Question

What must be added to the numbers 6,10,14 and 22 so that they are in proportion?

Answer 1

Hi ,

Let the number ' x ' should be added

to the given numbers so that they

are in proportional.

Therefore ,

( 6 + x ) , ( 10 + x ) , ( 14 + x ) , ( 22 + x )

are in proportion .

Product of extrems = product of
means

( 6 + x ) ( 22 + x ) = ( 10 + x ) ( 14 + x )

132 + 28x + x² = 140 + 24x + x²

28x - 24x = 140 - 132

4x = 8

x = 8/2

x = 2

Therefore ,

Required number = x = 2

I hope this helps you.

:)

Answer 2

Let the number to be added be x.

6 + x,10 + x, 14 + x, 22 + x are in proportion.

(6 + x) : (10 + x) :: (14 + x) : (22 + x)

$\frac{6 + x}{10 + x} = \frac{14 + x}{22 + x}$

On cross-multiplication, we get

(6 + x)(22 + x) = (10 + x)(14 + x)

132 + 6x + 22x + x^2 = 140 + 10x + 14x + x^2

132 + 28x + x^2 = 140 + 24x + x^2

28x - 24x = 140 - 132

4x = 8

x = 8/4

x = 2.

Therefore the number to be added so that they are in proportion = 2.


Verification:

(6  + x):(10 + x) :: (14 + x) : (22 + x)

= 8 : 12 :: 16 : 24

$\frac{8}{12} = \frac{16}{24}$

$\frac{2}{3} = \frac{2}{3}$


Hope this helps!