Vedanta Excel Maths Book 7

Unit Ratio, Proportion and Unitary Method

6

6.1 Ratio and Proportion – Looking back

Classroom - Exercise

1. Let's tell and write the ratios.

a) 'a is to b' = ................... and 'b is to a' = ...................

b) 'x is to y' = ................... and 'y is to x' = ...................

c) '3 is to 4' = ................... and '4 is to 3' = ...................
2. Let's tell and write the answers as quickly as possible.

a) Ratio of p to q = ................... b) Ratio of q to p = ...................

c) Ratio of 2 cm to 3 cm = ................... d) Ratio of 3 cm to 2 cm = ...................

e) In x:y the antecedent is = ............... f) In 5:7, the consequent is = .............

3. a) In the proportion a = c , means are ..................., .................. and extremes
b d
are ..................., ...................

b) If a : b = 2 : 1, then , ................... is double of b and b is ................... of a.

C) If m : n = 1 : 3 then, m is ................... of n and n is ................... of m.

We can compare two quantities of the same kind and in the same unit by using a
ratio. A ratio tells how many times a quantity is greater or smaller than another
quantity of the same kind.

Suppose, Hari has Rs 30 as his pocket money and Laxmi has Rs 15 as her pocket
money.

Here, the ratio of Hari's money to Laxmi's money = 30 :15 = 30 = 2 : 1
15
The ratio 2 : 1 tells that Hari's pocket money is two times that of Laxmi's.

Also, the ratio of Laxmi's pocket money to Hari's money = 15 : 30 = 15 = 1 : 2
30
The ratio 1 : 2 tells that Laxmi's pocket money is half of Hari's pocket money.

Thus, if a and b are any two quantities of the same kind and in the same unit, then,
the ratio of a to b = a : b and read as 'a is to b'.

Also, the ratio of b to a = b : a and read as 'b is to a'.

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Ratio, Proportion and Unitary Method

In a : b, the first term a is called antecedent and the second term b is called
consequent.
Similarly, in b : a, the first term b is antecedent and the second term a is consequent.

(i) A ratio compares two or more quantities of the same kind and in same unit.
(ii) A ratio is made by dividing one quantity by another quantity of the same unit.
(iii) A ratio does not have any unit.
(iv) A ratio is usually expressed in the simplest form of the terms.

Worked-out examples

Example 1: A table is 1 m long and 80 cm broad. Find the ratio of the length
and breadth of the table.

Solution: To make the ratio, the quantities should have the
Here, length (l) = 1 m = 100 cm same unit. So, 1 m is converted into 100 cm.
Breadth (b) = 80 cm

Now, ratio of length to breadth = 100 : 80 = 100 = 5: 4
80

Hence, the required ratio of length to breadth is 5 : 4.

Example 2 : Express 2 : 3 and 3 : 4 in the lowest common denominator and
compare them.

Solution:

2 : 3 = 2 and 3: 4 = 3
3 4

L.C.M. of the denominators 3 and 4 is 12.

Now, 32= 2 × 4 = 8 and 43= 3 × 3 = 9
3 × 4 12 4 × 3 12

So, 8 < 9 , i.e. 2 : 3 < 3 : 4
12 12

Example 3: If a : b = 3 : 4 and b : c = 8 : 9, find a : c.

Solution:

Here, a : b = 3 : 4 and b : c = 8 : 9

Now, multiplying these ratios, ba × b = 3 × 8923= 2
c 4 3
a 23 ,
? c = i.e. a : c = 2 : 3.

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Ratio, Proportion and Unitary Method

Example 4: The ratio of number of girls and boys in a class is 5 : 4. If there are
15 girls, find the number of boys.

Solution:

Let the number of boys be x.

Now, 15 = 5 5:4
x 4
or, 5x = 15 × 4
135 × 4
or, x = 51

or, x = 12 2×5:2×4 3×5:3×4

Hence, the required number of boys is 12.

Example 5: The ratio of the ages of a father and his son is 8 : 3. If the son is
15 years old, find the age of the father.

Solution:

Let the father’s age be x years.

Now, x = 8 5 5555555 555
15 3
or, 3x = 15 × 8 F FFFFFFF SSS

or, x = 155 × 8 = 40 Father’s age = 8 times 5 years Son’s age = 3 times 5 years
31
Hence, the father is 40 years old.

Example 6: Mrs. Rai has 15 sweets. If she divides these sweets between her
Solution: daughter and son in 3 : 2 ratio, how many sweets does each of
them get?

Here, the ratio of sweet received by her daughter and son is 3 : 2.

Let the daughter gets 3x and the son gets 2x sweets.

Now,3x + 2x = 15 3+2=5 6 + 4 = 10 9 + 6 = 15

or, 5x = 15
15
or, x = 5 = 3

? Number of sweets received by the daughter = 3x = 3 × 3 = 9

Number of sweets received by the son = 2x = 2 × 3 = 6

Example 7: The angles of a triangle are in the ratio 4 : 5 : 6. Find the size of
Solution: each angle of the triangle.

Here, the angles of the triangle are in the ratio 4 : 5 : 6. 5x°
Let the angles of the triangle are 4x°, 5x° and 6x°.

Now, 4x° + 5x° + 6x°= 180° 4x° 6x°
or, 15x°= 180°

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Ratio, Proportion and Unitary Method

or, x° = 180° = 12°
15
? The first angle of the triangle = 4x° = 4 × 12° = 48°

The second angle of the triangle = 5x° = 5 × 12° = 60°

The third angle of the triangle = 6x° = 6 × 12° = 72°

Example 8: If a : b = 2 : 1, find the value of 2a –+bb.
Solution: a

In a : b = 2 : 1, let, a = 2x and b = x

Now, 2a + b = 2 × 2x + x = 4x + x = 5x = 5
a–b 2x – x x x

Example 9: Two numbers are in the ratio 3 : 4. When 5 is added to each number,
the new ratio becomes 4 : 5. Find the numbers.

Solution:

Here, two numbers are in the ratio 3 : 4. Answer checking:
Let the required numbers be 3x and 4x.

According to the question, The required numbers that we
obtained are 15 and 20.
3x + 5 = 4 Ratio of 15 and 20 = 15 : 20 = 3 : 4
4x + 5 5 When 5 is added to each number,
(15 + 5) : (20 + 5) = 20 : 25 = 4 : 5
or, 5 (3x + 5) = 4(4x + 5) Which is given in the question.

or, 15x + 25 = 16x + 20

or, 15x – 16x = 20 – 25

or, – x = – 5

or, x = 5

? The first number = 3x = 3 × 5 = 15
The second number = 4x = 4 × 5 = 20

Example 10: A bag contains 1 rupee coins, 50 paisa coins and 25 paisa coins

in the ratio of 2 : 5 : 8. If the total amount of money in the bag is

Rs 65, find the number of each kind of coins.

Solution:

Let the number of coins of Re 1, 50 p and 25 p be 2x, 5x and 8x respectively.
5x
Then, 5x number of 50 p = Rs 2 = Rs 2.5 x 2 coins of 50 p = Re 1

8x number of 25 p = Rs 8x = Rs 2x 4 coins of 25 p = Re 1
4
According to the question,
Rs (2x + 2.5x + 2x) = Rs 65 Answer checking:

or, 6.5 x = 65 No. of Re 1 coins = 20 = Rs 20

or, x = 65 = 10 No. of 50 p coins = 50 = Rs 25
6.5 No. of 25 p coins = 80 = Rs 20
Now, the number of Re 1 coins = 2x = 2 × 10 = 20 ? Rs 20+Rs 25+Rs 20 = Rs 65
the number of 50 p coins = 5x = 5 × 10 = 50
the number of 25 p coins = 8x = 8 × 10 = 80 which is given in the question.

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Ratio, Proportion and Unitary Method

EXERCISE 6.1

General Section - Classwork

1. Let's tell and write the ratios as quickly as possible.

a) 2 m t0 3 m ......................... b) Rs 50 to Rs 40 .........................

c) 10 kg to 20 kg ......................... d) Re 1 to 50 p .........................

e) 70 cm to 1m ......................... f) 300 g to 1 kg .........................

2. Let's tell and write the answers as quickly as possible.

a) In p : q, antecedent is ....................... and consequent is .........................

b) In 3 : 8, antecedent is ...................... and consequent is .........................

c) If the antecedent is 9 and consequent is 5, the required ratio is .......................

3. a) If x : 2 = 3 :1, x = ..................... b) If 3 : x = 1 : 4, x = .....................

c) If 1 : 4 = 1 : y, y = ..................... d) If 5 : 1 = m : 1, m = .....................

Creative Section - A

4. Answer the following questions.

a) What is a ratio? Write with examples.

b) Is it possible to make a ratio between 2 kg and 3 m? Write with reason.

c) Define with examples, the antecedent, and consequent of a ratio.

d) The ratio of the number of girls and boys in a class is 1 : 2. What does it
mean? Write with examples.

e) The number of girls in a school is three times the number of boys. Express
it in a ratio.

5. Find the ratios and reduce them in their lowest terms.

a) 15 cm and 12 cm b) 8 cm and 8 mm c) 1 m and 75 cm

d) 9 months and 2 years e) 2 kg and 750 g f) 900 ml and 3 l.

6. Express the following ratios in the lowest common denominator and compare

them.

a) 1 : 2 and 3 : 4 b) 2 : 5 and 1 : 3 c) 2 : 3 and 5 : 6

d) 4 : 7 and 3 : 8 e) 7 : 9 and 5 : 8 f) 7 : 12 and 4 : 9

7. a) If a : b = 2 : 3 and b : c = 6 : 10, find a : c.
b) If x : y = 4 : 5 and y : z = 15 : 16, find x : z.
c) If p : q = 3 : 7 and q : r = 14 : 9, find p : r.

8. a) There are 40 students in a class and 16 of them are girls.

(i) Find the ratio of the girls and the total number of students.

(ii) Find the ratio of the boys and the total number of students.

(iii) Find the ratio of the girls and boys.

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Ratio, Proportion and Unitary Method

b) An alloy contains 120 g of copper and 150 g of zinc.

(i) Find the ratio of copper and zinc in the alloy.

(ii) Find the ratio of copper and the total weight of alloy.

(iii) Find the ratio of zinc and the total weight of alloy.

9. a) A ratio of two numbers is equal to 3 : 8. If the smaller number is 12, find the
greater number.

b) The ratio of two numbers is 5 : 2. If the smaller number is 14, find the greater
number.

c) The ratio of two numbers is 7 : 2. If the greater number is 56, find the smaller
one.

d) Two numbers are in the ratio 5 : 4. If the smaller number is 32, find the
greater one.

10. a) The ratio of the number of boys and girls in a class is 4 : 3. If there are 18
girls, find the number of boys.

b) The ratio of the ages of a mother and her son is 9 : 4. If the mother is 54 years
old, find the age of the son.

c) The ratio of monthly income to the monthly saving of a family is 9 : 2. If the
saving is Rs 4,320, find the income and expenditure of the family.

11. a) The ratio of length and breadth of a room is 3 : 2. If the room is 18 feet long,
find the following:

(i) the breadth of the room (ii) the perimeter of the room

(iii) the area of the floor of the room

b) The ratio of length and breadth of a piece of land is 5 : 3 and it's breadth is
48 m. Find the perimeter and area of the land.

12. a) Divide Rs 65 in the ratio of 2 : 3.

b) Divide 192 kg in the ratio of 7 : 5.

c) There are 32 students in a class. If the ratio of the number of boys and girls
is 5 : 3, find the number of boys and girls.

d) Mr. Yadav divides a sum of Rs 25,000 between his son and daughter in the
ratio of 2 : 3. Find the sum obtained by each of them.

e) Pratik and Debasis invested a sum of Rs 84,000 in a business. If the ratio of
their shares is 7 : 5, how much money did each of them invest?

f) There are 28 teachers in a school. If the ratio of the male and female teachers
is 4 : 3, find their numbers.

g) The population of a rural municipality is 9,966. If the ratio of the adult and
children population is 5 : 6, Find their numbers.

h) Electrum is an alloy of gold and silver. If the ratio of gold and silver in 200 g
of electrum is 11 : 9, find the weight of each metal in the alloy.

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Ratio, Proportion and Unitary Method

13. a) If a pair of complementary angles are in the ratio 2 : 3, find them.

b) If a pair of supplementary angles are in the ratio 3 : 7, find them.

c) The angles of a triangle are in the ratio 2 : 3 : 4. Find the size of each angle.

14. a) If a : b = 3 : 2, find the value of a + bb.
a –

b) If p : q = 1 : 3, find the value of 5p – q
p+q

c) If (5x + 3) : (7x + 3) = 3 : 4, find the value of x.

d) If (4x – 5) : (9x – 5) = 3 : 8, find the value of x.

Creative Section - B
15. a) Two numbers are in the ratio 5 : 7. When 3 is added to each number, the new

ratio becomes 3 : 4. Find the numbers.

b) The ratio of two numbers is 4 : 9. When 5 is subtracted from each of them,
the new ratio becomes 3 : 8. Find the numbers.

c) The ratio of the present age of father and son is 11 : 3. After 4 years, the ratio
of their age will be 3 : 1. Find their present age.

d) The ratio of the present age of Ram and Hari is 3 : 4. 5 years hence, the ratio
of their age is 4 : 5. Find their present age.

e) The ratio of the present age of mother and daughter is 3 : 1. Six years ago, the
ratio of their age was 6 : 1. Find their present age.

f) The present age of brother and sister are in the ratio 4 : 5. Three years ago,
the ratio of their age was 3 : 4. Find their present age.

16. a) The ratio of number of girls and boys in a class of 30 students is 7 : 8. If 5
new boy students admit in the class, what is the ratio of number of girls and
boys?

b) The ratio of milk and water in the mixture of 45 litres is 7 : 2. If 4 litres
more water is added in the mixture, what will be the new ratio of milk and
water?

17. a) A bag contains 1 rupee, 50 paisa and 25 paisa coins in the ratio 1 : 2 : 4. If the
total amount is Rs 90, find the number of each kind of coins.

b) Mrs. Sharma exchanged Rs 5,600 into the number of 10 rupee, 20 rupee and
50 rupee notes in the ratio of 5 : 4 : 3 at a bank for Dashain Tika, find the
numbers of each type of rupee notes.

It's your time - Project work!

18. a) How many teachers are there in your school? How many male teachers
and female teachers are there?

(i) Write the ratio of male teaches to the female teachers.
(ii) Write the ratio of male teachers to the total number of teachers.
(iii) Write the ratio of total number of teachers to the female teachers.

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Ratio, Proportion and Unitary Method

b) How many students are there in your class? How many boys and girls are
there in your class?
19. a)
b) (i) Find the ratio of the number of girls to boys.

(ii) Find the ratio of the number of girls to the total number of students.

(iii) Find the ratio of the total number of students to the number of boys.

(iv) If 2 more new girls and 3 more new boys are admitted in your class,
what is the new ratio of the number of boys to the girls?

Let's draw as many blue and white circles so that the total number of
circles are 18 and the ratio of blue to white circles is 4 : 5.

Let's draw as many triangles and rectangles so that the total number of
these plane shapes are 20 and the ratio of the number of triangles to the
number of rectangles is 3 : 2.

6.2 Proportion

Let’s take two ratios 2 : 3 and 6 : 9.

Here, 6 : 9 = 6 2 = 2 : 3
9 3

Thus, 2 : 3 and 6 : 9 are two equal ratios. The equality of two ratios is called a

proportion. Here, the terms 2, 3, 6, and 9 of the equal ratios are called proportional.

Again, let a, b, c and d are in proportion.

It is written as a : b = c : d or a : b :: c : d.

Here, the terms a, b, c, and d are called the first, second, third, and the fourth
proportional respectively.

Furthermore, the first and the fourth proportional are called extremes. The second
and third proportional are called means.

Extremes
Means

a:b c:d

1st 2nd 3rd 4th

Proportional Proportional

In a:b = c:d or a = c ; so, a×d=b×c
b d

? In a proportion, the product of extremes = the product of means

6.3 Types of proportions
There are two types of proportions- direct proportion and inverse proportion.

(i) Direct proportion
Suppose the cost of 1 pen = Rs 20
Then, the cost of 2 pens = 2 × Rs 20 = Rs 40.

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Ratio, Proportion and Unitary Method

Here, the ratio of the number of pens = 1 : 2
Also, the ratio of the cost of pens = 20 : 40 = 1 : 2

Thus, the ratio of the number of pens and the ratio of their cost are equal. So,
the ratios are in proportion. Furthermore, if the number of pens increases (or
decreases), their cost also increases (or decreases). Such type of proportion is
called direct proportion.
(ii) Inverse proportion

Suppose, 1 pipe can fill a water tank in 4 hours.
Then, 2 pipes can fill the tank in 2 hours.
Here, the ratio of the number pipes = 1 : 2
The ratio of time taken to fill the tank= 4 : 2 = 2 : 1

Thus, the ratio of the number of pipe and the ratio of the time taken to fill the
tank are exactly opposite. So, they are in proportion but inversely. Such type
of proportion is called inverse proportion.

Worked-out examples

Example 1: Test whether the ratios 4 : 5 and 12 : 15 are in proportion.

Solution:

Here, the product of extremes = 4 × 15 = 60
the product of means = 5 × 12 = 60

? Product of extremes = Product of means

? 4 : 5 = 12 : 15
Hence, 4 : 5 and 12 : 15 are in proportion.

Example 2: If the terms 3, 8, and 9 are in proportion, find the fourth proportional.

Solution:

Let, the fourth proportional be x.

Now, if 3, 8, 9, and x are in proportion,

3:8 =9:x If 3, 8, 9, and x are in proportion, the ratio of the first
two terms is equal to the ratio of the last two terms.
or, 3 = 9
8 x

or, 3x = 8 × 9

or, x= 8 × 93 = 24
31
Hence, the required fourth proportional is 24.

Example 3: If the terms 4, 6, and 18 are in proportion, find the third proportional.

Solution: We got it!
Let, the third proportional be x. To find third proportional
we should suppose the
Now, if 4, 6, x, and 18 are in proportion, third term as x!

4 : 6 = x : 18
4 x
or, 6 = 18

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Ratio, Proportion and Unitary Method

or, 6x = 4 × 18

or, x = 4 × 183 = 12
61

Hence, the required third proportional is 12.

Example 4: If the cost of 8 kg of apples is Rs 1,000, find the cost of 5 kg of
apples by proportion method.

Solution:

Let, the required cost of 5 kg of apples be Rs x.

Then, quantity of apples cost When the quantity of
8 kg Rs 1,000 apples decreases, cost
5 kg Rs x also decreases.

Since, the quantities of apples and their cost are in direct proportion,

8 : 5 = 1000 : x

or, 8 = 1000
5 x

or, 8x = 5 × Rs 1000

or, x = 5 × Rs 1010205 = 5 × Rs 125 = Rs 625
81
Hence, the required cost is Rs 625.

Example 5: If 20 workers can complete building a house in 24 days, in how
many days would 15 workers complete the same work?

Solution:

Let, the required number of working days be x.

Then, number of workers number of working days When the number of
20 24 workers decreases,
15 x working days increases.

Since the number of workers and their working days are in inverse proportion,
20 : 15 = x : 24

or, 20 4 = x
15 3 24

or, 3x = 4 × 24

or, x = 4 × 248 = 32
31

Hence, the required number of working days is 32 days.

Example 6: In a hostel, 50 students have food enough for 54 days. How many
students should be added in the hostel so that the food is enough
for only 45 days?

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Ratio, Proportion and Unitary Method

Solution:
Let, the required number of students to be added be x.
Then, total number students = (50 + x).

no. of students no. of days For fewer students, the food is enough
50 54 for longer days. For more students,
the food is enough for fewer days.

( 50 x ) 45

The number of students and food enough for number of days are in inverse

proportion.

So, 50 : (50 + x) = 45 : 54

or, 50 = 455
50 x 546

or, 5 (50 + x) = 50 × 6

or, 250 + 5x = 300

or, 5x = 300 – 250

or, 5x = 50

or, x = 5010
51

Hence, 10 students should be added.

EXERCISE 6.2

General Section – Classwork

\1. Let's tell and write the ratios which are in proportion.

a) 1 : 2 and 3 : 6 b) 2 : 3 and 4 : 9 Ratios which are in proportion

c) 2 : 5 and 1 : 2 d) 3 : 4 and 6 : 8

2. Let's tell and write whether these quantities are in direct or in inverse
proportion. Write 'Direct' or 'Inverse' in the blank spaces.

a) Number f books and their cost .....................................

b) Rate of cost of a pen and number of pens .....................................

c) Rate of cost of a pen and the total cost of pens .....................................

d) Speed of a vehicle and time taken to cover a .....................................
certain distance

3. Let's tell and write the values of x in these proportions.

a) x : 2 = 3 : 1, x = ............... b) 2 : x = 1 : 4, x = ....................

c) 5 : 1 = x : 3, x = .................... d) 1 : 5 = 5 : x , x = ....................

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Ratio, Proportion and Unitary Method

Creative Section - A
4. a) Define proportion with an example.

b) Write the types of proportions with examples.

c) Define extremes and means of a proportion with an example.

d) How do we say the rate of cost of rice and the total cost of rice are in direct
proportion write with an example.

e) How do we say the rate of cost of rice and the quantity of rice that we
purchase are in inverse proportion? Write with an example.

5. Let's test whether the following ratios are in proportion.

a) 2 : 3 and 8 : 12 b) 5 : 4 and 15 : 12 c) 6 : 8 and 24 : 16

6. Let's find the value of x in each of the following proportions.

a) 4 : 3 = 12 : x b) 5 : 7 = 20 : x c) 6 : 9 = x : 36

d) 8 : 10 = x : 50 e) 9 : x :: 36 : 48 f) 4 : x :: 28 : 42

g) x : 9 :: 30 : 27 h) x : 8 :: 35 : 40 i) 14 : 15 :: 42 : x

7. a) If the following terms are in proportion, find the fourth proportional.

(i) 2, 3, 8 (ii) 4, 7, 12 (iii) 9, 10, 36 (iv) 10, 15, 30

b) If the following terms are in proportion, find the third proportional.

(i) 4, 6, 18 (ii) 7, 3, 21 (iii) 5, 9, 45 (iv) 9, 10, 40

c) If the following terms are in proportion, find the second proportional.

(i) 6, 24, 36 (ii) 10, 30, 21 (iii) 12, 60, 40 (iv) 15, 45, 18

d) If the following terms are in proportion, find the first proportional.

(i) 5, 12, 30 (ii) 8, 21, 24 (iii) 12, 18, 24 (iv) 7, 24, 21

Let's solve these problems using proportion methods.
8. a) If the cost of 7 kg of rice is Rs 672, find the cost of 4 kg of rice.

b) If a bus covers 385 km in 7 hours, how many kilometres does it cover in
10 hours with the same speed?

c) If 6 packets of tea cost Rs 1,260, how many packets of tea can be purchased for
Rs 1,890?

d) The cost of 5 kg of oranges is same as the cost of 3 kg of apples. How many
kilograms of oranges are required to exchange 9 kg of apples?

9. a) If 10 pipes can fill a tank in 16 minutes, in how many minutes would 8
pipes fill the same tank?

b) If a group of 30 workers can complete a piece of work, in 21 days, in how
many days would 45 workers complete the same work?

c) When the rate of cost of rice is Rs 85 per kg, 18 kg of rice can be purchased
for a certain sum of money. How many kilograms of rice can be purchased
for the same sum if the rate is increased to Rs 90 per kg?

d) Mother can buy 15 kg of sugar at the rate of Rs 80 per kg. If the rate is
reduced by Rs 5 per kg, how much sugar can she buy for the same amount
of money?

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Ratio, Proportion and Unitary Method

10. a) 10 workers can complete a piece of work in 18 days. How many workers
are required to complete the work in 12 days?

b) 12 workers can complete a piece of work in 20 days. How many workers
are needed to complete the work in 16 days?

c) A garrison of 200 men has provisions for 45 days. For how many men
would the provisions last only for 30 days?

d) 60 boys in a hostel have food for 30 days. How long would the food last for
72 boys?

Creative Section - B

11. a) 15 labourers were employed to build a wall in 28 days. How many more
labourers should be employed to finish the construction in 21 days?

b) If 18 men can complete a piece of work in 42 days, how many more men
should be added to complete the work in 36 days?

12. a) 15 women can complete a piece of work in 16 days. How many women
should leave so that the work would be finished in 20 days?

b) In a hostel, 30 students have food enough for 40 days. How many students
should leave the hostel so that the food would be enough for 100 days?

13. a) A piece of work can be completed in 30 days working 8 hours a day. In
how many days would the work be completed working 6 hours a day?

b) A certain number of workers are employed to construct a road in
24 days working 9 hours a day. How many hours a day should they work
to complete the construction 6 days earlier?

c) A barrack has enough provisions to last 200 soldiers for 30 days. How
many soldiers must be transferred elsewhere to last the provisions for 40
days ?

It's your time - Project work!

14. a) Let's search and write any three pairs of quantities in our real life situations,
which are in direct proportion.

b) Let's search and write any three pairs of quantities in our real life situations,
which are in inverse (or indirect) proportion.

15. a) Let's write any three sets of four numbers such that the numbers in each
set are in proportion. Find the product of extremes and the product of
means in each proportion. Then, show that product of extremes = product
of means.

b) Let's write the rate of cost of few items such as rice, sugar, potatoes, milk,
etc. in your local market. Write short reports on the following cases.

(i) How does change in the rate of cost of these items affect the total cost
of each items that we purchase?

(ii) How does change in the rate of cost of these items affect the amount
of quantities of these items that we purchase?

111Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Ratio, Proportion and Unitary Method

6.4 Unitary method
Let's have some discussions on the following questions.
a) What are the rates of cost of the following items in your local areas?

(i) Milk per litre (ii) Rice per kg (iii) Sugar per kg (iv) Potatoes per kg
b) Now, will you please calculate the cost of the following quantities of these items?

(i) 2 l of milk (ii) 5 kg of rice (iii) 3 kg of sugar (iv) 10 kg of potatoes
c) How did you find the cost of these items? Discuss in the class.

Let the rate of cost of milk is Rs 45 per litre.
Then, the cost of 2 l of milk = 2 × Rs 45 = Rs 90
Here, 1 litre is the unit quantity and Rs 45 is the unit cost (or unit value). 2 l is more
quantity and Rs 90 is more value.
Thus, in the case of direct proportion, more value is obtained multiplying the unit
value by the given quantity.
Again, if the cost of 3 kg of sugar is Rs 240,
then, the cost of 1 kg of sugar = Rs 240 ÷ 3 = Rs 80
Thus, in the case of direct proportion, unit value is obtained dividing more value by
the given quantity.
Furthermore,
Let 1 pipe can fill a water tank in 60 minutes.
Then, 2 pipes of the same size can fill the tank in 60 ÷ 2 = 30 minutes.
Also, if 3 pipe can fill the tank in 20 minutes
Then, 1 pipe can fill the tank in 3 × 20 minutes = 60 minutes.
Thus, in the case of inverse (or indirect) proportion, more value is obtained by
division and unit value is obtained by multiplication.
In this way, the mathematical method that we apply to find the unit value or more
value is known as unitary method.

Worked-out examples

Example 1: The cost of 5 litres of petrol is Rs 540.

a) Find the cost of 12 l of petrol

b) How much petrol can be bought for Rs 972?

Solution:

a) The cost of 5 l of petrol = Rs 540 In direct proportion, unit
cost is obtained by division.
The cost of 1 l of petrol = Rs 540 = Rs 108
5

The cost of 12 l of petrol = 12 × Rs 108 = Rs 1,296

Hence, the required cost of 12 l of petrol is Rs 1,296.

Vedanta Excel in Mathematics - Book 7 112 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Ratio, Proportion and Unitary Method

b) Rs 540 is the cost of 5 l of petrol.
5
Re 1 is the cost of 540 l of petrol.

Rs 972 is the cost of 5 × 972 l of petrol = 9l
540

Hence, the required quantity of petrol is 9 l.

Example 2: The floor of a hall is 15 m long and 8.4 m broad. Find the cost of
carpeting the floor at Rs 85 per sq. m.

Solution:

Here, the length of the floor (l) = 15 m

the breadth of the floor (b) = 8.4 m

? Area of the floor = l × b = 15 m × 8.4 m = 126 m2

Now,the cost of carpeting 1 m2 is Rs 85.
the cost of carpeting 126 m2 is 126 × Rs 85 = Rs 10,710.

Hence, the required cost of carpeting the floor is Rs 10,710.

Example 3: A married person should pay Re 1 as the social security tax for the
Solution: annual income of Rs 100. How much tax does the person pay if his
annual income is Rs 3,60,900?

Here, when the annual income is Rs 100, the tax = Re 1

When the annual income is Re 1, the tax = Rs 1
100
1
When the annual income is Rs 3,60,900, the tax = Rs 100 × 3,60,900

= Rs 3,609.

Hence, the required social security tax is Rs 3,609.

Example 4: If 2 parts of a land costs Rs 45000, find the cost of 4 parts of the
3 5

land.

Solution:

Here, the cost of 2 parts of a land = Rs 45000
3
The cost of 1 (whole) land = Rs 45000 = 45000 × 3 = Rs 67,500.
2 2
3
4 4
The cost of 5 parts of the land = Rs 67,500 × 5 = Rs 54,000

Hence, the required cost of the land is Rs 54,000.

Example 5: A computer can download 200 megabyte (MB) of a movie file in 20
seconds.

a) Find the download speed of the internet in per second.

b) How long does the computer take to download the whole movie
file of size 2.4 gigabyte (GB)?

113Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Ratio, Proportion and Unitary Method

Solution:

a) Here, the internet speed = rate of download of the file

= downloaded size of file = 200 MB = 10 MB per second
time taken 20 seconds
Hence, the download speed of the internet is 10 MB per second (or 10 mbps)

b) Again, 10 MB of file is downloaded in 1 second.

1 MB of file is downloaded in 1 second.
10
2.4 GB or 2400 MB of file is downloaded in 1 × 2400 seconds = 240 seconds
10
= 240 minutes
60
= 4 minutes

Hence, the computer takes 4 minutes to download the whole movie file.

Example 6: At an average internet speed of 20 megabyte per second (mbps),
a computer takes 6 minutes to download an application file. How
long does the computer take to download the file if its speed slows
down to 15 mbps?

Solution:

a) Here, 20 mbps speed takes 6 minutes to download the file.
1 mbps speed takes 20 × 6 minutes to download the file.

15 mbps sped takes 20 × 6 = 8 minutes to download the file.
15
Hence, the computer takes 8 minutes to download the file.

Example 7: 24 students in a hostel had provisions enough for 30 days. If 16
more students join the hostel, how long would the provisions last?

Solution:

Here, after joining 16 more students, total number of students in the hostel = 24 + 16 = 40.

24 students had provisions for 30 days.

1 students had provisions for 24 × 30 days.

40 students had provisions for 24 × 30 days = 18 days.
40

Hence, the provisions would last for 18 days.

EXERCISE 6.3
General Section - Classwork

Let's tell and write the answer as quickly as possible.
1. a) Cost of 1 kg of sugar is Rs 80, cost of 2 kg of sugar is .....................

b) Cost of 3 kg of sugar is Rs 240, rate of cost of sugar is .....................

c) If the rate of cost of rice is Rs 110 per kg, the cost of 5 kg of rice is ................

Vedanta Excel in Mathematics - Book 7 114 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Ratio, Proportion and Unitary Method

2. a) 1 pipe can fill a tank completely in 2 hours. 2 pipes can fill in the tank
in ......................

b) 3 pipes can fill a tank completely in 2 hours. 1 pipe can fill the tank
in ......................

c) 2 workers can build a wall in 7 days; then, 1 worker can build it
in ......................

3. a) If 1 part of a sum is Rs 50, the whole (1) sum is ......................
2

b) If 1 part of a land worth Rs 20,000, the value of the whole land
3

is ......................

c) If 1 part of a tank is filled in 10 minutes, the whole tank is filled
4

in ......................

Creative Section - A
4. a) Define unitary method with examples.

b) Define unit quantity and unit value with an example.

c) How do we find unit value if the quantities are in direct proportion? Write
with an example.

d) How do we find unit value if the quantities are in inverse proportion? Write
with an example.

5. a) The cost of 9 litres of milk is Rs 765.
(i) Find the cost of 10 litres of milk.
(ii) How much milk can be bought for Rs 340?

b) A taxi covers 330 km distance in 6 hours.
(i) Find the speed of the taxi in km per hour.
(ii) How many kilometres does it travel in 8 hours at the same speed ?

c) A motorcycle covers 315 km distance with 7 l of petrol.
(i) How many kilometres does it covers with 12 l of petrol in the same
mileage?
(ii) How much petrol is needed to cover 270 km in the same mileage?

d) The rent of a room for 4 months is Rs 20,000.
(i) What is the rent for 7 months?
(ii) How long would a tenant hire the room for Rs 55,000 at the same rate
of rent?

6. a) The annual income of an individual is Rs 3,85,200. How much tax should
she pay at the rate of Re 1 per Rs 100 as social security tax?

115Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Ratio, Proportion and Unitary Method

b) Mrs. Nepali is a school teacher. She should pay Re 1 as the social security
tax for the annual income of Rs 100. If she pays Rs 3,348 as the annual tax,
find her annual income.

c) Mr. Rawal is a service man. His monthly salary is Rs 32,500. How much
social security tax should he pay in a year at Re 1 per Rs 100 annual income?

7. a) The interest of Rs 100 in one year is Rs 8. Find the interest of Rs 5,400
(i) in 1 year (ii) in 3 years.

b) The interest of Rs 100 in one year in a bank is Rs 10. Mr. Tharu borrowed
some loan from the bank and paid an interest of Rs 2,000 after 1 year. How
much loan did he borrow?

8. a) The floor of a rectangular room is 12m long and 7.5m broad.
(i) Find the area of the floor.
(ii) Find the cost of carpeting the floor at Rs 99 per sq. m.

b) A rectangular floor is 15m long and 10m broad. Find the cost of plastering it at
Rs 75.50 per sq.m.

9. a) If the exchange rate of 1 dollar is Rs 115.80, how many dollars can be
exchanged for Rs 5,790 ?

b) The cost of a bicycle is ICRs 7,450. If the exchange rate of Indian Currency
is ICRe 1 = NC Rs 1.60, find the cost of cycle in Nepali Currency.

10. a) I(35ifitp)haFerticnsodosftthaoefla47cnopdsatcrotosfsot23fR1psar3ro6tps,0ao0nf0it.lhaen(dlia)insFdRi.nsd2,t1h0e,0c0o0s,tfoinf dthtehewchooslteolfa32ndp.arts
b)

of the land. 2
5
c) Mr. Pandey spends Rs 4,800 on his children’s education which is part of
his monthly income. (i) Find his monthly income.
3
(ii) If he spends 8 parts of the income on food, how much does it amount to?

Creative Section -B

11. a) A computer can download 250 megabyte (MB) of a movie file in 25 seconds.

(i) Find the download speed of the internet in per second.

(ii) How long does the computer take to download the whole file of size
1.8 GB?

b) At an average internet speed of 30 megabyte per second (30 mbps), a
computer can download an application file in 3 minutes. Find the size of
the file in gigabyte (GB).

12. a) At an average internet speed of 15 mbps, a computer takes 4 minutes to
download an application file. If the speed slows down to 10 mbps, how
long does the computer take to download the file?

b) A computer can download 360 MB of a movie file from YouTube in
30 seconds. If the average internet speed increases by 8 mbps, how long
does the computer take to download the whole movie file of size 2.4 GB?

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Ratio, Proportion and Unitary Method

c) 5 machines of a noodle factory complete the required production of
noodles in 15 days. If 2 machines are shut down due to loadshadding, in
how many days would the remaining number of machines complete the
required production of noodles?

d) 2 printing machines of a printing house can complete the printing work of
the required number of Excel in Mathematics textbooks in 20 days. After
printing the books for 8 days, if 1 more machine was added, in how many
days would the remaining work be completed?

13. a) 50 students in a hostel had provisions for 60 days.

(i) How long would the provisions last for 1 student ?

(ii) How long would the provisions last for 40 students ?

b) A garrison of 60 soldiers had provisions for 45 days. If 15 more soldiers
joined the garrison, how long would the provisions last ?

14. a) With an average speed of 45 km per hour, a bus takes 10 hours to arrive
Kathmandu from Biratnagar. In how early does it arrive Kathmandu, if its
average speed is increased by 5 km per hour?

b) A contractor hired 18 workers to complete the construction of a road in 60
days. How many more workers should he hire to complete the construction
work 15 days earlier?

15. a) The transportation cost of 35 kg of potatoes is Rs 105.

(i) What is the transportation cost of 2.5 quintals of potatoes at the same rate
for the same distance?

(ii) How many quintals of potatoes can be transported for Rs 1,500 at the same
rate for the same distance?

b) Prabin types 765 words in 9 minutes. How many words would he type
in half an hour? How long does he take to type 12,750 words at the same
speed?

c) Dinesh is a mason and he takes Rs 3,250 for 5 days working everyday. If
he received only Rs 7,800 in two weeks, how many days was he absent in
his work?

It's your time - Project work!

16. a) Let's make groups of students and conduct a survey to find the rate of cost
of the following items in your local market.

(i) Milk per litre (ii) Rice per kg (iii) Sugar per kg (iv) Kitchen oil per litre

(v) Seasonal vegetables per kg (vi) Seasonal fruits per kg

Now, find the cost of three different quantities of each of the items and
discuss in the class.

b) Again, let's increase or decrease the rate of cost of the above surveyed items
and find the cost of different quantities of each item. Write a short report
about the impacts of rates of cost on the cost and quantities.

117Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Unit Percent and Simple Interest

7

7.1 Percent – Looking back

Classroom - Exercise

1. Let's tell and write the percents of these fractions and decimals.

a) 1 = ................ b) 1 = ................ c) 1 = ................
2 4 5

d) 1 = ................ e) 1 = ................ f) 1 = ................
10 20 50

g) 0.08 = ................ h) 0.15 = ................ i) 0.3 = ................

j) 0.66 = ................ k) 0.81 = ................ l) 0.99 = ..............

2. Let's tell and write fractions and decimal of each of the following percents.

a) 7 % = ........., ......... b) 33 % = ........., ......... c) 51 % = ........., .........
d) 69 % = ........., ......... e) 71 % = ........., ......... f) 99 % = ........., .........

Suppose, out of 100 people in a wedding ceremony, 60 are male and 40 are female.
Here 60 percent (60%) are male and 40 percent (40%) are female. Thus, percent
means per hundred or out of hundred.

A percentage is a fraction with the denominator always 100. It is important to
know that 'percentage' is not used with a number. For example, we do say or write
60 percentage, however, it is 60 percent (or 60%).

7.2 Operations on percent

Conversion of fraction or decimal into percent, percent into fraction or decimal,
finding the value of the given percentage of the given quantity, etc. are known as the
fundamental operations on percent.

(i) Conversion of fraction or decimal into percent

While converting a fraction or decimal into percent, we multiply it by 100 and affix

the symbol % to the product. For example:
3 3
5 = 5 × 100 % = 60 %, 0.75 = 0.75 × 100 % = 75 %

(ii) Conversion of percent into fraction or decimal

To covert a percent into fraction or decimal, we divide it by 100 and remove the

% symbol. For example:
25 1 72
25 % = 100 = 4 , 72 % = 100 = 0.72

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Percent and Simple Interest

(iii) To express a given quantity as the percent of whole quantity

In this case, we express the given quantity as the fraction of the whole quantity.

Then, we multiply the fraction by 100 %. For example:

12 as the percent of 60 = 12 × 100 % = 20 % Part of the whole quantity × 100 %
60 Whole quantity

45 as the percent of 180 = 45 × 100% = 25%
180

(iv) To find the value of the given percent of a quantity

In this case, we multiply the quantity by the given percent. Then, we convert the

percent into fraction and simplify. For example:
80
80 % of Rs 350 = 80 % × Rs 350 = 100 × Rs 350 = Rs 280.

(v) To find a quantity whose value of certain percent is given

In this case, we can consider the whole quantity by a variable such as x. Then,
we form an equation and by solving the equation, we can find the value of x. For
example:

If 15 % of a sum is Rs 45, find the sum.

Here, let the sum be Rs x. Alternate process

Now, 15 % of x = Rs 45 15 % of the sum = Rs 45.

or, 153 × x = Rs 45 ? The sum = Rs 45 = Rs 45 = Rs 45 × 100 = Rs 300
100 15% 15 15
20 3x
or, 20 = Rs 45 100

or, x = 20 × Rs 45 = Rs 300.
3

Hence, the required sum is Rs 300.

Worked-out examples

Example 1: If 3 of the number of students of a class are girls, find the percentage
5

Solution: of girls and boys.

Here, 3 = 3 × 100 % = 60 % To convert a fraction into percent
5 5 we should multiply by 100%.
? Percentage of girls = 60 %

Now, the percentage of boys = 100 % – 60 % = 40 %.

Example 2: If 25 % of the people in a village of Jumla have apple farming,

what fraction of the people do not have apple farming?

Solution: Alternate process
N?Hoe14rwe,,ofth2the5ef%rpae=cotpio1l2ne050hoaf=vpe41eaopppllee
Percentage of people who

farming. do not have apple farming
who do not
have apple = 100 % – 25 % = 75 %

farming = 1 – 1 = 3 ?75 % = 753 = 3
4 4 100 4
4

119Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Percent and Simple Interest

Example 3: Ganesh obtained 69 marks out of 75 full marks in Science. Express
his marks in percent.

Solution:
HHeenrec,e6, 9heasotbhtaeipneerdce9n2t%ofm75ar=ks.67591323 × 1040 % = 92 %

Example 4: The rate of cost of Sugar is increased from Rs 70 per kg to Rs 77 per
kg. Find the percent of increase in the price.

Solution:

Here, the initial rate of price = Rs 70 per kg.

The new rate of price = Rs 77 per kg.

? Increment in the rate of price = Rs 77 – Rs 70 = Rs 7

Now, the increment in percent = Increased price × 100 %
Initial price

= Rs 7 × 100 % = 10 %
Rs 70

Hence, the rate of price of Sugar is increased by 10 %.

Note: Percent of deducted price = Deducted price × 100 %
Initial price

Example 5: The monthly income of Mrs. Pariyar is Rs 36,000. She spends 20%
of her income on her children’s education, 25% on food, 30% on
house rent and the rest she saves in a bank.

a) Find her expenditure in each item.

b) How much does she save in the bank?

Solution:

Here, the total monthly income of Mrs. Pariyar = Rs 36,000

Expenditure on children’s education = 20 % of Rs 36,000 = 20 × Rs 36,000
= Rs 7,200 100

Expenditure on food = 25 % of Rs 36,000 = 25 × Rs 36,000
= Rs 9,000 100

Expenditure on rent = 30 % of Rs 36,000 = 30 × Rs 36,000
100
= Rs 10,800

? Her total expenditure = Rs 7,200 + Rs 9,000 + Rs 10,800 = Rs 27,000

Now, her saving in the bank = Rs 36,000 – Rs 27,000 = Rs 9,000.

Hence, she saves Rs 9,000 in the bank.

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Percent and Simple Interest

Example 6: There are 2475 males in a village which is 45 % of the total population.

a) Find the population of the village.

b) Find the number of females.

Solution: Alternative process:

Let, the total population of a village be x. 45% of total population = 2,475
a) Here, 45 % of x = 2475 2,475
or, total population = 45%

or, 45 9 × x = 2475 = 2,475
10020 45
100
or, 9x = 2475
20 = 2,475 × 100
20 × 2475 45
or, x = 9
= 5,500

= 5,500

Hence, the total population of the village is 5,500.

b) Again, the number of females = 5,500 – 2,475 = 3,025
Hence, there are 3,025 females in the village.

Example 7: Mr. Majhi spends 68 % of his monthly income to run the family

and the rest he saves in a bank. If he saves Rs 14,400 in the bank,

how much does he spend every month?

Solution:

Let his monthly income be Rs x.

Here, his saving percentage = 100 % – 68 % = 32 % The amount of saving is given. So,
we need to find the saving percent!
Now,32 % of x = Rs 14,400

or, 32 ×x = Rs 14,400
100

or, 8x = Rs 14,400
25

or, x = 25 × Rs 14,400 = Rs 45,000
8
? His monthly income = Rs 45,000

Now, his monthly expenditure = Rs 45,000 – Rs 14,400 = Rs 30,600.

Hence, he spends Rs 30,600 every month.

EXERCISE 7.1

General Section - Classwork

1. Let's tell and write the percent of these fractions and decimals.

a) 1 = ................., 3 = .............., b) 1 = .............., 2 = ..............
4 4 5 5

c) 1 = ................., 7 = .............., d) 1 = .............., 11 = ..............
10 10 20 20

121Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Percent and Simple Interest

e) 1 = .............., 27 = .............., f) 0.05 = .............., 0.5 = ................
50 50

g) 0.03 = .............., 0.3 = .............., h) 0.09 = .............., 0.9 = ................

2. Let's tell and write the fractions and decimals of the following percent.

a) 7% = ................, ................. b) 33 % = ................, .................

c) 59% = ................, ................. d) 97 % = ................, .................

3. 10% means 10 = 1 and 10% of 60 = 1 of 60 = 6. Apply this rule, tell and
100 10 10

write the values as quickly as possible.

a) 2 % of 100 = ............. b) 4 % of 50 = ............. c) 5 % of 120 = .............

d) 10 % of 150 = ............. e) 20 % of 50 = ............. f) 25 % of 80 = .............

Creative Section - A

4. Let’s express the following fractions and decimals in percent.
3 2 7
a) 4 b) 5 c) 10 d) 0.09 e) 0.3 f) 0.48

5. Let’s express the following percent in fractions and decimals.
a) 4% b) 18% c) 45% d) 64% e) 75% f) 98%

6. Lets express the following in percent.

a) 9 as percent of 36 b) Rs 55 as percent of Rs 550

c) 216 girls out of 450 students d) 60 marks out of 75 full marks

7. Let’s find the values.

a) 5% of Rs 360 b) 16% of 750 students c) 25% of 9,640 people
f) 99% of 900 villagers
d) 75% of 80 full marks e) 60% of 50 kg

8. Let’s find the whole quantity whose value of certain percentage is given below.

a) 12% is Rs 84 b) 10% is Rs 140 c) 25 % is 111 boys
d) 40% is 120 km e) 48% is 144 kg f) 85 % is 6,800 males

9. a) 3 parts of the total population of a village are literate. Find the percent of
5

(i) literate population. (ii) illiterate population

b) Mr. Joshi spends 3 of his monthly income every month.
10

(i) What percent does he spend ? (ii) What percent does he save ?

c) An alloy contains 35% zinc and the rest is copper.

(i) What fraction of the alloy is zinc? (ii) What fraction of the alloy is copper?

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Creative Section - B

10. a) Sunayana obtained 32 marks out of 40 full marks in mathematics. What
percent of marks did she obtain in mathematics ?

b) Mr. Jha earns Rs 25,000 in a month. He spends Rs 8,000 on food, Rs 7,000 on
rent, Rs 3,000 on his children’s education and the rest he saves in a bank.
(i) Express separately, the expenditure on food, rent and education in
percent.
(ii) What is his saving percent every month ?

11. a) There were 540 students in a school last year and 648 students this year.
Find the increased number of students in percent.

b) The rate of cost of petrol is decreased from Rs 110 per litre to Rs 104.50 per
litre. By how many percent is the cost decreased?

12. a) During the rapid diagnostic test of coronavirus among 250 people, 96%
were found negative. How many people were found negative result?

b) There are 640 students in a school and 55% of them are girls.

(i) Find the number of girls. (ii) Find the number of boys.

c) Debasis earns Rs 27,800 in a month. He spends 15% of his income on his
daughter’s education, 25% on food, and 10% on miscellaneous items.

(i) Calculate separately, his expenditure on education, on food and on
miscellaneous items.

(ii) How much is his total expenditure ? How much does he save every
month?

d) The population of a town increases by 4% every year. In the last year, the
population of the town was 1,10,000,

(i) Find the present population of the town.

(ii) Find the population of the town in the next year.

13. a) 52 % of the total number of students in a school are boys. If there are 260
boys,

(i) Find the total number of students in the school.

(ii) How many of them are girls ?

b) Mrs Gurung spends 75% of her income every month which amounts to
Rs 9000 and she saves the rest in a bank.

(i) Find her total income. (ii) How much money does she save in the bank?

c) 65% of the total number of students in a school are girls and there are 224
boys.

(i) What percent of the students are boys ?

(ii) Find the total number of students in the school.

(iii) Find the number of girls.

d) 45 % of the total population of a village are males and there are 11000
females. Find the male population of the village.

123Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Percent and Simple Interest

14. a) Bibek secured 18 marks out of 25 full marks in a unit test of mathematics.
Sunita secured 68% in the same test. Who did better in the test?

b) A monitor is elected from each class in a school. Bishwant got 28 out of 35
votes in his class and Sunayana got 34 out of 40 votes in her class. Who got
the greater percent of votes ?

It’s your time - Project work!

15. a) How many teachers are there in your school? How many are male teachers
and female teachers? Express the numbers in percent.

b) How many students are there in your class? How many are boys and girls
are there? Express the numbers in percent.

c) Let’s ask you parents, the rate of cost of rice five years ago. What is the rate
of cost of rice at present? Express the increased rate of cost percent.

7.3 Simple Interest – Review

When we deposit money in a bank for a certain interval of time, the bank pays
us some additional amount of money under a certain condition. Such additional
amount of money is called interest. Similarly, if we borrow money from a bank, we
should pay interest to the bank. We are already familiar about the following terms
that appear in the calculation of simple interest.

Principal (P) : It is the deposited or borrowed sum of money.

Interest (I) : It is the additional amount of money charged for the deposited
or borrowed sum of money.

Rate of interest (R): It is the condition under which the interest is charged. It is
usually expressed in percent.

Time (T) : It is the duration for which principal is deposited or borrowed.

Amount (A) : It is the total sum of principal and interest.

7.4 Calculation of simple interest

Let Rs P be principal, R % per year be the rate of interest, T years be the time and
Rs I be the interest.

Here, the rate R % per year means, Rs R is the interest of Rs 100 for 1 year.

So, the interest of Rs 100 for 1 year = Rs R

The interest of Re 1 for 1 year = Rs R
100
R
The interest of Rs P for 1 year = Rs 100 ×P

The interest of Rs P for T years = Rs R × P × T = Rs P×T×R
100 100
P×T×R
Thus, the simple interest (I) is calculated as: I= 100

Again, if I = P×T×R
100

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Percent and Simple Interest

P × T × R = I × 100

? P = I × 100 is the formula to calculate principal (P).
T×R

T = I × 100 is the formula to calculate time (T).
P×R

R = I × 100 is the formula to calculate rate (R).
P×T

Furthermore, amount (A) is the sum of the principal (P) and its interest (I).
? A = P + I is the formula to calculate amount (A).

P = A – I is the formula to calculate principal (P).
I = A – P is the formula to calculate interest (I).

Workedout examples

Example 1: Mr. Sherpa deposited a sum of Rs 7,800 in a bank at the rate of 6%
per year. Find the amount received by him at the end of 3 years.

Solution:

Here, Principal (P) = Rs 7,800

Rate (R) = 6 % per year

Time (T) = 3 years

Now, interest (I) = P×T×R
100

= Rs 7800 × 3 × 6 = Rs 1,404.
100

Again, amount (A) = P + I

= Rs 7,800 + Rs 1,404 = Rs 9,204.

Hence, he received an amount of Rs 9,204.

Example 2: In how many years does a sum of Rs 4500 amount to Rs 5400 at
5 % per year simple interest?

Solution:

Here, Principal (P) = Rs 4500

Amount (A) = Rs 5400

Rate (R) = 5 % per year

Now, Interest (I) =A–P

= Rs 5400 – Rs 4500 = Rs 900

Again, time (T) = I × 100 = 900 × 100 = 4 years
P×R 4500 × 5

So, the required time is 4 years.

125Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Percent and Simple Interest

EXERCISE 7.2
General Section A – Classwork

1. Let's tell and write the answers as quickly as possible.

(a) When P = Rs 100, T = 1 year, R = 5% per year, I = ......................

(b) When P = Rs 100, T = 1 year, R = 10 % per year, I = ......................

(c) When P = Rs 100, T = 1 year, I = Rs 8, R = ...................... per year.

(d) When P = Rs 100, T = 1 year, I = Rs 12, R = ...................... Per year.

(e) R = 7 % per year means P = ............, T = ................., I = .....................

(f) R = 9 % per year means, P = ..................., T = ................, I = ...........

Creative Section - A

2. Let’s find the simple interest and amount in the following cases.

a) P = Rs 1500, T = 2 years, R = 5 % per year

b) P = Rs 3600, T = 5 years, R = 10 % per year

c) P = Rs 4500, T = 4.5 years, R = 8 % year

d) P = Rs 8000, T = 3.5 years, R = 10.5 % per year

3. Let’s find the principal in the following cases. Also calculate the amount.

a) I = Rs 336, T = 2 years, R = 10.5 % per year

b) I = Rs 810, T = 4.5 years, R = 9 % per year

c) I = Rs 660, T = 2.5 years, R = 6 % per year

d) I = Rs 2856, T = 3.5 years, R = 8.5 % per year

4. Let’s find the rate of interest in the following cases.

a) P = Rs 1800, T = 3 years, I = Rs 324

b) P = Rs 2440, T = 4 years, I = Rs 488

c) P = Rs 3520, T = 2.5 years, I = Rs 792

d) P = Rs 8800, T = 3.5 years, I = Rs 2618

5. Let’s find the time in the following cases.

a) P = Rs 1400, R = 8 % per year, I = Rs 224

b) P = Rs 3600, R = 7 % per year, I = Rs 756

c) P = Rs 4400, R = 10.5 % per year, I = Rs 2310

d) P = Rs 9200, R = 12.5 % per year, I = Rs 2875

6. a) Shashwat deposited a sum of Rs 4,500 in a bank at the rate of 8 % per year for
2.5 years. (i) Find the interest of the sum. (ii) Find the amount received
by him at the end of 4 years.

Vedanta Excel in Mathematics - Book 7 126 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Percent and Simple Interest

b) Mr. Maharjan borrowed a sum of Rs 9,000 from an Agricultural Development
Bank at the rate of 12.5 %. How much amount should he pay to clear the
debt after 5 years?

7. a) Mrs. Karki deposited a sum of money in a bank at the rate of 5 % per year
simple interest. If she received an interest of Rs 4,740, at the end of 6 years,
how much money did she deposit in the bank?

b) Mr. Chaudhari borrowed a sum of money from a finance company at the
rate of 18 % per year. If he paid an interest of Rs 2,160 at the end of 5 years,
find the sum borrowed by him.

8 a) Dinesh lent a sum of Rs 4,200 to Dipesh at the rate of 7 % per year. If Dipesh
paid him an interest of Rs 588, how many years did Dipesh use his money?

b) In how many years would the simple interest of a sum of Rs 7,800 at 5.5 %
per year be Rs 1,716?

9. a) A services man deposited his saving of Rs 5,100 in a bank. At the end of
3 years if he received an amount of Rs 6,477 from the bank, find the rate of
interest.

b) At what rate of interest per year does a sum of Rs 6,900 amount to Rs 8,832
in 4 years?

Creative Section - B

10. a) Sita borrowed Rs 5,000 as a loan from Gita at the rate of 5% per annum for
4 years and she lent the same sum to Rita at the rate of 8% per annum for
3 years. How much profit or loss did Sita get in this transaction?

b) A money lender took a loan of Rs 8,000 from bank at the rate of 6% p.a. for
3 yeras and immediately he deposited the same sum in a cooperatives at the
rate of 9% for 2 years 6 months. How much did he gain in this transaction?

11. a) Shashwat has just celebrated his third birthday. On this occasion his father
deposited Rs 15,000 in a bank for him at the rate of 10% p. a. How much
amount does receive on his sixth birthday?

b) Mr. Bhandari borrowed Rs 24,000 from a bank in the beginning of 2073 BS.
If the bank fetched the interest at the rate of 8.5% per annum, how much
amount did he pay to clear the debt at the end of 2077 BS?

It's your time - Project work!

12. a) Ask your parents if they have any bank account. What type of account is
it: current saving, or fixed deposit? What is the rate of interest given by the
bank.

b) Let's visit the banks /cooperatives near your place. Record their rates of
interest and suggest your parents to deposit the money at the appropriate
bank or cooperative.

127Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Unit Profit and Loss

8

8.1 Profit and Loss – Looking back

Classroom - Exercise

1. Let's answer these questions as quickly as possible.

A stationer buys an exercise book for Rs 50 and sells it for Rs 60.

a) What is the cost price (C.P.) of the exercise book? .............................

b) What is the selling price (S.P.) of the exercise book? .............................
c) Does the stationer make profit or loss? .............................

How much and how did you find it? .............................

d) Which one between, C.P and should be greater S.P to make profit ? ...............

e) If the S.P. of the exercise book were Rs 48,
how much loss would he make? How did you find it? ..............................

f) Now, let's state the mathematical rule to find profit. Profit = ..........................

g) Also, let's state the mathematical rule to find loss. Loss = ..........................

2. Let's tell and write the answers as quickly as possible.

a) If C.P = Rs 40, S.P = Rs 50, profit = ..............................

b) If C.P = Rs 85, S.P = Rs 70, loss = ..............................

c) If C.P = Rs 50, profit = Rs 5, S.P. = ..............................

d) If S.P = Rs 75, profit = Rs 10, C. P = ..............................

e) If C.P = Rs Rs 120, loss = Rs 20, S.P. = ..............................

f) If S.P = Rs 250, loss = Rs 25, C.P = ..............................

3. a) If C.P = Rs = 100, Profit = Rs 15, profit percent = ..............................

b) If C.P = Rs 100, loss = Rs 20, loss percent = ..............................

When we purchase something, we need to pay the cost of the thing. This cost is
called cost price (C.P.) Similarly, the price at which something is sold is called
selling price (S.P.)

When selling price of an article is higher than its cost price, a profit is made.

Thus, when S.P. > C.P., profit = S.P. – C.P.
Also, S.P. = C.P. + Profit and C.P. = S.P. – Profit

However, if selling price of an article is less than its cost price, there is loss to the
seller.

Thus, when S.P. < C.P., loss = C.P. – S.P.
Also, S.P. = C.P. – Loss and C.P. = S.P. + Loss

Vedanta Excel in Mathematics - Book 7 128 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Profit and Loss

8.2 Profit and loss per cent
Let the C.P. of an article is Rs 100 and it is sold at a profit of Rs 10.
Here, the profit of Rs 10 is out of the C.P. of Rs 100. Therefore, it is called 10% profit.
Also, let the C.P. of an article be Rs 100 and it is sold at a loss of Rs 5.
Here, the loss of Rs 5 is out of the C.P. of Rs 100. Therefore, it is called 5% loss.
Thus, profit or loss is always on C.P. If C.P. is taken as Rs 100 and profit or loss is
calculated on it, it is called profit percent or loss percent.

Suppose, C.P. of a pen is Rs 40 Suppose, C.P. of a pen is Rs 40

and it is sold at a profit of Rs 8. Then, and it is sold at a loss of Rs 4. Then,

on C.P. of Rs 40, profit is Rs 8 on C.P. of Rs 40, loss is Rs 4

on C.P. of Re 1, profit is Rs 8 on C.P. of Re 1, loss is Rs 4
40 40
8 4
on C.P. of Rs 100, profit is Rs 40 × 100 on C.P. of Rs 100, loss is Rs 40 × 100

= Rs 20 = Rs 10

Here, Rs 20 is the profit out of the C.P. Here, Rs 10 is the loss out of the C.P.
of Rs 100. So, it is 20% profit. of Rs 100. So, it is 10% loss.

Now, from the above illustrations, we can generalise the following formulas to
calculate profit or loss percent.

Profit percent = Profit × 100 % S.P. – C.P. × 100 %
C.P. C.P.

Loss percent = Loss × 100 % C.P. – S.P. × 100 %
C.P. C.P.

Worked-out examples

Example 1: A shopkeeper buys a T-shirt for Rs 600 and sells it for Rs 678. Find
his profit percent.

Solution:
Here, C.P. of the T-shirt = Rs 678 and S.P. of the T-shirt = Rs 600

? Profit = S.P. – C.P. = Rs 678 – Rs 600 = Rs 78

Now, profit percent = Profit × 100 % = Rs 78 × 100 % = 13 %
C.P. Rs 600

Hence, the required profit percent is 13 %.

Example 2: Mr. Chamling bought 150 eggs at the rate of Rs 8 each. 10 of them
were broken and he sold the remaining at the rate of Rs 9 each.
Find his profit or loss percent.

129Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Profit and Loss

Solution:
Here, the remaining number of eggs = 150 – 10 = 140

C.P. of 150 eggs = 150 × Rs 8 = Rs 1,200

S.P. of 140 eggs = 140 × Rs 9 = Rs 1,260

? Profit = S.P. – C.P. = Rs 1,260 – Rs 1,200 = Rs 60

Now, profit percent = profit × 100% = Rs 60 × 100% = 5%
C.P. Rs 1200

Hence, his profit percent is 5%.

Example 3: A trader purchased a mobile for Rs 4,800 and sold it at a profit of
20%. How much profit did she make?

Solution:

Here, C.P. of the mobile = Rs 4,800

profit percent = 20% 20
100
? Amount of profit = 20% of C.P. = × Rs 4,800 = Rs 960

Hence, she made a profit of Rs 960.

8.3 Calculation of S.P. when C.P. and profit or loss per cent are given
In this case, C.P. is given and profit or loss percent is given. Then, we apply the
following processes to find S.P.:

Actual profit = profit percent × C.P.

Actual loss = loss percent × C.P.

Then, we calculate S.P. as:
S.P. = C.P. + Actual profit or S.P. = C.P. – Actual loss

Example 4: A shopkeeper bought a pair of shoes for Rs 1,620 and sold it at a
profit of 25%.

a) Find his profit amount. b) Find the selling price of the shoes.

Solution:

Here, C.P. of the shoes = Rs 1,620

Profit percent = 25 % 25
100
a) ? Actual profit = 25 % of C.P. = × Rs 1,620 = Rs 405

Hence, the required amount of profit is Rs 405.

b) Again, S.P. of the shoes = C.P. + profit = Rs 1,620 + Rs 405 = Rs 2,025
Hence, the required selling price of the shoes is Rs 2,025.

8.4 Calculation of C.P. when S.P. and profit or loss per cent are given
In this case, S.P. is given and profit or loss percent is given. Then, we apply the
following process to find C.P.

Vedanta Excel in Mathematics - Book 7 130 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Profit and Loss

Let's consider, C.P. = Rs x

Then, actual profit = P% of C.P. = P% of Rs x

Also, actual loss = L% of C.P. = L% of Rs x

Now, C.P. = S.P. – Profit OR C.P. = S.P. + Loss

or, x = S.P. – P% of Rs x x = S.P. + L% of x

Then, solving the equation, we find the value of x which is the required C.P.

Example 5: If S.P. = Rs 400 and loss percent is 20 %, find C.P.
Solution:

Let, the required C.P. be Rs x. Profit or loss percent is always out of C.P. So, actual
Here, actual loss = 20 % of C.P. profit or actual loss is always calculated from C.P.

= 20 × Rs x = Rs x
100 5

Now, C.P. = S.P. + loss x
or, 5
or, x = Rs 400 + Rs
x
or, x– 5 = Rs 400

or, 4x = Rs 400
5
5 × Rs 400
x = 4 = Rs 500

So, the required C.P. is Rs 500.

Example 6: Suntali sold a bag for Rs 832 at 4 % profit. At what price did she
purchase the bag?

Solution:

Here, S.P. of the bag = Rs 832

Profit percent = 4 %

Let the C.P. of the bag be Rs x. 4
100
Now, actual profit= 4 % of C.P. = × Rs x Alternative process:

= Rs x C.P. = 100 × S.P.
25 (100 + p)
Again, C.P. = S.P. – profit
= 100 × Rs 832
x (100 + 4)
or, x = Rs 832 – 25
= 100 × Rs 832
or, x+ x = Rs 832 104
25
26x = Rs 800
25
or, = Rs 832

or, x = 25 × Rs 832 = Rs 800
26

Hence, she purchased the bag for Rs 800.

131Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Profit and Loss

EXERCISE 8.1

General Section - Classwork
Tell and write the answers as quickly as possible.

1. a) If C.P = Rs 300, S.P. = Rs 330, then profit = .........................

b) If C.P. = Rs 420, S.P. = Rs 400, then loss = .........................

c) If C.P. = Rs 450, Profit = Rs 100, then S.P = .........................

d) If C. P.= Rs 200, Loss = Rs 20, then S.P = .........................

e) If S.P. = Rs 800, profit = Rs 100, then C.P = .........................

f) If S.P. = Rs 700, loss = Rs, 70, then C. P = .........................

2. a) If C.P. = Rs 100, profit = Rs 5, then profit percent = .........................

b) If C.P. = Rs 100, loss = Rs 10, then loss percent = .........................

c) If C.P. = Rs 100, S.P. = Rs 120, then profit percent = .........................

d) If C.P. = Rs 100, S.P. = Rs 80, then loss percent = .........................

3. a) If C. P = Rs 100, profit percent = 12% , then S.P. = ............................

b) If C.P = Rs 100, loss percent = 8% , then S.P. = ............................

Creative Section - A

Let's find the unknown variables in the following cases.
4. a) If C.P. = Rs 150 and profit = Rs 12, find profit percent.

b) If C.P. = Rs 440 and loss = Rs 22, find loss percent.
c) If C.P. = Rs 360, S.P. = Rs 396, find profit and profit percent.
d) If C.P. = Rs 700, S.P. = Rs 665, find loss and loss percent.
e) If C.P. = Rs 590 and profit = Rs 75, find S.P.
f) If C.P. = Rs 630 and loss = Rs 48, find S.P.
5. a) If C.P. = Rs 280 and profit percent = 10%, find profit amount.

b) If C.P. = Rs 810 and loss percent = 20%, find loss amount.

c) If C.P. = Rs 900 and profit percent = 8%, find profit amount and S.P.

d) If C.P. = Rs 1,500 and loss percent = 7%, find loss amount and S.P.

6. a) If S.P. = Rs 210, profit percent = 5%, C.P. = Rs x, find C.P.

b) If S.P. = Rs 315, loss percent = 10%, C.P. = Rs x, find C.P.

7. If S.P. = Rs 480, profit percent = 20%, then

C.P. = S.P. = Rs 480 = Rs 480 × 100 = Rs 400
120% 120 120
100

Vedanta Excel in Mathematics - Book 7 132 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Profit and Loss

If S.P. = Rs 240, loss percent = 20%, then

C.P. = S.P. = Rs 240 = Rs 240 × 100 = Rs 300
80% 80 80
100
Let’s apply the similar process and find C.P.

a) S.P. = Rs 132, profit percent = 10%, find C.P.

b) S.P. = Rs 325, profit percent = 25%, find C.P.

c) S.P. = Rs 264, loss percent = 12%, find C.P.

d) S.P. = Rs 425, loss percent = 15%, find C.P.

8. a) A stationer buys 1 dozen of pens at Rs 20 each and sells them at Rs 25 each.
Find his profit and profit percent.

b) A grocer purchased 5 dozen of eggs at Rs 8 each. 10 eggs were broken and
he sold the remaining at Rs 9 each. Find:
(i) his total profit or loss. (ii) Profit or loss percent.

c) A fruit seller sold 50 kg of oranges at the rate of Rs 80 per kg and gained
Rs 800. Calculate (i) his purchasing rate. (ii) Profit percent.

d) A vegetable seller purchased 1 quintal of potatoes at Rs 35 per kg and sold
at a loss of Rs 350. Find his (i) rate of selling price (ii) Loss percent.

9. a) Mrs. Shrestha bought a fancy item for Rs 720 and sold it at 10 % profit.

(i) Find her profit amount. (ii) Find the selling price of the item.

b) A retailer purchased a mobile for Rs 3,250 and sold it at 4% loss.

(i) Find his loss amount. (ii) At what price did he sell the mobile?

c) Anamol buys an old bicycle for Rs 4,500 and he spends Rs 500 to repair it.
If he sells it at 25% profit, find the selling price of the bicycle.

d) Ashma purchased a second hand scooter for Rs 1,52,000. She spent Rs 8,000
10. a) to repair it and sold it at 3% loss. At what price did she sell the scooter?
Mr. Agrawal has an electrical shop. He sold a tube light for Rs 240 at 20%
profit. Find the cost price of tube light.

b) Mrs. Sunuwar sells digital items. She sold a digital watch for Rs 1,260 and

made 10% loss. At what price did she purchase the watch?

Creative Section - B

11. a) A grocer bought 200 eggs at the rate of Rs 9 each. 20 eggs were broken and
he sold the remaining eggs at 10% profit. Find the rate of selling price of
each egg.

b) Suntali bought 50 kg oranges at the rate of Rs 80 per kg. She paid Rs 5 per
kg as the fare to bring the oranges to her shop. If she sold all the oranges
for Rs 4,675. Find her profit or loss percent.

133Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Profit and Loss

c) A stationer bought 100 books at Rs 80 each. He donated 10 books in a
school's library and sold rest of the books at Rs 88 each. Calculate his gain
or loss percent.

12. a) A shopkeeper purchased a bag for Rs 800 and an umbrella for Rs 400. If
she sold the bag at 15% profit and umbrella at 15% loss, find her profit or
loss percent in whole transaction.

b) A farmer bought a goat for Rs 15000 and a cow for Rs 35000. If he sold the
goat at 10% profit and cow at 20% loss, find his profit or loss percent in
whole transaction.

It’s your time - Project work!

13. a) Let’s become a problem maker and problem solver.

Write the values of the variables of your own. Then, solve each problem to
find unknown variable.

C.P. = ................... C.P. = ................... C.P. = ..........
S.P. = ................... S.P. = ................... Profit percent = ..........
Find profit percent Find loss percent Find S.P.

C.P. = .......... S.P. = .......... S.P. = ..........
Loss percent = .......... Profit percent = .......... Loss percent = ..........
Find S.P. Find C.P. Find C.P.

b) Let’s write appropriate amount of C.P. and S.P. to get the given profit or loss
percent.

Profit percent = 10% Loss percent = 5%
C. P. = ............................ C. P. = ............................
S. P. = ............................ S. P. = ............................

8.5 Discount
Have you ever asked a shopkeeper to get discount while buying some goods? If so,
what do you understand by discount?

A shopkeeper may label the price of her goods. This labelled price is called Marked
Price (M.P.). Later, the shopkeeper may reduce the marked price of his goods and
sells to the customers. The reduced amount of money from the marked price is
called discount. For example:

Suppose the marked price of a pen is Rs 65.

Let the shopkeeper reduces the marked price by Rs 5.
Here, Rs 5 is the discount given to the customer.

Vedanta Excel in Mathematics - Book 7 134 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Profit and Loss

In this case, S.P. of the pen = Rs 65 – Rs 5 = Rs 60

Thus, S.P. = M.P. – Discount
And, Discount = M.P. – S.P.

8.6 Discount per cent

Let the marked price of an article is Rs 100 and shopkeeper allows a discount of
Rs 10. Here Rs 10 is 10% discount.

Thus, when M.P. is considered as Rs 100 and discount is calculated from it, it is
called discount percent.

Discount percent = Discount u 100 %
M.P.

Discount amount = Discount percent of M.P.

Worked-out examples

Example 1: The marked price of an article is Rs 540. If the shopkeeper gives
some discount and sells it for Rs 513, find the discount percent.

Solution:

Here, M.P. of the article = Rs 540

S.P. of the article = Rs 513

? Discount = M.P. – S.P. = Rs 540 – Rs 513 = Rs 27

Now, discount percent = Discount × 100 % = Rs 27 × 100 % = 5 %
M.P. Rs 540

Hence, the required discount percent is 5 %.

Example 2: The marked price of a mobile is Rs 3,870 and the retailer allows
Solution: 10 % discount to the customer.

a) Find the discount amount.

b) How much should the customer pay for it?

Here, M.P. of the mobile = Rs 3,870

Discount percent = 10 % 10
100
a) ? Discount amount = 10 % of M.P. = × Rs 3,870 = Rs 387

b) Again, S.P.= M.P. – Discount = Rs 3,870 – Rs 387 = Rs 3,483

Hence, the customer should pay Rs 3,483.

8.7 Value Added Tax (VAT)

Value Added Tax (VAT) is known as a goods and services tax. Vat is charged on the
actual selling price of goods. VAT is calculated as a certain percent of selling price
(S.P.).

135Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Profit and Loss

VAT amount = VAT percent of S.P.
S.P. with VAT = S.P. + VAT percent of S.P.

Example 3: The selling price of a bag is Rs 1,200 and 13% VAT is charged on the
Solution: selling price.
a) Calculate the amount of VAT. b) Find S.P. with VAT.

Here, S.P. of the bag = Rs 1,200 and VAT = 13%

a) Now, amount of VAT = VAT % of S.P. 13
100
= 13 % of Rs 1,200 = × Rs 1,200 = Rs 156

b) Then, S.P. with VAT = S.P. + VAT amount = Rs 1,200 + Rs 156 = Rs 1,356
Example 4: The marked price of an article is Rs 2,000 and the shopkeeper

allows 20% discount.
a) Calculate the discount amount.
b) Find S.P. of the article after discount.
c) If 13% VAT is charged on S.P. after discount, calculate the VAT

amount.
d) Find S.P. of the article with VAT.
Solution:

Here, M.P. of the article = Rs 2,000, discount = 20%, VAT = 13%

a) Now, discount amount = discount% of M.P. = 20% of Rs 2,000

= 20 × Rs 2,000 = Rs 400
100

b) Then, S.P. of the article after discount = M.P. – discount amount

= Rs 2,000 – Rs 400 = Rs 1,600

c) Again, VAT amount = VAT% of S.P.

= 13% of Rs 1,600 = 13 × Rs 1,600 = Rs 208
100

d) Then, S.P. with VAT = S.P. + VAT amount = Rs 1,600 + Rs 208 = Rs 1,808

Example 5: The marked price of a tablet is Rs 7,000. If the shopkeeper allows

10 % discount, how much should a customer pay for it with 13 %

VAT? Shorter process:

Solution: S.P. = M.P. – discount % of M.P.
= Rs 7,000 – 10% of Rs 7,000
Here, M.P. of the tablet = Rs 7,000

Discount percent = 10 % = Rs 7,000 – 10 × Rs 7,000
VAT percent = 13 % 100
Now, discount amount = 10 % of M.P. = Rs 7,000 – Rs 700 = Rs 6,300

= 10 × Rs 7,000 = Rs 700
100

? S.P. = M.P. – Discount = Rs 7,000 – Rs 700= Rs 6,300

Vedanta Excel in Mathematics - Book 7 136 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Profit and Loss

Again, VAT amount = 13 % of S.P. Shorter process:

= 13 × Rs 6,300 S.P. with VAT = S.P. + VAT% of S.P.
100
= Rs 6,300 + 13% of Rs 6,300

= Rs 819 = Rs 6,300 + 13 × Rs 6,300
100
= Rs 6,300 + Rs 819 = Rs 7,119
? S.P. with VAT = S.P. + VAT amount

= Rs 6,300 + Rs 819 = Rs 7,119

Hence, the customer should pay Rs 7,119.

EXERCISE 8.2

General Section - Classwork
Tell and write the answers as quickly as possible.

1. a) M.P. = Rs 60, discount = Rs 5, S.P. = .............................

b) M.P. = Rs 450, discount = Rs 50, S.P. = .............................

c) M.P. = Rs 250, S.P. = Rs 225, discount = .............................

d) M.P. = Rs 500, S.P. = Rs 400, discount = .............................

e) S.P. = Rs 340, discount = Rs 10, M.P. = .............................

f) S.P. = Rs 660, discount = Rs 40, M.P. = .............................

2. a) M.P. = Rs 100, discount = Rs 12, discount percent= .............

b) M.P. = Rs 100, discount = Rs 20, discount percent= .............

c) M.P. = Rs 100, discount percent = 15%, discount = .............

d) M.P. = Rs 100, discount percent = 10%, discount = .............

e) M.P. = Rs 100, S.P. = Rs 80, discount percent= .............

f) M.P. = Rs 100, S.P. =Rs 95 discount percent= .............

g) M.P. = Rs 100, discount percent = 25%, S. P. = .............

h) M.P. = Rs 100, discount percent = 10%, S. P. = .............

3. a) S.P. = Rs 300, VAT = Rs 30, S.P with VAT = .........................

b) S.P. = Rs 400, VAT = Rs, 50, S.P. with VAT =.........................

c) S.P. = Rs 100, S.P with VAT = Rs 113, VAT =.........................

d) S.P. = Rs 200, S.P with VAT = Rs 226, VAT =.........................

e) S.P. = Rs 100, VAT = Rs 13, VAT Percent =.........................

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Profit and Loss

f) S.P. = Rs 100, VAT = Rs 10, VAT Percent =.........................

g) S.P. = Rs 100, VAT Percent = 13%, VAT =.........................

h) S.P. = Rs 100, VAT Percent = 10%, VAT =.........................

i) S.P. = Rs 100, VAT Percent = 13%, S.P. with VAT =.........................

j) S.P. = Rs 100, VAT Percent = 15%, S.P. with VAT =.........................

Creative Section - A
4. a) Define marked price and discount with examples.

b) Write the formula to find discount percent from marked price and discount
amount.

c) Write the formula to find S.P. from M.P. and discount amount.
d) What is the full form of VAT? What is the rate of VAT in Nepal at present?

Write the formula to find S.P. with VAT from S.P. and VAT amount.
Let's find the unknown variables in the following cases.
5. a) If M.P. = Rs 250, discount = 10%, find discount amount.
b) If M.P. = Rs 320, discount = 5%, find discount amount.
c) If M.P. = Rs 550, discount = 12%, find discount amount and S.P.
d) If M.P. = Rs 960, discount = 15%, find discount amount and S.P.

6. a) If M.P. = Rs 300, discount amount = Rs 12, find discount percent.
b) If M.P. = Rs 840, discount amount = Rs 42, find discount percent.
c) If M.P. = Rs 1,200, S.P. = Rs 1,080, find discount amount and discount
percent.
d) If M.P. = Rs 1,620, S.P. = Rs 1,296, find discount amount and discount
percent.

7. a) If S.P. = Rs 400, VAT = 13%, find the amount of VAT.
b) If S.P. = Rs 650, VAT = 12%, find the amount of VAT.
c) If S.P. = Rs 700, VAT = 13%, find S.P. with VAT.
d) If S.P. = Rs 1,080, VAT = 15%, find S.P. with VAT.

8. a) If M. P. = Rs 200, discount = 10%, find S.P. and S.P. with 13% VAT.
b) If M. P. = Rs 500, discount = 20%, find S.P. and S.P. with 10% VAT.
c) If M. P. = Rs 800, discount = 5%, find S.P. and S.P. with 15% VAT.
d) If M. P. = Rs 1,000, discount = 10%, find S.P. and S.P. with 13% VAT.

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Profit and Loss

9. a) The marked price of a fan is Rs 1,500 and the shopkeeper gives some
discount and sells it for Rs 1,200.
(i) Calculate the amount of discount.
(ii) Find the discount percent.

b) The marked price of a sunglasses is Rs 1,450. If the shopkeeper gives some
discount and sells it for Rs 1,334, find the discount percent.

10. a) The marked price of a mobile is Rs 5,600 and the shopkeeper allows 10%
discount.
(i) Find the amount of discount .
(ii) How much should a customer pay for it after discount ?

b) A supermarket announces a heavy discount of 40% to sell its old fashioned
items. How much should a customer pay for the following items ?
(i) Shoes – Rs 1580 (ii) T-shirt – Rs 860 (iii) Trousers – Rs 1220.

11. a) The selling price of a calculator is Rs 800 and 13% VAT is charged on the
selling price.

(i) Calculate the amount of VAT (ii) Find S.P. with VAT

b) The selling price of a digital watch is Rs 1,300. How much should a customer
pay for it with 15% VAT?

Creative Section - B
12. a) The marked price of a fan is Rs 1,600 and the shopkeeper allows 10%

discount.

(i) Calculate the discount amount. (ii) Find S.P. of the fan after discount.

(iii) If 13% VAT is charged on S.P. after discount, calculate the VAT amount.

(iv) Find S.P. of the fan with VAT.

b) The marked price of a jacket is Rs 4,000 and 20% of discount is given.

(i) Find S.P. of the jacket after discount.

(ii) If 15% of VAT is charged on S.P. after discount, calculate the VAT
amount.

(iii) Find S.P. of the jacket with VAT.

c) The marked price of a mobile is Rs 6,000 and shopkeeper allows 5%
discount.

(i) Find S.P. of the mobile after discount.

(ii) Find S.P. of the mobile with 13% VAT.

139Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Profit and Loss

d) The marked price of a television is Rs 10,000 and 10% discount is given.
Find the selling price of the television with 13% VAT.

e) A retailer marked the price of a camera as Rs 8,000 and he allows 15%
discount. How much should a customer pay for it with 15% VAT?

13. a) A supermarket announces an equal rate of discount in every item. If a
customer pays Rs 1,900 for a jacket with marked price Rs 2,000, how much
should the customer pay for a trousers marked with Rs 1,600?

b) The marked price of a dozen of copies is Rs 600. If the shopkeeper allows
20% discount and then charges 10% VAT, how many copies can be bought
for Rs 308?

c) Suppose you are going to buy a bicycle. The shop - A marked its price as
Rs 6,000 and allows 10% discount. Another shop - B fixed the price of same
model of bicycle as Rs 5,200 without giving any discount. In which shop
will you buy the bicycle? Give your reason with calculation.

It’s your time - Project work!
14. a) Let’s become a problem maker and problem solver.

Write the values of the variables of your own. Then, solve each problem to
find unknown variable.

M.P. = ................... M. P. = ........................
Discount % = .............. S. P. = ........................
Find S.P. Find discount percent

S. P. = ........................ S. P. with VAT = .........
VAT% = ........................ VAT% = ................
Find S.P. with VAT Find S.P. without VAT

b) Let’s search the rate of VAT of different countries in the available website.
Compare these VAT rates to the VAT rate of our country.

c) Let’s make group of 5 friends and visit your local market to find the selling
price of some electrical and electronic items. Calculate the S.P. of these items
with the current VAT rate system.

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Unit Algebraic Expressions

9

9.1 Algebraic terms and expressions – Looking back
Classroom - Exercise

1. Let's tell and write the answers as quickly as possible.

a) How many terms are there in each of the following expressions? Also write
the terms.

(i) In 2xy, number of terms: .................., terms are .......................................

(ii) In 2x + y, number of terms: .................., terms are ...................................

(iii) In 2+x+y, number of terms: .................., terms are ..................................

b) In 5x3 , coefficient is ............... base is .............. and power is ......................

c) If l = 3 and b = 2 then (i) 2 (l + b) = .................... (ii) l × b = ....................
2. a) The mathematical expression for the sum of 2x and 3y is .............................

b) The mathematical expression for the difference of 7ab and 3bc is ................

c) The mathematical expression for the product of 5p and 2q is .......................

3. Let's tell and write the sum, difference, or product.

a) 7x 2x = ......... 7x 2x = ......... 7x u 2x = .........

b) 5p2 2p2 = ......... 5p2 2p2 = ......... 5p2 × 2p2 = .........

x, 2x, 3ab, p2q, etc. are algebraic terms. An algebraic expressions is a collection of
one or more terms. Which are separated to each other by either addition (+) or
subtraction (–) sign.
For example: 3xyz, 7x – 2y, x + y – z, etc are algebraic expressions. We can represent
the terms and factors of the terms of an expression by a tree diagram.

For example: Expression (5x yz) (2xy 6)

Terms 5x yz 2xy 6
Factors 5x yz 2 x y 2 3

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Algebraic Expressions

9.2 Types of algebraic expressions

Algebraic expressions are categorized according to the number of terms contained
by the expressions. The table given below shows the types of expressions.

Monomial An algebraic expression with only one term is called a
Binomial monomial . For example : 5xy, – 8m, 9x2y, 11 etc are monomials.
Trinomial
An algebraic expression with two unlike terms is called a

binomial. For example : x + y, x – 4, 2xy + 3x, a2 – b2, pq – r,
x y
a b etc are binomials.

An algebraic expression with three unlike terms is called a trinomial.
For example : 2x +y – 1, a2 + ab + b2, xy + x + y etc are trinomial.

9.3 Polynomial

An algebraic expression with one or more terms and powers of the variables being
whole numbers in each term is called polynomial.

For example : x2y3z, x2 – 4, x + y + 7 etc. are polynomials.

However, x2 + 1 , x2/3 – y2/3 are not polynomials. Because the powers of the variables
x2
in these expressions are not whole numbers.

9.4 Degree of polynomials

Let's study the illustrations given in the table and learn about the degree of
polynomials.

Polynomials Degree of polynomials

2x Power of the variable x is 1. So, its degree is 1.

3y2 Power of the variable y is 2. So, its degree is 2.

x2yz The sum of the powers of the variables x, y, and z
= 2 + 1 + 1 = 4. So, it's degree is 4.

2p3 – 3p2 + 5 The highest power of the variable p is 3. So, its degree is 3.

a2b2 + 2a2b – 4ab2 The highest sum of the powers of ab = 2 + 2 = 4. So, its
degree is 4.

9.5 Evaluation of algebraic expressions
Let's take an algebraic expression 2x – 3y and evaluate it when x = 2 and y = 1.

Here, if x = 2 and y = 1, then 2x – 3y = 2 × 2 – 3 × 1 = 4 – 3 = 1

In this way, the process of finding the value of an algebraic expression by replacing
the variables with numbers is called evaluation.

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Algebraic Expressions

Worked-out examples

Example 1 : Which of the following expressions are polynomials? Write with

reason. 3
x p
a) 2 + 5 b) 3p2 – c) √ 5 y2 + 3 d) 2√ x – 1

Solution:
x
a) 2 + 5 is a polynomial because the power of is 1, which is a whole number.

b) 3p2 – 3 is not a polynomial because the power of the term 3 is –1, which is not
p p

a whole number.

c) √ 5 y2 + 3 is a polynomial because the power of y is 2, which is a whole number.

d) 2√ x – 1 is not a polynomial because the power of x is 1 , which is not a whole
number. 2

Example 2 : Find the degree of a) 4x2 b) 3x2y c) 7x5y2 9x2y3 4xy5
Solution: d) (xy)2 + x2 – y2

a) The degree of 4x2 is 2

b) The degree of 3x2y is 2 + 1 = 3

c) In 7x5y2 , the sum of powers of variables = 5 2 = 7

In 9x2y3, the sum of powers of variables = 2 3 = 5

In 4xy5, the sum of powers of variables = 1 5 = 6

Since the highest sum of powers of variables is 7,

the degree of 7x5y2 –9x2y3 + 4xy5 is 7.

d) Here, (xy)2 + x2 – y2 = x2y2 + x2 + y2

The highest sum of the powers of variables is 2 +2 = 4. So, its degree is 4.

Example 3 : If l = 5 and b = 3, evaluate 2(l + b).
Solution:

Here, when l = 5 and b = 3, then 2(l + b) = 2(5 + 3) = 2 × 8 = 16

EXERCISE 9.1

General Section - Classwork

1. Let's tick (— ) the correct answer.

a) The terms of expression 5x2 – 3xy are

(i) 5x2 and 3xy (ii) x2 and xy (iii) 5x2 and – xy (iv) 5x2 and –3xy

b) The number of terms in the expression x2 + y2 + z2 is
7

(i) 2 (ii) 3 (iii) 4 (iv) 7

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Algebraic Expressions

c) An algebraic expression with three unlike terms is called a

(i) Monomial ii) binomial (iii) trinomial (iv) all of these

d) Which of following expressions is a binomial?

(i) 3m2n – mn2 + m2n (ii) pq + qr + pr

(iii) √ 3 xy + z – 7 (iv) x2y –3x – x2y

e) Which one of the following expressions is not a polynomial? 4
x3 y3 x3 x2
(i) √ 5 x2y + z (ii) 2x3 – xy + y2 (iii) 2 3 (iv) 2

f) The degree of polynomial 2x4yz3 is

(i) 2 (ii) 8 (iii) 4 (iv) 3

g) The degree of polynomial 5x2y – xy + y2 is

(i) 2 (ii) 3 (iii) 5 (iv) 7

2. Let's tell and write the value of the expressions quickly.

a) If x = 3, y = 2, then (i) x + y = ........... (ii) x – y = ........... (iii) xy = ...........

b) If a = 2, b = 3, then (i) 2(a + b) = ........ (ii) a2 = ............ (iii) b2 = ............

Creative Section

3. a) Define algebraic expressions with examples.

b) What are monomial, binomial, and trinomial expressions? Write with
examples.

c) Is x – y + 2x a trinomial expression? Why?

d) What is a polynomial? Give an example of a polynomial.
1
e) Why is x2 a polynomial, but x2 is not a polynomial?

4. Let's identify and then classify the given expressions as monomial, binomial

or trinomial.

a) x2y + xy2 b) 9 – x2 c) XYZ d) pq + p + q

2

e) x2 + y2 f) a2 + a + 1 g) 3x2 + 7xy + 6y2 h) 3x + xy – 8y2

(i) x2 + x (j) – 6x2 k) 1 + x + xy l) 64
y2

5. Let's state with reason whether the given expressions are polynomials.

a) x3 + x2 b) 3 c) ab a b d) x2 + x –2
x2 3

e) √ 3 x2 – xy f) x1/2 + y1/2 g) √ x + 2x + 1 h) 5x3 – 4x2 + 6xy – 7

6. Let's find the degree of the following polynomials.

a) 3x2 b) –2xy c) 4x2yz

d) x2 + 5x + 6 e) 3y3 – 2y2 + 5y – 6 f) x2yz + xyz – 6

g) 2x2y2 + x2y – xy2 – 3xy + 4 h) x – x2y3 + (xy)3 i) (xy)2 + (xy)3 + (xy)4

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Algebraic Expressions

7. a) If l = 6, b = 4 and h = 2, evaluate the following expressions.

(i) l × b (ii) l × b × h (iii) 2(l + b)

(iv) 2h(l + b) (v) l2 (vi) 6l2

8. If x = 2 and y = 3, show that: a) (x + y)2 = x2 + 2xy + y2

b) (x – y)2 = x2 – 2xy + y2 c) x2 – y2 = (x + y) (x – y)

9.6 Addition and subtraction of algebraic expressions
x, 2x, 5x, etc. are the like terms.

The sum of x and 2x = x + 2x = 3x (1 + 2)x = 3x

The sum of 2x and 5x = 2x + 5x = 7x (2 + 5)x = 7x

The sum of x, 2x and 5x = x + 2x + 5x = 8x (1 + 2 + 5)x = 8x

The difference of 5x and 2x = 5x – 2x = 3x (5 – 2)x = 3x

The difference of 2x and x = 2x – x = x (2 – 1)x = x

Thus, when we add or subtract like terms, we should add or subtract the coefficients
of the like terms.

On the other hand, x, x2, y, 2y2 are unlike terms.

Sum of x and x2 = x + x2, difference of x and x2 = x – x2

Sum of x and y = x + y, difference of x and y = x – y

Thus, we do not add or subtract the coefficient of unlike terms.

Worked-out examples

Example 1: Add (i) 2a, 3a2, 4a and a2 (ii) 3x2y, 2x2y, 3xy2 and – 2xy2.
Solution:
(i) 2a + 3a2 + 4a + a2 = 2a + 4a + 3a2 + a2

= 6a + 4a2

(ii) 3x2y + 2x2y + 3xy2 + (– 2xy2) = 5x2y + 3xy2 – 2xy2
= 5x2y – xy2

Example 2: Add 4ab + 7bc – 5, 3bc – 8ab + 6 and 9ab – bc – 2.

Solution:

Addition by vertical arrangement Addition by horizontal arrangement

4ab + 7bc – 5 (4ab + 7bc – 5) + (3bc – 8ab + 6) + (9ab – bc – 2)

– 8ab + 3bc + 6 = 4ab + 7bc – 5 + 3bc – 8ab + 6 + 9ab – bc – 2

9ab – bc – 2 = 4ab 9ab 8ab 7bc 3bc – bc – 5 6 – 2
5ab + 9bc 1 = 5ab + 9bc 1
= 5ab + 9bc 1

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Algebraic Expressions

Example 3: Subtract 2a2 + 5ab – b2 from 5a2 – ab + 3b2.

Solution:

Subtraction by vertical arrangement Subtraction by horizontal arrangement
5a2 ab 3b2
± 2a2 ± 5ab b2 (5a2 – ab + 3b2) (2a2 + 5ab – b2)
= 5a2 – ab + 3b2 2a2 5ab b2)

3a2 – 6ab + 4b2 = 5a2 2a2 – ab 5ab + 3b2 b2
= 3a2 6ab + 4b2

Example 4: What should be added to 3a + 4x to get 7a – 2x ?

Solution:

Here, the required expression to be added is Let’s think, what should be
(7a 2x) (3a 4x) added to 3 to get 7.
= 7a – 2x – 3a – 4x It’s 4 and it is 7 – 3. It’s my
= 7a – 3a – 2x – 4x investigation to work out
= 4a 6x such problems.

Example 5: What should be subtracted from 8a – 5b + 2 to get 2a + 3b – 9?

Solution:

Here, the required expression to be subtracted is

(8a – 5b + 2) – (2a + 3b – 9) Let’s think, what should be
= 8a – 5b + 2 – 2a – 3b + 9 subtracted from to 8 to get 5.
= 8a – 2a 5b 3b 2 9 It’s 3 and it is 8 – 5. It’s my rule
= 6a 8b 11 to work out such problems.

EXERCISE 9.2
General Section - Classwork
1. Let's tell and write the sums or differences as quickly as possible.

a) 4x + 3x = ............... (b) 3xy + 6xy = ............. (c) 8a2 + 5a2 = .............

(d) 8p – 3p = ............. (e) 9ab2 – 6ab2 = ............. (f) 10x3 – 3x3 = .............

2. a) What is the sum of a2 and a ? ...............................
b) What is the difference of 2x and 3y? ...............................
c) What should be added to 3x2 to get 7x2? .............................

d) What should be subtracted from 11a3 to get 6a3? ...............................

Creative Section - A
3. a) What are like and unlike algebraic terms? Write with examples.

b) How do we add or subtract like terms? Write with examples.

c) Can we add or subtract the coefficients of unlike terms? Write with examples.

d) In what way the sum of x + x and the product of x × x different with each
other?

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Algebraic Expressions

4. Let's add.

a) 3x, 5y, 4x and 6y b) 6ab, 8bc, (–2ab) and (–3bc)

c) 7at2, (–2at), (–4at2) and at d) (–5p3q2), (–9p2q3), 6p3q2 and 10p2q3

e) 3t – 2tz + 4 and 5tz + 2t – 10 f) 2x + 3y – 6 and 5x – 4y + 1

g) 7a + 4b – 8 and 3b – 5a + 9 h) 5x2 + 3x + 4 and x2 + 2x – 7

i) 8a2 + 3ab – 2b2, 3a2 – ab + 5b2 and 5ab – 7a2 – b2

j) a2 + 3ab – bc, b2 + 3bc – ca and c2 + 3ca – ab

5. Let's subtract.

a) 5pq from 8pq b) 2x2y from 9x2y

c) –3ab from 5ab d) 2a2b3 from – 3a2b3

e) 4m + 5n from 6m + 9n f) 7p2 – 6q2 from 5p2 + 2q2

g) 2x2 – 3x + 6 from 4x2 + 5x – 3

h) y3 – 5y2 + y – 11 from 4y3 – 3y2 – y – 6

i) a3b3 – 2a2b2 + 3ab – 4 from –5a3b3 – a2b2 – 4ab – 7

j) 2.6x4 – 3.8x3 – 1.2x2 + 4.6x – 5.4 from 6.2x4 + 8.3x3 –2.1x2 + 6.4x – 4.5

6. Let's simplify. b) 9a – 2b – 4a + 7b
a) 3x + 2y + 5x – 9y d) 9a2 5a – 6a2 3a – 2
c) 4x2 – 2y2 + 2x2 – 5y2 f) 5x2 – (2x2 – y2) – 4y2
e) 7p2 + p + p2 – 6p + 3 h) 7p – 5q – (2p – 8q)
g) 10a + 4b – (3a + 2b) j) 13a2 – (3b2 – 4c2) + a2 – (8a2 – 5b2 + 7c2)
i) 12x – (5x + 4y) – (2y + 3x)

7. a) What should be added to 5xy to get 9xy ?
b) What should be added to ab bc ca to get ab bc ca ?
c) What should be subtracted from 9x2y to get 4x2y?
d) What should be subtracted from 3p2 + 2p – 1 to get p2 – 3p 4 ?

8. a) To what expression must 5a2 – 4a + 3 be added to make the sum zero?
b) From what expression must x2 + 5x – 7 be subtracted to make the difference
unity?

9. a) If a = x + y and b = x – y, show that (i) ( a + b)2 = 4x2 (ii) (a –b)2 = 4y2
b) If x = p + 2 and y = p – 3, show that (i) x + y + 1 = 2p (ii) x y – 5 = 0

It's your time – Project work

10. a) Let's write any three different pairs of like terms. Then, find the sum and
difference of each pair.

b) Write any three binomial algebraic expressions and denote them by A, B and C
respectively.

Then, find (i) A +B – C (ii) A – (B – C) (iii) (A + B) – (A – C)

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Laws of Indices Laws of Indices

Unit

10

10.1 Laws of indices (or exponents)

Let’s consider an algebraic term 2x3.

Here, 2 is called the coefficient, x is the base and 3 is the exponent of the base. The
exponent is also called the index of the base. The plural form of index is indices.
An index of a base shows the number of times the base is multiplied. For example:

x×x o x is multiplied two times = x2 (x squared)

x × x × x o x is multiplied three times = x3 (x cubed)

x × x × x × x o x is multiplied four times = x4 (x raise to the power 4)

While performing the operations of multiplication and division of algebraic
expressions we need to work out indices of the same bases under the certain rules.
These rules are also called laws of indices.

1. Product law of indices

Study the following illustrations and investigate the idea of the product law of
indices.

1 2 = 4 unit squares 123
3 4 2 = 21 × 21 = 21 +1 = 9 unit squares
22 32 4 5 6 3 = 31 × 31 = 31 +1

2 789

3

Similarly,

x2 x = x1 × x1 y2 y = y1 × y1
= x1 +1 = y1
+1

x y

Again,

3 = 8 unit cubes
2 = 27 unit cubes
14 = 31 × 31 × 31
2 = 21 × 21 × 21
23 7 8 = 21 + 1 + 1 3

56 2 33 = 31 + 1 + 1
2
3 3

x3 x = x1 × x1 × x1 y3 y = y1 × y1 × y1
= x1 + 1 + 1 = y1 + 1 + 1
y
xx y
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148 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur