Question

two numbers differ by 40 when each number is increased by 8, the bigger number becomes thrice the smaller number. If one number is x. Find the other number.

Answer 1

Hi ,

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

According to the problem given ,

x + 8 = 3 ( 48 - x )

x + 8 = 144 - 3x

x + 3x = 144 - 8

4x = 136

x = 136/4

x = 34

Therefore ,

The bigger number = x = 34

Smaller number = 40 - x

= 40 - 34

= 6

I hope this helps you.

:)

Answer 2

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• Let the bigger number = x
• smaller number = 40-x

Given:

Each number is increased by 8

New number =

= x+8

= 40-x +8

=48-x

ATQ,

$: \implies x + 8 = 3(48 - x) \\ : \implies x + 8 = 144 - 3x \\ : \implies \: x + 3x = 144 - 8 \\ : \implies \: 4x = 136 \\ : \implies x = \frac{ \cancel{136} {} \: ^{34} }{ \cancel4} \\ : \implies \: \blue{\underline{\boxed{x = 34}}}$

Therefore,

• Bigger number = 34
• Smaller number =

$40- x \\ 40 - 34 \\ : \implies 6$

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