Question

the sum of twice the 1st and thrice the 2nd is 92 and 4 times the 1st exceeds 7 times the 2nd by 2, then the numbers are (a) 25 and 20
(b) 25 and 14 (c) 14 and 22 (d) none of these ​

Answer 1

Answer:

option b is the correct answer

Step-by-step explanation:

Given :-

The sum of twice the first and thrice the second is 92, and four times the first exceeds seven times the second by 2.

To Find :-

The Numbers

Solution :-

Let the first number be x and the second number be y.

According to the Question

1st Equation

2x + 3y = 92 ……….(i)

2nd Equation

4x - 7y = 2 ………(ii)

On multiplying (i) by 7 and (ii) by 3,

14x + 21y = 644 ………..(iii)

12x - 21y = 6 ………..(iv)

On adding (iii) and (iv), we get

⇒ 26x = 650

⇒ x = 650/26

⇒ x = 25

Putting the x values Eq in (i)

⇒ 2x + 3y = 92

⇒ 2 × 25 + 3y = 92

⇒ 50 + 3y = 92

⇒ 3y = (92 – 50)

⇒ 3y = 42

⇒ y = 42/3

⇒ y = 14

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