Question

If two numbers are in the ratio 5: 6 and 8 is subtracted from each of the numbers they become in the ratio of 4: 5 than the
numbers are

Answer 1

Let the two numbers be 5x and 6x

When 8 is subtracted from 5x and 6x the ratio become 4x and 5x


A/q,

5x - 8 + 6x - 8 = 4x + 5x

11x - 16 = 9x

11x - 9x = 16

2x = 16

x = $\dfrac{16}{2}$

x = 8


1st number => 5x => 5 × 8 => 40

2nd number => 6x => 6 × 8 => 48



$\therefore{}$ $\text{The numbers are 40 and 48 respectively}$

Answer 2

$<b>Heya mate. (^_-). Solution below.$
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$<u>Your Question -</u>$

If two numbers are in the ratio 5:6 and 8 is subtracted from each of the numbers, they become in the ratio of 4:5. What are the numbers?

$<u>Solution -</u>$

Let the first number be '5x'
Then, second number = '6x'

Now, when 8 is subtracted from each, the numbers become

5x - 8 : 6x - 8

But, according to question, the numbers will become 4 : 5

Then,

5x - 8 : 6x - 8 = 4 : 5

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

Cross multiplying, we get

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

=> 24x - 32 = 25x -40

=> -32 + 40 = 25x - 24x

=> 8 = x

=> x = 8.

$<marquee>Hence, the first number is 5x = 5 × 8 = 40</marquee>$

$<marquee>And, second number is 6x = 6 × 8 = 48</marquee>$
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Thank you ;-)