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AMARCHAND SINGHVI INTERNATIONAL SCHOOL

QUESTIONS AND ANSWERS

CLASS – VII (MATHEMATICS)

CHAPTER 8 – RATIO AND PROPORTION

Note: All these questions and answers are to be done in the notebook.

Ex 8(A)

Q.1)Express each of the following ratios in the simplest form:

(i) 28:63

Answer

Now students in this we will be dividing numerator and denominator by HCF of 28 and 63 to

obtain simplest form. HCF of 28 and 63 = 7

28:63 =

=

=

= 4 : 9

(iii) 0.4 : 0.6

Answer

0.4 : 0.6 =

=

=

=

= 2 : 3

(v)

:

Answer

:

=

=

÷

=

= 9 : 4

(vii)

:

:

Answer

In this type of sums we will take the LCM of denominators first and then we will multiply the

LCM with each fraction to obtain simplest form. LCM of 4,8 and 10 = 40

:

:

=

:

:

x 40

=

:

:

=

:

:

=

=

:

:

= 10: 5 : 4

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(ix) 125 g : 1 kg

Answer

Now students we know that ratio can be taken of the same quantity and same units so here

one term is in gram and other is in kilogram so we will be converting kilogram into gram and

then we will be dividing numerator and denominator by HCF of 125 and 1000 to obtain

simplest form. HCF of 125 and 1000 = 125

125 g : 1 kg = 125 g : 1000 g

= 125 : 1000 =

=

=

= 1 : 8

(xi) 36 minutes to 1

hours

Answer

Now students we know that ratio can be taken of the same quantity and same units so here

one term is in minutes and other is in hours so we will be converting hours into minutes and

then we will be dividing numerator and denominator by HCF of 36 and 90 to obtain simplest

form. HCF of 36 and 90 = 18

1

hours =

hours =

x 60 minutes

=

minutes =

minutes

= 3 x 30 minutes = 90 minutes

36 minutes : 1

hours = 36 minutes : 90 minutes

= 36 : 90 =

=

=

= 2 : 5

Q.2) There are 2500 students in a school out of them 1100 are girls and the rest are boys find

the ratio of:

(i) number of boys to number of girls.

Answer

Number of Students = 2500

Number of Girls = 1100

Number of Boys = Number of Students Number of Girls = 2500 1100 = 1400

So, Number of boys to Number of girls = 1400 : 1100 =

=

14 : 11

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××(iii) number of students to number of boys.

Answer

So, Number of Students to Number of boys = 2500 : 1400 =

=

25 : 14

Q.4) If the ratio of the length to the breadth of a rectangle be 5:3 and its perimeter is 144 m,

find the length of the rectangle.

Answer

Ratio of length : breadth = 5:3

So here students we come to understand that this ratio is in simplest reduced form and the original

length and breadth of rectangle is not given to us. So there would surely be a number that would

have been a common factor in length and breadth which was eliminated in ratio by cutting and then

5:3 was left. So if we assume that common factor number as Y then

The length of rectangle = 5 x Y = 5Y and breadth of rectangle = 3 x Y = 3Y

Now, Perimeter of Rectangle = 2 ( Length + Breadth )

144 = 2 ( 5Y + 3Y )

144 = 2 ( 8Y )

144 = 2 x 8Y

144 = 16Y

144 ÷ 16 = Y

9 = Y

The length of rectangle = 5 x Y = 5 x 9 = 45 m

Q.8) (i) if a:b = 5:7 and b:c = 14:15, find a:c.

Answer

=

and

a : c =

x

=

x

=

=

=

= 2:3

Q.13) Rina lost her weight in the ratio 5:3. Her original weight was 80 kg. What is her new

weight?

Answer

Ratio = 5:3 =

Inverse of Ratio =

New weight = Inverse of Ratio x Original weight

=

x 80

=

=

= 3 x 16

= 48 kg

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So, rina’s new weight is 8 kg.

Q.16) if 3A=5B and 4B= 6C, then A:C equals.

a) 5:2 b) 3:5 c) 2:5 d) 4:5

Answer

3A = 5B

A =

=

and

4B = 6C

B =

=

A : C =

x

=

x

=

=

=

=

= 5 : 2

So the correct answer is option a) 5:2

Note: Students please note that question number 3, 5, 6, 7, 9, 10, 11, 12, 14,15, 17 and 18 of

Exercise 8(A) are omitted from the syllabus.

Ex 8 (B)

Q.1)

(i)

Answers

P

=

P

=

Now we can see that Product of extremes = Product of means

So we can say that yes

are in proportion.

(iii) 2

Answers

P

=

= 9

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P

=

Now we can see that Product of extremes is not equal to Product of means

So we can say that yes 2

are not in proportion.

Q.2)

(i) 0.9 : 0.6 : : X : 3

Answers

Here students it is already given that all these four numbers are in proportion.

So

Product of extremes = Product of means

0.9 x 3 = 0.6 x X

2.7 = 0.6 x X

2.7 ÷ 0.6 = X

= X

= X ( I have multiplied by 10 just to remove the decimal place)

= X

= X

4.5 = X

(iii) 1.6 : X : : 0.12 : 0.24

Answers

Here students it is already given that all these four numbers are in proportion.

So

Product of extremes = Product of means

1.6 x 0.24 = X x 0.12

0.384 = X x 0.12

0.384 ÷ 0.12 = X

= X

= X ( I have multiplied by 1000 just to remove the decimal place)

= X ( Do the cutting now by table of 12 )

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= X

3.2 = X

Q.3)

(i) 8, 32, 17

Answers

Let the fourth proportional be Y. Then we can say that 8, 32, 17, Y are in proportion.

That means 8:32::17:Y

Now we know that

Product of extremes = Product of means

8 x Y = 32 x 17

Y x 8 = 544

Y = 544 ÷ 8

Y = 68

So , the fourth proportional is 68.

(iii)

Answers

Let the fourth proportional be Y. Then we can say that

, Y are in proportion.

That means

: Y

Now we know that

Product of extremes = Product of means

x Y =

Y x

=

Y x

=

Y =

÷

Y =

x

Y =

Y =

So , the fourth proportional is

.

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Q.7) If 25 persons can dig a trench 36 m long in one day, then find the number of persons

required to dig a trench 108 m long in one day

Answers

Let the number of persons required to dig a trench 108 m long in one day be Y.

Now by cross multiplication method

Number of people Length of trench (m)

Y 108

25 36

=

=

= ÷ 36

=

So, 75 people are required to dig a trench of 108 m long in one day.

Q.8) The label on a flavoured milk bottle states that there is 1.5 g of fat in 200 mL. if you drink

500 mL of that milk, how much fat would you be getting?

Answers

Let the amount of fat in 500 mL be Y grams.

Now by cross multiplication method

Amount of fat (g) Amount of milk (mL)

Y 500

1.5 200

=

=

= ÷ 200

=

So, 3.75 grams of fat is there in 500 mL of milk.

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Q.9) Rajat can type 3200 words in one hour. How many words can he type in 15 minutes?

Answers

Let the number of words typed in 15 minutes be Y .

Now by cross multiplication method

Number of words Time (minutes)

Y 15

3200 60

=

=

= ÷ 60

=

So, Rajat can type 800 words in 15 minutes.

Q.12) Ayushree’s goal is to practice 20 minutes of Maths for every 60 minutes of class time.

How many minutes of Maths practice should she do if the class is 90 minutes long ?

a) 24 b) 32 c)25 d) 30

Answers

Let the number of minutes of Maths practice in 90 minutes be Y .

Now by cross multiplication method

Maths practice time (minutes) Class time (minutes)

Y 90

20 60

=

=

= ÷ 60

=

So, Ayushree should do 30 minutes of Maths practice in 90 minutes class.

So, the correct answer is Option d) 30

Q.13) A building casts a shadow that is 432m long. At the same time a person who is 2m long

casts a shadow 6m long . How high is the building?

a) 216 m b) c) 108 m d) m

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Answers

Let the Height of building be Y .

Now by cross multiplication method

Height (m) Length of Shadow (m)

Y 432

2 6

=

=

= ÷ 6

=

So, height of building is 144 m.

So, the correct answer is Option b) 144 m

Note: Students please note that question number 2(Part V), 4, 5, 6, 10, 11, 14 and 15 of

Exercise 8(B) are omitted from the syllabus.

Ex 8 (C)

Note: Students please note that Exercise 8(C) is omitted from the syllabus.