Question

Two numbers differ by 40. When each number is increased by 8 then the bigger becomes thrice of less number. If one number is x then other other number is (40-x)? true or false explain

Answer 1

Solution : FALSE

REASON -

Given-

one number = x

another number = (40-x)

Let

(40-x) > x

Then

According to the question,

$40 - x + 8 = 3(x + 8) \\ \\ \implies 48 - x = 3x + 2 \\ \\ \implies - x - 3x = 24 - 48 \\ \\ \implies - 4x = - 24 \\ \\ \implies \: x = - 24 \times \frac{ - 1}{4} \\ \\ x = 6$

Hence

one number is 6

and other number is (40-6) = 36

Now,

difference between numbers = 34-6 = 28 ≠ 40

This is not satisfy to the condition given in the question.

Answer 2

$\huge\boxed{\underline{\mathcal{\red{A}\green{N}\pink{S}\orange{W}\blue{E}\pink{R}}}}$

$\huge\green{\underbrace{\overbrace{\ulcorner{\mid{\overline{\underline{(40-x)\:is\:true}}}}}}}$

$\huge{\fbox{\fbox{\orange{\mathfrak{Explanation\:Required}}}}}$

Let, bigger number = x

Smaller number = 40 - x

If each number is increased by 8

The new numbers are ,

Bigger number = x + 8

& Smaller number = 40 - x + 8 = 48 - x

Acc to problem,

${\bf{\green{x+8=3(48-x)}}}$

${\bf{\green{x+8=144-3x}}}$

${\bf{\green{4x=136}}}$

${\bf{\green{x=136/4}}}$

${\bf{\green{x=34}}}$

So,

${\bf{\green{The bigger number = x = 34}}}$

${\bf{\green{The bigger number = x = 34}}}$

${\bf{\green{Smaller number = 40 - x}}}$

${\bf{\green{=40-34=>6}}}$