Arihant CBSE Class 11 Book

182 CBSE Term II Applied Mathematics XII

The following results are from independent sample taken from two populations.

Sample 1 Sample 2

n1 = 35 n 2 = 40

x1 = 13.6 x 2 = 10.1

S2 = 5.2 S2 = 8.5

Calculate test statistics and degrees of freedom for the t-distribution?

4. A manufacturer of a new, cheaper type of the light bulb claims that his product is better the higher priced

competitive light bulb. The average life of the other light bulb is known to be 5000 h. In a test to examine the
manufacturer’s claim, 100 of his bulbs are left on until they burn out. The average length of life in the sample is
5100 h. With a significance level of a = 0.05, is there enough evidence to support the manufacturer’s claim?
Assume that s = 500 h.

5. Naman borrowed ` 1000000 from a bank to purchase a house and decided to repay the loan by equal monthly

installments in 10 yr. If bank charges interest at 9% per annum compounded monthly, calculate the EMI.
(given (1.0075)120 = 2.4514 )

Or
Has set up a sinking fund in order to have ` 50000 after 12 yr for his children’s equation. How much should be
set aside bi-annually into on account paying 5% per annum compounded half-yearly? (given (1.025)24 = 1.809)

Section B (5 Marks Each)

This section consists of 4 questions of Long Answer Type.

6. Consider the following data

Dates in the month of April 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Number of units sold 2 5 0 12 13 25 45 13 31 18 11 2 3 1

Calculate 3 days moving average and display these and the original figures on the same graph.
Or

The production of a soft drink company in thousand of litres during each month of a year is as follows

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
1.2 0.8 1.4 1.6 1.8 2.4 2.6 3.0 3.6 2.8 1.9 3.4

Calculate the 5-monthly moving average and show these moving averages on a graph.

7. A manufacturing company has issued a bond having a face value of ` 10000 paying annual dividends at 8.5%

per annum. The bond will be redeemed at par at the end of 10 yr. Find the purchase price of this bond, if the
investor wishes a yield rate of 8% [given (1.08)-10 = 0.4631].

Or
A firm anticipates an expenditure of ` 500000 for plant modernization at end of 10 yr from now. How much
should the company deposit at the end of each year into a sinking fund earning interest 5% per annum?
(given (1.05)10 = 1.629)

8. (i) A machine costing ` 70000 has a useful life of years. The estimated scrap value is ` 20000 using straight line

method, find the annual depreciation.
(ii) The value of a car depreciates by 12.5% every year. By what percent will the value of the car decrease after

3 yr?

CBSE Term II Applied Mathematics XII 183

9. A manufacturing company makes two types of teaching aids A and B of Mathematics for

Class XII. Each type of A requires 9 labour hours of fabricating and 1 labour hour for finishing. Each type of B
requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the
maximum labour hours available per week are 180 and 30, respectively. The company makes a profit of ` 80 on
each piece of type A and ` 120 on each piece of type B. How many pieces of type A and type B should be
manufactured per week to get a maximum profit? Make it as an LPP and solve it graphically. What is the maximum
profit per week?

Or

A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in
min) required for each toy on the machines is given below

Types Machines

A I II III
B
12 18 6

60 9

Each machine is available for a maximum of 6 h per day. If the profit on each toy of type A is ` 7.50 and that the
each toy of type B is ` 5, then show that 15 toys of type A and 30 of type B should be manufactured in a day to
get maximum profit.

Section C (1 Mark Each)

This section consists of 1 Case Based comprises of 5 MCQs.

10. Consider the following equations of curve x 2 = y and y = x.

On the basis of above information, answer the following questions.

(i) The point of intersection of both the curves is

(a) (0, 0) and (2, 2) (b) (0, 0) and (1, 1) (c) (0, 0) and (- 1, - 1) (d) (0, 0) and (- 2, - 2)

(ii) The graph of the given curves is shown as

Y x2=y Y x2=y
y=x

(a) X¢ OX (b) X¢ OX

Y¢ y=x Y¢
Y y=x Y

(c) X¢ O X (d) X¢ OX

Y¢ x2=y x2=y Y¢ y=x

ò(iii) The value of integral 1 x dx is (c) 1 (d) 1
0 2 (d) 1
(a) 1 (b) 1 (d) 1
43
2
ò(iv) The value of integral 1 x2 dx is (c) 1
0 2
(a) 1 (b) 1
43

(v) The value of area bounded by the curves x2 = y and y = x is (in sq unit)

(a) 1 (b) 1 (c) 1
64 3

184 CBSE Term II Applied Mathematics XII

Solutions

1. Let I = ò dx 2. We have, MC = x
ax - x2
x x2 + 2500

Now, put x = 1 Þ dx = - 1 dt \ C( x) = ò x dx + k
t t2 x2 + 2500

æçè - 1 øö÷ dt - dt òPut x2 + 2500 = t 2 Þ x dx = t dt Þ C( x) = tdt + k
1 t2 1 at
\ I = ò = ò t
a 1 -1
t - t2 t2 × C( x) = ò dt + k = t + k = x2 + 2500 + k
t tt
Given that, when x = 0, then C( x) = 1000
=ò ò- dt = - ( at - 1 - 1 dt \ 1000 = 2500 + k = 50 + k Þ k = 950
2
at - 1 )
Thus, C( x) = x2 + 2500 + 950

( at - - 1 +1 and AC = 1 + 2500 + 950 êéëQAC = Cù
2 x2 x x ûú
= - 1) +
C
a × çæ - 1 + 1÷ö
è2 ø 3. Consider,
Null hypothesis H0 Mean life is same for both the batches.
= - 2 ( at 1 + C
Alternative hypothesis H1 Two batches have different
a - 1)2 mean lives.

= - 2 a× 1 -1 +C é t = 1ù We have,
ax êëQ x úû n1 = 10, n2 = 8, x1 = 750, x2 = 820, S1 = 12 and S2 = 14

= -2 a - x + C Test statistics t is given by
ax
t = x1 - x2 ´ n1n2 = n1S12 + n2S22
Or S n1 + n2 , where S n1 + n2 - 2

Let I = ò 1 + 3x - x2 dx = ò - ( x2 - 3x - 1) dx \ S= 10 ´ 144 + 8 ´ 196 = 3000
10 + 8 -2 16
çèæç x2 çèæ 3 øö÷ 2 çèæ 3 øö÷ 2 - 1÷øö÷
ò= 2 2
- - 3x + - dx = 188 = 13.711

é æçè 3 ÷øö 2 ù and t = 750 - 820 ´ 10 ´ 8
êadding 2 ú 13.711 10 + 8
ëê and subtracting úû

=ò - íìïïîèçæ x - 3÷ö 2 - 9 - 1 ïþïüý dx = -70 ´ 80 = -70 ´ 2 10
2ø 4 13.711 18 13.711 3

=ò ïìíïîçæè x 3 øö÷ 2 çèæ 9 + 4ö÷ø ïþïýü = -140 ´ 3.162 = - 10.762
2 4 41.133
- - - dx
So, degree of freedom = 10 + 8 - 2 = 16
=ò îíïïìçèæ x 3 ø÷ö 2 143 ïþýüï dx
2 Here, calculated|t| = 10.762 > 2.120 = calculated t16( 0.05).
- - - So, we reject the null hypothesis at 5% level of significance
and hence accept the alternate hypothesis.

Hence, the mean life for both the batches is not the same.

èçæç ö÷ø÷ 2 2 Or

ò= 13 - èçæ x - 3 ö÷ø dx We have, D = 0, n1 = 35, n2 = 40, x1 = 13.6,
2 2
x2 = 10.1, S1 = 5.2 and S2 = 8.5
Test statistics, t = ( x1 - x2 ) - D0 = (13.6 - 10.1) - 0
= æèç x - 23ø÷ö ì æèç x - 32 ø÷ö ü S12 2 (5.2)2 + ( 8.5)2
2 13 sin-1íïï ïïý + + S 2
1 + 3x - x2 + 4´2
ï 13 ï C n2 n2 35 40

îï 2 ïþ = 3.5 = 3.5

é ò a2 - x2 dx = x a2 - x2 + a2 sin-1 x + ù 27.04 + 72.25 0.77251 + 1.80625
êQ 2 2 a Cú 35 40
ë û
= 3.5 = 2.18
= 2x - 3 1 + 3x - x2 + 13 sin-1 çæ 2x - 3ö÷ + C 2.57876
4 8 è 13 ø
Test statistics = 2.18

CBSE Term II Applied Mathematics XII 185

æçèç S12 S 2 ÷øö÷ 2 = 0.025 ´ 50000
n1 2 (1 + 0.025)24 - 1
+
and degree of freedom = + n2 1250
( 1.025 ) 24
1 æçèç S12 ø÷ö÷ 2 1 èççæ S 2 öø÷÷ 2 =
n1 - 1 n1 n2 - 1 2
-1
n2
= 1250
ççæè ( 5.2) 2 ( 8.5 ) 2 ÷øö÷ 2 1.809 -1
35 40
+ = 1250
0.809
= èççæ 2 ø÷ö÷ 2 èççæ 2 ÷ø÷ö 2
= ` 1545.12
1 ( 5.2) + 1 ( 8.5 )
35 - 35 40 - 40
1 1 6. According to the question,

= ( 0.77251 + 1.80625)2 Dates in April Number of units sold
1 ( 0.77251)2 + 1 (1.80625)2 12 2
34 39 13 5
14 0
= 6.6503 15 12
1 ( 0.5969) + 1 16 13
34 39 ( 3.2625) 17 25
18 45
= 6.6503 = 65.64 19 13
0.0176 + 0.0837 20 31
21 18
Thus, degree of freedom = 65 22 11
4. We have, x = 5100; m = 5000, s = 500 and n = 100 23 2
24 3
The claim of the manufacturer that his product is better 25 1

than the competitor’s average life of 5000 h.

H0 : m = 5000; H1 : m > 5000
x -m
We know that, Z = s/ n

= 5100 - 5000 = 100 ´ 100
500 / 100 500

= 100 ´ 10 = 2.0
500

At a = 0.05, the table value is Z = 1.645. Let the 3 days moving average is m.

Since, the computed value is more than the table value, m1 = 2 + 5 + 0 = 7 = 2.33
we reject the null hypothesis. 3 3

Therefore, we conclude that the manufacturer’s claim is = 5 + 0+ 12 = 17 =
3 3
justified. m2 5.66

5. We have, P = ` 1000000, i = 9 = 0.0075 = 0 + 12 + 13 = 25 = 8.33
´ 100 3 3
12 m3

and n = 12 ´ 10 = 120 = 12 + 13 + 25 = 50 = 16.66
3 3
\ EMI = P´ i ´ (1 + i)n m4
(1 + i)n - 1
m5 = 13 + 25 + 45 = 83 = 27.66
3 3
= 1000000 ´ 0.0075 ´ (1 + 0.0075)120
(1 + 0.0075)120 - 1 m6 = 25 + 45 + 13 = 83 = 27.66
3 3
= 1000000 ´ 0.0075 ´ (1.0075 )120
(1.0075)120 - 1 45 + 13 + 31 89
m7 = 3 = 3 = 29.67

= 7500 ´ 2.4514 = 7500 ´ 2.4514 m8 = 13 + 31 + 18 = 62 = 20.67
2.4514 - 1 1.4514 3 3

= 12667.42 m9 = 31 + 18 + 11 = 60 = 20
3 3
Hence, EMI is ` 12667.42

Or m10 = 18 + 11 + 2 = 31 = 10.33
3 3
We have, S = ` 50000, n = 12 ´ 2 = 24 yr

and i= 5 = 0.025 m11 = 11 +2 + 3 = 16 = 5.33
´ 100 3 3
2

Now, R = iS m12 = 2 + 3 + 1 = 6 = 2
+ i)n -1 3 3
(1

186 CBSE Term II Applied Mathematics XII

On the basis of above data, we can draw the following graph. On the basis of above data, we can draw the following graph.
50 Daily units sold
45 Moving averages 4.0 Moving average
3.5 Production
40 3.0
2.5
35 2.0
1.5
30 1.0
0.5
25

20

15

10

5

0
12 13 14 15 16 17 18 19 20 21 22 23 24 25
Dates
Number of units sold
Production

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Or 7. We have, F = ` 10000
According to the question, n = Number of periodic dividend payments

= Period of redemption = 10,

Months Production i = 8 = 0.08
(in thousand litre) 100
Jan
Feb 1.2 and R = ` èçæ 8.5 ´ 10000øö÷ = ` 850
Mar 0.8 100
Apr 1.4 \ C = Redemption price or maturity value
May 1.6
Jun 1.8 = Face value = ` 10000. (bond is redeemed at par)
Jul 2.4
Aug 2.6 Let V be the purchase price of the bond.
Sep 3.0
Oct 3.6 We know that,
Nov 2.8
Dec 1.9 R íì1 - (1 + i) -n ü i)-n
3.4 î i ý
V = þ + C(1 +

\ V =` é 850 ççèæ 1 - (1 + 0.08)-10 öø÷÷ + 10000 (1 + 0.08)-10 ù
ê 0.08 ú
êë ûú

= ` é 85000 {1 - ( 1.08) -10 } + 10000 (1.08) -10 ù
ëê 8 ûú

= ` (10625 ´ 0.53680651 + 10000 ´ 0.46319349)

= ` (5703.57 + 4631.93) = ` 10335.50

Hence, the purchase price of the bond is ` 10335.50.

Let the 5 months moving average of the Jan-Dec production Or
is m.
We have, A = ` 500000, r = 5%,

m1 = 1.2 + 0.8 + 1.4 + 1.6 +1.8 = 6.8 i = 0.05 and n = 10 R = ?
5 5
So, = 1.36 We know that, A = R é (1 + i)n - 1 ù
ê i ú
0.8 + 1.4 +1.6 +1.8 + 2.4 = 8 = 1.6 ë û
m2 = 5 5
é ( 1.05 )10 - 1 ù
Þ 500000 = R ê 0.05 ú
1.4 + 1.6 + 1.8 + 2.4 + 2.6 9.8 ë û
m3 = 5 = 5 = 1.96
500000 ´ 0.05 25000
= 1.6 + 1.8 + 2.4 + 2.6 + 3.0 = 11.4 = Þ R = (1.629 - 1) = 0.629 = 39745.63
5 5
m4 2.28

m5 = 1.8 + 2.4 + 2.6 + 3.0 + 3.6 = 13.4 = 2.68 \ The company deposit ` 39745.63 every year into the
5 5 sinking fund.
8. (i) We have original cost of machine = ` 70000
m6 = 2.4 + 2.6 + 3.0 + 3.6 + 2.8 = 14.4 = 2.88
5 5 Scrap value of machine = ` 20000

m7 = 2.6 + 3.0 + 3.6 + 2.8 +1.9 = 13.9 = 2.78 Useful life = 5 yr
5 5 \Annual depreciation = 70000 - 20000 = 50000

m8 = 3.0 + 3.6 + 2.8 + 1.9 + 3.4 = 14.7 = 2.94 55
5 5 = 10000

Hence, annual depreciation is ` 10000.

CBSE Term II Applied Mathematics XII 187

(ii) Let the present value of the car be ` P. So, line passes through the points ( 0, 10) and ( 30, 0).
\ Value of the car after 3 years = P(1 - i)3 On putting (0, 0) in the inequality x + 3y £ 30, we get

= Pæç1 - 12.5 ö÷ 3 0 + 3( 0) £ 30 Þ 0 £ 30, which is true.
è 100 ø
So, the half plane is towards the origin.
= P æç1 - 125 ÷ö 3 Also, x ³ 0 and y ³ 0, so the region lies in the first quadrant.
è 1000ø On solving 3x + 4y = 60 and x + 3y = 30, we get the point of
intersection is B(12, 6).
3 3
Pæçè1 1 ø÷ö Pèçæ 7 ø÷ö The graphical representation of the lines is given below
= - 8 = 8

Decrease in the value of car = P - Pæç 7÷ö 3 Y
è 8ø 20

15 D (0,15)
x+3y = 30
= P éëê1 - 343 ù = ´ 169 C(0,10)
512 ûú P 512 10 3x+4y = 60

\ Percentage decrease in the value of the car after 5 B(12, 6)
A(20, 0)

çæ P ´ 169 ÷ö X¢ O 5 E(30, 0)
10 15 20 25 30 X
3 yr = ç 512 ´ 100÷ % (0, 0) Y ¢
èçç P ÷ø÷
\ Feasible region is OABCO.

= çæè 169 ´ 25 ÷öø % = 4225 % = 33 1 % The corner points of the feasible region are O( 0, 0),
128 128 128 A( 20, 0), B(12, 6) and C( 0, 10).

9. Let the number of pieces of two types of teaching aids A and Corner points Z = 80x + 120y
B be x and y, respectively. O(0, 0) Z = 80( 0) + 120( 0) = 0
A(20, 0) Z = 80( 20) + 120( 0) = 1600
We construct the following table B(12, 6) Z = 80(12) + 120( 6) = 1680
C(0, 10) Z = 80( 0) + 120(10) = 1200
Items Number Time on Time on Profit
of pieces fabricating finishing (in `)
(in hours) (in hours)

A x 9x x 80x Hence, the maximum value of Z is ` 1680.

B y 12y 3y 120y The manufacturer should produce 12 pieces of type A and 6
pieces of type B to earn maximum profit of ` 1680.
Total x+y 9x + 12y x + 3y 80x + 120y
Or
Availability 180 30
Let the manufacturer makes x toys of type A and y toys of
The profit on type A is ` 80 and on type B is ` 120. type B. We construct the following table.
Our objective is to maximise,
Time of Time of Time of
Z = 80x + 120y machine machine
...(i) Types Number machine Profit
subject to constraints, of toys II III (in `)
9x + 12y £ 180 or 3x + 4y £ 60 ...(ii) of toys I (in min) (in min)
x + 3y £ 30 ...(iii)
...(iv) (in min) 18x
and x ³ 0, y ³ 0 ...(v)
Table for line 9x + 12y = 180 is A x 12x 0y 6x 7.50x

x 0 20 B y 6y 18x + 0y 9y 5y

y 15 0 Total x + y 12x + 6y 6x + 9y 7.50x + 5y

Maximum 6´60= 360 6´60= 360 6´60= 360
time

requires

So, line passes through the points ( 0, 15) and ( 20, 0). Our problem is to maximise. ...(i)
On putting (0, 0) in the inequality 9x + 12y £ 180, we get Z = 7.50x + 5y
...(ii)
9( 0) + 12( 0) £ 180 Þ 0 £ 180, which is true. subject to constraints, ... (iii)
12x + 6y £ 360Þ 2x + y £ 60 ...(iv)
So, the half plane is towards the origin. ... (v)
Table for line x + 3y = 30 is 18x £ 360 Þ x £ 20

x 0 30 6x + 9y £ 360 Þ 2x + 3y £ 120
y 10 0
and x ³ 0, y ³ 0

188 CBSE Term II Applied Mathematics XII

Table for line 2x + y = 60 is 30 Therefore, feasible region is OABCDO.
0
x0 The corner points of the feasible region are O( 0, 0) A( 20, 0),
y 60 B( 20, 20), C(15, 30) and D( 0, 40).
The values of Z at the corner points are given below

So, the line passes through the points ( 0, 60) and ( 30, 0). Corner points Z = 7.50x + 5y
On putting ( 0, 0) in the inequality 2x + y £ 60, we get O( 0, 0) Z = 7.50( 0) + 5( 0) = 0
A( 20, 0) Z = 7.50( 20) + 5( 0) = 150
2 ´ 0 + 0 £ 60 Þ 0 £ 60, which is true. B( 20, 20) Z = 7.50( 20) + 5( 20) = 250
So, the half plane is towards the origin. Z = 7.50(15) + 5( 30) = 262.5
Table for line 2x + 3y = 120 is C(15, 30) Z = 7.50( 0) + 5( 40) = 200
D( 0, 40)
x 0 60
y 40 0

So, the line passes through the points ( 0, 40) and ( 60, 0). Thus, the maximum value of Z is at C(15, 30).
On putting ( 0, 0) in the inequality 2x + 3y £ 120, we get
Hence, the manufacturer should manufacture 15 toys of
2 ´ 0 + 3 ´ 0 £ 120 Þ 0 £ 120, which is true. type A and 30 toys of type B to maximise the profit.
10. (i) (b) We have, x2 = y and y = x
So, the half plane is towards the origin.
Draw the graph of the line x = 20. \ x2 = x Þ x2 - x = 0
On putting ( 0, 0) in the inequality x £ 20, we get Þ x( x - 1) = 0 Þ x = 0, 1
0 £ 20, which is true. Now, when x = 0, y= 0 and when x = 1, y = 1

So, the half plane is towards the origin. So, point of intersection is ( 0, 0) and (1, 1).
Also, x, y ³ 0, so the region lies in the first quadrant.
On solving equations 2x + y = 60 and 2x + 3y = 120, we get (ii) (a) The graph of the curves is given as
C(15, 30).
Similarly, solving the equations x = 20 and 2x + y = 60, Y x2=y
we get B( 20, 20). y=x

Y (1, 1)

80 X¢ X
(0, 0)

Y¢

70 ò(iii)(c)1xdx = é x2 ù1 = 1 - 0 = 1
60 (0, 60) ê 2 ú 2 2
50 0 ë û0
40 D(0, 40)
30 C(15, 30) ò(iv)(b)1x 2 dx = é x3 ù1 = 1 - 0 = 1
ê 3 ú 3 3
0 ë û0

ò ò(v) 1 1
(a)\Required area = 0 yLine dx - 0 yParaboladx

20 B(20, 20) ò ò= 1 x dx - 1 x2 dx
00
10
÷÷øö10 ÷÷øö10
X' A(20, 0) (60, 0) ççæè x2 çæèç x3 1 -1 1
2 3 2 3 6
O X = - = = sq unit
10 20 30 40 50 60 70 80
(0, 0) (30, 0)

Y' x = 20 2x + y = 60 2x + 3y = 120

CBSEATeprmpIIliAepdplieMd MaatthhemeamticsaXtII ics 189
Class 12th (Term II)

Practice Paper 2*

(Unsolved)

Ins tructions Tim e : 2 Hr
M a x. M a rks : 4 0
1. The ques tion paper contains three s ections A, B and C.
2. Section A has 5 ques tions w ith 3 internal choices .
3. Section B has 4 ques tions w ith 3 internal choices .
4 . Section C has 1 Cas e Bas ed MCQs com pris es of 5 MCQs .
5. There is no neg ative m arking .

As exact Blue-print and Pattern for CBSE Term II exams is not released yet. So the pattern of this
paper is designed by the author on the basis of trend of past CBSE Papers. Students are advised
not to consider the pattern of this paper as official.

Section A (3 Marks Each)

This section consists of 5 questions of Short Answer Type.

1. Evaluate ò dx .
1) 2x +3
( x -

òOr Evaluate 1 x(1 - x)n dx.
0

2. The average number, in lakhs, of working days lost in strikes during each year of the period (1981-90) was as

under.

1981 1982 1983 1984 1985 1986 1987 1088 1989 1990

1.5 1.8 1.9 2.2 2.6 3.7 2.2 6.4 3.6 5.4

Calculate the 3-yearly moving average and draw the moving average graph.

3. A machine being used by a company is estimated to have a life of 15 yr. At that time, a new machine would

cost ` 75000 and the scrap of the old machine would yield ` 9600 only. A sinking fund is created for
replacing the machine at the end of its life. What sum should be retained by the company at the end of
every year to accumulate at 6% per annum?

Or A bond of face value ` 1000 has coupon rate of 6% per annum with interest and matures in 5 yr. If the
bond is priced to yield 8% per annum, find the value of the bond.

4. Find the present value of perpetuity of ` 7800 payable at the begining of each year, if money is worth 6%

effective.

Or Find the purchase price of ` 1000 bond, redeemable at the end of 20 yr at 105 and paying annual
dividends at 5%, if the yield rate is to be 8% effective. (given (1.08)-20 = 0.21454821)

5. Shiv Suman took a housing loan of ` 800000 to be paid in 10 yr by equal monthly installments. The interest

charged is 10.5% compounded monthly. Find the monthly installments. (given (1.0075)-120 = 0.40793705)

190 CBSE Term II Applied Mathematics XII

Section B (5 Marks Each)

This section consists of 4 questions of Long Answer Type.

6. A random sample of size 16 has 53 as mean. The sum of the squares of the deviation taken from mean is 150, can

this sample be regarded as taken from the population having 56 as mean? (given t15 (0.01) = 2.95)

Or

Consider the following hypothesis test.
H0 : m £ 3530, Ha : m > 3530

A sample of 92 provided a sample mean x = 3740 and sample standard deviation S = 810.

(i) Compute the value of the test statistic and degrees of freedom.

(ii) Use t-distribution table to compute a range for the p-value.
(iii) At a = 0.01 what is your conclusion?

(iv) What is the rejection rule using critical value? What is your conclusion?

7. Fit a straight line trend by the method of least squares and tabulate the trend values from the following data.

Year 1980 1981 1982 1983 1984 1985 1986 1987

Production (in quintals) 346 411 392 512 626 640 611 796

8. A loan of ` 400000 at the interest rate of 6.75% per annum compounded monthly is to be amortized by equal

payments at the end of each month for 10 yr. Find
(i) the size of each monthly payment.
(ii) the principal outstanding at the beginning of 61st month.
(iii) the interest paid in 61st payment.
(iv) the principal contained in 61st payment.
(v) total interest paid.
(Given (1.005625)120 = 1.9603 and (1.005625)60 = 1.4001)

Or
Find the purchase price of a ` 50000, 6% bond, dividends payable semi-annually, redeemable at par in 10 yr,
if the yield rate is to be 5% compounded semi-annually. (given (1.025)-20 = 0.61027094)

9. A dietician wishes to mix two types of foods f1 and f 2 in such a way that the vitamin contents of the mixture

contain atleast 6 units of vitamin A and 8 units of vitamin B. Food f1 contains 2 units/kg of vitamin A and
3 units/kg of vitamin B while food f 2 contains 3 units/kg of vitamin A and 2 units/kg of vitamin B. Food f1 costs
` 50 per kg and food f 2 costs ` 75 per kg. Formulate the problem as an LPP to minimise the cost of mixture.

Or

An aeroplane can carry a maximum of 200 passengers. A profit of ` 400 is made on each first class ticket
and a profit of ` 300 is made on each economy class ticket. The airline reserves atleast 20 seats for first class.

However, atleast 4 times as many passengers prefer to travel by economy class to the first class. Determine,
how many each type of tickets must be sold in order to maximise the profit for the airline. What is the maximum
profit?

Section C (1 Mark Each)

This section consists of 1 Case Based comprises of 5 MCQs.

10. Consider the following curves x 2 + y2 £ 1 and x + y ³ 1.

On the basis of above information, answer the following questions.

(i) Both curves intersect at

(a) (1, 0) (b) (0, 1) (c) Both (a) and (b) (d) Do not intersect

CBSE Term II Applied Mathematics XII 191

(ii) Common region bounded by both curves

Y (0, 1) Y Y (0, 1)
(0, 1)

(a) X¢ (0, 0) X (b) X¢ (0, 0) X (c) X¢ (0, 0) X (d) None of these
Y¢ (1, 0) Y¢ (1, 0)
(1, 0) (d) None of these
Y¢ (d) x4 + 6x3 + x + C

(iii) Area of common region (in sq unit) 4
(d) 3p - 1
(a) p + 1 (b) p - 1 (c) 1 - p
42 42 24 44

(iv) ( x3 + 8) ( x -1) dx is equal to
x2 - 2x + 4

(a) x4 + 3x + C (b) x3 + x2 - 2x + C (c) x2 + 5x + C
3 32 4

(v) Area of uncommon region is (in sq unit) (c) p + 1
42
(a) 3p + 1 (b) 3p - 1
42 42

1. 1 log 2x + 3 - 5 +C Or 1 Answers
5 2x + 3 + 5 (n + 1) (n + 2)
3. ` 2810.89 Or ` 919.03
2. 1.73, 1.96, 2.23, 2.83, 2,83, 4.1, 4.07, 5.13
Y 4. ` 137800 Or ` 716.13

6.5 5. ` 10795
6 6. The sample is not taken from the population having 56 as mean.

5.5 Or (i) t = 2.49 and df = 91
5 Actual
(ii) 0.005< p -value < 0.01
(iii) Reject H0
(iv) Reject H0 , if ³ ta ; Reject H0
7. y = 541.75 + 59.62x

4.5 Year 1980 1981 1982 1983 1984 1985 1986 1987
Trend
Trend 333.08 392.7 452.32 511.94 571.56 631.18 690.8 750.42
4 values
3.5
8. (i) ` 4593
3 (ii) ` 233336.89
2.5 (iii) ` 1312.52
(iv) ` 3280.48
2 (v) ` 151160
1.5 Or ` 53897.29

1 Dotted curve represents 3-yearly moving average 9. Minimum cost of the mixture will be ` 150 at each point of the
0.5
line segment joining the points (3, 0) and èçæ 12 , 2 ÷öø.
X 5 5
0 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990
Or First class ticket = 40 and economy class ticket = 160;
maximum profit = ` 6400

10. (i) (c) (ii) (b) (iii) (b) (iv) (b) (v) (a)

192Applied Mathematics CBSE Term II Applied Mathematics XII
Class 12th (Term II)

Practice Paper 3*

(Unsolved)

Ins tructions Tim e : 2 Hr
M a x. M a rks : 4 0
1. The ques tion paper contains three s ections A, B and C.
2. Section A has 5 ques tions w ith 3 internal choices .
3. Section B has 4 ques tions w ith 3 internal choices .
4 . Section C has 1 Cas e Bas ed MCQs com pris es of 5 MCQs .
5. There is no neg ative m arking .

As exact Blue-print and Pattern for CBSE Term II exams is not released yet. So the pattern of this
paper is designed by the author on the basis of trend of past CBSE Papers. Students are advised
not to consider the pattern of this paper as official.

Section A (3 Marks Each)

This section consists of 5 questions of Short Answer Type.

ò ò1. If b x 3dx = 0 and, if b x 2dx = 2, find a and b.

a a3

Or

òEvaluate ex x 2 + 1 dx.

(x + 1)2

2. A 99% confidence interval for a population mean was reported to be 83 to 87. If s = 8, what sample size was

used in this study?

Or

Consider the following hypothesis test.

H0 : m ³ 45
Ha : m < 45

A sample of 36 is used. Compute a range for p-value and state your conclusion for the following sample.
(given as a = 0.01, x = 44 and S = 5.2)

3. A bond has a face value of ` 1000 matures in 5 years and present value is ` 1200. If coupon rate is 8%

per annum paid semi-annually, find the yield to maturity.

Or

A machine costing ` 30000 is expected to have useful life of 13 yr and final scrap value of ` 4000. Find the
annual depreciation charge, using the linear depreciation method.

4. A bond has a face value of ` 1000 matures in 4 yr and coupon rate is 4% per annum. The bond makes annual

coupon payments. If the yield to maturity is 4%, find the fair value of bond.

CBSE Term II Applied Mathematics XII 193

5. A factory manufactures two types of screws A and B, each type requiring the use of two machines, an automatic

and a hand-operated. It takes 4 min on the automatic and 6 min on the hand-operated machines to manufacture
a packet of screw ‘A’ while it takes 6 min on the automatic and 3 min on the hand-operated machine to
manufacture a packet of screw ‘B’. Each machine is available for atmost 4 h on any day. The manufacturer can
sell a packet of screw ‘A’ at a profit of 70 paise and screw ‘B’ at a profit of `1.

Assuming that he can sell all the screws he manufactures, how many packets of each type should the factory
owner produce in a day in order to maximize his profit? Formulate the above LPP and solve it graphically and
find the maximum profit.

Section B (5 Marks Each)

This section consists of 4 questions of Long Answer Type.

6. Find the area of the region bounded by curve y2 = 9x, x = 2, x = 4 and the X-axis in the first quadrant.

Or
Draw a rough sketch of the curve y =|x - 2|. Find the area under the curve and line x = 0 and x = 4.

7. The 9 items of a sample had the following values.

45, 47, 50, 52, 48, 47, 49, 53, 50

The mean is 49 and the sum of squares of deviations taken from means is 52. Can this sample be regarded as
taken form the population having 47 as mean? Also, obtain 95% and 99% confidence limits of the population
mean.

Or

Consider the following hypothesis test:
H0 : m =8
Ha : m ¹8

A sample of 120 provided a sample mean of 8.4. The population standard deviation is 3.2.

(i) What the value of the test statistic.

(ii) What is the p-value?
(iii) At a = 0.05, what is your conclusion?

(iv) Compute 95% confidence interval for the population mean. Does it support your conclusion?

8. The table given below shows the daily attendance (in thousand) at a certain exhibition over a period of two

weeks.

Week 1 52 48 64 68 52 70 72

Week 2 55 47 61 65 58 75 81

Calculate 7 day moving average and illustrate these and original information on the same graph using the same
scales.

Or
Assuming a four yearly cycle, calculate the trend by the method of moving average from the following data.

Year 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993

Value 12 25 39 54 70 87 105 100 82 65

9. Find the purchase price of a ` 20000 bond, redeemable at the end of 10 yr at 110 and paying annual dividends at

4%, if the yield rate is to be 5% effective (given (1.05)-10 = 0.61391325)

194 CBSE Term II Applied Mathematics XII

Section C (1 Mark Each)

This section consists of 1 Case Based comprises of 5 MCQs.

10. Mr. Ravi amortizes a loan of ` 1500000 for renovation of his house by 8 yr mortgage at the rate of 12%

per annum compounded monthly. (given (1.01)96 = 2.5993 and (1.01)57 = 1.7633)

On the basis of above information, answer the following questions.

(i) The equated monthly installment is

(a) ` 24379.10 (b) ` 24380.10 (c) ` 24378.10 (d) ` 24379.20
(d) ` 1055326.20
(ii) The principal outstanding at the beginning of 40th month is (d) ` 10543.26
(d) ` 13826.84
(a) ` 10563226.20 (b) ` 10554226.20 (c) ` 10553227.20 (d) ` 840395.60

(iii) The interest paid in 40th payment is

(a) ` 10554.26 (b) ` 10553.26 (c) ` 10553.46

(iv) The principal contained in 40th payment is

(a) ` 13823.84 (b) ` 13824.84 (c) ` 13825.84

(v) Total interest paid is

(a) ` 840393.60 (b) ` 840393.50 (c) ` 840393.30

Answers

1. a = - 1and b = 1

Or ex - 2e x + C 8. 60.85, 61.28, 61.14, 60.71, 61.28, 61.14, 61.85, 63.14
x+1

2. 106 Attendance (in thousand)
Or 0.10< p-value < 0.20; Do not reject H0 MoOvriingignaalveatrtaegnedagrnacpeh
88
3. 3.64% 84
Or ` 2000 80
76
4. ` 1000 72
68
5. Hence, the manufacturer should produce 30 packets of screw A 64
and 20 packets of screw B to get a maximum profit of ` 41. 60
56
6. 4(4 - 2 ) sq units 52
Or 4 sq units 48
44
7. 46.14 to 51.86
Or (i) 1.37 1 2 3 4 5 6 7 8 9 1011121314

(ii) 0.1706 Week 1 Week 2

(iii) Do not reject H0 Or
(iv) (7.83, 8.97); Yes 39.75, 54.75, 70.75, 84.75, 92, 90.75

9. ` 19683.48
10. (i) (a) (ii) (d) (iii) (b) (iv) (c) (v) (a)

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