Question

Find the sum of three consecutive largest positive integers such that the sum of one-third of first, one - fourth of second and one -fifth of third is atmost 20.

• A. 75
• B. 80
• C. 65
• D. 90

Answer 1

The correct option is A

75

Let the smallest positive integer = x

Then, second smallest integer = x +1

and the greatest integer = x +2

According to the condition,

$\Rightarrow \frac{x}{3}+\frac{x+1}{4}+\frac{x+2}{5}\le 20$

$\Rightarrow \frac{20x+15x+15+12x+24}{60}\le 20$

$\Rightarrow 47x+39\le 1200$

$\Rightarrow 47x\le 1161$

$\Rightarrow x\le \frac{1161}{47}$

$\Rightarrow x\le 24\frac{33}{47}$

Therefore, x = 24, x +1 = 25 and x+ 2 = 26

So, sum of the numbers = 24 + 25 + 26 = 75