One worker can make x^2x widgets in 6 hours, and a second worker can

Question

One worker can make x^2−x widgets in 6 hours, and a second worker can make 2x widgets in 3 hours. If the first worker works faster than the second worker, what must be true of x?

(A) x > 2
(B) x < 3
(C) x > 3
(D) x < 5
(E) x > 5

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Official Answer

E

aeforx · Answer

(x^2-x)/6 > 2x/3
x^2-x > 4x
x^2-5x > 0
x(x-5) > 0
x > 5 or x < 0
cuz x>=0 so x > 5

paragw · Answer

First worker's rate: (x^2 − x)/6 widgets/hour
Second worker's rate: 2x/3 widgets/hour
Given first worker is faster:
(x^2 − x)/6 > 2x/3
x^2 − x > 4x
x^2 − 5x > 0
x < 0 or x > 5
Since the number of widgets must be positive,
x > 1.
Hence, x > 5

toltango · Answer

bruh i thought it was x^(2-x). it is (x^2) - x.

Bunuel · Answer

If it were x^(2 - x) it would have been written that way. Without the brackets x^2 - x can only mean  x^2 - x

onlyPlanA · Answer

In 6 hours, first worker can make =  x^2 - x  widgets
So, in 1 hour, first worker can make =  \frac{(x^2 - x)}{6} widgets = Rate of first worker

In 3 hours, second worker can make =  2x  widgets
So, in 1 hour, second worker can make =  \frac{2x}{3} widgets = Rate of second worker

Given that, first worker works faster than second, so the rate of first worker will be greater than second,
Hence,  \frac{(x^2 - x)}{6}  >  \frac{2x}{3}

x^2 - x > 4x 
 x^2 - 5x > 0 
 x(x - 5) > 0 
 x > 5  or  x < 0

Answer is E.

One worker can make x^2x widgets in 6 hours, and a second worker can

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GMAT Problem Solving (PS) Questions

One worker can make x^2x widgets in 6 hours, and a second worker can

Prep Toolkit

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