = 2.6 + 3.7 + 2.2 = 2.83
3
(iii) The trend value for the year 2007, when x = 0

\ yt = 138.86 + 7.64 ´ 0 = 138.86 Þ m6 = 3.7 + 2.2 + 6.4 = 4.10
(iv) For year 2006, x = - 1 3

\ yt = 138.86 + 7.64 ´ ( - 1) Þ m7 = 2.2 + 6.4 + 3.6 = 4.07
= 138.86 - 7.64 3

= 131.22 Þ m8 = 6.4 + 3.6 + 5.4 = 5.13
For year 2009, x = 2 3
\ yt = 138.86 + 7.64 ´ 2
(i) Clearly, m1 = 1.73
= 138.86 + 15.28 (ii) Clearly, m4 = 2.83
(iii) Clearly, m3 + m5 = 2.23 + 2.83 = 5.06
= 154.14
\Sum of the trend value for year 2009 and 2006 (iv) Can not determine

= 154.14 + 131.22 = 285.36 (v) Moving average graph is as follows
(v) For year 2010, x = 3
7 Actual
\ yt = 138.86 + 7.64 ´ 3 6
= 138.86 + 22.92 5 Trend

= 161.78 4

For year 2005, x = - 2

\ yt = 138.86 + 7.64 ´( - 2) 3
= 138.86 - 15.28
2

= 123.58 1

\Difference of the trend value for the year 2010 and 0
2005 = 161.78 - 123.58 = 38.20 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990

Chapter Test 9. Give an example of least square method.
10. Write down two uses of the least square method.
Multiple Choice Questions
Long Answer Type Questions
1. The forecasts on the basis of a time series are 11. Write the limitations of least-square method.
12. Fit a straight line trend equation by the method of
(a) true to a great extent (b) 100% true
(c) never true (d) None of these least squares and estimate the trend value.

2. A fire in a factory delaying production for some weeks, Year 1991 1992 1993 1994 1995 1996 1997 1998
is Values 80 90 92 83 94 99 92 104

(a) secular trend 13. The profits of a paper bag manufacturing company
(b) cyclical trend (in ` lakh) during each month of a year are
(c) irregular trend
(d) seasonal trend Months Profit

3. Linear trend of a time series indicates towards Jan 1.2
Feb 0.8
(a) constant rate of growth Mar 1.4
(b) constant rate of change Apr 1.6
(c) change in geometric progression May 2.0
(d) None of the above Jun 2.4
Jul 3.6
4. For the given five values 16, 25, 19, 34, 43 the three Aug 4.8
years moving averages are Sep 3.4
Oct 1.8
(a) 22, 18, 25 (b) 20, 26, 32 Nov 0.8
(c) 24, 13, 15 (d) 25, 19, 33 Dec 1.2

5. If the trend line with 2006 as the origin is

y = 13.5 + 1.22x, the trend line with origin 2010 is

(a) y = 18.38 + 1.22x (b) y = 11.23 + 1.25x

(c) y = 21.34 + 1.9 x (d) y = 15.24 + 1.9 x

Case Based MCQs

6. The number of students in a college for the following
years shown as follows use 5-yearly moving average

Year 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 Plot the given data on a graph sheet. Calculate the
Number 332 317 357 392 402 405 510 427 405 438 four monthly moving averages and plot these on the
same graph sheet.
Based on the above information, answer the following
questions 14. The table below gives details of the electricity
generated in million kilowatt hours in each quarter for
(i) The average value for the year 1983 is the years

(a) 360 (b) 340

(c) 372 (d) 385 Quarter

(ii) The average value for the year 1981 is Year 1st 2nd 3rd 4th
8 76 9
(a) 347.10 (b) 352.24 2002 10 77 10
2003 11 78 10
(c) Can not determine (d) 323.52 2004

(iii) The average value for the year 1985 is

(a) 420 (b) 413.2

(c) 425.16 (d) 492.4

(iv) The average value for the year 1987 is Calculate the 4-quarterly moving averages and show
these moving averages on a graph.
(a) 429.8 (b) 434.4

(c) 462.5 (d) 472.2 Answers

(v) The average value for the year 1988 is 1. (a) 2. (c) 3. (b) 4. (b) 5. (a)

(a) 463 (b) 421 6. (i) (a) (ii) (c) (iii) (b) (iv) (a) (v) (d)

(c) 424 (d) 437

Short Answer Type Questions For Detailed Solutions
7. Describes the types of least squares problem. Scan the code
8. What are cyclic variations?

CBSE Term II Applied Mathematics XII 103

CHAPTER 07

Perpetuity, Sinking
Funds, Bonds & EMI

In this Chapter...

l Perpetuity
l Sinking Funds
l Bonds
l Equated Monthly Installment (EMI)

Perpetuity Type II When periodic payment is made at the beginning of
each payment period.
A perpetuity is a type of annuity that lasts forever. In
perpetuity, the stream of cash flows continues for an infinite The present value P of the perpetuity is given by
amount of time.
P = R + R + R +R + ....
1 + i (1 + i) 2 (1 + i) 3
Present Value of Perpetuity
çæ ö÷
Let us consider an annuity whose periodic payment is ` R for R ç 1 ÷ R
infinite periods, interest being r% per period or r = i per Þ P = R + 1+ ÷ = R + i
i ç 1 ÷
100 ç 1 - + ø
period per rupee. è 1 i

Type I When periodic payment is made at the end of each Þ P=R+ R
payment period. i

The present value P of the perpetuity is given by Hence, if money is worth i per period, then the present value

P = 1 R i + (1 R +R + .... of perpetuity of ` R payble at the beginning of each period is
+ + i) 2 (1 + i) 3 R + R.

çæ ÷ö i

Þ P = R i ç 1 ÷ = 1 R = R Present Value of a Growing Perpetuity
1+ ç -1 ÷ + i-1 i
èç 1 ÷ø
1+ i A growing perpetuity is a perpetuity in which periodic
payment do not remain fixed rather these payment keep on
Þ P = R, growing at the same constant rate of growth.
i
A growing perpetuity of `R , which grows at the rate of g per
where R is periodic payments of an annuity. rupee per period, payable at the end of each payment period.
When money is worth i per period is P = i -Rg.
Hence, the present value of ` R payable at the end of each
payment period when money is worth i period is R.

i

104 CBSE Term II Applied Mathematics XII

Sinking Fund (ii) Relative price approach

A sinking fund is a fund created to accumulate money over (iii) Arbitrage-free pricing approach
the years to discharge a future obligation like replacing an old
machine by a new machine, renovation of house, etc. (iv) Stochastic-Calculus approach
An annuity whose periodic payment is `R payable at the end We will study the one main approaches present
of each payment period of n period, interest beings r% per value approach.
period or i = r .
Present Value Approach
100
Then, the amount of obligation which can be discharged Let a bond of face value ` F matures in N yr has a coupon
rate of r% per annum, then each coupon payment
S = R[(1 + i) n - 1] C =F´ r .
i
100
Sinking Fund Payment
If the discount rate is d% per annum, then
The periodic payment of ` R required to accumulate a sum of
` S over n periods with interest charged at the rate of i per i= d
period is given by 100

R = i´S So, the present vale of bond is given by
(1 + i) n - 1
PV = C i + (1 C i) 2 +C + ....+ C +F
Bonds 1+ + (1 + i) 3 (1 + i) N (1 + i) N

A bond is a fixed income instrument that represents a loan = C i é - 1 ù + F
made by an invester to a borrower. 1+ êêë1 (1 + i) N úúû
A bond is characterised by the following terms
1 - 1 (1 + i) N
(i) Maturity Bond maturity is the time when the bond 1+
issuer must repay the original bond value to the bond i é a(1 - r n ) ù
holder. êëêQ using Sn 1 - r ûúú
=
(ii) Maturity date Maturity date is the date on which the
bond will mature and the bond issuer will pay the bond PV = C[1 - (1 + i) -N ] + F(1 + i) -N
holder the face value of the bond. i

(iii) Face value The face value of a bond is the price that the (i) If discount rate < coupon rate, then PV > face value.
issuer pays at the time of maturity, also referred to as (ii) If discount rate = coupon rate, then PV = face value
“Par value”.
(iii) If discount rate > coupon rate, then PV < face value.
(iv) Redemption price The price at which the issuing
company will repurchase the bond from investors before These results show that there is an inverse-relation between
its maturity date. yield and bond price.

(v) Dividend rate The rate at which a bond yields interest is Calculation of Yield To Maturity (YTM)
called the dividend rate or the nominal interest rate.
It is nearly impossible to calculate the yield to maturity using
(vi) Coupon rate A bond’s coupon rate denotes the annual the formula of present value of a bond. However,
interest rate paid by the bond issuer to the bond holder. approximate value of yield to maturity can be computed using
the following formula
(vii) Discount rate or Yield To Maturity (YTM) The rate of
interest used to discount the bond’s cash flows is known Approximate YTM
as discount rate or Yield to Maturity.
Coupon payment + Face value - Present value
Valuation of Bonds N
= Face value + Present value
There are mainly four approaches of valuation of bonds
2
(i) Present value approach
C + F - PV
N
i.e. Approximate YTM = (F + PV)

2

CBSE Term II Applied Mathematics XII 105

Equated Monthly Installment (EMI) where, P = Principal borrowed
i = Rate of interest per month
An equated monthly instalment (EMI) is the fixed amount n = Number of payments.
paid by a borrower every month to clear off the loan taken
from a lender. Amortization of Loans
EMI depends on the following factors
Amortization refers to the process of paying off a loan through
(i) Principal amount a scheduled, pre-determined sequence of equal payments
(ii) The loan tenure that includes
(iii) The interest rate
EMI can be calculated by two methods (i) interest on the outstanding loan and

Flat Rate Method (ii) repayment of part of the loan i.e. principal.

A loan is said to be amortized if each installment is used to
pay interest and the part of the principal.

In flat rate method the principal amount remains same Amortization Formulas
throughout the tenure and the interest is charged on it at a
constant rate throughout the loan tenure. While amortizing a loan, one should know the principal
outstanding at the beginning of any period, interest and
Suppose an amount of ` P is borrowed at flat rate of r per principal amounts in any EMI, total interest paid etc. The
rupee per month for a period of n months. formulas for the same are given below.

Then, Interest = P ´ i´ n, where i = r Principal outstanding at beginning of kth period
100
[(1 + i) n - k + 1 - 1]
\ EMI = Principal + Interest = EMI i(1 + i) n - k + 1
n

Þ EMI = P + P ´ i´ n or EMI(1 - (1 + i) k - (n + 1) )
n i

= Pçæ 1 + (i´ n) ÷ö = Pæèç i + 1 öø÷ Interest paid in kth payment = EMI [(1 + i) n - k + 1 - 1]
è n ø n (1 + i) n - k + 1

Reducing Balance Method or EMI(1 - (1 + i) k - (n + 1) )

In this method, EMI is calculated using the following formula Principal paid in kth payment
= EMI - Interest paid in kth period
EMI = - P ´i -n or P ´ i (1 + i) n
(1 + i) (1 + i) n - 1 Total interest paid = n ´ EMI - P

1

106 CBSE Term II Applied Mathematics XII

Solved Examples

Example 1. What sum of money is needed to invest We know that,

now, so as to get ` 6000 at the beginning of every R = (1 i ´S - 1
month forever, if the money is worth 9% per annum + i)n
compounded monthly?
= 0.0375 ´ 40000
Sol. Given, R = ` 6000, (1 + 0.0375)28 - 1

r = 9 % = 0.75 per month = 1500 = 1500
12 (1.0375)28 - 1 ( 2.80 - 1)

and i = 0.75 = 0.0075 = 1500 = 833.33
100 1.80

\ P=R+ R Thus, the required amount is ` 833.33
i
Example 4. A bond that matures in 6 yr has coupon rate
= 6000 + 6000
0.0075 of 12% per annum and has a face value of ` 15000.
Find the fair value of bond, if the yield to maturity
= 6000 + 800000 is 10%.

= ` 806000 Sol. Given, F = ` 15000, r = 12%

Thus, ` 806000 are needed to invest to get ` 6000 at the N = 6 yr and d = 10%
beginning of every month forever.
\ i = 10 = 0.1 é dù
Example 2. Sankar Sharma purchased a number of 100 ëêQ i = 100 úû

stocks of Tata power. At the end of first year, he Now, coupon payment, C = 15000 ´ 12
received a payment of ` 10000, which grows at a 100
rate of 5% per year and continues forever. If the
discount rate is 9%, find the present value of Sankar = ` 1800
Sharma’s investment. \ PV = C[1 - (1 + i)-N ] + F(1 + i)-N

Sol. Given, R = ` 10000, i = 9% and g = 5% i

Let P be the present value of Sankar Sharma’s investment. = 1800[1 - (1 + 0.1)-6 ] + 1800(1 + 0.1)-6
0.1
Then, P= R
i-g = 1800[1 - (1.1)-6 ] + 15000(1.1)-6
0.1
= 10000
(9 - 5) = 1800(1 - 0.56) + 15000( 0.56)
0.1

100 = 18000 ( 0.44) + 8400
= 10000 ´ 100
= 7920 + 8400
4
= 16320
= ` 250000
Hence, the fair value of bond is ` 16320.

Example 3. A company anticipates a capital Example 5. A bond of face value of ` 1200 has a coupon

expenditure of ` 40000 for a new equipment in 7 yr. rate 10% per annum paid semi-annually and
How much should be deposited quarterly in a matures in 6 yr. If the present value of the bond is
sinking fund carrying 15% per annum compounded ` 1560, find the yield to maturity.
quarterly to provide for the purpose?
(given (1.0375)28 = 2.80) Sol. Given, F = ` 1200,

Sol. Let R be deposited quarterly to accumulate ` 40000 in 7 yr. r = 10% = 10 = 5% per half yearly
2

Given, S = ` 40000, n = 7 ´ 4 = 28 yr \ C = ` èæç1200 ´ 5 öø÷ =` 60
100
15
and i = 4 ´ 100 = 0.0375 N = 6 yr = 12 half years

CBSE Term II Applied Mathematics XII 107

Since, PV = ` 1560 Sol. Given, P = ` ( 600000 - 200000) = ` 400000,

C + ( F - PV ) n = 20 ´ 12 = 240
N
Yield to maturity (YTM) = ( F + PV ) and i = 12 = 12 = 0.01
12 ´100 1200
2
P ´i
60+ (1200 - 1560) (i) \ The monthly installment (EMI) = 1 - (1 + i)-n

= 12 400000 ´ 0.01
1200 + 1560 - (1 + 0.01)-240
=
2 1
= 60+ ( - 30) = 60 - 30 = 0.0217
= 1 - 4000
2760 / 2 1380 ( 1.01 ) -240

\Approximate yield to maturity (YTM) = 0.0217 = 4000
- 0.092
or 2% per half year or 4% per annum. 1

Example 6. Neha Suman purchased a house from a = 4000
0.908
company for ` 600000 and made a down payment of
` 200000. She repays the balance in 20 yr by = ` 4405.29
monthly installments at 12% compounded monthly. (ii) \Total interest = n ´ EMI - P
(given (1.01)-240 = 0.092).
= 240 ´ 4405.29 - 400000
(i) What are monthly payments? = 1057269.6 - 400000
(ii) What is the total interest payment? = ` 657269.6

108 CBSE Term II Applied Mathematics XII

Chapter
Practice

PART 1 (a) ` 3914.81 (b) ` 2914.81
Objective Questions (c) ` 3614.81 (d) ` 3814.81

l Multiple Choice Questions 7. A machine costing ` 200000 has effective life of 7 yr
and its scrap value is ` 30000. What amount should
1. The present value of a perpetuity of ` 5000 payable the company put into a sinking fund earning 5% per
annum, so that it can replace the machine after its
at the end of each year, if money is worth 5% useful life? Assume that a new machine will cost
` 300000 after 7 yr (given (1.05)7 = 1.4071)

compounded annually is (a) ` 32169.53 (b) ` 34169.53
(c) ` 33161.38 (d) ` 36169.53
(a) ` 20000 (b) ` 100000

(c) ` 10000 (d) ` 25000 8. A company intends to create a sinking fund to
replace at the end of 20th year assets costing
2. The rate of interest will the present value of a ` 50000. Calculate the amount to be retained out
of profit every year, if the interest rate is 5%
perpetuity of ` 500 payable at the end of every
(given (1.05)20 = 2.6532)
6 months be ` 10000 is

(a) 5% (b) 8%

(c) 10% (d) 4% (a) ` 15122.18 (b) ` 15322.18
(c) ` 15422.18 (d) ` 15022.18
3. The present value of a perpetuity of ` 3120 payable

at the beginning of each year, if money is worth 6% 9. A bond has a face value of ` 1000, matures in 4 yr.
Coupon rate is 4% per annum. The bond makes
effective is annual payments. If the yield to maturity is 4%,
then the fair value of bond is
(a) ` 55120 (b) ` 56120 (given (1.04)-4 = 0.8551)

(c) ` 52120 (d) ` 52000

4. Ashwani holds a perpetual bond that generates an

annual of ` 50000 each year. He believes that the (a) ` 1500 (b) ` 1000
(c) ` 1200 (d) ` 1600
borrower is credit worthy and that an 8% interest

rate will be suitable for this bond. Then, present 10. A bond of face value of ` 1000 matures in 5 yr.
Interest is paid semi-annually and bond is priced to
value of this perpetuity is yield 8% per annum. If the present value of bond is
` 800, then the annual coupon rate is
(a) ` 625000 (b) ` 400000 (given (1.04)-10 = 0.6761)

(c) ` 600000 (d) ` 62500

5. Simran purchased a number of stocks of Amazon.

At the end of first year, she received a payment of (a) 3.06% (b) 4%
(c) 4.2% (d) None of these
` 8000, which grows at a rate of 3% per year and

continues forever. If the discount rate is 7%. Then,

the present value of Simran’s investment is 11. A company ABC Ltd has issued a bond having a
face value of ` 10000 paying annual dividend at 8%.
(a) ` 32000 (b) ` 20000 The bond will be redeemed at par at the end of
10 yr. The purchase price of this bond, if the
(c) ` 200000 (d) ` 40000 investor wishes a yield rate of 8% is
(given (1.08)-10 = 0.4632)
6. A person has set up a sinking fund in order to have
` 100000 after 10 yr for his children’s college (a) ` 8000 (b) ` 12000
education. How much amount should be set aside (c) ` 10000 (d) ` 9000
bi-annually into an account paying 5% per annum
compounded half-yearly? (given (1.025)20 = 1.6386)

CBSE Term II Applied Mathematics XII 109

12. A sinking fund is created for the redemption of 19. Mr. Malik borrowed ` 500000 from a bank to
debentures of ` 100000 at the end of 25 yr. How purchase a house and decided to repay the loan by
much money should be provided out of profits each equal monthly payment in 10 yr. If bank charges
year for the sinking fund, if the investment can earn interest at 7.5% per annum compounded monthly,
interest 4% per annum (given (1.04)25 = 2.6658) then EMI is (given (1.00625)120 = 2.1121)

(a) ` 2401.25 (b) ` 2501.25 (a) ` 5935 (b) ` 6380
(c) ` 2500 (d) ` 2700 (c) ` 7340 (d) ` 8520

13. A bond of face value ` 1000 has a coupon rate of l Case Based MCQs

10% per annum paid semi-annually and mature in

4 yr, if the present value of the bond is ` 1140, then 20. Aakriti invest in bond of face value of ` 1000 has a
coupon rate of 8% per annum with interest paid
the yield to maturity is semi-annually and mature in 5 yr.

(a) 5% (b) 6% (c) 4% (d) 8%

14. A bond of face value ` 1000 has a coupon rate of Based on above information, answer the following
6% per annum with interest paid semi-annually and questions.
matures in 5 yr. If the bond is priced to yield 8%
per annum, then the value of the bond is (i) If discount rate is 6% per annum, then present
(given (1.04)-10 = 0.6761) value of bond is

(a) ` 919.03 (b) ` 802.03 (a) greater than 1000
(c) ` 719.03 (d) ` 1019.03 (b) less than 1000
(c) equal to 1000
15. A bond of face value of ` 500 mature in 3 yr. (d) None of the above

Interest is paid half-yearly and bond is price to (ii) If discount rate is 8% per annum, then present

yield 8% annually. If the present value of bond is value of bond is

` 450, then annual coupon rate is (a) ` 1000 (b) ` 900

(a) less than 10% (b) equal to 10% (c) ` 1200 (d) ` 1100

(c) more than 10% (d) None of these (iii) If discount rate is 10% per annum, then present

16. Manish takes a loan of ` 300000 at an interest of value of bond is

10% compounded annually for a period of 3 yr, then (a) greater than 1000 (b) less than 1000

EMI by using flat rate method is (c) equal to 1000 (d) None of these

(a) ` 1083.33 (b) ` 1073.33 (iv) If the present value of bond is ` 1000, then

(c) ` 1093.33 (d) ` 1063.33 discount rate is

17. Mr. Ahuja borrowed ` 100000 from a bank to (a) 6% (b) 8%

purchase a car and decided to repay the loan by (c) 10% (d) 5%

equal monthly installments in 10 yr. If bank charges (v) If the present value of bond is ` 1100, then

interest at 9% per annum compounded monthly, discount rate is

then the value of EMI is (a) less than 8% (b) greater than 8%
(given (1.0075)120 = 2.4514)
(c) equal to 8% (d) None of these

(a) ` 1250 21. In year 2018, Mr. Verma took a loan of ` 250000 at
(b) ` 1266.74 the interest of 6% per annum compounded monthly
(c) ` 1300 is to be amortized by equal payment at the end of
(d) None of the above each of 5 yr.

18. A person buys a house for which he agrees to pay [given (1.005)60 = 1.3489 and (1.005)21 = 1.1104]
` 5000 at the end of each month for 8 yr. If money
is worth 12% converted monthly. Then, the cash Based on the above information, answer the
price of house is (given (1.01)-96 = 0.3847) following questions

(i) The number of payment is

(a) ` 307650 (b) ` 407650 (a) 10 (b) 12
(c) ` 507650 (d) None of these
(c) 30 (d) 60

110 CBSE Term II Applied Mathematics XII

(ii) The EMI is calculated using formula 7. A firm anticipates a capital expenditure of ` 80000
for a new equipment in 5 yr. How much should be
(a) EMI = Pi(1 + i)-n (b) EMI = P(1 + i)-n
(1 + i)-n - 1 i deposited quarterly in sinking fund carrying 12%
per annum compounded quarterly to provide for
(c) EMI = Pi(1 + i)n (d) EMI = Pi(1 + i)n the purchase? (given (1.03)20 = 1.8061)
(1 + i)n -1 i

(iii) The size of each monthly payment is 8. A company establish a sinking fund to provide for
payment of ` 250000 debt, maturing in 4 yr.
(a) ` 4832.69 (b) ` 4932.69 Contributions to the fund are made at the end of
every year. Find the amount of each annual deposit,
(c) ` 5000 (d) ` 4700 if the interest is 18% per annum.
(given (1.18)4 = 1.9387).
(iv) The interest paid in 40th payment is

(a) ` 500 (b) ` 600

(c) ` 480.48 (d) `490.48 9. A bond has face value of ` 10000 and maturity
period of 10 yr. The nominal interest rate is 6% per
(v) The total interest paid by Mr. Verma is annum. What should be the price of the bond to
yield an effective interest of 8%?
(a) ` 40000 (b) ` 50000 (given (1.08)-10 = 0.4631)

(c) ` 39961.40 (d) ` 40961.40

PART 2 10. A bond of face value of ` 1000 matures in 5 yr.
Interest is paid semi-annually and bond is priced to
Subjective Questions yield 8% per annum. If the present value of bond is
` 1100, find the annual coupon rate.
l Short Answer Type Questions (given (1.04)-10 = 0.6761)

1. Find the present value of a sequence of payments 11. Face value of bond is ` 1000, coupon rate 4.25% per
of ` 2000 made at the end of every 6 months and annum paid semi-annually and matures in 10 yr. If
continuing forever. If money is worth 5% per present value of bond is ` 918.23. What is the yield
annum compounded semi-annually. to maturity.

2. At what rate of interest will the present value of 12. A coupon bond has a ` 1000 face value and
` 300 payable at the end of each quarter be provides 10.5% semi-annual coupon for 14 yr. The
` 24000? discount rate is 9% per annum. What is the value of
the coupon bond. (given (1.045)-28 = 0.2915)
3. At 6% converted quarterly, find the present value of
a perpetuity of ` 45000 payable at the end of each 13. Mr. Taneja purchase a house of ` 4500000 with a
quarter.
down payment of ` 500000 and balance in EMI for
4. What sum of money is needed to invest now, as to
get ` 10000 at the beginning of every month 25 yr. If bank charges 6% per annum compounded
forever, if the money is worth 8% per annum
compounded quarterly? monthly. Calculate the EMI.
(given (1.005)300 = 4.4650).
5. The present value of a perpetual income of ` R at
the end of each 6 months is ` 42000. Find the value 14. A person buys a house for which be agrees to pay
of R, if money is worth 6% compounded ` 25000 at the end of each month for 8 yr. If money
semi-annually. is worth 12% converted monthly, what is the cash
price of house? (given (1.01)-96 = 0.3847)
6. A company intends to create a sinking fund to
replace at the end of 20th year assets costing 15. Mr. Kailash has taken a personal loan of ` 200000
` 500000. Calculate the amount to be retained out
of profits every year, if the interest rate is 5% per for 2 yr at an interest rate of 20% per annum, which
annum. (given (1.05)20 = 2.6532)
is to be paid back in equal monthly installments.

How much monthly installment Mr. Kailash will
÷ö-24
pay? æç given çæ 61 ø = 0.6725 øö÷÷.
çè è 60

CBSE Term II Applied Mathematics XII 111

l Long Answer Type Questions 18. Find the purchase price of a ` 20000 bond,
redeemable at the end of 10 yr at 110 and paying
16. A machine costs a company ` 52000 and its annual dividends at 4%, if the yield rate is to be 5%
effective life is estimated to be 25 yr. A sinking fund effective. (given (1.05)-10 = 0.6139)
is created for replacing the machine by a new
model at the end of its life time, when its scrap 19. Rajesh purchased a house from a company for
realizes a sum of ` 2500 only. The price of the new ` 2500000 and made a down payment of ` 500000.
model is estimated to be 25% more than the price He repays the balance in 25 yr by monthly
of present one. Find what amount should be set instalments at the rate of 9% per annum
aside at the end of each year out of the profit for the compounded monthly
sinking fund, if it accumulates at 3.5% compound
annually? (given (1.035)25 = 2.3632) (i) What are the monthly payment?

17. A machine is bought for ` 320000. Its effective life (ii) What is the total interest payment?
is 8 yr, after which its salvage value would be (given (1.0075)-300 = 0.1062)
` 25000. It is decided to create a sinking fund to
replace this machine at the end of its effective life 20. A loan of ` 400000 at the interest rate of 6.75% per
by making half yearly payments that will earn an annum compounded monthly is to be amortized by
interest of 8% per annum compounded half yearly. equal payment at the end of each month for 10 yr,
If it is known that the cost of machine increases by find
5% per annum. Calculate the amount of each (i) the size of each monthly payment
payment to the sinking fund.
[given (1.04)16 = 1.8730 and (1.05)8 = 1.4774] (ii) the principal outstanding at the beginning of 61st
month.

[given (1.005625)120 = 1.9603 and
(1.005625)60 = 1.4001]

SOLUTIONS

Objective Questions 5. (c) Here, R = ` 8000, i = 7% and growth of rate ( g) = 3%

1. (b) We know that, \ P = R = ( 8000 = 8000 ´ 100 = 200000
P= R i-g 7 - 3)% 4
i
\ Present value = ` 200000
Here, R = ` 5000 and i = 5%
\ P = 5000 = 5000 ´ 100 = 100000 6. (a) Let ` R be set a side bi-annually for 10 yr in order to have

5% 5 ` 100000 after 10 yr.
\Present value = ` 100000
2. (c) Here, P = ` 10000 and R = ` 500 Then, S = ` 100000, n = 10 ´ 2 = 20 yr

Let rate of interest be r % per annum and i = 5 = 0.025
\ i = çèæ 2r ø÷ö % (compounded semi-annually) 200

\ i= R \ R = (1 i ´S = 0.025 ´ 100000
P + i)n - 1 (1.025)20 - 1

Þ r = 500 = 2500 1
200 10000 1.6386 -

Þ r = 10% = 2500 = ` 3914.81
3. (a) Here, R = ` 3120 and i = 6 0.6386

100 7. (c) Cost of new machine = ` 300000
\ P = R + R = 3120 + 3120 ´ 100 Scrap value of old machine = ` 30000

i6 Hence, the money required for new machine after 7 yr
= 3120 + 52000 = ` 55120
4. (a) Here, R = ` 50000 and i = 8% = 300000 - 30000
\ P = R = 50000 ´ 100 = 625000
= ` 270000
i8
\ Present value = ` 625000 Here, S = ` 270000, n = 7 yr and i = 5% = 5 = 0.05
100

\ R = (1 i ´S = 0.05 ´ 270000
+ i)n - 1 (1.05)7 - 1

= 13500 = 13500 = ` 33161.38
1.4071 - 1 0.4071

112 CBSE Term II Applied Mathematics XII

8. (a) Here, S = ` 50000, i = 5% = 0.05 and n = 20 yr 13. (b) Here, F = ` 1000, r = 10%

\ R = (1 i ´S -1 = 0.05 ´ 50000 r = 10 = 5% = 0.05 (semi-annually)
+ i)n (1.05)20 - 1 2

= 25000 = 25000 and C = 1000 ´ 5 = ` 50
2.6532 - 1 1.6532 100

= ` 15122.18 N = 8 yr
9. (b) Here, F = ` 1000, r = 4%, N = 4 yr and d = 4% PV = ` 1140

Þ i = 4 = 0.04 é dù Approximate YTM
100 êëQ 100 ûú
i = C + çèæ F - PV øö÷ 50 + èæç 1000 - 1140 ø÷ö
N 8
So, Coupon payment, C = 1000 ´ 4 = ` 40 = =
100 F + PV 1000 + 1140

PV = C(1 - (1 + i)-N ) + F(1 + i)-N 2 2
i = 50 - 17.5 = 0.0304

= 40(1 - (1.04)-4 ) + 1000(1.04)-4 1070
0.04
Approximate YTM = 3% half yearly = 6% per annum
= 40(1 - 0.8551) + 1000( 0.8551)
0.04 14. (a) Here, F = ` 1000, r = 6 = 3% (semi- annually)
2
= 40( 0.1449) + 855.10
0.04 N = 5 ´ 2 = 10 yr and d = 8 = 4%
2
= 144.90 + 855.10 = ` 1000
or i = 4 = 0.04 é i = dù
100 ëêQ 100 ûú

10. (a) Let the annual coupon rate is r %. So, C =1000 ´ 3 = ` 30
100
Here, F = ` 1000, then C = 1000 ´ r = 5r
200 \ PV = C(1 - (1 + i)-N + F(1 + i)-N
i
d = 8% or i = 8 = 0.04 (semi-annually)
200 = 30(1 - (1.04)-10 ) + 1000 ´ (1.04)-10
0.04
N = 5 ´ 2 = 10 yr and PV = ` 800
= 30(1 - 0.6761) + 1000 ´ 0.6761
800 = 5r (1 - (1.04)-10 ) + 1000(1.04)-10 0.04
0.04
= 242.93 + 676.10 = ` 919.03
800 = 5r (1 - 0.6761) + 1000( 0.6761)
0.04 15. (a) Let annual coupon rate is r %

Þ 5r ( 0.3239) = 800 - 676.10 = 123.90 Here, F = ` 500, then C = 500 ´ r = 5 r
0.04 200 2

r = 123.90 ´ 0.04 = 3.06% i = 8 = 4% = 0.04, N = 3 ´ 2 = 6 yr and PV = ` 450
5 ´ ( 0.3239) 2

11. (c) Here, F = ` 10000, N = 10 yr, i = 8% = 0.08 \ PV = C(1 - (1 + i)-N ) + F(1 + i)-N
i
and C = 8 ´ 10000 = ` 800
100 450 = 5r (1 - (1.04)-6 ) + 500(1.04)-6
2 0.04
PV = C(1 - (1 + i)-N ) + F(1 + i)-N
i 450 = 5r (1 - 0.7903) + 500( 0.7903)
0.08
= 800(1 - (1.08)-10 ) + 10000(1.08)-10
0.08 500( 0.2097)r = 450 - 395.15 = 54.85
8
= 800(1 - 0.4632) + 10000( 0.4632)
0.08 r = 54.85 ´ 8 = 4.18
500 ´ 0.2097
= 10000( 0.5368) + 4632

= 5368 + 4632 = ` 10000 Hence, coupon rate is 4.18%, which is less than 10%.
16. (a) We have, P = Principal = ` 30000
12. (a) Here, S = ` 100000, n = 25 yr and i = 4% = 0.04

\ R = (1 i ´S = 0.04 ´ 100000 = 4000 i = 10 = 1 and n = 12 ´ 3 = 36
+ i)n - 1 (1.04)25 -1 2.6658 - 1 1200 120

= 4000 EMI = Pæèç i + 1 ÷øö = 30000 çèæ 1 + 316 ø÷ö
1.6658 n 120

= ` 2401.25 = 250 + 833.33 = ` 1083.33

CBSE Term II Applied Mathematics XII 113

17. (b) Given, P = ` 100000, i = 9 = 0.0075 i = 6 = 0.005 and n = 60
´ 100 1200
12

and n = 12 ´ 10 = 120 EMI = 250000 ´ 0.005(1.005)60 [from (ii)]
(1.005)60 - 1
So, EMI = P ´ i ´(1 + i)n
(1 + i)n -1 250000 ´ 0.005 ´ 1.3489
= 1.3489 - 1 = ` 4832.69

= 100000 ´ 0.0075 ´ (1.0075)120 (iv) (c) We know that, interest paid in kth payment
(1.0075)120 -1
[(1 + i)n - k + 1
= 750 ´ 2.4514 = ` 1266.74 = EMI (1 + i)n - k + 1 - 1]
2.4514 - 1

18. (a) Here, EMI = ` 5000, n = 12 ´ 8 = 96 and i = 12 = 0.01 So, interest paid in 40th payment
1200
= EMI[(1 + i)60 - 40 +1 - 1]
(1 + i)60 - 40 + 1
\ EMI = P ´i i)-n
1 - (1 +
= 4832.69 ´ ((1.005)21 -1)
5000 = P( 0.01) ( 1.005 ) 21
1 - (1.01)-96
= 4832.69 ´ (1.1104 - 1)
Þ P = 5000(1 - 0.3847) = 50000 ´ 0.6153 1.1104
0.01
= 4832.69 ´ 0.1104 = ` 480.48
= ` 307650 1.1104

19. (a) Here, P = ` 500000, i = 7.5 = 0.00625 (v) (c) Total interest paid by Mr. Verma
1200 \ Total interest = n ´ EMI - P
= 60 ´ ( 4832.69) - 250000
and n = 12 ´ 10 = 120 = 289961.40 - 250000
= ` 39961.40
EMI = P ´ i(1 + i)n
(1 + i)n - 1

EMI = 500000 ´ 0.00625 ´ (1.00625)120 Subjective Questions
(1.00625)120 - 1
1. The given annuity is perpetuity of first type in which
= 3125 ´ 2.1121 = 6600.3125 = ` 5934.99 R = ` 2000 and r = 5 % (semi-annually)
2.1121 - 1 1.1121 2

= ` 5935 (approx.) So, i = 5 % = 5
2 200
20. (i) (a) Discount rate < Coupon rate
Þ Present value > Face value We know that,
Þ Present value > 1000
P = R = 2000 = 2000 ´ 200 = ` 80000
(ii) (a) Discount rate = Coupon rate i 5 / 200 5
Þ Present value = Face value
Þ Present value =1000 Hence, the present value is ` 80000.

(iii) (b) Discount rate > Coupon rate 2. Let the rate of interest be r % per annum.
Þ Present value < Face value
Þ Present value < 1000 Then, i = r (quarterly)
400
(iv) (b) Present value =1000
Here, R = ` 300 and P = ` 24000
Here, Present value = Face value
\ Discount rate = Coupon rate = 8% We know that, P = R Þ i = R
iP
(v) (a) Present value =1100
Þ r = 300
Here, Present value > Face value 400 24000
\ Discount rate < Coupon rate
\ Discount rate is less than 8%. Þ r = 300 ´ 400 = 5%
21. (i) (d) Here, Time = 5 yr 24000

Hence, rate of interest is 5% per annum.

3. Let P be the present value of perpetuity.

Given, R = ` 45000 and i = 6 % = 6
4 400

\ Total number of payment = 5 ´ 12 = 60 \P = R = 45000 = 45000 ´ 400 =` 3000000
i 6 6
(ii) (c) EMI = P ´ i(1 + i)n
(1 + i)n - 1 400

(iii) (a) Given, P = ` 250000, Hence, the present value is ` 3000000.

114 CBSE Term II Applied Mathematics XII

4. Given, R = ` 10000 and r = 8 % = 2% = 0.02 Then, PV = C(1 -(1 + i)-N ) + F(1 + i)-N
4 i
So, i = 0.02
PV = 600 (1 - (1.08)-10 ) + 10000(1.08)-10
\ P = R + R = 10000 + 10000 0.08
i 0.02
= 600 (1 - 0.4631) + 10000( 0.4631)
= 10000 + 500000 0.08

= ` 510000 = 600 ´ 0.5369 + 4631
0.08
Hence, ` 510000 are needed to invest now to get ` 10000 at
the beginning of every quarter forever. = 322.14 + 4631 = 4026.75 + 4631 = ` 8657.75
5. Given, P = ` 42000 and i = 6 = 0.03 0.08

200 Hence, the price of bond is ` 8657.75.
\ R = P ´ i = 42000 ´ 0.03 = ` 1260
Hence, R = ` 1260 10. Let the annual coupon rate is 6%
6. Given, S = ` 500000 and i = 5 = 0.05 and n = 20 yr Given, F = ` 1000, then C = 1000 ´ r = ` 5r
200
100 d = 8% per annum or 4% per half year

We know that,

R = i ´S - 1 i = 4 = 0.04
(1 + i)n 100

= ( 0.05) (500000) = ( 0.05)(500000) N = 5 ´ 2 = 10 yr and PV = ` 1100
(1 + 0.05)20 - 1 (1.05)2 - 1
\ PV = C(1 - (1 + i)-N ) + F(1 + i)-N
= 25000 = 25000 i
2.6532 - 1 1.6532
1100 = 5r (1 - (1.04)-10 ) + 1000(1.04)-10
= ` 15122.18 0.04

Thus, ` 15122.18 are retained out of profits every year for 1100 = 5r (1 - 0.6761) + 1000( 0.6761)
0.04
20 yr to accumulate ` 500000.
1100 = 5r ´ 0.3239 + 676.10
7. Given, S = ` 80000, i = 12 = 0.03 and n = 5 ´ 4 = 20 yr 0.04
400
0.04 (1100 - 676.10) 0.04 ´ 423.90
\ R = i ´S = ( 0.03)( 80000) r = 5 ´ 0.3239 = 5 ´ 0.3239
+ i)n - 1 (1 + 0.03)20 - 1
(1 r = 10.47

= 2400 Hence, coupon rate is 10.47%.
(1.03)20 - 1 11. Given, F = ` 1000, r = 4.25%

= 2400 = 2400 and i = 4.25 % = 4.25 (semi-annually)
1.8061 - 1 0.8061 2 200

= ` 2977.29 So, C = ` 1000 ´ 4.25 = ` 21.25
200
Thus, ` 2977.29 deposited quarterly in a sinking fund.

8. Here, S = ` 250000, n = 4 yr and i = 18 = 0.18 N = 10 ´ 2 = 20 yr and PV = ` 918.23
100
C + ( F - PV )
Now, R = i ´S = ( 0.18) ´( 250000) N
(1 + i)n - 1 (1 + 0.18)4 - 1 Approximate Yield to Maturity (YTM) = F + PV

= 45000 = 45000 = 45000 2
(1.18)4 - 1 1.9387 - 1 0.9387
21.25 + 1000 - 918.23
= ` 47938.64 20
= 1000 + 918.23

Hence, the required amount is ` 47938.64. 2
9. We have, F = Face value of bond = ` 10000
= 21.25 + 4.0885 = 25.3385 = 0.0264
N = Number of period = 10 959.115 959.115
C = 8 = 0.08
Approximate YTM = 2.64% semi-annually = 5.28% per annum
100
C = Coupon payment = Annual dividend ´ 10000 Hence, yield to maturity is 5.28 per annum.
12. Given, F = ` 1000, C = 1000 ´ 10.5 = ` 52.5 and d = 9%
= 6 ´ 10000 = ` 600
100 200
So, i = 9 = 0.045 (semi-annually)
F = Maturity value = Face value = ` 10000
200
Let PV be the price of the bond. and N = 14 ´ 2 = 28 yr

CBSE Term II Applied Mathematics XII 115

PV = C(1 - (1 + i)-N ) + F(1 + i)-N Cost of machine = ` 52000
i
Cost of new machine after increasing the 25% of the cost of
= 52.5 (1 - (1.045)-28) + 1000(1.045)-28 machine,
0.045 i.e. 52000 + 25% of 52000

= 52.5 (1 - 0.2915) + 1000 ( 0.2915) = 52000 çæè1 + 12050÷øö
0.045
= 52000 ´ 5 = ` 65000
= 52.5 ´ 0.7085 + 291.50 4
0.045
Scrap value of the present machine = ` 2500
= 826.58 + 291.50 = ` 1118.08

Hence, the value of coupon bond is ` 1118.08. So, net amount required at the end of 25 yr to purchase the
new model = ` ( 65000 - 2500) = ` 62500
13. Cost of house = ` 4500000
We know that,
Down payment = ` 500000 i ´S
+ i)n
\ Balance = ` ( 4500000 - 500000) = ` 4000000 R = (1 - 1

So, P = ` 4000000, i = 6 = 0.005 Here, S = ` 62500, n = 25 yr and i = 3.5 = 0.035
1200 100

and n = 25 ´ 12 = 300 ( 0.035) ´ ( 62500)
(1 + 0.035)25 - 1
\ EMI = P ´ i(1 + i)n Þ R =
(1 + i)n - 1
2187.5
= 4000000 ´ 0.005 ´ (1.005)300 = (1.035)25 - 1
(1.005)300 - 1
2187.5
= 20000 ´ 4.4650 = 89300 = ` 25772 = 2.3632 - 1
4.4650 - 1 3.4650
= 2187.5 = ` 1604.68
Hence, EMI of Mr. Taneja is ` 25772. 1.3632
14. Given, EMI = ` 25000, n = 12 ´ 8 = 96 and i = 12 = 0.01
Thus, ` 1604.68 are set aside each year out of the profit to
1200 purchase the new model of the machine.

Let P be the cash price of house. 17. Let each semi-annually deposit in the sinking fund of ` R.
Since, the cost of new machine is increases by 5% per
Then, P = EMI(1 -(1 + i)-n) annum the cost of present.
i Cost of machine at present = ` 320000

= 25000 (1 - (1.01)-96 ) Cost of machine after increasing 5% per annum after 8 yr
0.01
320000çèæ1 5 ø÷ö 8
= 25000(1 - 0.3847) 100
0.01 = + = 32000 ( 1.05 ) 8

= 25000 ´ 0.6153 = ` 1538250 = 320000 ´ 1.4774 = ` 472768
0.01 Salvage value of present machine = ` 25000

Hence, the cash price of house is ` 1538250. So, net amount required at the end of 8 yr to purchase the

15. Given, P = ` 200000 n = 12 ´ 2 = 24 and i = 20 = 1 new model is ` ( 472768 - 25000) = ` 447768
1200 60
We know that, R = i ´S
200000 ´ 1 (1 + i)n - 1
P ´i 60
\ EMI = - (1 + i)-n = -24 Here, S = ` 447768, n = 8 ´ 2 = 16 yr and i = 8 = 0.04
1 æèç1 1 ø÷ö 200
1 - + 60

= 200000 = 200000 \ R = ( 0.04) ´( 447768)
60(1 - 0.6725) (1 + 0.04)16 - 1
60çæèç1 -24 ÷ö÷ø
æç 61 ÷ö 17910.72
- è 60 ø = (1.04)16 - 1

= 200000 =` 10178 = 17910.72
60 ´ 0.3275 1.8730 - 1

Hence, EMI of Mr. Kailash is ` 10178. = 17910.72 = ` 20516.28
0.8730
16. Let ` R be the amount set aside each year.
Since, the cost of new machine is 25% more than the cost of Thus, ` 20516.28 deposited half yearly out of the profit to
present. purchase the new model of the machine.

116 CBSE Term II Applied Mathematics XII

18. We have, (ii) We have, EMI = ` 16782.27

n = Number of period = 10, i = yield rate = 5 = 0.05 n = 300 and P = 2000000
100 \ Total interest = n ´EMI - P

R = Annual dividend = 4% of face value = ` çæè 4 ´ 20000øö÷ = 300 ´ 16782.27 - 2000000
100 = 5034681 - 2000000
= ` 3034681
= ` 800

The bond is redeemed at 110. \Total interest = ` 3034681.

Therefore, the redemption price of the bond is 110% of its 20. Given, P = ` 400000, n = 12 ´ 10 = 120

face value. and i = 6.75 = 0.005625
1200
Thus, C = Redemption price =` èçæ 20000 ´ 110 ÷öø =` 22000
100
(i) Size of monthly payment

Let V be the purchase price of the bond. Then, EMI = P ´ i(1 + i)n
(1 + i)n - 1
ìí1 - (1 + i) -n ü i)-n
î i ý
V = R þ + C(1 +

ÞV =` é 800ìí1 - (1 + 0.05 ) -10 ü + 22000(1 + 0.05 ) -10 ù = 400000 ´ 0.005625 (1 + 0.005625)120
ê ý ú (1 + 0.005625)120 - 1

ëê î 0.05 þ úû = 2250 ´ 1.9603
1.9603 - 1
= ` [16000{1 - (1.05)-10} + 22000(1.05)-10 ]

= ` [16000(1 - 0.6139) + 22000( 0.6139)] = 4410.675 = ` 4593
= ` [16000( 0.3861) + 13505.80] 0.9603
= ` ( 6177.60 + 13505.80)
= ` 19683.40 \ Monthly payment = ` 4593
\The purchase price of bond is ` 19683.40
19. Cost of house = ` 2500000 (ii) We know that,
Down payment = ` 500000
\ Principal amount = ` ( 2500000 - 500000) = ` 2000000, Principal outstanding at beginning of kth period

n = 25 ´ 12 = 300 = EMI[(1 + i) n- k+1 - 1 ]
and i = 9 = 0.0075 i(1 + i) n- k+1

1200 So, principal outstanding at beginning of 61st month

= EMI ((1 + i)120 - 61 + 1 - 1)
i(1 + i)120 - 61 + 1

(i) We know that, = EMI ((1 + i)60 - 1)
i(1 + i)60
P ´i
EMI = 1 - (1 + i)-n = 4593 ((1.005625)60 - 1)
0.005625 ( 1.005625 ) 60
= 2000000 ´ 0.0075 = 15000
1 - (1 + 0.0075)-300 1 - (1.0075)-300 = 4593 (1.4001 - 1)
0.005625 ( 1.4001 )
= 15000
1 - ( 0.1062) = 4593 ´ 0.4001 =
0.005625 ´ 1.4001 ` 233336.89

= 15000 = ` 16782.27 Principal outstanding at beginning of 61st Month is
0.8938 ` 233336.89

\Monthly payment = ` 16782.27

Chapter Test (iv) The principal paid by Mr. Jain in 50th month

(a) ` 91053 (b) ` 16753

(c) ` 14273 (d) ` 15273

(v) Total interest paid by Mr. Jain

Multiple Choice Questions (a) ` 840321 (b) ` 905310

(c) ` 105316 (d) None of these

1. A fund which is created to accumulate money over the [given (1.01)-96 = 0.3847 and (1.01)-47 = 0.6265]
years to discharge a future obligation is called ……… .
Short Answer Type Questions
(a) Perpetuity (b) Sinking fund
(c) Bonds (d) EMI 9. Find the present value of a perpetuity of ` 1200
payable at the end of each year, if money worth is 4%
2. The present value of a perpetual of ` R at the end of per annum.
each of the 6 months is ` 144000 at 6% compounded
semi-annually, then the value of R is 10. What sum of money is needed to invest now, so as to
get ` 6000 at the beginning of every month, of the
(a) ` 4000 (b) ` 4200 money is worth 8% per annum compounded monthly.
(c) ` 4320 (d) ` 4230
11. A firm anticipates a capital expenditure of ` 10000 for a
3. If the discount rate is greater than coupon rate, then new equipment in 5 yr. How much should be
the present value is deposited quarterly in a sinking fund earning 10% per
annum compounded quarterly to provide for the
(a) greater than face value (b) less than face value purchase? (given (1.025)20 = 1.6836).
(c) equal to face value (d) None of these

4. A bond of the face value ` 1000 in 7 yr, coupon rate is 12. A bond that matures in 5 yr has coupon rate of 10%
8% paid semi-annually. If the discount rate is 8% per per annum and has a face value of ` 10000. Find the
annum, then the present value of the bond is fair value of bond, if yield to maturity is 8%.
(given (1.08)-5 = 0.6808)
(a) ` 950 (b) ` 1048
(c) ` 1000 (d) ` 1150 13. Mrs. Sharma took a housing loan of ` 800000 to be
paid in 10 yr by equal monthly installments. The
5. The present value of a perpetuity of ` 500 payable at interest charged is 10.5% per annum compounded
the end of each quarter, if money is worth 8% per monthly. Find her monthly installment.
annum is (given (1.00875)-120 = 0.3515)

(a) ` 25000 (b) ` 20000
(c) ` 30000 (d) None of these

6. The present value of a perpetuity of ` 7800 payable at Long Answer Type Questions
the beginning of each year. If money is worth 6%
effective is 14. A machine costs a company ` 52000 and its effective
life is estimated to be 12 yr. A sinking fund is created
(a) ` 138800 (b) ` 137800 for replacing the machine by a new model at the end
(c) ` 136800 (d) ` 135800 of its life time, when its scrap realizes a sum of ` 5000
only. The price of new model is estimated to be 25%
7. At what rate of interest will the present value of a higher than the price of the present one. Find what
amount should be set aside at the end of each year,
perpetuity of ` 500 payable at the end of each quarter out of the profit, for the sinking fund, if it accumulates
at 10% effective. (given (1.1)12 = 3.1384)
be ` 40000?
15. A person amortizes a loan of ` 150000 for a new home
(a) 1.25% per annum (b) 2.5% per annum by obtaining a 10 yr mortgage at the rate of 9% per
annum compounded monthly. Find
(c) 5% per annum (d) 6% per annum
(i) the monthly payment (ii) the total interest paid
Case Based MCQs (given (1.0075)-120 = 0.4079)

8. In year 2020, Mr. Jain amortizes a loan of ` 1500000 for
renovation of his house by 8 yr mortgage at rate of
12% per annum compounded monthly.

Based on the above information, answer the following
questions.

(i) The equated monthly installment paid by Mr. Jain was Answers

(a) ` 24378.35 (b) ` 30379.10 1. (b) 2. (c) 3. (b) 4. (c) 5. (a) 6. (b) 7. (c)
8. (i) (a) (ii) (b) (iii) (c) (iv) (d) (v) (a) 9. ` 30000 10. ` 906000
(c) ` 35691.10 (d) None of these 11. ` 365.71 12. ` 10798 13. ` 10794.14
14. ` 2805.85 15. (i) ` 1900 (ii) ` 78000
(ii) The interest paid by Mr. Jain in 50th month
For Detailed Solutions
(a) ` 11000 (b) ` 9105.31 Scan the code

(c) ` 10055.30 (d) ` 12135.40

(iii) The principal outstanding at the beginning of 50th

month

(a) ` 110541 (b) ` 10341

(c) ` 910531 (d) ` 1213540

118 CBSE Term II Applied Mathematics XII

CHAPTER 08

Stocks, Shares
and Debentures

In this Chapter...

l Shares
l Dividend
l Debentures
l Stocks

Shares (vi) At discount When shares are issued at a value which is
lower than the face value of shares, the shares are said to
A company’s capital is divided into small equal units of a have been issued at discount, i.e. issue price is less than
finite number. Each unit is known as a share. In simple face value, e.g. share with a face value of ` 10 is issued at
terms, a share is a percentage of ownership in a company or a ` 8, here ` 2 being discount.
financial asset. e.g. If the market capitalization of a company
is ` 10 lakh, and a single share is priced ` 10, then the Important Formulae Related to Share
number of shares to be issued will be 1 lakh.
(i) Total face value = Number of shares ´ Nominal value
Some Basic Definitions of one share

(i) Nominal/Face value The original value of a share is (ii) Money invested = Number of shares ´ Market value
called its nominal (printed or face) value, which is fixed of one share
by the company.
Dividend
(ii) Shareholder An individual who purchases/possesses the
share(s) of the company is called a shareholder of the Dividend is sharing of profit of a company with its
company. shareholder. Company usually share a part of their profit with
shareholder in the form of dividend which is paid on a per
(iii) Market value The price of a share at any time in the share basis.
market is called market value of the share. The market
value of a share is not fixed, it varies from time-to-time. 1. The dividend is always expressed as the percentage of
the face value of the share, which is called rate of
(iv) At par When shares are issued at their face value, the dividend.
shares are said to have been issued at par, e.g. shares i.e. Dividend on one share = d ´ Nominal value,
with a face value of ` 10 is issued at ` 10. 100

(v) At premium When shares are issued at a value which is where d is the rate of dividend.
higher than the face value of shares, the shares are said to
have been issued at premium, i.e. issue price is more 2. The dividend is always given (by the company) on the
than face value, e.g. share with a face value of ` 10 is face value of the share irrespective of the market value of
issued at ` 12, here ` 2 being premium. the share.

CBSE Term II Applied Mathematics XII 119

Types of Shares payments at fixed periods until that time. The money paid to
company or government for buying such bond is called stock.
There are two types of shares However, even before redemption date, the stock can be sold
and purchased in open market and the rate varies, just like
(i) Preferred Shares shares. Note that stock can be bought or sold in fraction,
unlike shares.
Preferred shares are shares in a company that are owned by
people who have the right to receive part of the company’s Types of Stocks
profit before the holders of ordinary shares are paid.
Preferred share holder are paid dividend only, if the company There are mainly two types of stocks,
has profit after paying working expenses and taxes.
(i) Common Stock
(ii) Common Shares or Ordinary Shares
Common stock holder have to right of voting in shareholder’s
Common or ordinary shares are those shares which are paid meeting. They receive dividends from the company on
dividend only when profits are left after preferred share regular intervals.
holders have been paid dividend at specified rate.
(ii) Preferred Stock
Debentures
Preferred stock holders don’t have voting right. However,
A company may borrow money from the public/shareholders they receive dividends from the company in preference of
by issuing debentures. They promise a fixed rate of interest common stock holders.
for a fixed period. The main difference between stock/share
and debentures is that debentures given a fixed return. Brokerage
Whether the company in losses or profit. Also note that
shares form a part of the capital of the company whereas Stocks and shares are sold and purchased in share market
debentures are debt taken by company. through stockbrokers and their charges are called broker’s
commission or brokerage.Brokerage is usually calculated as
Stock percentage of face value (not market value or sale/purchase
value), unless given otherwise.
Join stock companies or the government can also raise loan
from the market by issuing bond or promisory notes. They Note
promise to pay a fixed amount on a future date and interest
(i) Brokerage is added to market value while purchasing stock/shares.
(ii) Brokerage is deducted from market value while selling.

120 CBSE Term II Applied Mathematics XII

Solved Examples

Example 1. Find, what a buyer would have to pay for Now, total dividend paid to 6% preference shareholders
= ` ( 2 lac ´ 6)
800 shares of ` 12 each quoted at ` 65. What should = ` 12 lac
be the gain to the share-holder, if he had purchased
the share at par? \ Dividend paid to ordinary shareholders
= ` (15 lac - 12 lac)
Sol. Given, face value of a share = ` 12 = ` 3 lac
Market value of share = ` 65
Amount paid by the buyer for 800 shares = ` ( 65 ´ 800) \ Dividend paid to ordinary shareholders = 3 lac %
= ` 52000 2 lac
Since, gain from one share = ` ( 65 - 10) = ` 55
\ Gain from 500 shares = ` (55 ´ 500) = 1.5%
= ` 27500
Hence, the dividend received by a person holding 200
Example 2. A man buys ` 10 shares at a premium of preference shares and 300 ordinary shares

` 6 per share. If the company pays 8% dividend, = ` ( 200 ´ 6 + 300 ´ 1.5)
= ` (1200 + 450)
then find the percentage return on his investment = ` 1650

in buying 400 shares. Example 4. A man has 25 debentures of a company

Sol. Given, face value of one share = `10 and receives an interest of ` 40 per annum quarter.
If the return on his investment is 10% per annum,
and premium = ` 6 find the per value of debenture.

Investment on one share = ` (10 + 6) = ` 16

\Investment in buying 400 shares = ` ( 400 ´ 16) Sol. Given, man earns ` 40 per quarter from 25 debentures.

= ` 6400 \ Man annual income from 25 debentures = ` 160

Also, given rate of dividend = 8% Let par value of debenture is ` x.

Now, annual income on 1 share = 8% of ` 10 Since, income from 25 debentures giving 10% annual return

= æèç 8 ´ 10ø÷ö = 4 on investment.
100 5
` ` Þ æçè 25 ´ 10 ´ 10x0÷øö =`5x
2
\ Annual income on 400 shares = ` èæç 400 ´ 54÷öø
Since, we have
= ` 320 5 x = 160
2
Now, percentage return on his investment
Þ 5x = 320
= èæç Annual income ´ 100ø÷ö % Þ x = 64
Total investment
Hence, the par value of a debenture is ` 64.
æèç 320 ´ 100÷øö %
= 6400 Example 5. A man invested ` 14400 in ` 100 shares of a

= 5% company at 20% premium. If the company declares
5% dividend at the end of the year, then how much
Example 3. A company has a total capital of ` 4 crores does he got.

divided equally into preference shares of 6% and Sol. Market value of share = ` 120
ordinary shares, each of face value ` 100. It gives an
annual dividend of ` 15 lac. If a person holds ` 200 Total number of shares = 14400 = 120
preference shares and 300 ordinary shares, how 120
much dividend does he receive?
Now, income of one share = ` 5
Sol. Given, total capital = ` 4 crores, \ Income on 120 shares = `(120 ´ 5)

The number of preference shares = ` 600
= Number of ordinary shares
= 2 crores = 2 lac Example 6. A man buys ` 25 shares in a company which
100
pay 9% dividend. The money invested in such that,
and the number of ordinary shares is 2 lac. if gives 10% an investment. Find the price at which
he bought the shares.

CBSE Term II Applied Mathematics XII 121

Sol. Let cost of one share = ` x Example 9. Rohan holds 50 shares of par value ` 600

Face value of share = ` 25 each of a company which pays dividend at the rate

Dividend on one share = 9% of ` 25 = ` çèæ 9 ´ 25 øö÷ of 5% per annum. When the price of a share rises
100
to ` 784, the shares are sold an he invests half the

= ` 2.25 sale proceed in 7% stock at ` 98 and the other

It is given that 10% profit on the market price half in 8% debenture at ` 100. Find the change in
\ 10x = 2.25
his income.
100
Þ x = ` (10 ´ 2.25) Þ x = ` 22.50 Sol. We have, face value of a share = ` 600

Total value of 50 shares = `(50 ´ 600)

Example 7. ` 100 share of a company is quoted ` 750 in = ` 30000

the market. How much does Ankit pay to purchase Rate of dividend = 5%
600 such shares, if the brokerage paid by him 1%?
What is his gain percent on this investment, if the Dividend on ` 30000 = ` èçæ 5 ´ 30000ö÷ø
company pays a dividend of 20%? 100

= ` 1500

Sol. Given, the brokerage is paid on the market value of a share. Since, each share sold of ` 784.
The market value of a share is ` 750.
\Total sale proceeds of 50 shares = `(50 ´ 784) = ` 39200

\Brokerage on one share = 1% of ` 750 It is given that half of the sale proceeds in 7% stock at ` 98
and other half in 8% debenture at `100.
= ` èçæ 1 ´ 750öø÷
100 \Income on investment in 7% stock

= ` 7.5 = ` æèç 7 ´ 392200 øö÷ = ` æçè 7 ´ 19600øö÷ = ` 1400
98 98
\ Total brokerage on 600 shares = ` ( 600 ´ 7.5)

= ` 4500 Income on investment in 8% stock

Dividend on one share = 20% of ` 100 = ` çæè 20 ´ 100÷öø = ` 20 = ` èæç 8 ´ 392200 öø÷ = ` çèæ 8 ´ 19600÷øö = ` 1568
100 100 100

Dividend on 600 shares = ` ( 600 ´ 20) \Total income = ` (1400 + 1568) = ` 2968

= ` 12000 \ Change in income = ` ( 2968 - 1500) = ` 1468

Total investment = ` ( 600 ´ 750 + 4500) Example 10. Amit invested ` 4444 in the shares of face

= ` 454500 value ` 100 each of a company. At the end of the
year. The company declared dividend at 15% which
\ Gain percentage = çèæ 12000 ÷øö % ´ 100 = 2. 64% gave him an income of ` 600. At what price was the
454500 share quoted, if the brokerage was 1%?

Example 8. Find the money invested in 4 1 % stock at Sol. Let the share was quoted to ` x.

2

` 98 to produce the some annual income as would Let the total number of share bought be y.
be obtained by investing ` 7395 in 6 3 % stock
Face value of a share = ` 100
4
Dividend on 1 share = 15% of ` 100 = ` 15

at ` 105. \Dividend on y shares = ` 15y

Sol. By investing ` 105, the income = `6 3 = ` 27 It is given that the income = ` 600
44
\ 15y = 600 Þ y = 600 = 40
\ By investing ` 7395, the income = ` èæç 27 ´1 ´ 7395öø÷ 15
4 105
Brokerage = 1%

= ` 13311 Market value of 40 shares = ` 40x
28 \Brokerage = 1% of 40x = ` 4x

Now, for the income ` 9, investment required = ` 98 10
2
\ Total investment =` æèç 40x + 4x øö÷ = ` 404x
10 10
\ For the income of ` 13311, investment required
28 But the total investment is given to be ` 4444.
\ 404x = 4444
= ` æç 98 ´ 13311 ÷ö
è 9 / 2 28 ø 10
Þ x = 4444 ´ 10 = 110
= ` çæè 98 ´ 2 ´ 13311 ø÷ö
9 28 404

= ` 10353 Hence, the share was quoted is ` 110.

122 CBSE Term II Applied Mathematics XII

Chapter
Practice

PART 1 8. A man purchases 600 shares of face value of ` 40 at
Objective Questions
par. If a dividend of ` 1680 was received at the end

of the year, the rate of dividend is

(a) 5% (b) 6%

l Multiple Choice Questions (c) 7% (d) 8%

9. The money needed to invest ` 24250 of 11% stock

1. If Rakhi invests ` 11200 on ` 40 shares at a at ` 97 is

premium of 25%, then the number of shares she (a) ` 25000 (b) ` 24000

buys is (c) ` 24500 (d) ` 25500

(a) 224 (b) 225 10. The value of stock that can be bought by investing
` 5500 in a stock at 8 2 premium, brokerage being
(c) 280 (d) 264 3
1 1 is
2. The investment in buying 525 shares of `100 each 3

at 12% premium is

(a) ` 52500 (b) ` 46200

(c) ` 58800 (d) ` 53550 (a) ` 6050 (b) ` 5500

3. A company declared an annual dividend of 10%. (c) ` 5250 (d) ` 5000

Then, the annual dividend of Manoj owning 1500 11. The money is obtained by selling ` 30000 of 15%

shares of the company of par values of ` 10 each is stock at 40 premium, brokerage being ` 2 is

(a) ` 1000 (b) ` 1500 (a) ` 41400 (b) ` 45000

(c) ` 1800 (d) ` 2000 (c) ` 42600 (d) ` 42000

4. A man invest ` 24000 on ` 60 shares at a discount of 12. A man has 20 debentures of a company and

20%. If the dividend declared by the company is receives an interest of ` 30 per quarter. If the

10%, then his annual income is return on his investment is 8% per annum. The par

(a) ` 3000 (b) ` 2880 value of a debenture is

(c) ` 1500 (d) ` 1440 (a) ` 80 (b) ` 75 (c) ` 60 (d) ` 50

5. Ravi bought 500 shares of a company quoted at 13. The percent income of buyer on 9% debenture of

` 280.The amount spent by him on this purchase of face value ` 100 available in the market of ` 150 is

the brokerage is 1% is (a) 3% (b) 4% (c) 5% (d) 6%

(a) ` 140000 (b) ` 151000 14. By investing ` 3450 in a 4 1 % stock, a man obtains
2
(c) ` 141400 (d) ` 154000

6. The price of 15% share with face value of ` 100, an income of ` 150. The market price of the stock is

that gives 12% income is (a) ` 110 (b) ` 105

(a) ` 120 (b) ` 125 (c) ` 103.50 (d) ` 107.50

(c) ` 112 (d) ` 115 15. Isha wants to secure an annual income of ` 1500 by

7. The investment required to purchase ` 125000 of investing in 15% debentures of face value of ` 100

8% stock at ` 92 is each and available for ` 104 each. If the brokerage

(a) ` 125000 (b) ` 100000 is 1%, then sum of money he should invest is

(c) ` 120000 (d) ` 115000 (a) ` 10504 (b) ` 10540

(c) ` 15000 (d) ` 19642

CBSE Term II Applied Mathematics XII 123

16. A company declared an annual dividend of 10%. (iii) Income of Anurag from 4% stock is

The annual dividend received by Ravi owning 400 (a) ` 2100 (b) ` 2000

shares of the company having a par value of ` 100 (c) ` 1600 (d) ` 1500

each is (iv) Income of Anurag from 7% stock is

(a) ` 440 (b) ` 4400 (a) ` 1600 (b) ` 2100

(c) ` 4000 (d) ` 3600 (c) ` 2000 (d) ` 1500

17. The annual yield percent on 15% debentures of face (v) Total income of Anurag is

value of ` 100 each and available at 20% discount is (a) ` 3600 (b) ` 3700
(c) ` 3500 (d) ` 3800
(a) 18.75% (b) 18%

(c) 16% (d) 18.50%

l Case Based MCQs PART 2

18. Mr. Gupta purchase 500 shares from ABC Ltd Subjective Questions
Company from the market. The market price of `10
shares of a company is ` 16. l Short Answer Type Questions

Based on above information, answer the following 1. Shyam invested ` 10846 in buying the shares of a
questions. company at ` 17 each. If the face value of the each
share be ` 10 and company paid 15% dividend at
(i) The investment required by Mr. Gupta is the end of the year. Find the earned by him.

(a) ` 5000 (b) ` 8000 2. How much should a man invest in ` 25 shares
(c) ` 58000 (d) ` 6000 selling at ` 38 as to obtain an income of ` 450, if the
dividend declared is ` 5%?
(ii) Annual income of Mr. Gupta from his investment,
if the company declared 7.5% per annum is 3. A man invests ` 15840 in buying shares of face
value ` 24 each and selling at 10% premium. Find
(a) ` 375 (b) ` 500 the number of shares of bought by him.
(c) ` 600 (d) ` 800
4. A man invests ` 8800 in a company paying 6%
(iii) The percentage will be obtain on his investment, dividend, when a share of face value ` 100 is selling
at ` 60 premium. What is the percentage return on
dividend at 7.5% per annum is his investment?

(a) 7.5% (b) 75% 5. A man by ` 80 shares of a company which pays 15%
16 dividend. He buys the shares at such a prize that
his profit is 20% on his investment. At what price
(c) 75 % (d) None of these did he buy the shares.
8
6. A ` 100 share of a company is quoted at ` 899 in the
(iv) The number of shares buy, if he invest ` 7200 is market. How much does Manish pay to purchase
500 such shares, if the brokerage paid by him is
(a) 720 (b) 300 1%? What is the gain percent on this investment, if
(c) 250 (d) 450 the company pays a dividend of 25%?

(v) The number of shares purchased by Mr. Gupta, if 7. Manoj invested a certain sum in 18% debenture of
he invests ` 3248, of the brokerage is 1% face value of ` 100 available at ` 90 and earned an
annual income of ` 8100. Find the amount invested
(a) 150 (b) 200 by him, if the brokerage is 1%.
(c) 100 (d) 300
8. A man has 30 debentures of a company and
19. Anurag invested ` 75000 partly in 4% stock at 90 receives an interest of ` 30 per quarter. If the
and the balance in 7% stock at 130. If the income return on his investment is 8% per annum. Find the
from second investment is ` 500 more than the par value of a debenture.
income from the first investment.

Based on above information, answer the following
questions

(i) Investment by Anurag in 4% stock is

(a) ` 35000 (b) ` 36000
(c) ` 39000 (d) ` 40000

(ii) Investment by Anurag in 7% stock is

(a) ` 39000 (b) ` 36000
(c) ` 40000 (d) ` 45000

124 CBSE Term II Applied Mathematics XII

9. Ram has 200 shares of par value of ` 100 each of a 18. A man invests part of ` 8370 in 9% stock at ` 96
company paying dividend at 6% per annum.He also and the remainder in 12% stock at ` 120. A income
has 50 debentures of par value of ` 100 each, on from each investment is same. Find the amount
which the company pays on annual interest of 10%. invested in each stock.
Find the total income of Ram from investment in the
company. l Long Answer Type Questions

10. What amount of money does Rashmi get on selling 19. A company has a total capital of ` 500000 divided
14% debentures worth ` 16000 at 10% discount, the into 1000 preferred shares of 6% dividend, with
face value of each being ` 100 and brokerage 1.5%? par value of ` 100 each and 4000 ordinary shares of
par value of ` 100 each. The company declares an
11. Find the market value of a 8% stock when a man get annual dividend of ` 40000. Find the dividend
9% return on his investment after paying a tax of received by a person having 100 preferred shares
paise 12 per rupee on his income. and 200 ordinary shares.

12. A man buys ` 3000 stock at ` 90 and sells when its 20. A man invest ` 13500 partly in shares paying 6% at
price rises to ` 92 1. Find the gain. ` 140 and partly in 5% at ` 125. If his total income
4 is ` 560. How much has he invested in each?

13. What amount of 3 1 % stock will produce an annual 21. A person invests ` 1365 in 3% at ` 91. He sells out
2 ` 1000 stock when they have risen to ` 93 1 and
2
income of ` 77 after paying an income tax of 8 1 the remainder when they have fallen to ` 85. How
3 much did he lose or gain by this transaction?

paise a rupee. 22. Ravi and Rakhi invest ` 12000 each in buying
shares of two companies. Ravi buys 15% ` 100
14. Ravish invests ` 20500 in 4% stock at 125 while Ravi shares at a discount of ` 20, while Rakhi buys ` 25
invest ` 24600 in another stock at 120. If the ratio of shares at a premium of 20%. If both receive equal
the incomes of Ravish to that of Ravi is 4 : 5, at what dividends at the end of the year. Find the rate
rate has the annual dividend been paid to Ravi? percent of the dividend declared by Rakhi’s
company.
15. Which is better investment 4% stock at ` 90 with
income tax 10% or 5% stock at ` 110 without income 23. A man invests ` 25000 partly in 8% stock at ` 89.75
tax? and the remaining in 6 1 % stock at ` 129.75. If
2
16. A man invested a certain sum in 7 1 % stock at ` 103 after paying 10% income tax. His net annual
2 income is ` 1755, how much stock of each kind
does he hold? If brokerage is 0.25%.
and sold when the price rose to ` 107 gaining
there by a sum of ` 36. What sum did he invest?
How much stock did he buy?

17. Which is better investment; 16%, ` 100 shares at
` 132 or 18% ` 100 shares at ` 140.

CBSE Term II Applied Mathematics XII 125

SOLUTIONS

Objective Questions Þ By investing ` 24250, stock purchase on = ` çæè 100 ´ 24250÷øö
97
1. (a) Market value of shares = ` æçè 40 + 25 ´ 40÷øö
100 = ` 25000

= ` ( 40 + 10) = ` 50 10. (d) At 8 2 premium means that for a ` 100 stock, investment
\ Total number of shares = 11200 = 224 3

50 will be ` 108 2.
3
2. (c) Market value of shares = 100 + 12% of ` 100
Since, the brokerage is added to the cost of stock.
= ` (100 + 12) = ` 112
Investment = Number of shares ´ Market values of shares, \ Investment of ` 100 stock = ` èçæ108 + 2 + 4 öø÷ = ` 110
3 3
= ` (525 ´ 112) = ` 58800
Þ Investment of ` 1 stock = ` 110
3. (b) Total investment = Number of shares 100
´ Market values of shares
Þ Investment of ` 5500 = ` çèæ 100 ´ 5500 öø÷
= ` (1500 ´ 10) 110
= ` 15000 (market value at par)
= ` 5000
Rate of dividend = 10%
11. (a) At 40 premium means the cost of `100 stock is ` 140.
çæè15000 10 ø÷ö
Total dividend = ` ´ 100 =` 1500 Since, the brokerage is subtracted from the selling price.

4. (a) Market value of shares = 60 - 20% of ` 60 \Net selling price of `100 stock = ` (140 - 2)

èæç 60 20 60÷öø = ` 138
100
= ` - ´ Þ SP of ` 30000 stock = ` çèæ 138 ´ 30000öø÷
100

= ` ( 60 - 12) = ` 48 = ` 41400

Total number of shares = 24000 = 500 12. (b) A man earns ` 30 per quarter, his annual income from
48 20 debentures = ` 120

Dividend on one shares = ` æçè 10 ´ 60ö÷ø = `6 Now, assume that par value of debenture is `x.
100
\Income from 20 debentures giving 8% annual return on

Dividend on 500 shares = ` (500 ´ 6) = `3000 investment =` æçè 20 ´8 ´ 10x0÷øö = ` 8x
5
5. (c) Market value of 500 shares = ` (500 ´ 280) = `140000

Brokerage paid by Ravi = 1% of 140000 = ` 1400 By given condition, 8x = 120
Total amount spent by Ravi = ` (140000 + 1400) = ` 141400 5
6. (b) Let the market value of each shares = `x
Þ x = 120 ´ 5 = 75
The given statement means that divided on a share of ` 100 8
is ` 15.
Q Income = `15 Hence, the par value of a debenture is ` 75.

This is 12% of the investment of `x. 13. (d) Here, market value of a debenture = ` 150
\ 12x = 15 Þ x = 1500 = ` 125
\Income on ` 150 is ` 9
100 12
7. (d) The market value of ` 100 stock = ` 92 Þ Income on ` 100 = ` çèæ 9 ´ 100öø÷ = ` 6
150

Thus, there is 6% income on the debentures.

Market value of ` 125000 stock 14. (c) Let the market value of 41% stock be `x.
2
= èçæ 92 ´ 125000øö÷ =
` 100 ` 115000 Then, investment of ` x yields income = ` 41 = ` 9
22
8. (c) Total investment = ( 600 ´ 40) = ` 24000
= èçæ 9 ´ 3450÷øö
Let the rate of dividend = r % Investment of ` 3450 will yield income ` 2x
\ r % of 24000 = 1680
= ` 15525
r = 1680 ´ 100 = 7 x
2400
\ 15525 = 150
Hence, rate of dividend is 7% x
9. (a) By investing ` 97, stock purchased = ` 100
Þ x = 15525 = 103.50
\By investing ` 1, stock purchased = ` 100 150
97
Hence, the market value of the stock is ` 103.50.

126 CBSE Term II Applied Mathematics XII

15. (a) Amount invested = Market value of debenture In 7% stock at 130 yields income = ` 7
+ brokerage
Þ Investment of `( 75000 - x) will yield income
= ` (104 + 1% of 104)
= ` çèæ 7 ´ ( 75000 - x)÷öø
= ` (104 + 1.04) = ` 105.04 130

Income obtained = Interest on ` 100 = ` 15 Since, the income from second investment is ` 500 more

Thus, for a annual income of ` 18. than the income from first investment.

Amount invested = ` 105.04 \ 7 ( 75000 - x) - 4x = 500
130 90
\For an income of ` 1500,

Investment mode = ` æèç 105.04 ´ 1500ø÷ö = ` 10504 Þ 52500 - 7x - 4x = 500
15 13 130 90

16. (c) We have, Þ 7x + 4x = 52500 - 500
Total par value of 400 shares = ` ( 400 ´ 100) = ` 40000 130 90 13
Rate of dividend = 10% per annum
\Total dividend of Ravi = ` æèç 40000 ´ 11000÷øö = `4000 Þ 63x + 52x = 52500 - 6500
1170 13

17. (a) It is given that, Þ 115x = 46000 Þ x = 36000
1170 13

Face value of a debenture = ` 100 (i) (b) Investment in 4% stock is ` 36000.

Market value of a debenture = ` 80 [Q discount at 20%] (ii) (a) Investment in 7% stock is ` ( 75000 - 36000)

Rate of interest on debentures = 15% = ` 39000
(iii) (c) Income on 4% stock is ` èæç 4 ´ 9306000öø÷ = ` 1600
Now, by investing ` 80, annual interest = ` 15

\By investing ` 100, annual interest = ` æçè 15 ´100÷øö = `18.75 æèç 7 ´ 39000öø÷ =
80 130
(iv) (b) Income on 7% stock is ` ` 2100

Hence, annual yield is 18.75%. (v) (b) Total income = ` (1600 + 2100) = ` 3700

18. (i) (b) Market value of share = ` 16

Total investment of 500 shares = ` (16 ´ 500) = ` 8000 Subjective Questions

(ii) (a) Rate of dividend = 7.5% 1. Market value of one share = ` 17
Face value of one share = ` 10
Total money invested by Shyam = ` 10846
Face value of 500 shares = ` 5000
çèæ 7.5 ´ 5000ø÷ö \Number of shares bought = 10846 = 638
Dividend received by Mr. Gupta = ` 100 17

= ` 375 Face value of 638 share = ` ( 638 ´ 10) = ` 6380

(iii) (b) Investment of ` 8000, Mr. Gupta gain is ` 375. Rate of dividend paid by company = 15%

èçæ 375 ´ 100÷öø % 375 75 \ Dividend received by Shyam = 15% of ` 6380
8000 80 16
Percentage gain = = % = % = ` æèç 15 ´ 6380ö÷ø = ` 957
100
(iv) (d) Market value of one share = ` 16
2. Dividend on one share = 15% of ` 25
Total investment = ` 7200
Number of share = 7200 = 450 = ` èçæ 15 ´ 25 ÷øö = ` 15
100 4
16
(v) (b) Market value of one share = ` 16 Since, the total income is ` 450.

Brokerage percentage = 1.5% \The number of shares bought = 450 = èçæ 450 ´ 4 ÷öø = 120
15 / 4 15
Brokerage on one share = 1.5% of = ` 16 = ` çèæ 1.5 ´16÷øö
100
Since, market value of one share = ` 30
= ` 0.24
\The sum of money invested by man = ` ( 30 ´ 120) = ` 3600
Net cost on the purchase of one share
= ` (16 + 0.24) = ` 16.24 3. Given, total investment = ` 15840

Total investment = ` 3248 and face value of one share = ` 24
\ Number of share bought = 3248 = 200
Now, premium on one share = 10% of ` 24
16.24
= ` çæè 10 ´ 24øö÷ = ` 2.40
100
19. Let the investment in 4% stock be ` x and in 7% stock be
` ( 75000 - x). and investment on one share = ` ( 24 + 2.40) = ` 26.40

In 4% stock at ` 90 \ Number of shares bought = Total investment
Investment of ` 90 yields income = ` 4 Investment on one share
Þ Investment of ` x will yield income = ` 4x
= 15840 = 600
90 26.40

CBSE Term II Applied Mathematics XII 127

4. Market value of one share of ` 60 premium of face value 9. Annual dividend on a share = ` æçè 6 ´ 100öø÷ = ` 6
` 100 = ` (100 + 60) = ` 160 100

Total investment = ` 8800 \ Annual dividend on 200 shares = ` ( 200 ´ 6)
= ` 1200
Number of shares bought = 8800 = 55
160 Par value of a debenture = ` 100

Rate of dividend paid by company = 6% \ Par value of 50 debenture = ` (50 ´ 100) = ` 5000

Face value of 55 shares = ` (55 ´ 100) = ` 5500 çèæ 10 5000öø÷
100
\ Dividend received = æèç 6 ´ 5500ø÷ö = 330 \ Annual interest on 50 debentures = ` ´
100
` `

æçè 330 ´ 100ø÷ö % = ` 500
8800
Percentage return of investment = = 3.75% Hence, total income of Ram = ` (1200 + 500)
= ` 1700
5. Dividend on one share of ` 80 = 15% of ` 80
10. Total worth of debentures = `16000
= ` æèç 15 ´ 80öø÷ = ` 12 Face value of one debenture = ` 100
100 \Number of debenture = 16000 = 160
100
Let the man buy one share for ` x. Market value of debenture after 10% discount = ` 90
His profit on one share 20% of x = 20x = x. Market value of 160 debenture have = ` (160 ´ 90) = ` 14400

100 5 Brokerage = 1.5%
\ Total brokerage = ` çæ1.5 ´ 14400÷ö = ` 216
Since, the dividend paid by the company on one share = ` 12
\ x = 12Þ x = 60 è 100 ø

5
\ The boys each share at ` 60.

6. The brokerage is paid on the market value of a share it is Thus, amount received by Rashmi = ` (14400 - 216)
= ` 14184
given that the market value of share is ` 899.

\ Brokerage on one share = 1% of ` 899 = 899 11. Income from ` 100 stock = ` 8
100

Total brokerage on 500 shares = ` æèç500 ´ 899 ö÷ø = ` 4495 Income tax on ` 8 income = ` 8 ´ paise 12 per rupee
100
= ` çæè 8 ´ 12 øö÷ =` 0.96
100
Dividend on one share = 25%of `100 = ` èçæ 25 ´ 100ö÷ø = ` 25
100 \ Net income = ` ( 8 - 0.96) = `7.04

Dividend on 500 share = ` ( 25 ´ 500) = ` 12500 However, we are given the return on investment is 9%.

Total investment = ` (500 ´ 899 + 4495) = ` 453995 When income is ` 9, investment = ` 100

æçè 12500 ´ 100ø÷ö % = \ When income is ` 7.04, investment
453995
\ Gain percentage = 2.75% = ` èæç100 ´ 7.04 öø÷ = ` 704 = ` 78 2
9 9 9
7. Amount invested = Market value of debenture + Brokerage
\ The market value of ` 100 stock is ` 78 2
= ` èçæ 90 + 1 ´ 90öø÷ = ` ( 90 + 0.90) = ` 90.90 9
100
12. CP of ` 100 stock = ` 90
Income obtained = Interest on ` 100 = 18% of ` 100
SP of ` 100 stock = ` 921
= ` çèæ 18 ´ 100öø÷ = ` 18 4
100
\ Gain on 100 stock = æèç 92 1 - 90øö÷ =` 9
Thus, for an annual income of ` 18 amount invested = ` 90.90 ` ` 4 4

\ For an income of ` 8100, Þ Gain on ` 3000 stock = ` æçè 9 ´ 1 ´ 3000öø÷
4 100
Investment = ` èçæ 90.90 ´ 8100ø÷ö = ` 40905
18
= ` 67.50

Hence, the amount invested by Manoj is ` 40905. 13. Income on ` 100 stock = ` 31 = ` 7
22
8. A man earn ` 30 per quarter form 30 debentures.
\ Annual income from 30 debentures = ` ( 30 ´ 4) = ` 120 Income tax on ` 1 = 81 paise = 25 paise
33
Let the par value of debenture is ` x.
çæè 25 1 ö÷ø 1
\ Income from 30 debentures giving 8%, =` 3 ´ 100 =` 12

Annual return on investment = ` æèç 30 ´ 8 ´ xö÷ø =` 24x \ 7 = æèç 1 ´ 72 ÷öø = 7
100 10 2 12 24
Income tax on ` ` `

By given condition, 24x = 120Þ x = 120 ´ 10 = 50 \ Net income on ` 100 stock = ` èçæ 7 - 7 ÷öø =` 77
10 24 2 24 24

Hence, the par value of a debenture is ` 50.

128 CBSE Term II Applied Mathematics XII

Now, to produce annual income of ` 77, the amount of stock \By investing ` 100, annual dividend = ` çèæ 18 ´ 100÷øö
24 140

required = ` 100 = ` 12.85

\ To produce annual income of ` 77, the amount of stock Clearly, second investment yields more interest.
çèæ 77øö÷
required = ` 100 ´ Hence, it is better investment.
77 / 24
18. Let the investment in 9% stock at `96 be x.
= ` (100 ´ 24) = ` 2400
\Investment in 12% stock at ` 120 = ` ( 8370 - x)
14. Let the rate of dividend paid to Ravi be x%. Then,
Investment of ` 96 yields income = ` 9
= çæè x ´ 24600ö÷ø =
Ravi’s income ` 120 ` 205 x Investment of ` 120 yields income = ` 12

Ravish’s income =` çèæ 4 ´ 20500øö÷ =` 656 Investment of ` x will yields income = ` çæè 9x ö÷ø …(i)
125 96

It is given that Investment of ` ( 8370 - x) will yield income
Ravish’s income : Ravi’s income = 4 : 5 = ` 12 ( 8370 - x) …(ii)
Þ 656 = 4 120

205x 5 It is given income from each investment is same.
Þ 820x = 656 ´ 5 = 3280 \ 9x = 12 ( 8370 - x)
Þ x = 3280 = 4
96 120
820 Þ 30x = 267840 - 32x
Þ 62x = 267840
Hence, the rate of dividend on Ravi’s stock is also 4%. Þ x = 267840 = 4320

15. Let the man invest ` 100 in each stock. 62
\Investment in 9% stock = ` 4320
For first stock, on investment of ` 90 earns on income of ` 4 Investment in 12% stock = ` ( 8370 - 4320)
which is subject to income tax of 10%.
\Income tax on investment of ` 90 = 10% of ` 4 = 40 paise = ` 4050
Hence, net income on investment of ` 90 = ` (4 - 0.40 ) 19. The par value of 1000 preferred shares = ` (1000 ´ 100)

= ` 3.60 = ` 100000
\ Net income on investment of ` 100 = ` æèç 3.60 ´ 19000÷øö = `4
Since, dividend on preferred shares is paid at the rate of 6%.

For second stock, investment of ` 110 earns income ` 5 \Total dividend paid to preferred share holders

\Investment of ` 100 earns on income of ` æèç5 ´ 100 ø÷ö =` 4.54 = ` æçè 6 ´ 100000öø÷
110 100

Clearly, second stock gives better return. = ` 6000

16. CP of ` 100 stock = ` 103 Since, there are 1000 preferred shares
\Dividend for 1 preferred shares = ` ( 6 ´ 100)
SP of ` 100 stock = ` 107
= ` 600
\ Gain on ` 100 stock = SP - CP = ` (107 - 103) = ` 4
Remaining dividend
When gain is ` 4, investment = ` 103 = Total dividend - Dividend paid to preferred shares
= ` ( 40000 - 6000) = ` 34000
When gain is ` 36, investment = ` çæè 103 ´ 36ö÷ø = `927
4 \Dividend for one ordinary share = ` èæç 344000000÷öø = ` 8.50

Now, by investing ` 103, stock purchased = ` 100

\By investing ` 927, stock purchased = ` çæè 100 ´ 927ø÷ö Þ Dividend on 200 ordinary shares = ` ( 8.50 ´ 200)
103 = ` 1700

= ` 900 \Total dividend received by person
= Dividend received on preferred shares
17. Face value of each share = ` 100 + Dividend received on ordinary shares
= ` ( 600 + 1700) = ` 2300
Market value of share = ` 132

Rate of dividend = 16%

By investing ` 132, annual dividend = ` 16 20. Let the investment of the man in shares paying 6% of ` 140
be ` x, then an investment in shares paying 5% at ` 125
\By investing ` 100, annual dividend = ` çèæ 16 ´ 100÷öø = ` (13500 - x)
132 Income on one share of ` 140 = ` 6
\ Income on ` x = ` 6x = ` 3x
= ` 12.12 140 70
\ Income on one share of ` 125 = ` 5
Now, on 18% ` 100 share of ` 140.
Market value = ` 140 and rate of dividend = 18%
By investing ` 140, annual dividend = ` 18

CBSE Term II Applied Mathematics XII 129

\ Income on ` (13500 - x) = ` 5 (13500 - x) Rakhi purchased ` 25 shares at premium of 20%.
125
\Market value of one share purchased by Rakhi
-
= ` çæè 13500 x øö÷ = èæç1 + 12000 ø÷ö of ` 25 = ` æçè 6 ´ 25÷øö =` 30
25 5

But the total income of the man is ` 560. Investment by Rakhi = ` 12000

\ 3x + èçæ 13500 - x ø÷ö = 560 \Number of shares purchased by Rakhi = 12000 = 400
70 25 30

Þ 15x + 14(13500 - x) = 560 Let r % be the rate of dividend declared by Rakhi company,
350 then

Þ 15x + 14 ´ 13500 - 14x = 560 ´ 350 Annual dividend = Number of shares held by Rakhi

Þ x = 560 ´ 350 - 14 ´ 13500 ´ rate of dividend of Rakhi company

Þ x = 196000 - 189000 ´ face value of shares

Þ x = 7000 Þ 2250 = 400 ´ r ´ 25
100
\ ` (13500 - 7000) = ` 6500
2250 ´ 100
\Investment in 6% shares at ` 140 = ` 7000 Þ r = 400 ´ 25 = 22.5

and investment in 5% shares at ` 125 = ` 6500.

21. A person invest ` 1365 in 3% at ` 91. Hence, the rate percent is 22.5%.

By investing ` 91, stock bought = ` 100 23. Let the investment in 8% stock be `x, so that investment
in 61 % stock is ` ( 25000 - x).
\By investing ` 1365, stock bought = ` æèç 100 ´ 1365÷öø 2
91 Now, investment of ` ( 89.75 + 0.25) = ` 90

= ` 1500 ` 90 in 8% stock fetches a gross income = `8

Now, CP of ` 100 stock = ` 91 So, investment of ` x fetches on income of ` 8 ´ x
90
SP of ` 100 stock = ` 931
2 = ` 8x = ` 4x
90 45
\Gain = SP - CP = ` èæç 93 1 - 91 ö÷ø = ` 2.5
2
Similarly, investment of ` (129.75 + 0.25) = `130
Thus, gain on ` 100 stock = ` 2.5
` 130 in 61 % stocks fetches a gross income = ` 6.50
\Gain on ` 1000 stock =` æèç 2.5 ´ 1000ø÷ö =` 25 2
100
So, investment of ` ( 25000 - x) fetches on income of

Remaining stock = ` (1500 - 1000) = ` 500 ` æèç 6.50 ´ ( 25000 - x)ö÷ø = ` ( 25000 - x)
130 20
Since, the remaining stock is sold ` 85.

\CP of ` 100 stock = ` 91 \ Total gross income = ` çèæ 8x + 25000 - x öø÷
90 20
SP of ` 100 stock = ` 85

\Loss = CP - SP = ` ( 91 - 85) = ` 6 = æçè 16x + 225000 - 9x ö÷ø
180
Thus, loss on ` 100 stock = ` 6 `

\Loss on ` 500 stock =` èçæ 6 ´ 500øö÷ =` 30 =` çæè 7x + 225000 ÷øö =` èçæ 7x + 1250ö÷ø
100 180 180

Hence, the net loss in the transaction = ` ( 30 - 25) = ` 5 Since, income tax is 10%.

22. Market value of one share purchased by Ravi = ` (100 - 20 ) Net income = ` çæè 7x + 1250÷øö ´ 90
180 100
= ` 80

Investment by Ravi = ` 12000 Since, net income is given to be ` 1755, we get

\Number of shares purchased by Ravi = 12000 = 150 çèæ 7x + 1250÷øö ´ 90 = 1755
80 180 100

\Annual dividend received by Ravi Þ 7x + 1250 = 175500 = 1950
180 90
= Number of shares held by Ravi ´ rate of dividend
Þ 7x = 1950 - 1250
´ face value of one share held by Ravi 180

= ` é æçè150 ´ 15 ø÷ö ´ 100ùûú = ` 2250 Þ 7x = 180( 700)
ëê 100

\Annual dividend received by Rakhi Þ x = 18000
= Annual dividend received by Ravi
Hence, investment in 8% stock is ` 18000.

\Annual dividend received by Rakhi = ` 2250 and investment in 61 % = ` 7000.
2

Chapter Test (iii) Annual income of Alok from invested in shares is

(a) ` 400 (b) ` 500 (c) ` 600 (d) ` 300

(iv) Annual income of Alok from invested in debentures is

(a) ` 500 (b) ` 300 (c) ` 400 (d) ` 600

Multiple Choice Questions (v) The total annual income of Alok from invested in

shares and debentures is

1. If the market value of a share is less than face value, (a) ` 1000 (b) ` 400 (c) ` 800 (d) ` 1200
then the share is said to be
Short Answer Type Questions
(a) at par (b) above par
(c) below par (d) None of these 10. Ramesh invested ` 5500 in the shares of a company.
At the end of the year, the company declared 20%
2. If the selling price of `100 stock is ` 100 cash, the stock dividend which gave him on income of ` 500. At what
is said to be premium were the share purchased?

(a) at par (b) at premium 11. A man sells 400 shares of ` 10 each at ` 15 each. The
(c) at a discount (d) None of these company declares dividend on 21% on these shares
with the money be gets, he buys shares of ` 10 at ` 8
3. Shekhar invested ` 4455 in ` 10 shares quoted at each of some other company which declares a
` 8.25. If the rate of dividend 12% his annual income is dividend of 18%. Find the difference in his income.

(a) ` 648 (b) ` 668 12. Which is better investment. 10% debentures at 10%
(c) ` 655.60 (d) ` 534.60 discount or 20% debenture at 10% premium?

4. At what price should I buy a share the value of which is 13. A man invest a certain sum in 5% stock at ` 95 1 and
` 100, paying a dividend of 8%, so that my yield is 11%. 2

(a) ` 84 (b) ` 75 sold it out when the price rose to ` 100 3. In the
4
(c) ` 70 (d) ` 72.72
transaction he gained sum of ` 90, What sum did he
5. Rahul transferring ` 6000 of 13 1 % stock at ` 90 to invest? How much stock did he buy (brokerage 1.5)?
2 14. How much Raju invest in 3 1 % stock at ` 96 9 to get a

10 1 % stock at par. The change in his income is 2 10
2 net income of ` 532 after paying on income for 5 paise
in a rupee of brokerge is ` 1 1 .
(a) ` 245 (b) ` 243
(c) ` 240 (d) ` 250 10

6. A ` 100 shares of a company are sold at 20% discount. Long Answer Type Questions
If the return on investment is 15%, then the rate of
dividends is 15. A man invested ` 9000 in 9% stock at ` 90. He sold the
stock when price rose to ` 105 and invested the sale
(a) 12% (b) 10% proceeds in 15% at some price. By doing this, his
(c) 16% (d) 14% income increased by ` 675. At what price did he
purchase the second stock?
7. The percent income of buyer on 9% debentures of face
to value of ` 100 available in the market for ` 120. 16. Meenakshi sells 5000 ordinary shares of company A,
for value ` 50 each, paying 15% dividend at ` 75 per
(a) 7% (b) 7.5% (c) 8% (d) 6% share. She invested the sale proceed in ordinary shares
of a company B, with the par value of ` 50 each and
8. A person has deposited ` 13200 in a bank which pays paying a dividend of 12%. If the market value of share
14% interest. He withdraws the money and invest in of company B is ` 60. Find
` 100 stock at ` 110 which pays dividend of 15%. How
much does he gain or lose? (i) The number of shares of company B bought by

(a) losses ` 48 (b) losses ` 312 Meenakshi.
(c) gain ` 48 (d) gain ` 132
(ii) Change in the dividend income of Meenakshi.
Case Based MCQs
9. Alok invested a part of ` 5300 in 13 1 % debenture of Answers
3
face value of ` 100 each available at ` 110. The rest he 1. (c) 2. (a) 3. (a) 4. (d) 5. (b) 6. (a) 7. (b) 8. (a)
invested in ` 100 shares available at par. At the end of 9. (i) (c) (ii) (b) (iii) (a) (iv) (c) (v) (c) 10. 120% 11. ` 510
the year, he received 20% dividend on his investment 12. Second 13. ` 4000 14. ` 15680
in debentures. 15. second stock was bought at par 16. (i) 6250 (ii) No change

Based on above information, answer the following For Detailed Solutions
questions. Scan the code

(i) The amount invested in shares is

(a) ` 3000 (b) ` 3300

(c) ` 2000 (d) ` 2300

(ii) The amount invested in debentures is

(a) ` 3500 (b) ` 3300

(c) ` 1800 (d) ` 3000

CBSE Term II Applied Mathematics XII 131

CHAPTER 09

Return, Growth
and Depreciation

In this Chapter...

l Return on Investment
l Compound Annual Growth Rate
l Depreciation

Return on Investment Let a stock is held for one year, then rate of return can be
calculated as,
When a business desires to measure the efficiency of its
investment. It uses a financial calculator known as return on Rate of Return = D1 + ( P1 - P0 ) ´ 100
investment. P0

In simple words, return on investments estimates what you where, P0 = Price of shares/stocks at the time of investment
receive back as compared to what you invest. P1 = Price of shares/stocks of the end of the year
D1 = Dividend earned during the year
Return on Investment Formula
ROI = Net profit / Loss Nominal Rate of Return
Cost of Investment

Another Formula The nominal rate of return is the amount of money generated
by an investment before factoring in expenses such as taxes,
Final value of Investment investment fees and inflation.

ROI = - Initial value of Investment The formula for the nominal rate of return is
Cost of Investment
Nominal Rate of Return percent
It can also be expressed as a percentage, then it is said to be
rate of return. æç Current Market Value of Investment ÷ö
ç - Original Investment Value ÷÷ ´
Rate of return is the gain or loss of an investment over a = ç Original Investment Value ÷ 100
certain period of time. Rate of return can be calculated by
using formula, ç
èø
Rate of Return = Net Profit / Loss ´ 100
Cost of Investment Effective Rate of Return

Rate of return may be positive or negative, if rate of return is The effective rate of return is the rate of interest on an
negative, it means we are in loss and, if rate of return is investment when compounding occurs more than once.
positive, it means we are in profit.
The formula for the effective rate of return is
r ö÷m
Investment in share market involves the receipt of dividends Effective Rate of Return, re = æç1 + mø - 1.
and a capital gain or loss. è

132 CBSE Term II Applied Mathematics XII

Here, r stands for the annual interest rate and m stands for Methods of Computing Depreciation
the number of compounding periods.
There are several methods of computing depreciations.
If the interest is compounded continuously, then Following are some commonly used methods.
m ® ¥ Þ re = er - 1
(i) Straight line method or linear method of depreciation.
Compound Annual Growth Rate
(ii) Reducing balance method
The Compound Annual Growth Rate (CAGR) is the rate of
return that would be required for an investment to grow from (iii) Sum of the years digits method
its beginning balance to its ending balance, assuming the
profits were re-invested at the end of each period of the (iv) Double declining balance method
investments life span.
(v) Units of production method.
Formula and Calculation of the Compound Annual Growth
Rate (CAGR) Linear or Straight Line Method

1 The linear or straight line depreciation method is the simplest
and most widely used method. In this method, the value of an
CAGR = æç EV ö÷ n - 1 asset is reduced uniformly over each period until it reaches to
è BV ø its Scrap or Salvage value. The formula of calculating annual
depreciation by linear method is given below
where, EV = Ending value Annual depreciation
BV = Beginning value
n = number of years = Original cost of an asset - Scrap value
Useful life of asset in years
Depreciation
Reducing Balancing Method
Depreciation is defined as the reduction of original cost of
fixed asset such as buildings, furniture, machinery, vehicles, The reducing balance method of depreciation results
etc over a period of time due to wear and tear. Original cost is in declining depreciation with each accounting period.
the total price at which the asset was purchased. The amount of depreciation reduces as the life of the asset
progress.
l Useful life The useful life of an asset is an accounting estimate The formula of calculating the depreciated value or scrap
of the number of years it is likely to remain in service for the value of an asset is given below
purpose of cost effective revenue generation.
S = P(1 - i) n ,
l Scrap value/Salvage value The value of a depreciable asset at
the end of its useful life is called its scrap value. where S = Depreciated or Scrap value of an asset
P = Present value of an asset
l Total Depreciation The difference between the original value i= r ,
of an asset and the scrap value is the total depreciation. 100

l Book value The book value of an asset on a given data is the where r = constant rate of depreciation
original value of the asset minus the accumulated depreciation n = number of years.
at time.