Ratio and Proportions – Exercise 2.4.2 – Class IX

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

Hire and Purchase - Exercise 2.4.2 - Class IX

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 30out of 500 components for a compter were found defective. At this aret 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 defecrive components in 1600 components

then 30:500 :: x :1600

30×1600 = 500x

x = 30×1600/500 =96


 

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