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Isnt this a risky question

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by [email protected] » Mon Oct 28, 2013 9:48 pm
The positive value of x that satisfies the equation (1 + 2x)^5 = (1 + 3x)^4 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

We can assume a lot of values no?


Ans C
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Source: — Problem Solving |

by Uva@90 » Mon Oct 28, 2013 11:22 pm
[email protected] wrote:The positive value of x that satisfies the equation (1 + 2x)^5 = (1 + 3x)^4 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

We can assume a lot of values no?


Ans C
Shibriz,
You can do by Trail and error method.

Start with X=1, for easy calculation
(1 + 2x)^5 = 243
(1 + 3x)^4 = 256

You can see there is slight difference between the two values.
So X should e slightly greater than 1

Hence you will go with C

Regards,
Uva.
Known is a drop Unknown is an Ocean
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by [email protected] » Tue Oct 29, 2013 11:58 am
Hi shibsriz,

Uva@90 has offered a nice approach to this question, and I'm going to add a few details.

First, the wording and set-up for this question imply that there's just 1 answer, not several, so you shouldn't feel too overwhelmed.

Second, the answers are numbers, so we should TEST THE ANSWERS. Since the "math" in this question could get ugly, I'm going to stick to integers.

X = 1 is a great place to start. Plug that value into both calculations and you'll end up with...

3^5 and 4^4
243 and 256

So the numbers are real close; the answer has to be close to 1. The question is do we pick B or C? Notice that the SECOND value is BIGGER.

For the next test, you can decide between a few options (.5, 1.5 or 2 if you prefer using an integer).

X = 2 gives us...

5^5 and 7^4
3025 and (49)(49) = approx. (50)(50) = 2500

Notice now that the FIRST value is BIGGER.

This tells me that increasing the value of X makes the first part of the equation "grow" faster than the second part, so I would have to choose an answer that was a little bigger than 1

Final Answer: C

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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by Matt@VeritasPrep » Wed Oct 30, 2013 10:45 pm
shibsriz wrote:We can assume a lot of values no?
By the Fundamental Theorem of Algebra (which you aren't expected to know, mercifully!) any polynomial has as many roots as its degree - so since this equation is of degree 5 (its highest term is x�), it has five solutions. Not all of these are real solutions though: this one, for instance, has 0, a positive solution (~1), a negative solution (~-.5), and a couple of complex solutions that are out of scope here.
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