8
RATIO AND PROPORTION
1. Find the ratio of each of the following in simplest form :
(i) Rs 3.25 to Rs 2.7
(ii) 15 hours to 1 day
(iii) 4 months to 3 years (iv) 35 paise to Rs 7.80
(v) 75 kg to 2 quintals
(vi) 15 g to 2 kg
Ans. (i) Rs 3.25 to Rs 2.75 =
Rs 3.25 3.25 13
=
=
= 13 : 11
Rs 2.75 2.75 11
15 hours 15 hours 5
= =5:8
=
1 day
24 hours 8
4 1
= =1:9
(iii) 4 months to 3 years = 4 months to 3 × 12 months =
36 9
(ii) 15 hours to 1 day =
(iv) 65 paise to Rs 7.80 =
65
1
65 p aise
65 p aise
=
=
=
= 1 : 12
Rs 7.80
7.80 × 100 p aise 780 12
(v) 75 kg to 2 quintals =
75 kg
75 kg
3
= =3:8
=
2 quintals
2 × 100 kg 8
(vi) 150 g to 2 kg =
150 g
150 g
3
=
=
= 3 : 40
2 kg
2 ×1000 g 40
2. Express each of the following ratios in simplest form :
(i) 49 : 35 (ii)
Ans. (i) 49 : 35 =
(iii)
146 : 365
49 7
= =7:5
35 5
(iii)
(ii)
1 1 1
(vi) 2 : 3 : 5
3 4 6
3 4 5
: :
4 5 6
146 : 365 =
146 2
= =2:5
365 5
3 4 5
4
3
5
: : =  × 60  :  × 60  :  × 60  = 45 : 48 : 50
4 5 6
4
 5
 6

7 13 31
1 1 1
LCM of 3, 4 and 6 = 12
(iv) 2 : 3 : 5 = : :
3 4 6
3 4 6
7 13 31  7
  13
  31

: :
=  × 12  :  × 12  :  × 12  = 28 : 39 : 62
3 4 6
3
 4
  6

1
Question Bank
Class - VI Mathematics
3. In a school, there are 420 boys and 180 girls. Find the ratio of:
(i) girls to total number of students (ii) boys to total number of students
Ans. Number of boys = 420 and number of girls = 180
⇒ Total number of students = 420 + 180 = 600
(i) Ratio of girls to total number of students = 180 : 600 =
180 3
=
= 3 : 10
600 10
(ii) Ratio of boys to total number of students = 420 : 600 =
420 7
=
= 7 : 10
600 10
4. The population of a village is 3540. If the number of males is 2065, find the ratio of
males to females.
Ans. Total population of a village = 3540; Number of males = 2065
Number of females = 3540 – 2065 = 1475
2065 7
= =7:5
Thus, ratio of males to females = 2065 : 1475 =
1475 5
5. A line segment 32 cm long is divided into two parts in the ratio 3 : 5 Find the
length of each part.
Ans. Total ratio = 3 + 5 = 8, Length of one part =
3
× 32 = 12 cm , Length of another
8
5
× 32 = 20 cm
8
6. The ratio of tin and zinc in an alloy is 3 : 4. How much tin is there in 10.5 g of the
alloy ?
part =
3
× 10.5 g = 3 × 1.5 g = 4.5 g
7
7. Divide Rs 6655 between Amit and Kunal in the ratio 7 : 4.
Ans. Sum of ratio = 3 + 4 = 7. Quantity of tin in alloy =
Ans. Sum of terms of ratio = 7 + 4 = 11 Amit’s share = Rs
7
× 6655 = Rs 7 × 605 = Rs 4235
11
4
× 6655 = Rs 4 × 605 = Rs 2420
11
8. Divide Rs 9002 among A, B, C in the ratio 3 : 4 : 7.
Kunal’s share = Rs
Ans. Sum of ratio = 3 + 4 + 7 = 14, A’s share = Rs
3
× 9002 = 3× 643 = Rs 1929
14
4
× 9002 = 4 × 643 = Rs 2572
14
7
C’s share = Rs
× 9002 = 7 × 643 = Rs 4501.
14
2
Class - VI Mathematics
B’s share = Rs
Question Bank
9. Divide 182 into three parts in the ratio
1 1 1
: : .
10 15 20
Ans. LCM of 10, 15 and 20 = 60
1 1 1
1
 1
  1

: :
=  × 60  :  × 60  :  × 60  = 6 : 4 : 3
10 15 20  10
  15
  20

6
× 182 = 6 × 14 = 84
Total ratio = 6 + 4 + 3 = 13, First’s part =
13
4
3
× 182 = 3 × 14 = 42
Second’s part = × 182 = 4 × 14 = 56, Third’s part =
13
13
10. Divide Rs 3010 among A, B, C in such way that A gets double of what B and B gets
double of what C gets.
Ans. Let C gets = Rs 1, B gets = Rs 2 × 1 = Rs 2, A gets = Rs 2 × 2 = Rs = 4
∴ A : B : C = 4 : 2 : 1, sum of terms of ratio = 4 + 2 + 1 = 7
∴
2
4
× 3010 = 4 × 430 = Rs 1720, B’s share = Rs × 3010 = 2 ×
7
7
1
430 = Rs 860, C’s share = Rs × 3010 = Rs 430.
7
The sides of a triangle are in the ratio 2 : 3 : 4. If its perimeter is 54 cm, find the
lengths of sides of the triangle.
Sum of the terms of ratio = 2 + 3 + 4 = 9, Sum of side i.e. perimeter = 54 cm
2
4
3
1st side = × 54 = 12 cm; 2nd side = × 54 = 18 cm; 3rd side = × 54 = 4 ×6
9
9
9
= 24 cm.
The angles of a triangle are in the ratio 7 : 5 : 3. Find the measure of each angle of
the triangle.
Sum of the terms of ratio = 7 + 5 +3 = 15
We know that the sum of angles of triangle = 180°
A’s share = Rs
11.
Ans.
12.
Ans.
7
5
× 180 = 7 × 12 = 84°, 2nd angle =
× 180 = 5 × 12 = 60°,
15
15
3
× 180 = 3 × 12 = 36°.
3rd angle =
15
13. Find the ratio of the price of coffee to that of tea, when coffee costs Rs 24 per 100
gm and tea costs Rs 80 per kilogram.
∴ Ist angle =
24
24 1000 3
×
Ans. The ratio of the price of coffee to tea = 100 =
= =3:1
80
1
100 80
1000
3
Question Bank
Class - VI Mathematics
14. An office opens at 9.30 A.M. and closes at 5.30 P.M.
Official lunch time is from 12.30 P.M. to 1.00 P.M. What is the ratio of
(i) lunch interval to total office hours ?
(ii) lunch interval to working hours ?
Ans. Opening time = 9.30 A.M. Closing time = 5.30 P.M. or 17.30
Total office hours = (17.30 – 9.30) = 8.00 hours
∴ Lunch interval = 1.00 P.M. – 12.30 P.M. = (13.00 – 1230) = 0.30 hours
30 1
= hr.
60 2
1
1 8
1 1 1
(i) Lunch Interval : Total office hour. = : ⇒ 2 = × =
= 1 : 16
8 2 8 16
2 1
1
1 16 − 1 15
=
(ii) Working hours = 8 − =
2
2
2
1
1 15
1 2
1
Lunch interval: working hours = : = 2 = × =
= 1 : 15
2 2 15 2 15 15
2
India has a fleet of 28 naval ships, Pakistan has 14 naval ships and Bangladesh has
2 naval ships. Find the ratio of the sizes of three fleets, in lowest terms.
India has naval ship = 28, Pakistan has naval ship = 14, Bangladesh has naval ship
=2
Required ratio = 28 : 14 : 2 = 14 : 7 : 1
A pole of length 1m 80 cm is divided into two parts such that their lengths are in
the ratio 5 : 7, find the length of each part of the pole.
Total length of pole = 1 m 80 cm or 180 cm
(∵ 1 metre = 100 cm)
Ratio of length = 5 : 7; Sum of terms of ratio = 5 + 7 = 12
=
15.
Ans.
16.
Ans.
7
5
= 75 cm; Length of IInd part = 180 ×
= 105 cm
12
12
17. Divide Rs. 560 between Ramu and Munni in the ratio 3 : 2.
Ans. Total amount to be divided = Rs 560; Sum of the terms of ratio = 3 + 2 = 5
Length of Ist part = 180 ×
3
2
3
; Fraction of Munni = ; Part of Ramu = 560 × =
5
5
5
2
Rs 336; Part of Munni = × 560 = Rs 224.
5
18. Arnav and Prerna are aged 8 years and 10 years respectively. Their mother divides
Rs 90 in the ratio of their ages. How much does each get ?
4
Question Bank
Class - VI Mathematics
Fraction of Ramu =
Ans. Ratio of ages = 8 : 10 = 4:5. Sum of terms of ratio = 4 + 5 = 9
Total money = Rs 90, Part of Arnav = 90 ×
Part of Prerna = 90 ×
4
= Rs 40
9
5
= Rs 50
9
19. An amount of hundred rupees is divided among two persons in the ratio
1 1
: .
10 15
How much money does each get ?
1
1 1
1 15 3
Ans. Ratio =
:
⇒ 10 = × = = 3 : 2; Sum of ratio = 3 + 2 = 5
1 10 1 2
10 15
15
3
Total amount = Rs 100; Part of first person = 100 × = Rs 60
5
2
Part of second person = 100 × = Rs 40
5
20. The lengths of the sides of a triangle are in the ratio 2 : 3 : 4. If the perimeter of the
triangle is 63 cm, find the lengths of the sides of the triangle.
Ans. Ratio of sides of triangle = 2 : 3 : 4; Sum of ratio of sides of triangle = 2 + 3 + 4 = 9
If the perimeter = 63 cm; First side = 63 ×
63 ×
2
= 7 × 2 = 14 cm; Second side =
9
3
4
= 7 × 3 = 21 cm; Third side = 63 × = 7 × 4 = 28 cm.
9
9
1
kg. If in the alloy, the ratio of zinc to
2
copper is 1 : 4, find the weight of copper in it.
21. An alloy of zinc and copper weighs 12
1
25
kg. Given ratio = 1 : 4; Sum of the terms of the
Ans. Weight of alloy = 12 kg =
2
2
4 25
kg ; = 2 × 5 = 10 kg
ratio = 1 + 4 = 5; Weight of copper = ×
5 2
22. Divide a sum of Rs 1,350 among A, B and C such that B gets equal to one - third
of A and C gets equal to half of B.
Ans. Let the share A = 1; Share of B =
1
1
1
1
rd of A = × 1 = ; Share of C =
of B =
3
3
3
2
1 1 1
× =
2 3 6
Class - VI Mathematics
5
Question Bank
1 1
1
1
∴ Given ratio = 1: : = 1× 6 : × 6 : × 6 = 6 : 2 : 1 (∵ LCM of 3 and 6 = 6)
3 6
3
6
Sum of the terms of the ratio = 6 + 2 + 1 = 9; Total sum = Rs 1350
A’ share =
2
6
× Rs 1350 = 6 × Rs 150 = Rs 900; B’ share = × Rs 1350 = Rs 300
9
9
1
× Rs 1350 = 1 × Rs 150 = 150
9
23. A mixture weighing 155 kg contains three substances A, B and C in the ratio
1 1 1
: : . Find the quantity of each substance in the mixture.
2 5 3
C’ share =
Ans. Weight of mixture = 155 kg. Given ratio =
= 15 : 6 : 10
1
1
1
1 1 1
: : = × 30 : × 30 : × 30
2
5
3
2 5 3
Sum of the terms of the ratio = 15 + 6 + 10 = 31
Quantity of substance A in mixture =
15
× 155 kg = 15 × 5 kg = 75 kg
31
Quantity of substance B in mixture =
6
× 155 kg = 6 × 5 kg = 30 kg
31
10
× 155 kg = 10 × 5 kg = 50 kg.
31
24. Mr.Gupta divides Rs 4,050 among his three children Ashok, Mohit and Geeta in
such a way that Ashok gets equal to four times of what Mohit gets and Mohit gets
equal to 2.5 times of what Geeta gets. Find the share of each.
Ans. Let the share of Geeta = 1. Share of Mohit is (2.5 times of Geeta) = 2.5
Share of Ashok is (4 times of Mohit) = 4 × 2.5 = 10.
Ratio = 1 : 2.5 : 10 = 1 × 2 : 2.5 × 2 : 10 × 2 = 2 : 5 : 20
Sum of the terms of ratio = 2 + 5 + 20 = 27
Quantity of substance C in mixture =
Share of Geeta =
2
× Rs 4050 = 2 × Rs 150 = Rs 300
27
Share of Mohit =
5
× Rs 4050 = 5 × Rs 150 = Rs 750
27
Share of Ashok =
20
× Rs 4050 = 20 × Rs 150 = Rs 3000.
27
Class - VI Mathematics
6
Question Bank
25. In a proportion, the first, second and fourth terms are 32, 112 and 217 respectively.
Find the third term.
Ans. Let the third term of the proportion be x
32
x
32 × 217 2 × 217
=
=
= 2 × 31 = 62.
⇒ x=
112 217
112
7
26. The ratio of length to breadth of a rectangular playground is 13 : 10. If the length
of the playground is 260 meters, find its breadth.
Ans. Let the breadth be x. Length = 260 meters.
Length : Breadth = 13 : 10 ⇒ 260 : x = 13 : 10
∴ 32 : 112 : : x : 217
⇒
13 × x = 260 × 10 ⇒ x =
260 × 10
= 200
13
Hence, breadth = 200 metre.
27. The ratio between weights of copper and zinc in an alloy is 9 : 7. If the weight of zinc
in the alloy is 9.8 kg. find :
(i) the weight of copper in the alloy.
(iii) the ratio of weight of copper to that of alloy.
Ans. Weight of Zinc = 9.8 kg Copper : Zinc = 9 : 7
(i) Let the weight of copper be x, then x : 9.8 = 9 : 7
⇒
7 × x = 9 × 9.8 = x =
9 × 9.8 88.2
=
= 12.6
7
7
Hence, weight of copper = 12.6 kg.
12.6 126 9
=
=
= 9 : 16
22.4 224 16
27. Find the fourth term of the proportion whose first, second and third terms are 18,
27 and 32 respectively.
Ans. Let the fourth term be x. ∴ 18 : 27 : : 32 : x ⇒ 18 × x = 27 × 32
(ii) Weight of copper : weight of alloy ⇒ 12.6 : 22.4 =
⇒ x=
27 × 32
= 3 × 16 = 48
18
Thus fourth term = 48
29. The ratio of the length to the width of a school ground is 2. Find its length, if width
is 40 metres.
Ans. Let the length = x m, width = 40 m; The ratio of length to width = x : 40
As per condition 5 : 2 = x : 40 ⇒ 2 × x = 20 × 5 x =
∴
40 × 5
= 20 × 5 = 100 m
2
Length of school ground = 100 m.
Class - VI Mathematics
7
Question Bank
30. On a map, a 4 cm tall building is drawn and in front of it a 1.5 cm tall tree is drawn.
If the actual height of the building is 12 metre, find the actual height of the tree.
Ans. Let the actual height of tree be x; In map height of building = 4 cm. Height of tree
= 1.5 cm. Actual height of building = 12 m.
4 : 15 = 12 : x ⇒ 4 × x = 1.5 × 12 ⇒ x =
1.5 ×12
= 4.50 metre.
4
∴ Actual height of tree is 4.5 metre.
Class - VI Mathematics
8
Question Bank