9th mathematics exercise question with answer

# Ratio and Proportion – Full Chapter – Class IX

After studying the chapter Ratio and Proportion and their general form; to understand and differentiate between different types of proportion; to acquire skills of writing proportion; to solve problems on time and work involving proportions; to apply proportion in day to day life situations.

## 2.4.1 Introduction to Ratio and Proportion

In a ratio a : b, the first term a is called the antecedent and the second term b is called the consequent. Ratio is an abstract quantity and has no unit. Ratio tells how many times the first term is there in the second term. Example 1: In the adjacent figure, find the ratio of the shortest side of the triangle to the longest side.

Solution:

We see that the shortest side is of length 5 cm and the longest side is of length 13 cm. Hence the ratio is 5:13

Example : Suppose the ratio of boys to girls in a school of 720 students is 7 : 5. How many more girls should be admitted to make the ratio 1 : 1?

Solution:

For every 7 boys there are 5 girls. Thus out of 12 students, 7 are boys and 5 are girls. Hence the number of boys is 7/12 x 720 = 420.

The number of girls is 720 – 420 = 300. Now we want the ratio of boys to girls to be 1:1. This means the number of boys and girls must be same. Since the deficiency of  girls is 420 – 300 = 120, the school must admit 120 girls to make the ratio 1:1.

Example 5: Consider the ratio 12:5 If this ratio has to be reduced by 20% which common number should be added to both the numerator and denominator?

Solution:

Consider 12/5. This has to be reduced by 20%. This means we have to consider 80% of this number. Thus, we must get,

12/5 x 80/100 = 48/25

We have to find a such that,

12+a/5+a = 48/25

Cross multiplying

25(12 + a) = 48(5 + a)

48a – 25a = (25 x 12) – (48 x 5) =

23a = 60

a = 60/23

If we add 60/23 to both terms of 12 : 5 we get a ratio which is 20% less than the original ratio.

### Ratio and Proportion – Exercise 2.4.1

1.Write each of these ratios in the simplest form.

(i) 2:6

(ii)24:4

(iii) 14:21

(iv) 20: 100

(v) 18:24

(vi) 22:77

Solution:

(i) 2:6 = 1:3 (dividing both by 2)

(ii) 24:4 = 6:1 (dividing both by 2)

(iii) 14:21 = 2:3 (dividing both by 7)

(iv) 20:100 = 1:5 (dividing both by 2)

(v) 18:24 = 3:4 (dividing both by 6)

(vi) 22:77 = 2:7 (dividing both by 11)

2. A shop-keeper mixes 600 ml of orange juice with 900 ml of apple juice to make a fruit drink. Write the ratio of orange juice to apple juice in the fruit drink in its simplest form.

Solution:

Ratio of volumes of

Orange juice and apple juice O:A

= 600:900

= 6:9

= 2:3

3. a builder mixes 10 shovels of cement with 25 shovels of sand. Write the ratio of cement to sand.

Solution:

Ratio of cement to sand = 10 shovels :25 shovels

4.In a school there are 850 pupils and 40 teachers. Write the ratio of teachers to pupils.

Solution:

Number of teachers : Number of pupils

= 40 : 850 = 4:85

5. On a map, a distance of 5cm represent an actual distance of 15km. Write the ratio of the scale of the map.

Solution:

Let x be the number to be added them

(49 + x) = (68 + x) = 3:4

4(49+X) = (68 + X)3

196+4X = 204 + 3X

4X – 3X = 204 – 196

x = 8

## 2.4.2 Proportion – Ratio and Proportion

### Ratio and Proportion – Exercise 2.4.2

1. In the adjacent figure, two triangles are similar find the length of the missing side Solution:

Let the triangles be ABC and PQR

BC/QR = AC/PR

5/X  = 13/39

13X = 5 X 39

X = 5X39/13  = 5 X3 = 15

1. What number is to 12 is 5 is 30?

Solution

Let x be the number

x:12 :: 5 : 30

30x = 12×5

x = 12×5/30   = 2

1. Solve the following properties:

(i). x : 5 = 3 : 6

(ii) 4 : y = 16 : 20

(iii) 2 : 3 = y : 9

(iv) 13 : 2 = 6.5 : x

(v) 2 : π = x : 22/7

Solution:

(i). x : 5 = 3 : 6

6x = 5 x 3

6x = 15

x = 15/6

(ii) 4 : y = 16 : 20

4×20 = 16y

y = 4×20/16

y = 5

(iii) 2 : 3 = y : 9

2×9 = 3y

y = 2×9/3 = 2×3 = 6

(iv) 13 : 2 = 6.5 : x

13x = 2 x 6.5

13x = 13

x = 13/13 = 1

(v) 2 : π = x : 22/7

2x22/7 = πx

x = (2x22/7) /π =(2x22/7) /(22/7)

x = 2

1. find the mean proportion to :

(i) 8, 16

(ii) 0.3, 2.7

(ii)162/3 , 6

(iv)  1.25, 0.45

Solution:

(i) 8, 16

Let x be the mean proportion to 8 and 16

Then 8/x = x/16

x2 = 8 x 16 = 128

x = √128 = √(64×2) = 8√2

(ii) 0.3, 2.7

Let x be the mean proportion to 0.3 and 2.7

Then 0.3/x = x/2.7

x2 = 0.3 x 2.7 = 0.81

x = √(0.81) = 0.9

(ii)162/3 , 6

Let x be the mean proportion to 162/3  and 6

Then (162/3) /x = x/6

x2 = 162/3  x 6 = 50/3 x 6 = 100

x = √100 = 10

(iv)  1.25, 0.45

Let x be the mean proportion to 1.25 and 0.45

Then 1.25/x = x/0.45

x2 = 1.25 x 0.45

x = √(1.25 x 0.45) = √(1.25 x 0.45)x√(100×100)/ √(100×100)

= √(125×45)/√(100×100) = √(25x5x5x9)/√(100×100)  = 5x5x3/10×10  = 3/4

1. Find the fourth proportion for the following:

(i) 2.8,  14, 3.5

(ii) 31/3, 12/3, 21/2

(iii)15/7, 23/4, 33/5

Solution:

(i) 2.8,  14, 3.5

Let x be the  fourth proportion

Then, 2.8 : 14  :: 3.5  : x

2.8x = 14×3.5

x  = 14×3.5/2.8 = 17.5

(ii) 31/3, 12/3, 21/2

Let x be the  fourth proportion

Then, 31/3: 12/3  :: 21/2: x

10/3 :5/3 : : 5/2 : x

10/3x = 5/3 x 5/2

10/3x = 25/6

x  = 25/6 x 3/10 = 75/60  = 15/12 = 5/4

(iii)15/7, 23/14, 33/5

Let x be the  fourth proportion

Then, 15/7: 23/14:: 33/5: x

12/7 :31/14 : : 18/5 : x

12/7 x = 31/14 x 18/5

12/7 x = 31×18/14×5

x  = 31×18/14×5 x 7/12 = 31×3/5×4 = 93/20 = 413/20

1. Find the third proportion to:

(i) 12, 16

(ii) 4.5, 6

(iii) 51/2 , 161/2

Solution:

(i) 12, 16

Let x be the third proportion

Then 16:12 :: x : 16

12x = 16 x16

x = 16×16/12 = 64/3  = 211/3

(ii) 4.5, 6

Let x be the third proportion

Then 6:4.5 :: x : 6

4.5x = 6 x6

x = 6×6/4.5 = 36/4.5  = 360/45  = 8

(iii) 51/2 , 161/2

Let x be the third proportion

Then 161/2: 51/2:: x : 161/2

11/2 x = 33/2 x 33/2

x = 33/2 x 33/2 x 2/11= 33×3/2  = 99/2  = 491/2

1. In a map 1/4 cm represents 25km, if two cities are 21/2c apart on the map, what is the actual distance between them?

Solution:

Let 21/2 cm represemts x km

1/4cm: 25km :: 21/2cm : x km

1/4 x x = 25 x 21/2

x/4 = 25 x5 /2

x = 25×5/2 x 4 = 25x5x2 = 250km

1. Suppose 30 out of 500 components for a computer were found defective. At this rate how many defective components would he found in 1600 components?

Solution:

Number of defective components in 500 components = 30

Let x be the number of defective components in 1600 components

then 30:500 :: x :1600

30×1600 = 500x

x = 30×1600/500 =96

## 2.4.3 Time and Work – Ratio and Proportion

### Ratio and Proportion – Exercise 2.4.3

1. Suppose A and B together can do a job in 12 days, while B alone can finish a job in 24 days. In how many days can A alone finish the work?

Solution:

Number of days in which A and B together can finish the work = 12 days

Number of days in which B alone can finish the work = 30

1/T = 1/m + 1/n

1/12 = 1/m + 1/30

1/m = 1/301/12 = 5-2/60 = 3/60 = 1/20

A can finish the work in 20 days.

Suppose A is twice as good a workman as B and together they can finish a job in 24 days. How many days A alone takes to finish the job?

Solution:

A is twice as good a workman as B

i.e if B can finish a work in t days A can finish it in 1/2 days

1/T = 1/m + 1/n

1/24 = 1/t/2 + 1/t = 2/t + 1/t = 3/t

1/24 = 3/t

t = 24 x 3 = 72

i.e, B takes 72days to finish the job

A takes 72/2 = 36 days to finish it

1. Suppose B is 60% more efficient them A. if A can finish a job in 15 days how many days B needs to finish the same job?

Solution:

A can finish a work in 15 days.

Work done A in 1 day = 1/15

B is 60% more efficient

Work done by B in 1 day

1/15 + 1/15 x 60/100

= 1/15 (1 + 60/100)

= 1/15 ( 8/5)

= 8/75

Number of days in which B alone can finish the work = 1/(8/75)  = 75/8 = 93/8 days

1. Suppose A can do a piece of work in 14 days while B can do it in 21 days. They begin together and worked at it for 6 days. Then A fell ill B had to complete the work alone. In how many days was the work completed?

Solution:

M = 14 days

N = 21 days

Part of work done in 6 days

= (1/14 + 1/21)6

= 6(3+2/42) = 5×6/42 = 5/7

Remaining part of the work = 1-5/7 = 2/7

Days taken by B to finish

2/7 part of the work = (2/7)/(1/21) = 2/7 x 21/7 = 6 days

Total number of days in which the work is completed = 6+6 = 12 days

1. Suppose A takes twice as much time as B and thrice as much time as C to complete a work. If all of them work together they can finish the work in 2 days. How much time B and C working together will take to finish it?

Solution:

If A alone takes to t1 days to do the work , B finishes it in t1/2  and C is t1/3 days

1/T = 1/t1 +1/t2 + 1/t3

= 1/t1 +1/(t1/2) + 1/(t1/3)

= 1/t1 +2/t1 + 3/t1

= 6/t1

1/T = 1/2

1/2 = 6/T1

i.e. t1 = 12 dyas

B takes 12/2 = 6days

C takes 12/3 = 4 days

Part of work done by B

In one day = 1/6

Part of work done by C in one day = 1/4

If B and C together takes t days to finish the work 1/T = 1/6 + 1/4 = 2+3/12 = 5/12

T = 12/5 = 2.4 days