Question

Two numbers are in the ratio of 3:4 if 10 be subtracted from each of them the remainder are in the ratio of 1:3 find the numbers

Answer 1

Let the two numbers be x and y
so
x/y=3/4
x=3y/4
Now according to second condition
(x-10)/(y-10)=1/3
3x-30=y-10
3(3y/4)-30=y-10
9y/4 - 30=y-10
(9y-120)/4=y-10
9y-120=4y-40
5y=80
y=16
x=3(16)/4
x=12
The two numbers are 12 and 16
Hope it helps
:)

Answer 2

Let the first number be 3x.
Then, second number = 4x

Now, subtracting 10 from both sides we get,
First number = 3x - 10
Second number = 4x - 10

Now, according to the question,

$\frac{3x - 10}{4 x - 10} = \frac{1}{3}$

Cross multiplying, we get,

$3(3x - 10) = 1(4x - 10)$

$9x - 30 = 4x - 10$

$= > 9 x - 4x = - 10 + 30$

$= > 5x = 20$

$= > x = \frac{20}{5}$

$= > x = 4$

Hence,
First number = 3x = 3 × 4 = 12
Second number = 4x = 4 × 4 = 16

That's your answer.

Hope it'll help.. :-)