If 2/x^3

Question

If  \frac{2}{x^3}< 0.02  what is the least positive integer value of x?

A. No such least value exists.
B. 101
C. 100
D. 5
E. 4

EgmatQuantExpert · Answer

Solution

Given:  
 •  \frac{2}{x^3} < 0.02

To find:  
 • The least integer value of x

Approach and Working:  
We can rewrite the given expression in the following manner:
 •  \frac{2}{x^3} < 0.02 
Or,  \frac{2}{x^3} < \frac{2}{100}

If this is true, then  x^3  must be greater than 100
 • If x = 4, then  4^3  = 64 < 100
• If x = 5, then  5^3  = 125 > 100

Therefore, the least integer value of x must be 5

Hence, the correct answer is option D.

Answer: D

KSBGC · Answer

Note: we are asked to determine the least value of x.

what if  x^3  = 100

\frac{2}{100} = 0.02

then we can take care x = 5

x^3  = 125

it is clear that

\frac{2}{125}  < 0.02..........as we know \frac{2}{100}= 0.02

Thus the best answer is D.

Qoofi · Answer

By substitution,

2/64 = 1/32 > 0.02 - Incorrect
2/125 < 0.02 - Correct.

JeffTargetTestPrep · Answer

We can test the values given in the answer choices, starting with 4 (choice E):

2/4^3 < 0.02 ?

2 < 0.02 x 4^3 ?

2 < 0.02 x 64 ?

2 < 1.28 ? → No!

Now let’s test 5:

2/5^3 < 0.02 ?

2 < 0.02 x 5^3 ?

2 < 0.02 x 125 ?

2 < 2.5 ? → Yes!

Answer: D

kpxu08 · Answer

What if x is negative?

Insiya · Answer

What if we take the value of X to be negative ?
Can anyone please clear my doubt? 
Thanks!

Sent from my vivo 1611 using  GMAT Club Forum mobile app

Bunuel · Answer

You are right. The word "positive" was missing in the stem.

GMATGuruNY · Answer

Since x must be a positive integer, both sides of the inequality can be multiplied by  x^3

\frac{2}{x^3} < 0.02

2 < 0.02x^3

200 < 2x^3

100 < x^3

x^3 > 100

In the resulting inequality, the least positive integer value for x is 5.

D

chetan2u · Answer

\frac{2}{x^3}< 0.02...............2<x^3*0.02......100<x^3 
Now x is integer, 4^3=64 and 5^3=125
So minimum is 5

D

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If 2/x^3 < 0.02 what is the least positive integer value of x?

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