Ratio and Proportion ICSE Class-7th Concise Selina

Ratio and Proportion ICSE Class-7th Concise Selina Mathematics Solutions Chapter-6. We provide step by step Solutions of Exercise / lesson-6 Ratio and Proportion for ICSE Class-7 Concise Selina Mathematics. Our Solutions contain all type Questions with Exe-6 A and Exe-6 B, to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7.

Ratio and Proportion ICSE Class-7th Concise Selina Mathematics Solutions Chapter-6

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Exe-6 A,

Exe-6 B,

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Exercise – 6 A

Ratio and Proportion ICSE Class-7th Concise Selina Mathematics

Question 1.

Express each of the given ratio in its simplest form :

Answer-1.

(i)

(HCF of 22 and 66 = 22)

= 1 : 3

(ii)

1.5 : 2.5 = 1.5/2.5 = 15/25

=(15÷5)/(25÷5)=3/5 (HCF of 15, 25 = 5)

= 3 : 5

(iii)

(iv)

40 kg : 1 quintal = 40 kg : 100 kg  …(1 quintal = 100 kg)

= 40/100 = (40÷20)/(100÷20)

=25

 (HCF of 40, 100 = 20)

= 2: 5

(v)

(vi)

(vii)

(viii)

6 months: 1 (1/3) years = 6 months : (4/3)×12 months

= 6 months : 16 months

(ix)

Question 2.

Divide 64 cm long string into two parts in the ratio 5 : 3.

Answer-2.

Sum of ratios = 5 + 3 = 8

∴ first part = 5/8 of 64 cm = 40 cm

Second part = 3/8 of 64 cm = 24 cm

Question-3.

Rs. 720 is divided between x and y in the ratio 4:5. How many rupees will each get?

Answer-3.

Total amount = Rs. 720 Ratio between x, y = 4 : 5

Sum of ratios = 4 + 5 = 9

x’s share = 4/9 of Rs. 720 = Rs. 320

y’s share = 5/9 of Rs. 720 = Rs. 400

Question-4.

The angles of a triangle are in the ratio 3 :2 : 7. Find each angle.

Answer-4.

Ratio in angles of a triangle = 3:2:7

Sum of ratios = 3 + 2 + 7=12

Sum of angles of a triangle = 180°

∴ First angle = (3/12)×180°=45°

Second angle = (2/12) x 180°= 30°

Third angle =(7/12) x 180°= 105°

Question-5.

A rectangular field is 100 m by 80 m. Find the ratio of
(i) length to its breadth
(ii) breadth to its perimeter.

Answer-5.

(i)

Length of field (l) = 100 m

Breadth (b) = 80 m

∴Perimeter = 2 (l + b) = 2 (100 + 80) m

= 2 x 180 = 360 m

Ratio between length and breadth

= 100 : 80 = 5 : 4
(Dividing by 20, the HCF of 100 and 80)

(ii)

Length of field (l) = 100 m

Breadth (b) = 80 m

∴Perimeter = 2 (l + b) = 2 (100 + 80) m

= 2 x 180 = 360 m

Ratio between breadth and its perimeter

= 80 : 360 = 2 : 9
(Dividing by 40, the HCF of 80 and 360)

Question-6.

The sum of three numbers, whose ratios are 3  : 4  : 6  is 4917.Find the numbers.
Answer-6.

Sum of three numbers = 4917

Ratio between them = 3(1/3) : 4(1/5) : 6(1/8)

= 10/3:21/5:49/8

= (400:504:735)/120  (LCM of 3, 5, 8 = 120)

= 400 : 504 : 735

Sum of ratio’s = 400 + 504 + 735 = 1639

∴ First number = 400/1639 of 4917 = 1200

Second number = 504/1639 of 4917 = 1512

and third number = 735/1639 of 4917 = 2205

The ratio between two quantities is 3 : the first is Rs. 810, find the second.
Answer-7.

Ratio between two quantities = 3 : 4

Sum of ratio = 3+4 = 7

∴ Second quantity = Rs. 810x/43

= Rs. 270 x 4 = Rs. 1080

Question-8.

Two numbers are in the ratio 5 : 7. Their difference is 10. Find the numbers.
Answer-8.

Question-9.

Two numbers are in the ratio 10 : 11. Their sum is 168. Find the numbers.
Answer-9.

Question 10.

A line is divided in two parts in the ratio 2.5 : 1.3. If the smaller one is 35T cm, find the length of the line.
Answer-10.

Ratio between two parts of a line

= 2-5 : 1-3 =25 : 13

Sum of ratios = 25 + 13 = 38

Length of smaller part = 35.1 cm 38

Now length of line = (38/13)×35.1 cm

= 38 × 2.7 cm = 102.6 cm

Question-11.

In a class, the ratio of boys to the girls is 7:8. What part of the whole class are girls.
Answer-11

Ratio between boys and girls = 7:8

Sum of ratios = 7 + 8 = 15

∴ Girls are 8/15 of the whole class.

Question-12.

The population of a town is ’ 50,000, out of which males are  of the whole population. Find the number of females. Also, find the ratio of the number of females to the whole population.
Answer-12

Total population = 180,000

Population of males = 1/3 of 180,000 = 60,000

∴ Population of females = 180,000 – 60,000 = 120,000

Ratio of females to whole population

= 120,000 : 180,000 = 2:3

Question-13.

Ten gram of an alloy of metals A and B contains 7.5 gm of metal A and the rest is metal B. Find the ratio between :
(i) the weights of metals A and B in the alloy.
(ii) the weight of metal B and the weight of the alloy.
Answer-13.

 

Total weight of A and B metals = 10 gm A’s weight = 7.5 gm B’s weight = 10 – 7.5 = 2.5 gm

(i) Ratio between A and B = 7.5 : 2.5

=75/10 : 25/10 = 3:1

(ii) Ratio between B and total alloy

= 2.5:10 = (25/10):10

⇒ 25 : 100 = 1 : 4

Question-14.

The ages of two boys A and B are 6 years 8 months and 7 years 4 months respectively. Divide Rs. 3,150 in the ratio of their ages.
Answer-14

A s age = 6 years 8 months

= 6 x 12 + 8 = 72 + 8 = 80 months

B’s age = 7 years 4 months = 7 x 12 + 4 = 84 + 4 = 88 months

∴ Ratio between them = 80 : 88 = 10 : 11

Amount = Rs. 3150

Sum of ratios = 10 + 11 =21

∴ A’s share = (3150×10)/21 = 1500 =

Rs. 1500

B’s share = (3150×11)/21=1650 =

Rs. 1650

Question-15.

Three persons start a business and spend Rs. 25,000; Rs. 15,000 and Rs. 40,000 respectively. Find the share of each out of a profit of Rs. 14,400 in a year.
Answer-15

A’s investment = Rs. 25000

B’s investment = Rs. 15000

C’s investment = Rs. 40000

∴ Ratio between their investment

= 25000 : 15000 : 40000

= 5 : 3 : 8

Sum of ratios = 5 + 3 + 8=16 Total profit = ₹ 14400

∴ A’s share = (14400/16)×5=₹4500

B’s share = (14400/16)×3 = ₹ 2700

C’s share = (14400/16)×8 = ₹ 7200

Question-16.

A plot of land, 600 sq m in area, is divided between two persons such that the first person gets three-fifth of what the second gets. Find the share of each.
Answer-16.

Area of plot of land = 600 sq. meter

Let second’s share = x

Then first share = (3/5)x

∴ Ratio between them (3/5)x:x

= (3/5):1 = 3:5

Sum of ratios = 3 + 5 = 8

∴ Share of first person = (600/8)×3 = 225 sq.m

and second share = (600/8)×5=375 sq.m.

Question-17.

Two poles of different heights are standing vertically on a horizontal field. At a particular time, the ratio between the lengths of their shadows is 2 :3. If the height of the smaller pole is 7.5 m, find the height of the other pole.
Answer-17.

Question-18.

Two numbers are in the ratio 4 : 7. If their L.C.M. is 168, find the numbers.
Answer-18.

Given, Ratio in two numbers = 4:7

and their L.C.M. = 168

Let first number = 4x

and second number = 7x

Now, L.C.M. of 4x and 7x

= 4 x 7 x x = 28x

∴ 28x = 168

x = 168/28

x = 6

∴ Required numbers = 4x and 7x = 4 x 6 = 24

and 7 x 6 = 42

Question-19.

is divided between A and B in such a way that A gets half of B. Find :
(i) the ratio between the shares of A and B.
(ii) the share of A and the share of B.
Answer-19.

Question-20.

The ratio between two numbers is 5 : 9. Find the numbers, if their H.C.F. is 16.

Answer-20.

Let the first number be 5x and second number be 9x

H.C.F. of 5x and 9x = Largest number common to 5x and 9x = x

Given H.C.F. = 16 ⇒ x = 16

∴ Required numbers = 5x and 9x = 5×16 and 9×16 = 80 and 144

Question-21.

A bag contains ₹ 1,600 in the form of ₹10 and ₹20 notes. If the ratio between the numbers of ₹10 and ₹20 notes is 2 : 3; find the total number of notes in all.
Answer-21.

Total amount in the bag = 1600

It contains notes in the denomination of ₹10 and 20

Ratio between the number of ₹10 and 20 notes is = 2 : 3

Let number of ₹10 note = x

and number of ₹ 20 notes = y

According to condition,

10x + 20y = 1600   ….(i)

and x = 2/3y         ….(ii)

Now, substitute the value of x in eq (i)

10×(2/3)y+20y=1600

⇒(20/3)y+20y=1600

⇒(20+60)/3 y=1600

⇒(80/3)y=1600

⇒ y = (1600×3)/80

∴ y = 60

Now, substitute the value of y in eq (ii), we get

x = 2/3 × 60=40

Total number of notes in all = x + y

= 60 + 40 = 100 notes

Question-22.

The ratio between the prices of a scooter and a refrigerator is 4 : 1. If the scooter costs ₹45,000 more than the refrigerator, find the price of the refrigerator.
Answer-22.

Ratio between the prices of scooter and a refrigerator = 4:1

Cost price of scooter = ₹45,000

Let the cost of scooter = 4x

Cost of refrigerator = 1x

According to condition,

Cost of scooter > Cost of refrigerator

⇒ 4x- 1x = 45000

⇒ 3x = 45000

x = 45000/3

⇒ x = 15000

∴ Price of refrigerator = ₹15000

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Exercise – 6 B

Solutions of Ratio and Proportion for ICSE Class-7th Concise Selina Mathematics 

Question-1.

Check whether the following quantities form a proportion or not ?

Answer-1.

(i)

3x, 7x, 24 and 56

If these are in proportion, then

3x × 56

= 7x × 24

⇒ 168 x

= 168 x

which is true.

Hence 3x, 7x, 24 and 56 are in proportion.

(ii)

0.8, 3, 2.4 and 9 are in proportion.

if 0.8 × 9

= 3 × 2.4

⇒ 7.2

= 7.2

which is true

Hence 0.8, 3, 2.4 and 9 are in proportion.

(iii) 

(iv)

0.4, 0.5, 2.9 and 3.5 are in proportion

if 0.4 × 3.5

= 0.5 × 2.9

⇒ 1.40

= 1.45

which is not true

Hence 0.4, 0.5, 2.9

and 3.5 are not in proportion.

(v) 

proportion.

Question-2.

Find the fourth proportional of

Answer-2.

(i) 

Fourth proportional of 3, 12 and 4

(12 x 4)/3 = 16

(ii)

(iii)

(iv)

(v)

Fourth proportional of 4 hours 40 minutes, 1 hour 10 minutes and 16 hours.

4 hours 40 minutes = 4 × 60 + 40

= 240  40 = 280

1 hour 10 minutes = 1 × 60 + 10

= 60 + 10 = 70 minutes

16 hours = 16 × 60 = 960 minutes

∴ Fourth proportional = (70×960)/280

= 240 minutes = 240/60 = 4 hours

Question-3.

Find the third proportional of

Answer-3.

(i)

Third proportional of 27 and 9

(9 x 9)/27 = 3

(ii) 

Third proportional to 2 m 40 cm and 40 cm or 240 cm and 40 cm

(40×40)/240

=203

=623 cm

(iii) 

(iv) 

(v)

Question-4.

Find the mean proportional between

Answer-4.

(i)

Mean proportional between 16 and 4

(ii)

Mean proportional between 3 and 27

(iii)

Mean proportional between 0.9 and 2.5

(iv)

Mean proportional between 0.6 and 9.6

(v)

Question-5.

Answer-5.

(i)

A : B = 3 : 5

=3/5 : 1

(Dividing by 5)

and B : C = 4 : 7

= 1: 7/4    …

(Dividing by 4)

∴ A : B : C = 3/5 : 1 : 7/4

= 12 : 20 : 35    …

(Multiplying by 5 × 4 = 20)

(ii)

x : y = 2 : 3

= 2/3 : 1

(Dividing by 3)

and y : z = 5 : 7

=1 : 7/5    …

(Dividing by 5)

∴ x : y : z = 2/3 : 1 : 7/5

= 10 : 15 : 21    …

(Multiplying by 3 × 5 = 15)

(iii)

m : n = 4 : 9

(iv)

(v)

a : b = 1.5 : 3.5

⇒ a : c = 5 : 14

(vi)

∴ p : r = 45 : 98

Question-6.

If x: y – 5 :4 and 2 : x = 3 :8, find the value of y.
Answer-6.

Question-7.

Find the value of x, when 2.5 : 4 = x : 7.5.

Answer-7.

2.5 : 4 = x : 7.5

4 × x = 2.5 × 7.5

Question-8.

Show that 2, 12 and 72 are in continued proportion.

Answer-8.

Three numbers a, b and c are in continued

proportion if, a : b : : b : c

The numbers are 2, 12 and 72

 

— End of Ratio and Proportion Solutions :–

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