Question

Three consecutive integers are such that when they are taken in increasing order and

multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.

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Answer 1

Let the integers be x, x+1 , x+2.

Now, we multiply them by 2,3,4 as we are told in the question to get 74 as total.

Therefore according to question

2 (x) + 3 (x+1) + 4 ( x+2) = 74

2x + 3x +3 +4x +8 = 74

9x + 11 = 74

9x = 63

x = 63÷9 = 7

1st integer = x = 7

2nd integer = x + 1 = 7+1 = 8

3rd integer = x + 2 = 7 + 2 = 9

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Answer 2

Answer:

$\huge \mathbb \color{blue}{ANSWER \implies}$

Let the three consecutive integers be x, x + 1 and x + 2.

As per the condition, we have

2x + 3(x + 1) + 4(x + 2) = 74

⇒ 2x + 3x + 3 + 4x + 8 = 74

⇒ 9x + 11 = 74

⇒ 9x = 74 – 11 (transposing 11 to RHS)

⇒ 9x = 63

⇒ x = 63 ÷ 9

⇒ x = 7 (transposing 7 to RHS)

Thus, the required numbers are 7, 7 + 1 = 8 and 7 + 2 = 9, i.e., 7, 8 and 9.

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