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The Quest for 700: Weekly GMAT Challenge (Answer)

Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:

The two expressions in question, (1 + 2x)5 and (1 + 3x)4, could in theory be expanded, but you’ll wind up doing a lot of algebra, only to find an equation involving x5, x4, x3, x2, 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 35 =?= 44.
Compute the two sides: 35 = 243, while 44 = 162 = 256. So the right side is bigger.
Now, we need another benchmark. Try x = 2. The “equation” becomes 55 =?= 74.
Compute or estimate the two sides. 54 = 252 = 625, and multiplying in another 5 gives you something larger than 3,000 (3,125, to be precise). Meanwhile, 74 = 492 < 502 = 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 45 =?= (11/2)4
45 = 210 = 1,024 (it’s good to know your powers of 2 this high)
For the other side, first compute the denominator: 24 = 16. Now the numerator: 114 = 121×11×11. 121×11 = 1,331 (this is quick to do longhand)
1,331×11 = 14,641 (also quick to do longhand)

So (11/2)4 = 14,641/16 < 1,000. The right side is bigger.

Let’s recap:
(1 + 2x)5 = (1 + 3x)4 at what value of x?
(1 + 2x)5 < (1 + 3x)4 when x = 1
(1 + 2x)5 > (1 + 3x)4 when x = 1.5
(1 + 2x)5 > (1 + 3x)4 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 (x2, x3, etc.), and this simplifies the algebraic expansion:
For tiny positive x,
(1 + 2x)5 = 1 + 5(2x) + higher powers of x ? 1 + 10x
(1 + 3x)4 = 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:
(1 + 2x)5 = 25 = 32
(1 + 3x)4 = (5/2)4 = 2.54 = 6.252 > 36 (= 62) > 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.




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