Real Number System
4.8 Multiplication and division of irrational numbers
In the case of multiplication and division of irrational numbers, we multiply or divide
their rational coefficients as well as the irrational numbers. For example,
(i) 2 2 u 2 = (2 × 1) 2 × 2 = 2 4 = 2 × 2 = 4 We got it !
(ii) 2 2 u 5 3 = (2 × 5) 2 × 3 = 10 6
(iii) 9 10 ÷ 3 2 = (9 ÷ 3) 10 ÷ 2 = 3 5 We should simply multiply or
divide rational numbers and
irrational numbers separately.
(iv) 4 15 ÷ 2 5 = (4 ÷ 2) 15 ÷ 5 = 2 3
4.9 Rationalisation
Let’s consider an irrational number 5 .
When it is multiplied by 5 , the product = 5 u 5 = 25 = 5 and it is a rational
number. The process of changing an irrational number into a rational number is called
rationalisation. Study the following examples.
2 u 2 = 4 =2 2 is the rationalising factor of 2 .
2 3 u 3 =2 9 =2×3 3 is the rationalising factor of 2 3 .
Thus, if the product of two irrational numbers is a rational number, each of them is
called the rationalising factor of the other.
4.10 Conjugate
Let’s consider a binomial irrational number 3 + 2 .
Here, ( 3 + 2 ) u ( 3 – 2 ) = ( 3 )2 – ( 2 )2 (a + b) (a – b) = a2 – b2
=3–2=1
Thus, when 3 + 2 is multiplied by 3 – 2 , the product is a rational number. So,
3 – 2 is the rationalising factor of 3 + 2 . In this case, the rationalising factor
3 – 2 is called the conjugate of 3 + 2 or 3 + 2 is the conjugate of 3 – 2 .
Study a few more examples given below.
Conjugate of 5 + 3 is 5 – 3 , conjugate of 7 – 2 is 7 + 2 and so on.
Therefore, in a + b , it's conjugate is a – b or vice versa.
Worked-out examples
Example 1: Simplify a) 3 3 + 5 3 – 4 3 b) 5 8 – 2 18 + 50
Solution:
a) 3 3 + 5 3 – 4 3 = 8 3 – 4 3 = 4 3
49Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8
Real Number System
b) 5 8 – 2 18 + 50 8 = 4 × 2 o 4 is a perfect square
= 5 4 × 2 – 2 9 × 2 + 25 × 2 18 = 9 × 2 o 9 is a perfect square
=5u2 2 –2×3 2 +5 2
= 10 2 – 6 2 + 2 50 = 25 × 2 o 25 is a perfect square
= 11 2 – 6 2
=5 2 Thus, one factor is a perfect square and
another is 2 in each case.
Example 2: Simplify a) 2 3 u 3 6 b) 15 24 ÷ 3 8
Solution:
I understood !
a) 2 3 u 3 6 = 2 u 3 3 × 6 In 18, I should factorise 18 in
= 6 18 such a way that one of the factors
=6 9×2 should be a perfect square.
= 6 u 3 2 = 18 2
b) 15 24 ÷ 3 8 = (15 ÷ 3) 24 ÷ 8 or 15 24 =5 3
3 8
=5 3
Example 3: Simplify ( 5 – 2 ) ( 5 + 2 ).
Solution:
( 5 – 2 ) ( 5 + 2 ) = ( 5 )2 – ( 2 )2 = 5 – 2 = 3
Example 4: Rationalise the denominator of a) 3 b) 10 6 c) 2
3 35 5+ 3
Solution:
a) Multiplying the numerator and denominator by the rationalising factor 3 .
3 = 3× 3 = 3 3 = 3
3 3× 3 3
b) Multiplying the numerator and the denominator by the rationalising factor 5 .
10 6 = 10 6 × 5 = 10 30 = 10 30 =2 30
35 3 5× 5 3×5 15 3
c) Multiplying the numerator and denominator by the rationalising factor 5 – 3 .
2= 2 ( 5 – 3)
5+ 3 ( 5+ 3) ( 5 – 3)
= 2( 5 – 3 ) = 2( 5 – 3 ) =2 5– 3) = 5– 3
( 5 )2 – ( 3 )2 5–3 2
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Real Number System
Example 5: Simplify a) 3 – 1 b) 2 – 6 + 3 18 – 32
23 28
Solution:
a) 3 – 1 = 3 – 1× 3 = 3 – 3 = 3 3 –2 3 = 3
2 3 2 3× 3 2 3 6 6
b) 2 – 6 + 3 18 – 32 = 22 – 6 + 3 9×2– 16 × 2
2 8 2× 2 4×2
= 22 – 6 +9 2 –4 2
2 22
= 2– 3 +5 2
2
=6 2– 3× 2 =6 2 – 32 = 12 2 –3 2) = 92
2× 2 2 2 2
EXERCISE 4.3
General Section - Classwork
1. Let's say and write the rationalising factors of the following surds.
Surds Rationalising factors Surds Rationalising factors
a) 2
b) 2 3 e) 3 + 5
c) 3 5 f) 3 – 2
g) 1
d) 1
2 3+ 2
h) 1
5 –2
2. Let's simplify the following operations on surds.
a) 3 + 3 = ................... b) 5 + 5 = ..................
c) 2 2 + 3 2 = ................. d) 2 3 – 3 = ................
e) 5 6 – 2 6 = .................. f) 2 × 3 = ..................
g) 7 × 2 = ................... h) 2 3 × 3 5 = ...................
i) 6 ÷ 2 = ................... j) 6 15 ÷ 2 3 = ...................
51Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8
Real Number System
3. Let's rationalise the denominators.
a) 1 = ................... b) 1 = ................... c) 1 = ...................
2 3 5
Creative Section - A
4. Let's simplify.
a) 3 2 + 4 2 b) 7 3 – 5 3 c) 3 + 3 3 + 2 3
d) 4 7 – 2 7 + 7 e) 9 6 – 3 6 – 6 f) 2 + 8
g) 2 2 + 3 18 h) 4 3 – 27 i) 4 8 + 3 18 + 32
j) 6 3 + 5 75 – 2 48 k) 9 32 – 6 50 – 2 2 l) 3 20 – 2 45 + 80 – 5
5. Let's simplify.
a) 2 2 × 3 b) 3 5 × 2 c) 2 2 × 4 5
d) 4 3 × 3 5 e) 2 × 6 f) 6 × 3
g) 2 × 2 × 2 h) 10 ÷ 2 i) 12 30 ÷ 6 6
j) 8 27 ÷ 3 12 k) 9 24 ÷ 18 l) 12 35 ÷ 2 28
6. Let's simplify.
a) 5 × ( 2 + 8 ) b) 4 3 × (2 12 – 3 3 )
c) ( 3 + 2 ) ( 3 – 2 ) d) ( 5 – 2 ) ( 5 + 2 )
e) ( 7 + 3 ) ( 7 – 3 ) f) (2 2 – 3 ) (2 2 + 3 )
g) (3 3 + 5 ) (3 3 – 5 ) h (4 6 – 2 7 ) (4 6 + 2 7 )
7. Let's rationalise the denominators of the following irrational numbers and simplify.
2 3 c) 3 d) 3 2 e) 5
a) 2 b) 3 22 23 56
g) 2
f) 1 h) 7 i) 2 5 3 2 j) 3 2
2 +1 3 –1 3+ 2 3 – 2 6 +3 2
Creative Section - B
8. Let's rationalise the denominators as per necessary and simplify.
a) 3 + 1 b) 5 – 1 c) 4 + 6 d) 21 – 5 7
2 3 2 5 3 72
4 f) 3 + 5 g) 32 – 6 +5 2 h) 6 –2 5 – 53
e) 3 2 – 2 52 2 3 4
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Real Number System
9. Simplify:
a) 42 + 60 –2 20 + 2 175 b) 16 – 24 +2 50 –3 8
28 45 2 32
6+
c) 98 – 40 – 48 + 2 72 d) 2 24 + 2 46 + 3 12
50 128 12
e) 15 – 30 + 18 + 20 f) 72 – 48 – 45 +2 98
5 45 27 50 128
10. a) Find the perimeter of a rectangle with width 18 cm and length 50 cm.
b) Find the perimeter of a triangle whose sides are 75 cm, 48 cm, and 27 cm.
c) Find the area of rectangular ground of length 25 3 m and breadth 20 6 m.
d) The base of a triangular field is 30 2 m and its height (h) is 24 2 m. Find its
area using the formula, area of triangle = 1 × b × h.
2
11. a) Find the perimeter and area of each of the following rectangles.
p = 27cm(i) (ii) (2 + 5) cm (iii) (7 + 2) m
( 3 + 3) cm
( 3 + 5) cm (4 + 5 cm (10 + 8 m
b) Find the area of each triangle. (Area of triangle =12 × base (b) × perpendicular (p))
(i) (ii) (iii) b = 5 5 cm
p = 4 18cm p = 3 20cm
b = 12 cm b = 6 8 cm
It's your time - Project work!
12. a) Let's write any four fractions with irrational denominators. Then, rationalise
each denominator and simplify.
b) Let's write any four pairs of irrational numbers that give the following sums or
difference after addition or subtraction.
(i) sum is 9 2 (ii) sum is 12 3 (iii) difference is 5 (iv) difference is 2 7
c) Let's write any four pairs of irrational numbers that give the following products
or quotients after multiplication or division.
(i) product is 24 (ii) product is 30 (iii) quotient is 2 (iv) quotient is 3
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Unit Ratio, Proportion and Unitary Method
5
5.1 Ratio – Looking back
Classroom - Exercise
1. Let’s say and write the answers as quickly as possible.
a) Ratio of 11 boys to 17 girls is ……………. and 17 girls to 11 boys is …………….
b) Ratio of 1 m to 70 cm is ………….. and 70 cm to 1 m is …………..
c) In 4 : 5, the antecedent is ………….. and consequent is …………..
d) In w : x = y : z, the first, second, third and fourth proportional are …………...
………….. respectively.
e) In a : b : : c : d, extremes are ………………, and means are ………………
f) Are 1, 2, 3 and 6 in proportion ? ………………
g) Are 1, 2, 3 and 4 in proportion? ………………
h) If x : 2 = 3: 1, the value of x is ………………
i) If 3 : y = 1 : 2, the value of y is ………………
2. Let’s say and write which one is in direct or inverse proportions.
a) Amount of rice and its cost ………………….
b) Number of workers and their working days ………………….
c) Speed of a bus and time taken to cover a certain distance ………………….
d) Speed of a bus and distance covered in an interval of time ………………….
e) Number of workers and amount of work done by them ………………….
Let Ram has Rs 50 and Sita has Rs 100. 50
100
Here, the ratio of Ram’s to Sita’s money = = 1 : 2
It shows that Ram’s money is half of Sita’s money.
Similarly, the ratio of Sita’s to Ram’s money = 100 = 2 : 1
50
It shows that Sita’s money is two times of Ram’s money.
Thus, a ratio compares two or more quantities of the same kind. It shows how many
times a quantity is less or more than another quantity of the same kind.
Vedanta Excel in Mathematics - Book 8 54 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Ratio, Proportion and Unitary method
Let’s take a ratio a:b. We read it as ‘a is to b’.
Here, the first term a is called the antecedent and the second term b is called the
consequent. For example,
In 2 : 3 (read as ‘2 is to 3’), 2 is antecedent and 3 is consequent.
In 3 : 2 (read as ‘3 is to 2’), 3 is antecedent and 2 is consequent.
Remember that while making a ratio, the quantities must be of the same kind and they
should have the same unit. The ratio itself does not have any unit.
5.2 Compounded ratio
Let’s take any two ratios a : b and c : d. a c
b d
The compounded ratio of a : b and c : d = a : b × c : d = × = ac:bd
Thus, a ratio which is obtained by multiplying two or more ratios is called a compounded
ratio. For example,
If 1:2 and 3:4 are any two ratios,
the compounded ratio = 1 × 3 = 3 = 3:8
2 4 8
Workedout examples
Example 1: Find the ratio of 75 cm and 1 m and reduce it in the lowest terms.
Solution:
Here, 1 m = 100 cm 1 m is converted into 100 cm to make
the units of both quantities same.
? Ratio of 75 cm and 1 m = 75:100
75
= 100
= 3 = 3:4
4
Example 2: Find the compounded ratio of a) 3:2 and 1:4 b) 1:2, 4:5 and 10:3.
Solution: = 3 × 1 = 3 =3:8
The compounded ratio of 3:2 and 1:4 2 4 8
The compounded ratio of 1:2, 4:5 and 10:3 = 1 × 4 × 10 = 4:3
2 5 3
Example 3: If x:y = 2:3 and y:z = 6:5, find a) x:z b) x:y:z
Solution:
Here, x:y = 2:3 and y:z = 6:5
Now, x u y = 2 u 6 = 4
y z 3 5 5
or, x:z = 4:5
Again, in the first ratio, y = 3 and in the second ratio, y = 6.
The L.C.M. of 3 and 6 = 6
55Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8
Ratio, Proportion and Unitary method
Now, x = 2 = 2 × 2 = 4
y 3 3 × 2 6
And, y = 6
z 5
Therefore, x:y:z = 4 : 6 : 5
Example 4: The ratio of the number of boys and girls in a school is 2:3. If there are
360 girls, find the number of boys.
Solution:
Let the number of boys be x.
According to the question,
x = 2
360 3
It’s easier!
or, 3x = 360 u 2 boys : girls = 2 : 3
So, x : 360 = 2 : 3
or, x = 360 × 2 = 240
3
Hence, the required number of boys is 240.
Example 5: An alloy contains copper and zinc in the ratio 3:2. Find the mass
of the metals in 450 g of alloy.
Solution:
Let the mass of copper be 3x g and that of zinc is 2x g.
According to the question,
3x + 2x = 450 Here, if copper is 3g, zinc is 2g
If copper is 2 × 3g, zinc is 2 × 2g
or, 5x = 450 If copper is 3 × 3g, zinc is 3 × 2g
If copper is x × 3g, zinc is x × 2g
or, x = 450 = 90 g So, we consider copper is 3x g and
5 zinc is 2x g.
? Mass of copper= 3x = 3 × 90 g = 270g
Mass of zinc = 2x = 2 × 90 g = 180 g
Example 6: A, B, and C invested a sum of Rs 2,70,000 on a business in the ratio
of 2:3:4.
(i) Find the sum invested by each of them.
(ii) At the end of the year, if they gained Rs 90,000, divide the profit
among them in the ratio of their shares.
Solution:
(i) Let the sums invested by A, B, and C are Rs 2x, Rs 3x and Rs 4x respectively.
According to the question,
2x + 3x + 4x = Rs 2,70,000
or, 9x = Rs 2,70,000
or, x = Rs 270000 = Rs 30,000
9
? The sum invested by A = 2x = 2 u Rs 30,000 = Rs 60,000
The sum invested by B = 3x = 3 u Rs 30,000 = Rs 90,000
The sum invested by C = 4x = 4 u Rs 30,000 = Rs 1,20,000
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Ratio, Proportion and Unitary method
(ii) Again, the profit = Rs 90,000 and the ratio of sharing profit is 2:3:4
Now, 2x + 3x + 4x = Rs 90,000
or, 9x
= Rs 90,000
or, x
= Rs 90000 = Rs 10,000
9
? The profit received by A = 2x = 2 u Rs 10,000 = Rs 20,000
The profit received by B = 3x = 3 u Rs 10,000 = Rs 30,000
The profit received by C = 4x = 4 u Rs 10,000 = Rs 40,000
Example 7: The monthly income of X is double than that of Y and the monthly
income of Y is treble than that of Z. If the total monthly income of these
three persons is Rs 1,00,000, find the monthly income of each person.
Solution:
3
Here, the ratio of the income of Y and Z = Y : Z = 3 : 1 = 1
the ratio of the income of X and Y = X : Y = 2 : 1 = 2 = 2 × 3 = 6
1 1 × 3 3
? The ratio of X, Y and Z = X : Y : Z = 6 : 3 : 1
Let, the monthly incomes of X, Y and Z be Rs 6x, Rs 3x, and Rs x respectively.
According to the question,
6x + 3x + x = Rs 1,00,000
or, 10x = Rs 1,00,000
or, x = Rs 100000 = Rs 10,000
10
? The income of X = 6x = 6 u Rs 10,000 = Rs 60,000
The income of Y = 3x = 3 u Rs 10,000 = Rs 30,000
The income of Z = x = Rs 10,000.
Example 8: Last year, 50 students in a school appeared SEE. Among them, 16
students secured grade C, 4 secured grade D, and the rest of them
secured grades A and B in the ratio 2 : 3.
(i) Find the number of students who secured grades A and B.
(ii) Find the ratio of the number of students who secured the grades
A, B, C, and D.
Solution:
i) Here, the number of students who secured grades A and B = 50 – (16 + 4)
= 30
Let the number of students who secured grade A is 2x and grade B is 3x.
Now, according to the question,
2x + 3x = 30 or, 5x = 30 or, x = 6
Now, the number of students who secured grade A = 2x = 2 × 6 = 12
The number of students who secured grade B = 3x = 3 × 6 = 18
(ii) Again, the ratio of the number of students who secured grades A, B, C, and D
= 12 : 18 : 16 : 4 = 6 : 9 : 8 : 2
57Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8
Ratio, Proportion and Unitary method
Example 9: Two numbers are in the ratio 3:4. When 9 is subtracted from each of the
numbers, the ratio becomes 2:3. Find the numbers.
Solution:
Let the required numbers be 3x and 4x.
According to the question, I got it!
3x – 9 = 2 9 is subtracted from 3x = 3x – 9
4x – 9 3
9 is subtracted from 4x = 4x – 9
2
or, 3 (3x – 9) = 2 (4x – 9) And the new ratio = 3x – 9 = 3
or, 9x – 27 = 8x – 18 4x – 9
or, x = 9
Now, the required first number = 3x = 3 u 9 = 27
The second number = 4x = 4 u 9 = 36
Example 10: The ratio of the present age of a father and his son is 4:1. Four years
hence, the ratio of their ages will be 3:1. Find their present ages.
Solution:
Let the present age of the father be 4x years and that of the son is x years.
4 years hence, age of the father = (4x + 4) years.
4 years hence, age of the son = (x + 4) years.
According to the question,
4x + 4 = 3
x+4 1
or, 4x + 4 = 3x + 12
or, x = 8
Now, the present age of the father = 4x = 4 u 8 years = 32 years
The present age of the son = x = 8 years
Example 11: If the ratios of the angles of a quadrilateral are 1:2:3:4, find the size of
each angle.
Solution:
Let the sizes of the angles of the quadrilateral be xq, 2xq, 3xq and 4xq respectively.
Now, x + 2x + 3x + 4x = 360q The sum of the angles of a quadrilateral is 360°.
or, 10x = 360q
or, x = 360°
10
= 36°
Now, the first angle = x° = 36°, the second angle = 2x° = 2 × 36° = 72°
the third angle = 3x° = 3 × 36° = 108° and fourth angle = 4x° = 4 × 36° = 144°
Hence, the required angles of the quadrilateral are 36°, 72°, 108°, and 144°.
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Ratio, Proportion and Unitary method
EXERCISE 5.1
General Section - Classwork
Let’s say and write the answers as quickly as possible.
1. a) Ratio of 3 to 4 is …………, antecedent is ………, consequent is ………
b) Ratio of 1 kg to 500 g is ………, antecedent is ………, consequent is ………
c) The compounded ratio of p : q and r : s is ……………..
d) The compounded ratio of 1 : 4 and 3 : 1 is ……………..
e) If a = 1 , and b = 2 , then a = ……………..
b 2 c 3 c
f) If x : y = 1 : 2 and y : z = 4 : 3, then x : z = ……………..
2. a)Two numbers are in the ratio 1 : 2. If the smaller one is 5, the greater one is …......
b)Two sums are in the ratio 3 : 1 and the first sum is Rs 30, the second sum is …......
c)Two weights are in the ratio 1 : 4 and the second weight is 20 kg, the first weight
is …......
d) x : y = 2 : 1. If x = 10, then y = ………. e) a : b = 1:5. If a = 5, then b = …………
Creative Section - A
3. Answer the following questions.
a) The ratio of the incomes of A and B is 1 : 2. Define its meaning with an example.
b) The ratio of length and breadth of a rectangle is 3 : 1. Define its meaning with an
example.
c) The weight of an object X is two-fifth times the weight of object Y. Express it in a
ratio and define it’s meaning.
d) Define antecedent and consequent with an example.
4. Let’s find the ratios in the lowest terms.
a) 20 girls and 15 boys b) 1.5 m and 75 cm
c) 800 g and 1 kg d) 750 m and 1.5 km
5. a) There are 15 teachers in a school and 10 of them are male teachers.
(i) Find the ratio of the number of male and female teachers.
(ii) Find the ratio of the number of female teachers and the total number of
teachers.
b) One of the angles of a pair of complementary angles is 50° . Find the ratio of the
pair of complementary angles.
6. Find the compounded ratios. c) 7 : 9 and 6 : 5 d) 8 : 3 and 15 : 16
a) 1:2 and 4:3 b) 6:5 and 1 : 4
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Ratio, Proportion and Unitary method
7. a) If p:q = 1:2 and q:r = 4:3, find (i) p:r (ii) p:q:r
b) If a:b = 2:3 and b:c = 6:5, find (i) a:c (ii) a:b:c
8. Let’s find the value of x in the following proportions.
a) x : 3 = 4 : 6 b) 2 : x = 6 : 9 c) 5 : 4 = x : 12 d) 4 : 7 = 8 : x
9. a) Two numbers are in the ratio of 2 : 5 and the bigger number is 30. Find the
smaller number.
b) The ratio of length and breadth of a rectangular ground is 4 : 3 and the breadth
is 75 m. Find the length of the ground.
c) There are 18 girls in a class and the ratio of the number of girls and boys in the
class is 3 : 2. Find the number of boys.
d) The ratio of the monthly income and expenditure of a family is 7:4. If the income
of the family in a month is Rs 21,000, find:
(i) the monthly expenditure.
(ii) The saving in a month.
10. a) When a sum of Rs 100 is divided in the ratio of 2 : 3, find the share of each part.
b) Mr. Karki divides a sum of Rs 1,80,000 between his son and daughter in the ratio
of 4:5. Find the sum obtained by each of them.
c) An alloy contains copper and zinc in the ratio 5:3. Find the mass of the metals
in 960 g of alloy.
11. a) What is the actual distance between two places which is represented by 4.5 cm
b) on a map if the map is drawn to the scale 1:100000 ?
The distance between two places is 15 km. If the map scale is 1:500000, find the
distance between the places in centimetres drawn in the map.
12. a) If the angles of a triangle are in the ratios 1:2:3, find the size of each angle.
b) If the ratios of the angles of a quadrilateral are 2:3:4:6, find the size of each
angle.
c) If a pair of complementary angles are in the ratio 3:2, find the angles.
d) If a pair of supplementary angles are in the ratio 1:2, find the angles.
e) The acute angles of a right angled triangle are in the ratio 4:5, find the angles.
Creative Section - B
13. a) The cost of a book is two times more than the cost of a box and the cost of the
box is two times more than the cost of a pen. If the total cost of these three items
is Rs 350, find the cost of each item.
b) The monthly income of A is double than that of B and the monthly income of B
is treble than that of C. If the total income of three persons is Rs 80,000, find the
monthly income of each of person.
c) Last year 60 students of a school appeared SEE. Among them 8 students secured
grade C, 4 students secured grade D and the rest of them secured grades A and
B in 3 : 5 ratio.
(i) Find the number of students who secured grades A and B.
(ii) Find the ratio of the number of students who secured the grades A, B, C and D.
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Ratio, Proportion and Unitary method
14. a) Ram, Hari and Shyam invested a sum of Rs 1,80,000 on a business in the ratio
of 2:3:4. Find the sum invested by each of them.
b) Four businessmen invested a sum of Rs 4,50,000 in the ratio of 1:2:3:4 to start a
new business.
(i) Find the sum invested by each of them.
(ii) At the end of the year, if they gained Rs 1,80,000, divide the profit among
them in the ratio of their shares.
15. a) Two numbers are in the ratio 4:5. When 6 is added to each term, their ratio
b) becomes 5:6. Find the numbers.
Two numbers are in the ratio 7:5. When 10 is subtracted from each term, their
ratio becomes 3:2. Find the numbers.
16. a) The ratio of the present age of a father and his son is 3:1. Five years hence, the
ratio of their age will be 5:2. Find their present age.
b) The ratio of the present ages of A and B is 4:7. After three years, the ratio of their
age will be 5:8. Find the present age of A and B.
c) The present age of Bishwant and Sunayana are in the ratio 4:3. Three years ago,
the ratio of their age was 7:5. Find their present age.
d) The ratio of the present age of a mother and her daughter is 15:4. Four years
ago, the ratio of their age was 13:2. Find the present age of the mother and the
daughter.
17. a) The ratio of the length and breadth of a rectangle is 2:1. If the rectangle is 14 cm
b) long, find its breadth.
c)
A rectangular field is 16 m broad and the ratio of the length and breadth is 3:2.
d) (i) Find the length of the field. (ii) Find the perimeter of the field.
(iii) Find the area of the field.
The ratio of the length and breadth of a rectangular ground is 4:3 and its
perimeter is 126 m.
(i) Find the length and breadth of the ground.
(ii) Find the area of the ground.
The ratio of the length and breadth of a rectangle is 5:4 and its area is 180 cm2.
(i) Find its length and breadth (ii) Find its perimeter
It’s your time - Project work
18. a) How many students are there in your class? How many girls and boys are there?
Let’s make the following ratios of these numbers.
(i) Ratio of the total number of students to the number of girls.
(ii) Ratio of the total number of students to the number of boys.
(iii) Ratio of the number of girls to the number of boys.
b) Let’s measure the length and breadth of the following objects and write the ratio
of length and breadth.
(i) Maths book (ii) desk or table (iii) floor of the classroom
61Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8
Ratio, Proportion and Unitary method
5.3 Proportion
Let’s take any two ratios 6 : 8 and 9 : 12.
6 3 9 3
Here, 6 : 8= 8 = 4 = 3 :4 and 9 : 12 = 12 = 4 = 3 :4
Thus, the ratios 6 : 8 and 9 : 12 are equal ratios.
The equality of ratios is called a proportion.
6 : 8 and 9 : 12 are said to be in proportion. Here, the terms 6, 8, 9, and 12 are called
proportional.
Suppose , the terms a, b, c and d are in proportion. We can write it as a : b = c : d or
a : b :: c : d and we read it as ‘a is to b as c is to d.’
Here, the terms a, b, c, and d are the first second, third and fourth proportional
respectively.
In a proportion, the first and the fourth proportional are called extremes. The second
and the 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
We can use this fact to test whether the given ratios are in proportion or not.
5.4 Types of proportions
There are two types of proportions: direct proportion and inverse proportion.
(i) Direct proportion
Suppose, the cost of 1 pen = Rs 30
Then, the cost of 2 pens = 2 u Rs 30 = Rs 60
Here, the ratio of the number of pens = 1:2
Also, the ratio of the cost of pens = 30:60 = 1:2
Thus, the ratio of the number of pens and the ratio of the cost of pens are equal.
Therefore, these two ratios are in proportion. Here, if we increase or decrease the
number of pens in a ratio, the cost also increases or decreases in the same ratio.
Therefore, such proportion is called a direct proportion.
Thus, in a proportion, if one ratio is increased (or decreased), the corresponding
ratio also increases or decreases, such proportion is said to be a direct proportion.
(ii) Inverse proportion
Suppose 1 pipe can fill a tank in 12 hours.
Then, 2 pipes of the same size can fill the tank in 6 hours.
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Ratio, Proportion and Unitary method
Here, the ratio of the number of pipes = 1:2
The ratio of time taken to fill the tank = 12:6 = 2:1
Thus, the ratio of the number of pipes and the ratio of time taken to fill the tank are
oppositely equal. Therefore, these two quantities are in proportion with oppositely
(inversely) equal ratios. In this case, the proportion is called an inverse proportion.
In this way, in a proportion, if one ratio is increased (or decreased), the
corresponding ratio decreases (or increases) inversely, such proportion is said to
be an inverse proportion.
Worked-out examples
Example 1: Test whether the terms 5, 6, 10, and 12 are in proportion.
Solution:
Here, the ratio of the first two terms is 5:6 and the last two terms is 10:12
Now, the product of extremes = 5 × 12 = 60
The product of means = 6 u 10 = 60
? Product of extremes = Product of means
Hence, 5, 6, 10 and 12 are in proportion.
Example 2: If 6, 9, and 42 are the terms of a proportion, find the fourth proportional.
Solution:
Let, the fourth proportional be x.
Now, 6, 9, 42 and x are in proportion. Answer checking
6
? 6:9 = 42:x Here, 6 : 9 = 9 = 2 : 3
or, 6 = 42 42 : 63 = 42 = 2 : 3
9 x 63
? 6 : 9 = 42:63
or, 2x = 126
126 6 : 9 = 42 : x
or, x = 2 = 63
Hence, the required fourth proportional is 63.
Example 3: If 4, 7, and 35 are the terms of a proportion, find the third proportional.
Solution:
Let, the third proportional be x. Alternative process
In, 4 : 7 = x : 35,
Now, 4, 7, x and 35 are in proportion.
? 4:7 = x:35
or, 4 = x product of means = product of extremes
7 35 7x = 4 × 35
or, 7x = 140 or, x = 4 × 35 = 20
7
oHre, nce,xthe=re1q74u0ire=d
20 proportional is 20.
third
Example 4: If the ratios 5:x and 15:24 are in direct proportion, find the value of x.
Solution:
Here, 5:x and 15:24 are in direct proportion.
? 5:x = 15:24
63Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8
Ratio, Proportion and Unitary method
or, 5 = 15 I got it!
x 24
When 5 : x and 15 : 24 are
or, 15x = 5 × 24 in direct proportion, these
5 × 24 ratios are directly equal.
or, x = 15 =8
Example 5: If the ratios 4:3 and x:12 are in inverse proportion, find the value of x.
Solution:
Here, 4:3 and x:12 are in inverse proportion.
? 4:3 = 12:x It’s easier!
4 12 When 4 : 3 and x : 12 are in
or. 3 = x inverse proportion, the ratios
are equal but oppositely.
or, 4x = 3 × 12 So, 4 : 3 = 12 : x
3 × 12
or, x = 4 =9
Example 6: If the cost of 9 kg of onions is Rs 360, how much onions can be purchased
for Rs 480?
Solution:
Let x kg of onions can be purchased for Rs 480.
Here, the ratio of the quantity of onions = 9:x
The ratio of the cost of onions = 360:480
The quantity of onions and the cost are in direct proportion.
? 9:x = 360:480 Answer checking
360
or, 9 = 480 Cost of 9 kg of onions = Rs 360
x
Cost of 1 kg of onions = Rs 360
9 3 9
or, x = 4
= Rs 40
or, 3x = 9 × 4 Cost of 12 kg of onions= 12 × Rs 40
9×4 = Rs 480
3
or, x = = 12 Which is given in the question.
Hence, the required quantity of onions is 12 kg.
Example 7: If 9 workers can complete a piece of work in 20 days, how many workers
should be added to complete the work in 15 days?
Solution:
Let, x number of workers can complete the work in 15 days.
Here, the ratio of the number of workers = 9:x Answer checking
9 workers complete the work in 20 days
The ratio of the number of working days = 20:15
The number of workers and their working days 1 worker complete the work in 9 × 20 days
are in inverse proportion.
12 workers complete the work in 9 × 20
? 9:x = 15:20 12
= 15 days
15 3
or, 9 = 20 = 4 Which is given in the question.
x
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Ratio, Proportion and Unitary method
or, 3x = 9 × 4
or, x = 9×4 = 12
3
Hence, the required additional number of workers = 12 – 9 = 3.
Example 8 : 40 students of a hostel have food enough for 30 days. If 10 more students
join the hostel after 5 days, how long days the remaining food last ?
Solution :
Let, the remaining food will last for x days.
Here, the total number of students after 5 days = 40 + 10 = 50
The remaining number of days = 30 – 5 = 25 days
Here, the number of students and the number of days are in inverse proportion.
40 : 50 = x : 25
or, 4 = x
5 25
or, x = 4 × 25 = 20
5
Hence, the remaining food will last for 20 days.
EXERCISE 5.2
General Section - Classwork
1. Let’s say and write whether these quantities are in direct or inverse proportions.
a) Rate of cost and total cost of item ..................…………….
b) Rate of cost and number of purchased items ..................…………….
c) Interval of time and distance covered ..................…………….
d) Speed of a bus and time taken to cover certain distance ..................…………….
e) Number of workers and number of working days ..................…………….
2. Let’s say and write the terms which are in proportion.
a) 1, 2, 5, 10 b) 3, 4, 9, 15 Terms which are in proportion
c) 2, 3, 7, 9 d) 4, 5, 8, 10
3. Let’s say and write the values of x as quickly as possible.
a) x : 2 = 2 : 1, x = …………… b) 3 : x = 1 : 3, x = ………………….
c) 4: 1 = x : 2, x = …………….. d) 1 : 5 = 2 : x, x = ………………….
65Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8
Ratio, Proportion and Unitary method
Creative Section - A
4. Let’s find the products of extremes and means. Which of the following terms are in
proportion?
a) 2, 5, 4, 10 b) 3, 7, 6, 15 c) 4 : 5 and 12 : 15 d) 6 : 7 and 21 : 12
d) x : 7 : : 18 : 14
5. Find the values of x in the following proportions.
a) 3:2 = 9 : x b) 4 : 5 = x : 15 c) 5 : x :: 10 : 12
6. a) If the following terms are in proportion, find the fourth proportional.
(i) 3, 4, 6 (ii) 2, 3, 8 (iii) 4, 6, 12 (iv) 5, 7, 15
b) If the following terms are in proportion, find the third proportional.
(i) 2, 5, 10 (ii) 5, 6, 18 (iii) 8, 3, 6 (iv) 9, 7, 28
c) If the following terms are in proportion, find the second proportional.
(i) 3, 12, 28 (ii) 6, 24, 32 (iii) 7, 21, 15 (iv) 10, 30, 27
d) If the following terms are in proportion, find the first proportional.
(i) 4, 15, 12 (ii) 16, 5, 4 (iii) 7, 30, 35 (iv) 6, 24, 18
7. a) If the following ratios are in direct proportion, find the value of x.
(i) 3:5 and x : 15 (ii) 9 : x and 18:16 (iii) x :10 and 28:40
b) If the following ratios are in inverse proportion, find the value of x.
(i) 5:6 and x:20 (ii) 12:9 and 36:x (iii) x:20 and 60:45
Creative Section - B
Let’s use the rules of direct proportion and inverse proportion to workout the following
problems.
8. a) If the cost of 5 kg of is Rs 450, find the cost of 9 kg of rice.
b) The cost of 10 pens is Rs 250. How many pens can be purchased for Rs 450?
c) If a bus can travel 240 km in 4 hours, how many hours does it take to travel
420 km with the same speed?
9. a) The bus fare of 8 passengers is Rs 640. If the fare is decreased by Rs 5 per
b) passenger, how many passengers can travel for Rs 900?
10. a) The bus fare of 12 passengers is Rs 540. If the fare is increased by Rs 5 per
passenger, how many passengers can travel for Rs 750?
If 8 workers can complete a piece of work in 18 days, how many workers are
required to complete the work in 12 days?
b) 20 labourers complete the construction of a building in 24 days. In how many
days would 16 labourers complete the construction?
c) A bus completes its journey in 14 hours with an average speed of 40 km per
hour. How long does the bus take to complete the journey if its average speed is
increased to 56 km per hour?
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Ratio, Proportion and Unitary method
d) 15 workers were employed to complete a piece of work in 36 days. How many
more workers should be added to complete the work in 30 days?
e) 20 students in a hostel have provisions for 60 days. If 10 more students are
admitted to the hostel, for how many days would the provisions be enough?
f) A garrison of 60 men had provisions for 90 days. If 15 men left the garrison,
how long would the provisions last?
g) 60 students in a hostel have food enough for 30 days. If 20 more students join
the hostel after 10 days, how long will the remaining food last?
It’s your time - Project work!
11. a) Let’s write any three sets of four numbers such that each set of four numbers are
in proportion. Find the product of extremes and means and verify that each set
of numbers are in proportion.
b) Let’s find the rate of cost of the following items in your local market.
(i) rice (ii) sugar (iii) tomatoes (iv) onions (v) apples
Let’s find the cost of different quantities of each of these items and verify that
the cost and quantities are in direct proportion.
c) Again, increase or decrease the rate of cost of items yourself. Then show that
the rate of cost and number of quantities are in inverse proportion.
5.5 Unitary method
Suppose the cost of 1 kg of vegetables is Rs 40.
Here, 1 kg of vegetable is unit quantity and Rs 40 is the unit cost (or unit value).
Therefore, the value of unit number of quantity is called the unit value.
In the case of direct proportion (or direct variation), unit value is obtained by division
and the value of more number of quantity is obtained by multiplication. For example,
The cost of 2 kg of vegetables is Rs 80.
The cost of 1 kg of vegetables is Rs 80 . Unit value is obtained by division.
2
80
The cost of 5 kg of vegetables is Rs 2 × 5. More values is obtained by multiplication.
However, in the case of inverse proportion (or inverse variation), unit value is obtained
by multiplication and the value of more number of quantity is obtained by division.
For example,
2 pipes can fill a cistern in 6 hours.
1 pipe can fill the cistern in 2 × 6 hours Unit values is obtained by multiplication.
3 pipe can fill the cistern in 2×6 hours. More values is obtained by division.
3
In this way, the method of finding the unit value or the value of more number of quantity
by the process of division or multiplication is known as unitary method.
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Ratio, Proportion and Unitary method
Worked-out examples
Example 1: The cost of 4 packets of DDC milk is Rs 160. Find the cost of 10 packets of
milk. How many packet of milk can be bought for Rs 280?
Solution:
Cost of 4 packets of milk = Rs 160
Cost of 1 packet of milk = Rs 160 = Rs 40 Unit cost is obtained by division.
4 More cost is obtained by multiplication.
Cost of 10 packets of milk = 10 × Rs 40 = Rs 400
Again,
Rs 40 is the cost of 1 packet of milk.
Re 1 is the cost of 1 packet of milk.
40
1
Rs 280 is the cost of 40 × 280 packets of milk = 7 packets of milk.
Hence, the cost of 10 packets of milk is Rs 400 and 7 packets of milk can be bought for
Rs 280.
Example 2: When the average speed of a bus is 50 km per hour, it covers 300 km in a
certain interval of time. If the speed of the bus is increased to 60 km per
hour, find the distance covered by the bus in the same interval of time.
Solution:
When the speed is 50 km/hr, the bus covers 300 km
When the speed is 1 km/hr, the bus covers 300 km = 6 km
50
When the speed is 60 km/hr, the bus covers 60 u 6 km = 360 km
Hence, the bus covers a distance of 360 km with the average speed of 60 km/hr.
Example 3: If a pipe can fill 1000 l of water in 50 minutes, how much water does it
fill in 1 hour?
Solution:
In 50 minutes a pipe can fill 1000 l of water.
In 1 minute, the pipe can fill 1000 = 20 l of water
50
In 60 minutes (1 h) the pipe can fill 60 × 20 l = 1200 l of water.
Hence, the pipe can fill 1200 l of water in 1 hour.
Example 4: If the cost of 2 parts of a land is Rs 1, 80,000, find the cost of 4 parts of
the land. 3 5
Solution: 2
3
The cost of parts of a land = Rs 1,80,000 The cost whole land is definitely
The cost of 1 land (whole land) = Rs 1,80,000 more than the cost of 2 part of the
2 land. 3
3
= 1,80,000 u 3 = Rs 2,70,000
2
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Ratio, Proportion and Unitary method
The cost of 4 parts of the land = 4 u Rs 2,70,000 = Rs 2,16,000.
5 5
Hence, the required cost of 4 parts of the land is Rs 2,16,000.
5
Example 5: The cost of 5 exercise books and 2 pens is Rs 250. If the cost of 1 exercise
book is Rs 40, find the cost of 6 pens.
Solution:
Here, the cost of 1 exercise book = Rs 40
? The cost of 5 exercise books = 5 u Rs 40 = Rs 200
The cost of (5 exercise books + 2 pens) = Rs 250
or, Rs 200 + cost of 2 pens = Rs 250
or, cost of 2 pens = Rs 250 – Rs 200 = Rs 50
? cost of 1 pen = Rs 50 = Rs 25
2
And cost of 6 pens = 6 u Rs 25 = Rs 150
Hence, the required cost of 6 pens is Rs 150.
Example 6: If 16 workers can construct a road in 45 days, how many additional
numbers of workers should be employed to finish the construction in
30 days?
Solution:
In 45 days, 16 workers can construct the road.
On 1 day, 45 u 16 workers can construct the road.
In 30 days, 45 × 16 workers can construct the road.
30
= 24 workers can construct the road.
Hence, the additional number of workers = 24 – 16 = 8.
Example 7: 32 students of a hostel had provisions for 30 days. If 8 more students
were admitted after 5 days, how long would the remaining provisions
last?
Solution:
Here, the remaining number of days = 30 – 5 = 25 days.
Total number of students with 8 more students = 32 + 8 = 40
32 students had provisions for 25 days.
1 student had provisions for 32 u 25 days.
40 students had provisions for 32 × 25 days
40
= 20 days.
Hence, the remaining provisions would last for 20 days.
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Ratio, Proportion and Unitary method
EXERCISE 5.3
General Section - Classwork
1. Let’s say and write the answers in the blank spaces.
a) Cost of 1 kg of onions is Rs 50, cost of 3 kg of onions is ……………
b) Cost of 5 exercise books is Rs 200, cost of 1 exercise book is …………….
c) Cost of 3 pens is Rs 75, cost of 1 pen is ……….., cost of 4 pens is ……………..
d) A bus covers 120 km in 3 hrs, it covers ……. km in 1 hr and ……… km in 6 hrs.
2. a) 1 worker can do a work in 6 days, 2 workers can do it in ………….
b) 3 pipes can fill a cistern in 2 hours, 1 pipe can fill it in ……………..
c) 2 labourers can do a work in 6 days, 6 labourers can do it in …………….
3. a) 1 part of a sum is Rs 10, (i) the sum is …… (ii) 1 part of the sum is ………
2 4
b) 1 part of a distance is 4 km, (i) the distance is … (ii) 1 part of the distance is …
3 4
Creative Section - A
4. a) The cost of 5 kg of rice is Rs 400.
(i) Find the cost of 1 kg of rice and 12 kg of rice.
(ii) How many kilograms can be bought for Rs 240?
b) A bus completed its journey of 320 km with an average speed of 40 km per hour.
(i) How long did it take to complete the journey ?
(ii) What would be the speed of the car if it covered only 240 km in the same
interval of time?
c) 5 litres of petrol is needed for a motorbike to cover 200 km.
(i) What is the mileage of the bike?
(ii) How much petrol is needed to cover 320 km?
5. a) A worker can do a piece of work in 14 days.
(i) How much work does he do in 1 day ?
(ii) How much work does he do in 7 days ?
(iii) If he works for 2 days and leaves, how much work is left to finish?
b) A factory completes the work of production of items in 12 days.
(i) How much work does it do in 9 days.
1
(ii) In how many days would it finish 3 part of the work?
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Ratio, Proportion and Unitary method
c) A computer can download 1 part of an application file in 40 minutes.
3
(i) In how many hours would it complete to download the whole file?
(ii) How much file does it download in 1 minute?
(iii) How much file is left to download if it downloaded only for 90 minutes?
d) A pipe can fill a water tank in 1 hour.
(i) What part of the tank is filled by the pipe in 20 minutes ?
(ii) If the capacity of the tank is 1500 , how many litres of water does the pipe
fill in 20 minutes ?
6. a) 3 parts of the distance between two places is 36 km.
4
(i) Find the distance between the places.
(ii) Find 2 parts of the distance between the places.
3
b) A man spends Rs 7,500 every month to run his family, which is 1 part of his
3
monthly income.
(i) Find his monthly income.
(ii) If he spends 3 parts of his income on his children’s education, how much
25
money does he spend on education?
Creative Section - B
7. a) The cost of 2 kg of sugar and 5 kg of rice is Rs 710. If the cost of 1 kg of sugar is
Rs 80, find the cost of 8 kg of rice.
b) The cost of 3 kg of apples and 2 kg of oranges is Rs 540. If the cost of 1 kg of
oranges is Rs 90, find the cost of 6 kg of apples.
8. a) 3 pieces of the same size can fill a cistern in 90 minutes. How long would 5
pipes take to fill the cistern?
b) 4 machines of a factory can finish the required amount of production in 6 days.
If one machine had machinery defect and stopped functioning, in how many
days would the remaining machines finish the same quantity of production at
the same rate of production?
c) 6 workers complete a piece of work in 15 days. In how many days would 9
workers complete the same work?
d) 12 workers were employed to complete the construction of a building in 60 days.
How many additional numbers of workers should be employed to complete the
construction in 45 days?
e) 60 students of a hostel had provisions for 40 days. If 20 more students were
admitted, how long would the provisions last?
71Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8
Ratio, Proportion and Unitary method
f) 120 students of a hostel have food enough for 60 days. If 30 more students join
the hostel after 15 days, how long does the remaining food last?
g) A garrison of 110 people had provisions for 30 days. If 22 people leave the
garrison after 10 days, how long does the remaining provisions last?
9. a) A computer can finish to download an application file in 4 minutes at the rate
of 600 MB per minute.
(i) Find the size of the application file.
(ii) How long does it take to download the file when the download rate increases
to 800 MB per minute?
b) The size of a movie file in YouTube is 1.8 gigabyte (GB) and the rate of download
of the file is 900 megabyte (MB) per minute.
(i) How long does it take to download the file?
(ii) Find the rate of download of the file per second.
It’s your time - Project work!
10. Let’s make a survey and find the rate of cost of the following items in your local
market.
(i) rice (ii) potatoes (iii) cooking oil (iv) milk
a) Estimate the cost of these items consumed by your family per day.
b) How much budget do your family allocate on these items for a month and for
one year?
11. Let’s search the sizes of some mobile application files. Calculate and estimate the
time taken to download these files in your available internet speed.
12. Let’s visit to the available website and search today’s exchange rate of foreign
currencies with our Nepali currency. Then, find how much Nepali currency is
needed to exchange the following foreign currencies?
a) £ 1,500 b) $ 1,500 c) 50,000 d) AUD 1,500
Vedanta Excel in Mathematics - Book 8 72 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Unit Percent and Simple Interest
6
6.1 Percent – Looking back
Classroom - Exercise
1. Let’s say and write these fractions and decimals in percent.
a) 1 = ……….. b) 1 = ……….. c) 1 = ……….. d) 3 = ………..
2 4 5 10
e) 0.04 = ……….. f) 0.63 = ……….. g) 0.7 = ……….. h) 0.99 = ………..
2. Let’s say and write the following percent in fractions and decimals.
a) 9% = ……., …… b) 43% = ………., …….. c) 71% = ………, ………
d) 53 % = ……, ……. e) 87% = ……, ……….. f) 90% = ………, ……….
3. a) Principal (P) = Rs 100, Time (T) = 1 year, Interest (I) = Rs 8, Rate (R) = .............
b) Principal (P) = Rs 100, Time (T) = 1 year, Rate (R) = 9% p.a., Interest (I) = ........
c) Rate (R) = 10% p.a. Time (T) = 1 year, Interest (I) = Rs 10, Principal (P) = .........
d) Principal (P) = Rs 100, Rate (R) = 12% p.a., Interest (I) = Rs 12, Time (T) = ......
Percent means per hundred or out of hundred. 40 percent (40%) of the number of
students in a class are boys means, out of 100 students 40 are boys. Similarly, 80 marks
out of 100 full marks means 80 percent (80%) marks.
A percentage is simply a fraction in which the denominator is always 100.
Note that ‘percentage’ is not used with a number, while ‘percent’ is. For example, we do
not say or write 40 percentage, however, it is 40 percent.
6.2 Fundamental Operations on Percent
Study and learn the following fundamental operations on percent.
Operation Process Examples
To express a fraction Multiply the given 2 = 2 × 100% = 40%
or decimal as fraction or decimal by 5 5
1. percent 100%.
0.75 = 0.75 × 100% = 75%
To express a percent Divide the given percent 75% = 75 = 3 or 0.75
as a fraction or by 100 and remove % 100 4
2. decimal sign.
60% = 60 = 3 or 0.6
100 5
To find the value of Multiply the quantity by 10% of 50 = 10 × 50 = 5
the given percent of the given percent. Express 100
3. a given quantity the percent in fraction and
simplify. 75% of 60 = 75 × 60 = 45
100
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Percent and Simple Interest
To express a part of Make the ratio of the part 16 as the percent of 80
a whole quantity as to the whole quantity and 16
4. the percent of the multiply it by 100%. = 80 × 100% = 20%
whole quantity
To find a whole Divide the given value by The number whose 40% is 10
quantity when its the given percent. 10 10 100
5. value of certain = 40% = 40 = 10 × 40
percent is given = 25 100
Alternative process of ‘operation -5’: Consider the whole quantity as x (a variable) and by
making an equation and solving it, we can find the value of x. For example,
Let the required number be x.
40
Now, 4% of x = 10 or, 100 × x = 10 or, x = 25
Worked-out examples
Example 1: 3 parts of the number of students of a class are girls. Find the percent
4
of girls and boys.
Solution: 3 To express a fraction into percent,
4 multiply by 100%
Here, the percent of girls = × 100%
= 3 × 25%
= 75% Whole number of students is 100%. So,
Again, the percent of boys = 100% – 75% percent of boys = 100% – percent of girls.
= 25%
Hence, there are 75% girls and 25% boys in the class.
Example 2: Sarita got 720 marks out of 800 full marks in the Basic Level Examination
of grade 8. How many percent of marks did she obtain?
Solution: 720
800
Here, 720 as the percent of 800 = × 100%
= 90%
Hence, she obtained 90% marks.
Example 3: If the price of petrol is increased from Rs 102 per litre to Rs 112.20 per
litre, by how many percent is the price increased?
Solution:
Here, the initial price = Rs 102 per litre
The new price = Rs 112.20 per litre
? The increase of price = Rs 112.20 – Rs 102 = Rs 10.20
Now, the percent of increase of price = Increase price × 100% = 10.20 × 100% = 10%
Initial price 102
Hence, the price of petrol is increased by 10%.
Remember, Percent of reduction = amount of reduction × 100%
Initial amount
Vedanta Excel in Mathematics - Book 8 74 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Percent and Simple Interest
Example 4: The tagged price of mobile is Rs 3,600. If the shopkeeper allows 20%
discount, how much should a customer pay for it?
Solution:
Here, the tagged price of the mobile (M.P.) = Rs 3,600
Discount percent = 20% 20
100
? Amount of discount = 20% of Rs 3,600 = × Rs 3,600 = Rs 720
Now, the payment for the mobile = Rs 3,600 – Rs 720 = Rs 2,880
Hence, the customer should pay Rs 2,880.
Example 5: The monthly income of Mrs. Magar is Rs 25,000. She spends 25% of her
income on food, 20% on house rent and 10% on her children’s education.
a) Find her expenditure on each item.
b) If her miscellaneous expenditure is 15%, find her saving in a month.
Solution:
Here, the monthly income of Mrs. Magar = Rs 25,000
a) Expenditure on food = 25% of Rs 25,000 = 25 × Rs 25,000 = Rs 6,250
100
Expenditure on house rent = 20% of Rs 25,000 = 20 × Rs 25,000 = Rs 5,000
100
Expenditure on education = 10% of Rs 25,000 = 10 × Rs 25,000 =Rs 2,500
100
15
b) Miscellaneous expenditure = 15% of Rs 25,000 = 100 × Rs 25,000 =Rs 3,750
? Total expenditure in a month = Rs (6,250 + 5,000 + 2,500 + 3,750) = Rs 17,500
Now, her monthly saving = Rs 25,000 – Rs 17,500 = Rs 7,500
Example 6: The population of a village was 18,000 in the beginning of 2075 B.S.
It was increased by 5% in 2076 and again increased by 10% in 2077.
Find the population of the village at the end of 2077 B.S.
Solution:
Here, the population at the end of 2075 B.S. = 18,000 + 5% of 18,000
= 18,000 + 5 × 18,000
100
= 18,000 + 900 = 18,900
Again, the population at the end of 2077 B.S. = 18,900 + 10% of 18,900
= 18,900 + 10 × 18,900
100
= 18,900 + 1890
= 20,790
Hence, the population of the village at the end of 2077 B.S. would be 20,790.
75Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8
Percent and Simple Interest
Example 7: 40% of the total number of students of a school are girls. If there are
216 girls,
a) find the total number of students. b) Find the number of boys.
Solution: Alternative process
a) Let the total number of students be x.
40% of students = no. of girls
Here, 40% of x = 216
40% of students = 216
40 216
or, 100 ×x = 216 ? No. of students = 40%
or, 2x = 216 = 216
5 40
100
216
216 × 5 = 40 × 100 = 540
2
or, x = = 540
Hence, the total number of students of the school is 540.
b) Again, the number of boys = 540 – 216 = 324.
Example 8: Mr. Thapa spends 30% of his monthly income and saves Rs 22,400. How
much does he spend in a month?
Solution: Alternative process
Let his monthly income be Rs x. 22,400
Here, his percent of saving = 100% – 30% = 70% Monthly income = Rs 70%
= 22,400
70
Now, 70% of x = Rs 22,400 100
100
or, 70 ×x = Rs 22,400 = Rs 22,400 × 70
or, 100 7x = Rs 32,000
or,
10 = Rs 22,400
x = Rs 22,400 × 5 = Rs 32,000
7
? His monthly income = Rs 32,000
Now, his monthly expenditure = Rs 32,000 – Rs 22,400 = Rs 9,600.
Example 9: In an examination, 40% marks is needed to pass the examination in
grade C. If an examinee got 260 marks and unable to obtain grade C by
20 marks, find the full marks of the examination.
Solution:
Let the full marks of the examination be x. Alternative process
Here, the marks necessary for grade C = 260 + 20 = 280 280
Now, 40% of x = 280 Full marks = 40%
or, 40 ×x = 280 = 280
100 40
100
2x 100
or, 5 = 280 = 280 × 40
= 700
5
or, x = 280 × 2 = 700
Hence, the full marks of the examination is 700.
Vedanta Excel in Mathematics - Book 8 76 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Percent and Simple Interest
Example 10: Before last year, when the population of a municipality was increased
by 5%, it became 1,22,850. Last year, again the population was increased
by 10%. Find the increased number of population in two years.
Solution:
Let, the population of the municipality before 2 years be x.
Then, x + 5% of x = 1,22,850
or, x + 5x = 1,22,850
100
x
or, x + 20 = 1,22,850
or, 21x = 1,22,850
20
or, x = 1,17,000
? The increased number of population before last year = 1,22,850 – 1,17,000
= 5,850
Again, the increased number of population in last year = 10% of 1,22,850
= 12,285
Now, the total increased number of population in 2 years = 5,850 + 12,285
= 18,135
Hence, the increased number of population in two years is 18,135.
EXERCISE 6.1
General Section - Classwork
1. Let’s say and write these fractions and decimals in percent.
a) 1 = ………. b) 1 = ………. c) 1 = ………. d) 1 = ……….
4 5 20 25
e) 0.009 = ………. f) 0.05 = …….. g) 0.7 = ………. h) 0.99 = ……….
2. Say and write the following percent in fractions and decimals.
a) 0.7% = ………, …..…. b) 7% = …….., ………. c) 70% = ……., ……
3. 25% means 25 = 1 and 25% of 40 = 1 of 40 = 10. Using this rule, tell and write
100 4 4
the values as quickly as possible.
a) 10% of 50 = …….. b) 20% of 60 = ………… c) 25% of 20 = …………
4. If 10% of a number is 25, the required number =1205% = 25 × 100 = 250. Using this
10
rule, say and write the required numbers as quickly as possible.
a) 5% of a number is 15, the required number is ……………
b) 20% of a number is 30, the required number is ……………
c) 50% of a number is 50, the required number is ……………
77Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8
Percent and Simple Interest
Creative Section - A
5. Let’s find the values of the following numbers or quantities.
6. a) 8% of Rs 500 of the total ndbu))m136b3%e31r of Rs 2,500 given number or
c) 40% of 680 students % of 1296 people
Let’s find what percent or quantities is the
quantities?
a) 30 students are 3 boys? b) Rs 750 is Rs 150 ?
c) 2 km is 600 m ? d) 1840 people are 1380 male?
7. Let’s find the whole quantity from the given value of given percentage of quantity.
a) 5% is Rs 240 b) 17% is Rs 8500 c) 60% is 180 kg d) 75% is 900 km
8. a) 2 parts of the earth’s surface is covered with water.
3
(i) Express it in percent. (ii) How many percent is the continental part ?
b) If 4 part of the total number of students in a class are girls, how many percent
5
are boys?
9. a) Out of 800 full marks, Shivani obtained 680 marks in an examination. Express
her marks in percent.
b) Mother earns Rs 27,000 and spends Rs 10,800 every month.
(i) Express her expenditure in percent. (ii) Calculate her saving percent.
10. a) The price of petrol is increased from Rs 104 per litre to Rs 109.20. By how many
percent is the price increased?
b) The number of students of a school was 640 last year and it is 800 this year. By
how many percent did the number increase?
11. a) The marked price of a mobile is Rs 3,250 and the shopkeeper allows 10%
discount.
(i) Calculate the discount amount. (ii) How much should a customer pay for it?
b) On the occasion of Tihar festival, a super-market announces a discount of 20%
on every item. If the tagged price of a T-shirt is Rs 1,230, how much should a
customer pay for it?
c) 870 g of brass contains 40% zinc and the rest is copper. Find the mass of zinc
and copper in the alloy.
d) The monthly salary of Mr. Gurung is Rs 31,500. He spends his income in every
month in the following ways. Food – 20%, House rent – 25%, fuel – 10%,
miscellaneous -15% (i) Find his monthly expenditure on each item.
(ii) How much does he save every month ?
Creative Section - B
12. a) The monthly salary of Mr. Nepali was Rs 18,500 before last year and it was
increased by 10% last year. Again, it was increased by 20% this year.
(i) How much was his salary last year ? (ii) How much is he drawing this year?
Vedanta Excel in Mathematics - Book 8 78 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Percent and Simple Interest
b) In 2018 A.D., the number of tourist who visited Nepal was 12,60,000. In 2019, it
was decreased by 10% and in 2020 it was increased by 40%. How many tourists
visited Nepal in 2020?
c) The population of a town is 80,000. If the rate of growth of population of the
town is 4% per year, estimate the population of the town at the end of two years.
d) Recently Mrs. Yadav bought a new scooter for Rs 2,10,000. If its cost price is
depreciated by 10% per year, find its cost after 2 years.
13. a) 35% of the students of a school are boys which is 259.
(i) Find the total number of students in the school.
(ii) Find the number of girls.
b) Ganesh spends 60% of his income every month and saves Rs 11,120 in a month.
Find his monthly income and expenditure.
c) Father gave 35% of his money to his son, 25% to his daughter and the remaining
Rs 27,000 to his wife.
(i) Find the total percent of his money received by his son and daughter.
(ii) How many percent of money did his wife receive?
(iii) How much money did father have in the beginning?
(iv) Calculate the amount of money received by his son and daughter.
d) Sahayata spends 30% of her income on food, 15% on rent and 5% on other
items in a month. If she saves Rs 15,500 in a month, find her monthly income
and expenditures on each item.
14. a) In an examination, an examinee needs 40% marks to pass the examination in
grade C. If Kishan got 350 marks and unable to get grade C by 10 marks, find the
full marks of the examination.
b) In an examination, an examinee needs 60% marks to secure grade A. If Rinky
got 472 marks and could not secure grade A by 8 marks, find the full marks of
the examination.
c) Two years ago, when the number of students of a school was increased by 25%,
it became 500. Last year, again the number was increased by 20%. How many
new students were enrolled in two years?
d) Two years ago, when a Publication House increased the monthly salary of its
employees by 10%, an employee started to draw Rs 23,100 per month. Again, the
Publication House increased the salary by 20% last year. Calculate the increased
amount of monthly salary of the employee in two years.
It’s your time - Project work!
15. a) How many students are there in your class? Express the number of boys and
girls in percent.
b) How many teachers are there in your school? Express the number of
(i) mathematics teachers (ii) male teachers (iii) female teachers in percents.
c) Let’s study the fluctuation of rate of cost of some vegetables in your local market
within a week. Express the increased or decreased rate of cost in percent.
79Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8
Percent and Simple Interest
6.3 Simple Interest
What do we do when we are living in a rented house? Of course, we should pay the
rent to the house-owner for the use of the house. Similarly, when we use other’s money
we should pay some charge for the use of money. This charge (an additional amount of
money) is called interest.
In this way, when we borrow money from a bank, we should pay interest to the bank.
When we deposite money in a bank, the bank pays interest to us.
The interest which is calculated from the original borrowed (or deposited) sum is called
simple interest.
Let’s review the following terms which are needed to calculate simple interest.
Principal (P) – It is the deposited or borrowed sum of money.
Rate of interest (R) – It is the interest of Rs 100 for 1 year. It is expressed as the percent
per year (or per annum, p.a.)
Time (T) – It is the duration for which principal is deposited or borrowed.
Interest (I) – It is the the simple interest on Rs P at R% p.a. in T years.
Amount (A) – It is the total sum of principal and interest. So, A = P + I
6.4 Rate of interest
The interest on the principal of Rs 100 for 1 year is called rate of interest (R). Rate of
interest (R) is usually expressed in percent per year or per annum.
For example,
If the interest of Rs 100 in 1 year is Rs 5, then R = 5% p.a.
If the interest of Rs 100 in 1 year is Rs 12, then R = 12% p.a. and so on.
6.5 Formula of simple interest
According to the definition of the rate of interest,
On Rs 100, for 1 year, interest (I) = R
On Re 1, for 1 year, interest (I) = R
100
On Rs P, for 1 year, interest (I) = P×R
100
P uTu R
On Rs P, for T years, interest (I) = 100
Thus, the formula to calculate simple interest is (I) = PTR
100
PTR
Again, if I = 100 , then,
PTR = I u 100
? P = Iu 100 o is the formula to calculate principal (P)
T uR
Vedanta Excel in Mathematics - Book 8 80 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Percent and Simple Interest
T = Iu 100 o is the formula to calculate time (T)
P uR
R = Iu 100 o is the formula to calculate rate (R)
P uT
Also, from the definition of amount (A), A = P + I
or, A = P + PTR
100
or, A =P 1+ TR
100
or, A = P 100 + TR
100
or, P = A u 100 which is used to calculate principal when time, rate
100 + TR and amount are given.
Worked-out examples
Example 1: A farmer borrowed a sum of Rs 50,000 from an Agricultural Development
Bank at the rate of 7.5% per year to promote his poultry farming.
Calculate the amount required to clear the loan at the end of 5 years.
Solution:
Here, Principal (P) = Rs 50,000
Rate (R) = 7.5% per year
Time (T) = 5 years
Now, interest (I) = PTR = Rs 50,000 × 5 × 7.5 = Rs 18,750
100 100
Again, amount (A) = P + I = Rs 50,000 + Rs 18,750 = Rs 68,750
Hence, the required amount is Rs 68,750.
Example 2: At what rate percent per annum does a sum of Rs 9,450 amount to
Rs 12,474 in 4 years?
Solution:
Here, principal (P) = Rs 9,450 Answer checking
Amount (A) = Rs 12,474 P = Rs 9,450, R = 8%, T = 4 years
? Interest (I) = A – P = Rs 12,474 – Rs 9,450 I= PTR = Rs 9,450 × 4 × 8
= Rs 3,024 100 100
= Rs 3,024
Time (T) = 4 years A = Rs 9,450 + Rs 3,024
Now, rate (R) = I u 100 = 3,024 × 100 = 8% = Rs 12,474 which is given in the
PuT 9,450 × 4 question.
Hence, the required rate of interest is 8% per year.
81Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8
Percent and Simple Interest
Example 3: Find the sum that amounts to Rs 9,144 in 3 years at 9% per year simple
interest.
Solution:
Here, amount (A) = Rs 9,144
Time (T) = 3 years
Rate (R) = 9% per year 9,144 u 100
100 + 3 u 9
Now, principal (P) = A u 100 =
100 + TR
9,14,400
= 127
= Rs 7,200
Hence, the required sum is Rs 7,200.
EXERCISE 6.2
General Section - Classwork
1. The table given below shows the known variables. Tell and write the formulae to
find the unknown variables.
No. Known variables Unknown variables
a) Principal (P), Time (T), Rate (R%) Interest (I) =
b) Interest (I), Time (T), Rate (R%) Principal (P) =
c) Principal (P), Interest (I), Time (T) Rate (R) =
d) Principal (P), Interest (I), Rate (R%) Time (T) =
e) Amount (A), Time (T), Rate (R%) Principal (P) =
f) Principal (P), Interest (I) Amount (A) =
2. Let’s tell and write the answers as quickly as possible.
a) When P = Rs 100, T = 1 year, I = Rs 5, then R = …………
b) When P = Rs 100, T = 1 year, I = Rs 9, then R = ………….
c) R = 10% p.a. means, P = …………, T = ………….. I = …………..
d) R = 12% p.a means, P = …………, T = ………… I = …………..
3. a) If P = Rs 1,000, I = Rs 100, A = ………………
b) If A = Rs 2,000, P = Rs 1,800, I = ………………
c) If A = Rs 3,000, I = Rs 300, P = ………………
Vedanta Excel in Mathematics - Book 8 82 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Percent and Simple Interest
Creative Section - A
Let’s calculate the variables which are not given.
4. a) P = Rs 1,000, T = 2 years, R = 4% per year , find I.
b) I = Rs 525, T = 3 years, R = 7% per annum, find P.
c) I = Rs 1,200, P = Rs 3,000, R = 10% p.a., find T.
d) I = Rs 2,700, P = Rs 4,500, T = 5 years, find R.
e) A = Rs 1,800, T = 2 years, R = 10%, find P and I.
f) A = Rs 5,250, T = 5 years, R = 15% , find P and I.
5. a) Mrs. Chapagain borrowed Rs 9,900 from a bank at the rate of 8% per annum.
How much amount should she pay at the end of 2 years?
b) Mr. Shah deposited a sum of Rs 15,000 in a bank at the rate of 10% per year.
Find the amount received by him at the end of 3 years 6 months.
c) Mrs. Rai borrowed a sum of Rs 10,800 from Mr. Yadav at the rate of 13% per
year. Calculate the amount received by Mr. Yadav in 9 months.
6. a) Find the rate of interest per annum at which a sum of 2,700 yields an interest of
Rs 594 in 2 years?
b) If a sum of Rs 5,600 amounts to Rs 8,400 in 5 years, find the rate of interest.
c) At what rate percent per year will a sum of Rs 12,000 amount to Rs 12,720 in 6
months?
7. a) If the interest of a sum at 7.5% p.a. in 4 years is Rs 1,050, find the sum.
b) Find the principal that yields an interest of Rs 2,240 at the rate of 10% p.a. in 3
years 6 months.
c) In how many years will the interest of a sum of Rs 3,600 at the rate of 5.5% per
year be Rs 594?
Creative Section - B
8. a) Mrs. Joshi borrowed a sum of money from a bank at 12% p.a. simple interest. If
she paid an interest of Rs 1,920 in 2 years, find the sum borrowed by her.
b) A farmer borrowed a sum of Rs 10,000 from a Rural Development Bank at the
rate of 7.5% p.a. If he/she paid an amount of Rs 15,250 to clear the debt, how
long did he/she use the sum?
c) A businessman deposited Rs 60,000 in a Commercial Bank at 4.5% p.a. How
long should he deposit the sum to receive an amount of Rs 61,350?
9. a) Mr. Gharti borrowed a loan from Mrs. Gupta at 10% p.a. At the end of 2 years,
he paid an amount of Rs 3,600 and cleared the debt. Find the sum borrowed by
Mr. Gharti.
b) What sum of money amounts to Rs 5,454 at 8% p.a. in 30 months?
c) Pradeep lent a sum to Santosh at the rate of 12 1 % p.a. After 4 years, Santosh
2
paid him Rs 7500 and clear the debt, what sum did Pradeep lend to Santosh.
83Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8
Percent and Simple Interest
10. a) Mr. Gurung borrowed a loan of Rs 25,000 from Mr. Pandey at 12% p.a. At the
b) end of 3 years, if he agreed to pay Rs 26,300 with a goat to clear his debt. Find
the cost of goat.
11. a) Shashwat lent a sum of Rs 44,000 to his friend Rahul at 10% p.a. After 221 year,
his friend paid him Rs 40,000 with a cow. What was the cost of the cow?
b)
Sajina deposited Rs 20,000 at the rate of 8% p.a. in her saving account. After
12. a) 2 years, she withdrew Rs 5,000 and the total interest of 2 years. How long
b) should she keep the remaining amount to get total interest of Rs 6,800 from the
beginning?
Ramesh took a loan of Rs 50,000 from Urmila at the rate of 10% p.a. If he paid a
half of the principal and all the interest at the end of 3 years, in how many years
should he pay the remaining amount with total interest of Rs 20,000 from the
beginning?
A man took a loan of Rs 12,000 with simple interest for as many years as the rate
of interest per year. If he paid Rs 3,000 as interest at the end of loan period, what
was the rate of interest?
A teacher deposited Rs 66,000 in his/her saving account for as many years as the
rate of interest per annum. If he/she received Rs 10,560 as interest at the end of
saving period, find the time duration and rate of interest.
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.
P = ................ I = ................ I = ................ P = ................ A = ................
T = ................ T = ................ P = ................ I = ................ T = ................
R = ................ R = ................ R = ................ T = ................ R = ................
Find I and A Find P and A Find T Find R Find P and I
b) Let’s make the groups of your friends and visit nearby banks or finance company.
Collect the various information about the rate of simple interest in different
types of accounts such as current account, saving account and fixed deposit
account. In which account are they providing more interest?
c) Ssk 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?
Vedanta Excel in Mathematics - Book 8 84 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Unit Profit and Loss
7
7.1 Profit and loss - Looking back
Classroom - Exercise
Let’s say and write the correct answers.
1. a) What is the formula to find profit from the given C. P. and S. P.? ……………..
b) What is the formula to find loss from the given C. P. and S. P. ? ……………..
c) What is the formula to find profit or loss percent? ……………..
d) What is the formula to find S.P. from the given M.P. and discount? ……………….
2. a) If C.P. = Rs 200, S. P. = Rs 240, profit = ……………….
b) If C. P. = Rs 180, S. P. = Rs 160, loss = ……………….
c) If C. P. = Rs 100, profit = Rs 15, profit percent = ……………….
d) If M. P. = Rs 500, discount = Rs 50, S. P. = ……………….
e) If M. P. = Rs 100, discount = Rs 10, discount percent = ……………….
Let a shopkeeper buys a calculator for Rs 900 and he/she sells it for Rs 1,080. Here,
Rs 900 is the price paid to buy the calculator called cost price (C. P.) and Rs 1,080 is the
price at which the calculator is sold called selling price (S. P.).
In this case, the shopkeeper makes a profit of Rs 180 which is Rs 1,080 – Rs 900.
Thus, when S. P. is higher than C. P., there is profit and if S. P. is lower than C. P., there
is loss.
Hence, when S. P. > C. P. , Profit = S. P. – C. P.
Also, when S. P. < C. P., Loss = C. P. – S. P.
Furthermore,
Profit = S. P. – C. P., then C. P. = S. P. – Profit and S. P. = C. P + Profit
Loss = C. P. – S. P., then C. P. = S. P. + Loss and S. P. = C. P. – Loss
7.2 Profit and loss percent
Let C. P. of an article be Rs 100 and profit = Rs 20, then profit percent is 20%.
Again, C. P. of the article is Rs 100 and loss = Rs 15, then loss percent is 15%.
Thus, when profit or loss is expressed out of C. P. of Rs 100, it is called profit or loss
percent.
Again, let’s consider the above given example.
C. P. of the calculator = Rs 900 and S. P. of the calculator = Rs 1,080
Now, profit = S. P. – C. P. = Rs 1,080 – Rs 900 = Rs 180
85Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8
Profit and Loss
When C. P. is Rs 900, profit is Rs 180.
When C. P. is Re 1, profit is Rs 180
900
When C. P. is Rs 100, profit is Rs 180 × 100 = Rs 20
900
Here, Rs 20 is the profit of C. P. of Rs 100. Therefore, profit percent is 20%.
Thus, profit percent = profit × 100% S. P. – C. P. × 100%
C. P. C. P.
And, loss percent = loss × 100% C. P. – S. P. × 100%
C. P. C. P.
7.3 To find S.P. when C.P. and profit or loss percents are given
We know that, S.P. = C.P. + Profit or C.P. – Loss. So, to calculate S.P., at first we should
calculate profit amount or loss amount from C.P. For example,
If C.P. = Rs 400 and profit percent = 20%, find S.P. Direct process
20 S. P. = (100 + 20)% of C. P.
100
Here, profit amount = 20% of C.P. = u Rs 400 = 120% of Rs 400
= Rs 80 = 120 × Rs 400 = Rs 480
100
Now, S.P. = C.P. + Profit = Rs 400 + Rs 80 = Rs 480 Direct process
Again, if C. P. = Rs 400 and loss percent = 12%, find S. P. S. P. = (100 – 12)% of C. P.
12 = 88% of Rs 400
100
Here, loss amount = 12% of C. P. = × Rs 400 = Rs 48 = 88 × Rs 400
100
Now, S. P. = C. P. – Loss = Rs 400 – Rs 48 = Rs 352
= Rs 352
7.4 To find C.P. when S.P. and profit or loss percent are given
Profit or loss percent is always calculated on C.P. But, here, C.P. is unknown. So, we
should consider C.P. as Rs x. Then, the profit or loss amount is calculated from Rs x.
Study the following examples:
If S.P. = Rs 550 and prfit percent = 10%, find C.P.
Let, the required C.P. be Rs x 10 x
100 10
Here, profit amount = 10% of Rs x = u Rs x = Rs
Now, C.P. = S.P. – profit
or, x = 550 – x Direct process
or, 10
x+ x = 550 (100 + 10)% of C. P. = S.P.
10
or, 110% of C. P. = Rs 550
110
or, 11x = 550 or, 100 × C. P. = Rs 550
10
550 × 10 or, C. P. = Rs 550 × 100
11 110
or, x = = Rs 500
or, C. P. = Rs 500
Hence, the required C.P. is Rs 500.
Vedanta Excel in Mathematics - Book 8 86 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Profit and Loss
Summary of calculation of different variables in profit and loss.
Calculation of Given Process
Profit or loss C. P. and S. P. Profit = S. P. – C. P. and Loss = C. P. – S. P.
Profit or loss C. P. or C. P. and Profit% = profit × 100% and
percent S. P. C. P.
Loss
Loss% = C. P. × 100%
Profit or loss Profit % or Loss % Profit = Profit% of C. P. and
amount
and C. P. Loss = Loss% of C. P.
S. P. Profit % or Loss % S. P. = C. P. + Profit% of C. P. and
and C. P. S. P. = C. P. – Loss% of C. P.
C. P. + profit % of C. P. = S. P. and
C. P. – Loss % of C. P. = S. P.
C. P. Profit % of Loss % or, x+Profit% of x = S.P. or, x–Loss% of x=S.P.
and S. P. Direct process:
C. P. =(100+SP. rPo. fit)% and C. P. =(100–SP.rPo.fit)%
Worked-out examples
Example 1: A shopkeeper bought a watch for Rs 1250 and sold it for Rs 1500. Find
her profit percent.
Solution: Direct process
Profit = S. P. – C. P.
Here, C.P. of the watch = Rs 1,250
S.P. of the watch = Rs 1,500 Profit % = S. P. – C. P. × 100%
C. P.
? Profit = S.P. – C.P. = Rs 1,500 – Rs 1,250 = Rs 250 = 1500 – 1250 × 100%
Now, profit percent=PCro.Pf.it u 100%=1225500u100%=20% 1250
= 250 × 100% = 20%
1250
Hence, the required profit percent is 20%.
Example 2: A grocer bought 300 eggs at Rs 8 each. 25 eggs were broken and he sold
the remaining eggs at Rs 9.60 each. Find his/her profit or loss percent.
Solution:
Here, the remaining number of eggs = 300 – 25 = 275
C.P. of 300 eggs = 300 u Rs 8 = Rs 2,400
S.P. of 275 eggs = 275 u Rs 9.60 = Rs 2,640
? Profit = S.P. – C.P. = Rs 2,640 – Rs 2,400 = Rs 240
Now, profit percent = Profit u 100% = 240 u 100% = 10%
C.P. 2400
Hence, his profit percent is 10%.
87Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8
Profit and Loss
\Example 3: A stationer bought 12 gel pens for Rs 192 and sold 8 gel pens for
Rs 176. If he/she sold the remaining pens at the same rate of cost price,
find her profit or loss percent.
Solution:
Here,C.P. of 12 gel pens = Rs 192
C.P. of 1 pen = Rs 192 = Rs 16
12
S.P. of 4 pens= 4 × Rs 16 = Rs 64
S.P. of 8 pens = Rs 176
Total S.P.= Rs 64 + Rs 176 = Rs 240
Profit= S.P. – C.P. = Rs 240 – Rs 192 = Rs 48
? profit percent = Profit u 100% = 48 u 100% = 25%
C.P. 192
Hence, her profit percent is 25%.
Example 4: Mr. Shakya purchased a gold item for Rs 32,400. If the gold is slightly
devaluated and he sold it at 5% loss, find the selling price of the item.
Solution:
Here, C.P. of the gold article = Rs 32,400 Direct process
Loss percent = 5% S. P. = (100 – 5)% of C. P.
5 = 95% of Rs 32,400
100
Now, Loss amount = 5% of C.P. = u Rs 32,400 = 95 × Rs 32,400
100
= Rs 1,620
= Rs 30,780
? S.P. of the gold item = C.P. – Loss
= Rs 32,400 – Rs 1,620 = Rs 30,780
Hence, the selling price of the gold item is Rs 30,780.
Example 5: Mrs. Tamang sold a cosmetic item for Rs 522 at 16% profit. At what
price had she purchased the item?
Solution:
Let she had purchased the cosmetic item for Rs x.
So, C. P. = Rs x Direct process
Now, C.P. = S.P. – profit (100 + 16)% of C. P. = S.P.
or, 116% of C. P. = Rs 522
or, x = Rs 522 – 16% of x Rs 522
or, C. P. = 116%
or, x = Rs 522 – 16 ux = Rs 522
or, = Rs 522 – 140x0 116/100
or, x 25 100
4x = Rs 522 × 116
x+ 25 = Rs 522
= Rs 450
Vedanta Excel in Mathematics - Book 8 88 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Profit and Loss
or, 29x = Rs 522
or, 25
= Rs 522 u 25 = Rs 450
x 29
Hence, the cost price of the cosmetic item is Rs 450.
Example 6: Suntali bought a mobile phone and sold to Dhurmus at 10% loss.
Dhurmus again sold it for Rs 6,750 at 25% profit.
(i) Find the cost price of mobile for Dhurmus.
(ii) Find the cost price of mobile for Suntali.
Solution:
(i) Let the cost price of the mobile for Dhurmus be Rs x.
Now, C.P. = S.P. – profit Direct process
or, x = Rs 6,750 – 25% of x (100 + 25)% of C. P. = S.P.
or, 125% of C. P. = Rs 6,750
or, x = Rs 6,750 – 25 ux
100 or, C. P. = Rs 6,750
x 125%
or, x = Rs 6,750 – 4
Rs 6,750
or, x+ x = Rs 6,750 = 125/100
4
Rs 6,750 × 100
or, 5x = Rs 6,750 = 125
4
6,750 u 4 = Rs 5,400
or, x = Rs 5 = Rs 5,400
?C.P. of Dhurmus = Rs 5,400
(ii) Again, C.P. of Dhurmus = S.P. of Suntali = Rs 5,400
Let C.P. of Suntali be Rs y.
Now, C.P. = S.P. + Loss Direct process
or, y = Rs 5,400 + 10% of y (100 – 10)% of C. P. = S.P.
10 or, 90% of C. P. = Rs 5,400
100 Rs 5,400
or, y = Rs 5,400 + uy or, C. P. = 90%
or, y = Rs 5,400 + y = Rs 5,400
10 90/100
y
or, y– 10 = Rs 5,400 = Rs 5,400 × 100
= Rs 5,400 90
or, 9y
10 = Rs 6,000
Rs 5,400 10
or, y = 9 u
= Rs 6,000
? C.P. of Suntali is Rs 6,000.
89Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8
Profit and Loss
EXERCISE 7.1
General Section - Classwork
Let’s say and write the answers as quickly as possible.
1. a) C.P. = Rs 200, S.P. = Rs 224, profit = ………….
b) C.P. = Rs 400, S.P. = Rs 380, loss = ………….
c) C.P. = Rs 500, profit = Rs 60, S.P. = ………….
d) C.P. = Rs 700, loss = Rs 150, S.P. = ………….
e) S.P. = Rs 600,
f) S.P. = Rs 800, profit = Rs 100, C.P. = …………..
2. a) C.P. = Rs 100,
b) C.P. = Rs 100, loss =Rs 50, C.P. = …………..
c) C.P. = Rs 100,
d) C.P. = Rs 100, S.P. = Rs 120, profit percent = …………..
S.P. = Rs 85, loss percent = …………..
profit percent = 18%, S.P. = …………..
loss percent = 5% , S.P. = …………..
e) S.P. = Rs 110, profit percent = 10% C.P. = …………..
f) S.P. = Rs 88, loss percent = 12% C.P. = …………..
Creative Section - A
3. a) If C.P. = Rs 480, S.P. = Rs 528, find profit and profit percent.
b) If C.P. = Rs 640, S.P. = Rs 608, find loss and loss percent.
c) If C.P. = Rs 750, loss percent = 8% find S.P.
d) If C.P = Rs 1,560, profit percent = 20%, find S.P.
e) If S.P. = Rs 2,065, profit percent = 18%, find C.P.
f) If S.P. = Rs 1,869, loss percent = 11% find C.P.
4. a) A grocer purchased 80 kg of rice at Rs 90 per kg and sold at Rs 99 per kg.
(i) Find his profit. (ii) Find his profit percent.
b) A stationer purchased 5 dozen of pens at Rs 25 per piece and sold at Rs 24 per
piece.
(i) Find her loss. (ii) Find her loss percent.
c) Sunayana bought second-hand bicycle for Rs 4,350 and spent Rs 450 to repair
it. If she sold it for Rs 5,280, find her profit or loss percent.
d) A trader bought 1000 glass tumblers at Rs 8 each. 100 glass tumblers were
broken and she sold the rest at Rs 10 each. Find her gain or loss percent.
e) A grocer purchased 200 eggs at Rs 9 each. 20 of them were broken and he/she
sold the rest at Rs 9.50 each. Find her profit or loss percent.
f) A book-seller bought 7 story books for Rs 525 and sold 4 books for Rs 360. If
he sold the remaining books at the same rate of cost price find his/her profit or
loss percent.
Vedanta Excel in Mathematics - Book 8 90 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Profit and Loss
g) A shopkeeper bought 10 kg of tomatoes for Rs 240 and sold 7 kg of it for Rs 210. If
he sold the remaining quantity of tomatoes at the rate of cost price, find his gain
or loss percent.
5. a) A trader bought a pen drive for Rs 460 and sold it at 5% profit.
(i) Find the profit amount. (ii) Find the selling price of the pen drive.
b) A Shopkeeper bought an electric fan for Rs 1,520 and sold it at 10% loss.
(i) Find the loss amount. (ii) Find the selling price of the item.
c) Parbati buys a mobile for Rs 6,300 and sells it to Laxmi at 15% profit. How much
does Laxmi pay for it?
d) Shiva bought a second-hand motorcycle for Rs 1,53,200. He spent Rs 1,200 to
repair it and sold it to Bishnu at 8% loss. How much did Bishnu pay for it?
e) A stationer bought 80 gel pens at Rs 30 each. 20 gel pens were stolen and he/she
sold the remaining pens at 5% loss. Find the selling price of each gel pen.
f) A retailer purchased 1000 LED bulbs at Rs 80 each. 100 bulbs were damaged
and he sold the remaining bulbs at 21.5% profit. Find the selling price of each
bulb.
6. a) A trader sold a bag for Rs 440 at 10% profit. Find the cost price of the bag.
b) A shopkeeper sold a T-shirt for Rs 665 at a loss of 5%. At what price did she buy
the T-shirt?
c) A glassware seller bought 1000 glass tumblers. 100 of them were broken and
she sold the remaining tumblers at Rs 80 each. If she made a loss of 4%, at what
price did she purchase each tumbler?
Creative Section - B
7. a) Anita bought a bag for Rs 4,000 and sold to Shova at 20% profit. Again, Shova
sold it to Laxmi at 10% profit.
(i) Find the S. P. of Anita (or C. P. of Shova).
(ii) Find the S. P. of Shova (or C. P. of Laxmi)
b) A dealer bought a laptop for Rs 45,000 and sold it to a retailer at 12% profit. The
retailer again sold it to a customer at 25% profit. How much did the customer
pay for the laptop?
8. a) Sunayana bought a mobile phone and sold to Pratik at 8% loss. Pratik sold it to
Debashis for Rs 11,040 and made 20% profit.
(i) Find the cost price of mobile to Pratik
(ii) Find the cost price of mobile to Sunayana
b) Sayad bought a watch and sold to Dakshes at 10% profit. Dakshes again sold it
to Sahayata for Rs 6,050 at 10% profit.
(i) Find the cost price of Dakshes.
(ii) Find the cost price of Sayad.
c) A wholesaler sold an electric heater to a retailer at 20% profit. The retailer sold
it for Rs 2,052 to a customer at 5% loss.
(i) How much did the retailer pay for it?
(ii) How much did the wholesaler pay for it?
91Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8
Profit and Loss
It’s your time - Project work!
9. 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. = ...................
Profit percent = ...................
S.P. = ................... S.P. = ................... Find S.P.
Find Find
profit percent loss percent
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. = ............................
7.5 Discount and discount percent
The price on the label of an article or product is called the marked price or list price.
When a shopkeeper reduces the marked price (M.P.) of any articles and sells them to the
customer, the amount of reduction in the price is called discount. A shopkeeper allows
discount from the marked price (M.P.) of any article. When M.P. is considered as Rs 100
and discount is calculated from it, it is called discount percent.
Selling price (S.P.) = M.P. – Discount Discount percent = Discount u 100%
Discount = M.P. – S.P. M.P.
Discount amount = Discount % of M.P.
7.6 Value Added Tax (VAT)
Value Added Tax (VAT) is a tax which is charged at the time of consumption of goods
and services. VAT is levied on the selling price of goods. So, VAT is changed as a certain
percentage of S. P.
VAT amount = VAT percent of S.P.
S.P. with VAT = S.P. + VAT% of S.P.
Vedanta Excel in Mathematics - Book 8 92 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Profit and Loss
Summary of calculation of different variables in discount and VAT.
Calculation of Given Process
S. P. M. P. and discount
M. P. and discount % S. P. = M. P. – discount
Discount amount
M. P. and discount% Discount amount = Discount % of M. P.
S. P.
M. P. and discount Discount amount = discount % of M. P.,
Discount % S. P. and VAT % then, S. P. = M. P. – discount amount
VAT amount S. P. and VAT
Direct process
VAT %
S. P. = (100 – discount)% of M. P.
Discount % = discount × 100%
M.P.
VAT amount = VAT % of S. P.
VAT % = VAT × 100%
S.P.
S. P. with VAT S. P. and VAT % S. P. with VAT = S. P. + VAT % of S. P.
Direct process
S. P. with VAT = (100 + VAT)% of S. P.
Worked-out examples
Example 1: The marked price of a fan is Rs 1,620 and the shopkeeper allows some
discount to sell it for Rs 1,539. Find the discount amount and discount
percent.
Solution:
Here, M.P. of the fan = Rs 1,620
S.P. of the fan = Rs 1,539
? Discount amount = M.P. – S.P. = Rs 1,620 – Rs 1,539 = Rs 81
Also, discount percent = Discount amount u 100% = 81 u 100% = 5%
M.P. 1,620
Hence, the discount amount is Rs 81 and discount percent is 5%.
Example 2: The marked price of a mobile is Rs 5,650 and the shopkeeper allows
20% discount to the customer.
(i) Calculate the discount amount.
(ii) How much should a customer pay for it?
Solution:
Here, M.P. of the mobile = Rs 5,650
Discount percent = 20%
(i) Now, discount amount = 20% of M.P.
20
= 100 u Rs 5,650 = Rs 1,130
93Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8
Profit and Loss
(ii) Again, S.P. of the mobile = M.P. – Discount amount
= Rs 5,650 – Rs 1,130 = Rs 4,520
Hence, the discount amount is Rs 1,130 and the customer should pay Rs 4,520.
Example 3: The marked price of an article is Rs 3,200 and 10% discount is allowed
on it. Then, 13% VAT is levied on it while selling?
(i) Find the discount amount (ii) Find S. P. after discount
(iii) Find VAT amount
(iv) How much should a customer pay for it with VAT.
Solution:
Here, M.P. of the article = Rs 3,200
Discount percent = 10% 10
100
(i) Discount amount = 10% of Rs 3,200 = u Rs 3,200 = Rs 320
(ii) S.P. of the article = M.P. – Discount = Rs 3,200 – Rs 320 = Rs 2,880
(iii) Again VAT amount = 13% of S.P. = 13 u Rs 2,880 = Rs 374.40
100
(iv) Now, S.P. with VAT = S.P. + VAT amount
= Rs 2,880 + Rs 374.40 = Rs 3,254.40
Hence, the customer should pay Rs 3,254.40.
Example 4: A retailer bought a woolen jacket for Rs 4,000 and fixed its price 30%
above the cost price. He/she then allows 10% discount. How much
should a customer pay for it with 13% VAT?
Solution: Direct process
Here, C.P. of the jacket = Rs 4,000 M. P.= (100 + 30)% of C. P.
= 130% of Rs 4,000
? M.P. of the jacket = Rs 4,000 + 30% of Rs 4,000
30
= Rs 4,000 + 100 u Rs 4,000 = 130 × Rs 4,000
100
= Rs 4,000 + Rs 1,200 = Rs 5,200
= Rs 5,200
Again, discount amount = 10% Rs 5,200
10
= 100 u Rs 5,200
= Rs 520 Direct process
? S.P. of the jacket = M.P. – Discount
S. P. = (100 – discount)% of M. P.
= Rs. 5,200 – Rs 520 = (100 – 10)% of Rs 5,200
= Rs 4,680 = 90% of Rs 5,200
90
Now, VAT amount = 13% of S.P. = 100 × Rs 5,200
13 = Rs 4,680
100
= u Rs 4,680
= Rs 608.40
Vedanta Excel in Mathematics - Book 8 94 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Profit and Loss
? S.P. with VAT = S.P. + VAT amount Direct process
= Rs 4,680 + Rs 608.40
= Rs 5,288.40 S. P. with VAT = (100 + 13)% of S. P.
113
Hence, the customer should pay Rs 5,288.40. = 100 × Rs 4,680
= Rs 5,288.40
Example 5: Bijaya marked the price of a watch as Rs 2,400. He sold it allowing a
discount of 15% and made a profit of Rs 240. Find the cost price of the
watch.
Solution:
Here, M.P. of the a watch = Rs 2,400
Discount percent = 15%
Profit amount = Rs 240
C.P. of the watch = ?
Now, discount amount = Discount % of M. P.
15
= 100 × Rs 2400
= Rs 360 Direct process
? S.P. of watch = M. P. – Discount amount S. P. = (100 – 15)% of M. P.
= Rs 2,400 – Rs 360 = 85% of Rs 2,400
= Rs 2,040
Again, C. P. of watch = S. P. – Profit amount = Rs 2,040
= Rs 2,040 – Rs 240
= Rs 1,800
Hence, the cost price of the watch was Rs 1,800.
Example 6: After allowing 20% discount on the marked price of a computer, 13%
Value Added Tax (VAT) was levied on it. If the computer was sold for
Rs 48,816, calculate the marked price.
Solution:
Discount percent = 20% , VAT percent = 13% and S. P. with VAT = Rs 48,816
Now, at first let’s find S.P. without VAT. Direct process
(100 + 13)% of S.P. = S.P. with VAT
S.P. + 13% of S.P. = Rs 48,816
or, S.P. + 13 × S.P. = Rs 48,816 or, 113% of S.P. = Rs 48,816
100
113 S.P. Rs 48,816
or, 100 = Rs 48,816 or, S.P. = 113%
or, S.P. = Rs 48,816 × 100 = Rs 43,200 = Rs 48,816
113 113%
100
Again,
M. P. – D% of M.P. = S.P. without VAT = Rs 48,816 × 100
113
or, M. P. – 20% of M. P. = Rs 43,200
= Rs 43,200
20
or, M.P. – 100 of M.P. = Rs 43,200
95Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8
Profit and Loss
or, 4 M.P. = Rs 43,200
5
Rs 43,200 × 5
or, M.P. = 4 = Rs 54,000
Hence, the marked price of the computer was Rs 54,000
EXERCISE 7.2
General Section
1. Say and write the answers as quickly as possible.
a) M.P. = Rs 200, discount = Rs 30, S.P. = …………….
b) M.P. = Rs 400, S.P = Rs 375, discount = …………….
c) S.P. = Rs 550, discount = Rs 60, M.P. = …………….
2. a) M. P. = Rs 100, discount = Rs 25, discount percent = …………….
b) M.P. = Rs 100, S.P. = Rs 90, discount percent = …………….
c) M.P. = Rs 100, discount percent = 10% , discount amount = …………….
d) M.P. = Rs 100, discount percent = 20% , S. P. = …………….
3. a) S. P. = Rs 100, VAT amount = Rs 13, VAT percent = …………….
b) S. P. = Rs 100, VAT percent = 15%, VAT amount = …………….
c) S.P. = Rs 100, S.P with VAT = Rs 113, VAT percent = …………….
d) S.P. = Rs 100, VAT percent = 13% , S.P. with VAT = …………….
Creative Section - A
4. a) If M.P. = Rs 720, discount = Rs 36, find the discount percent.
b) If M.P. = Rs 1,530, S.P. = Rs 1,377, find the discount and discount percent.
c) If M.P. = Rs 940, discount percent = 5%, find the discount and S.P.
d) If S.P. = Rs 450, S.P. with VAT = Rs 495, find VAT percent.
e) If S.P. = Rs 800, VAT percent = 12%, find VAT amount and S.P with VAT.
5. a) The marked price of a bag is Rs 960 and the shopkeeper allows 10%
discount to the customer while selling.
(i) Find the amount of discount. (ii) Find its selling price after the discount.
b) The marked price of a mobile is Rs 2,500 and the shopkeeper allows 15%
discount. How much should a customer pay for buying it?
c) On the occasion of ‘Dashain festival’ , 15% discount is given to every item in a
supermarket. If the labeled price of a pair of shoes is Rs 3,200, find its selling
price.
Vedanta Excel in Mathematics - Book 8 96 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Profit and Loss
6. a) Let the marked price be Rs x, S.P. is Rs 180 and discount percent is 10%. Find
the marked price. [Hints: x – 10% of x = 180]
b) If a shopkeeper allows 25% discount and sells an umbrella for Rs 450, find the
marked price of the umbrella.
Creative Section - B
7. a) A shopkeeper marked the price of a calculator as Rs 1,600 and she/he sold it
allowing a discount of 10%.Then, she made a profit of Rs 140.
(i) Find the amount of discount.
(ii) Find the selling price after allowing discount.
(iii) Find the cost price of the calculator.
b) Mr. Muhammad fixed the marked price of a cycle as Rs 6,000. He sold it after
allowing a discount of 15% and made a profit Rs 500, find the cost price of the
cycle.
c) Kamala marked the price of a cosmetic item as Rs 400. She offered her customers
a discount of 20% and made a loss of Rs 30, what was the actual cost of the item
to her?
8. a) The marked price of a fan is Rs ,1500. The shopkeeper allows 20% discount on
it and then 13% VAT is charged.
(i) Find the amount of discount.
(ii) Find the selling price after giving discount.
(iii) Find the VAT amount.
(iv) Find the selling price with VAT.
b) The price of a rice cooker is marked as Rs 2,500. If 40% discount is allowed and
then 13% VAT is charged, find the selling price of the cooker with VAT.
c) The marked price of a jacket is Rs 5,000. What is the price of the jacket if 10%
VAT was levied after allowing 20% discount?
9. a) A retailer bought a watch for Rs 800 and fixed its price 25% above the cost price.
He then allows 10% discount.
(i) Find the marked price of the watch.
(ii) Find the discount amount.
(iii) Find the selling price of the watch.
(iv) How much should a customer pay for it with 13% VAT?
97Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8
Profit and Loss
b) A shopkeeper bought an electric fan for Rs 2,000 and fixed its price 20% above
the cost price. If he then gives 15% discount, how much should a customer pay
for it with 15% VAT?
c) A dealer marks the price of his mobiles 40% above the cost price and allows
20% discount. If his purchase price of a mobile is Rs 7,500, how much should a
customer pay for it with 13% VAT?
10. a) A gift house allowed 20% discount on the marked price of a doll, 13% VAT was
levied on it. If the doll was sold at Rs 1,808, what was its marked price?
b) After allowing 16% discount on the marked price of a watch, 13% Value Added
Tax (VAT) was levied on it. If the watch was sold for Rs 4,746, calculate the
marked price of the watch.
It’s your time - Project work!
11. 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 Find
S.P. discount percent
S. P. = ........................ S. P. with VAT = .........
VAT% = ........................ VAT% = ................
Find S.P. Find S.P.
with VAT without VAT
b) Let’s search the rate of VAT of different countries in the available website.
Compare these VAT rates with 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.
Vedanta Excel in Mathematics - Book 8 98 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
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