The sum of four numbers is 64. If 3 is added, subtracted, multiplied and divided to the first, second, third and fourth number respectively, then all the results are equal. What is the difference between the largest and the smallest of the original numbers?

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Solution

The correct option is C
32

Let the four numbers be A, B, C and D. Then, A+3 = B–3 = 3C = $\frac{D}{3}$ = x

Then, A = x-3, B = x+3, C = $\frac{x}{3}$ and D = 3x.

A + B + C + D = 64

$\Rightarrow (x - 3) + (x + 3) + \frac{x}{3} + 3 x = 64$

$\Rightarrow 5 x + \frac{x}{3} = 64$

$\Rightarrow x = 12$

Thus, the numbers are 9, 15, 4 and 36.

$∴$ Required difference = (36 - 4) = 32

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The sum of four numbers is 64. If 3 is added, subtracted, multiplied and divided to the first, second, third and fourth number respectively, then all the results are equal. What is the difference between the largest and the smallest of the original numbers?

The correct option is C 32 Let the four numbers be A, B, C and D. Then, A+3 = B–3 = 3C =D3 = x Then, A = x-3, B = x+3, C = x3 and D = 3x. A + B + C + D = 64 ⇒(x−3)+(x+3)+x3+3x=64 ⇒5x+x3=64 ⇒x=12 Thus, the numbers are 9, 15, 4 and 36. ∴ Required difference = (36 - 4) = 32