Question : Find the three consecutive numbers such that twice the first, three times the second and four times the third together make 191.
Option 1: 19, 20 and 21
Option 2: 21, 22 and 23
Option 3: 20, 21 and 22
Option 4: 21, 23 and 24
Correct Answer: 20, 21 and 22
Solution :
Let the three consecutive numbers be $x, x+1$ and $x+2$.
$⇒ 2x + 3(x+1) + 4(x+2) = 191$
$⇒ 2x + 3x + 3 + 4x + 8 = 191$
$⇒ 9x + 11 = 191$
$⇒ 9x = 180$
$⇒ x = 20$
Hence, the three consecutive numbers are 20, 21 and 22.
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