Question

two numbers are in the ratio of 5:6. if 8 is subtracted from each of the numbers, they become in the ratio of 4:5. find the numbers. somebody please solve it.

Answer 1

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★Answer:

Let x and y two numbers such that:

x/y =5/6

=6x-5y=0....(1)

IF 8 IS SUBTRACTED FROM BOTH THE NUMBERS , THE RATIO IS 4:5

x-8/y-8=4/5

=5(x-8)=4(y-8)
=5x-40=4y-32

=5x-4y=8.......(2)

MULTIPLYING EQUATION (1) by 4 and equation (2) by 5, we have ,

24x - 20y =0.....(3)

25x - 20y =40....(4)

SUBTRACT EQUATION (3rd) FROM (4th),
WE HAVE ,
X =40

SUBSTITUTING THE VALUE OF X= 40 ,IN EQUATION (1) we Have,

6×40 -5y =0
240=5y

Y =48

★THUS, THE NUBERS ARE 40 and 48.....

Hope' This Helps....
:)

Answer 2

First Method :

Let , x is included in the ratio.

So, numbers = 5x and 6x

According to question,

=> ( 5x - 8 ) / ( 6x - 8 ) = 4 / 5

=> 5( 5x - 8 ) = 4( 6x - 8 )

=> 25x - 40 = 24x - 32

=> 25x - 24x = 40 - 32

•°• x = 8

Now,

Numbers = 5x = 5 × 8 = 40

= 6x = 6 × 8 = 48

Verification :

R.H.S = ( 4/5 )

L.H.S = ( 40 - 8 ) / ( 48 - 8 )

= ( 32 / 40 )

= 4/5

= R.HS

Verified !!

Second Method :

Let the numbers are x and y.

Given,

=> ( x/y ) = ( 5/6 )

=> 6x = 5y

•°• 6x - 5y = 0 -------- ( 1 )

According to question,

=> ( x - 8 ) / ( y - 8 ) = ( 4/5 )

=> 5( x - 8 ) = 4( y - 8 )

=> 5x - 40 = 4y - 32

=> 5x - 4y = 40 - 32

•°• 5x - 4y = 8 -------- ( 2 )

Multiplying by 4 in ( 1 ) and by 5 in ( 2 ),

=> 4( 6x - 5y ) = 0 × 4

=> 24x - 20y = 0 ------ ( 3 )

And,

=> 5( 5x - 4y ) = 8 × 5

=> 25x - 20y = 40 --------- ( 4 )

Subtracting ( 3 ) from ( 4 ),

=> 25x - 20y - ( 24x - 20y ) = 40 - 0

=> 25x - 20y - 24x + 20y = 40

=> x = 40

Substitute the value of x in ( 1 ),

=> 6x - 5y = 0

=> ( 6 × 40 ) - 5y = 0

=> 240 - 5y = 0

=> 240 = 5y

=> ( 240 ÷ 5 ) = y

=> 48 = y

•°• y = 48

Hence, numbers are 40 and 48.

Verification :

R.H.S = 4/5

L.H.S = ( 40 - 8 ) / ( 48 - 8 )

= 32 / 40

= 4/5

= R.H.S

Verified !!

Hope it helps !!