Question

Three times of a number is 122 less than the five times of another number. If the sum of both the numbers is 74, find the two numbers.

Answer 1

Let the required numbers satisfying the given equation are a and b.

As given in the question, three times of a number is 122 less-than five times the another number.


Now, according to the question :

= > three times of a = ( 5 times of b ) - 122

= > 3 x a = ( 5 x b ) - 122

$= > 3a = 5b - 122 \: \: \: \: \: \: \textit{...(i)}$


It is also given that the sum of both the numbers is 74.
So,
Sum of a and b = 74

= > a + b = 74

Now, multiply by 3 on both sides,
= > 3a + 3b = 3 x 74

= > 3a + 3b = 222


Then, substituting the value of 3x from $\textit{(I)}$

= > 5b - 122 + 3b = 222

= > 8b = 222 + 122

= > 8b = 344

= > b = 344 / 8

= > b = 43


Then, substituting the value of b in $\textit{(I)}$

= > 3a = 5b - 122

= > 3a = 5( 43 ) - 122

= > 3a = 215 - 122

= > a = 93 / 3

= > a = 31



Therefore the numbers required here are 43 and 31.