Manhattan GMAT Challenge Problem of the Week – 23 April 2012
Here is a new Challenge Problem! If you want to win prizes, try entering our Challenge Problem Showdown. The more people that enter our challenge, the better the prizes!
Question
The positive value of x that satisfies the equation
=
is between
A. 0 and 0.5
B. 0.5 and 1
C. 1 and 1.5
D. 1.5 and 2
E. 2 and 2.5
Answer
The two expressions in question,
and
, could in theory be expanded, but you’ll wind up doing a lot of algebra, only to find an equation involving
,
,
,
, and x. Solving for the positive value of x that makes the equation true is nearly impossible.
So how on Earth can you answer the question? Notice that the question does not require you to find a precise value of x; you just need a range. So plug in good benchmarks and track the value of each side of the equation.
Start with x = 1. The “equation” becomes
=?=
.
Compute the two sides:
= 243, while
=
= 256. So the right side is bigger.
Now, we need another benchmark. Try x = 2. The “equation” becomes
=?=
.
Compute or estimate the two sides.
=
= 625, and multiplying in another 5 gives you something larger than 3,000 (3,125, to be precise). Meanwhile, 74 =
<
= 2,500, so the left side is now bigger. Somewhere between x = 1 and x = 2, then, the equation must be true.
The only benchmark left to try is 1.5, or 3/2. Plugging in, you get
=?= 
=
= 1,024 (it’s good to know your powers of 2 this high)
For the other side, first compute the denominator:
= 16. Now the numerator:
= 121×11×11. 121×11 = 1,331 (this is quick to do longhand)
1,331×11 = 14,641 (also quick to do longhand)
So
= 14,641/16 < 1,000. The right side is bigger.
Let’s recap:
=
at what value of x?
<
when x = 1
>
when x = 1.5
>
when x = 2
So the value at which the two sides are equal must be between 1 and 1.5.
The correct answer is C.
Extra points:
As x grows past 2, the larger power (5) on the left takes over, so you can see that the left side will always be bigger.
What about when x is between 0 and 1? Well, first notice that at x = 0, the two sides are again equal. If x is a tiny positive number (say, 0.001 or something), then you can ignore higher powers of x (
,
, etc.), and this simplifies the algebraic expansion:
For tiny positive x,
= 1 + 5(2x) + higher powers of x ˜ 1 + 10x
= 1 + 4(3x) + higher powers of x ˜ 1 + 12x
So right away, the right side is bigger than the left side. You can also check x = ½:
=
= 32
=
=
=
> 36 (=
) > 32
Again, the right side is bigger than the left side. For every x between 0 and 1, in fact, the right side is larger than the left side. This fact isn’t particularly easy to prove, but you don’t need to do so.
Special Announcement: If you want to win prizes for answering our Challenge Problems, try entering our Challenge Problem Showdown. Each week, we draw a winner from all the correct answers. The winner receives a number of our our Strategy Guides. The more people enter, the better the prize. Provided the winner gives consent, we will post his or her name on our Facebook page.


4 comments
Ravisankar Vemuri on April 26th, 2012 at 3:09 am
Its a unique quant method which takes advantage of the answer options. Thanks for the great content
showbiz on April 26th, 2012 at 8:15 am
I have a quicker way of doing it.
Remove the 1 from both brackets as it is a very small number in the general scheme of things.
2x^5 = 3x^4
Do basic exponential arithmentic to get
x = 3/2
that number is between 1 and 1.5.
Biingo!
Kalpana Sharma on August 23rd, 2012 at 2:28 pm
Great and easy solution by showbiz. Actually ignoring the 1 is a trick that is used in many places. Thanks for reminding to use it.
gaurav on April 29th, 2012 at 9:43 pm
I agree with showbiz. I opted for same approach. With RHS, LHS manipulations, I got the value of x. Can anyone confirm whether it is fine to proceed in such a manner as described by us.