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by infiniti007 » Sat Sep 12, 2015 3:46 pm
Is 4^(x+y) = 8^(10)?

1.) x - y = 9
2.) y/x = 1/4

I've seen the solution for the above problem, but I'm confused why I could not solve for x and y using each statement separately?

Once you get down to the point that you're solving:
Does x + y = 15?

At this point, why could I not use Statement 1 as follows:
x - y = 9
So, x = 9 - y

Therefore, 9 - y + y = 15
9 = 15 False.

Statement 2:
y = 1/4x
So, x = 4y

Therefore, 4y + y = 15
5y = 15
y = 3

Plug back into Statement 2:
x = 4(3) = 12

Does x + y = 15? 12 + 3 = 15 True

I feel like I'm missing something very basic here.
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Source: — Data Sufficiency |

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by Brent@GMATPrepNow » Sat Sep 12, 2015 3:52 pm
Does 4^(x+y) = 8^10?
1) x - y = 9
2) y/x = 1/4
This is a great candidate for REPHRASING the target question.
Aside: We have a free video with tips on rephrasing the target question: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100

Target question: Does 4^(x+y) = 8^10?
Take 4^(x+y) = 8^10 and rewrite each side with the same base of 2 to get:
(2^2)^(x+y) = (2^3)^10
Simplify to get: 2^(2x + 2y) = 2^30 [power of a power rule]
For this equation to hold true, it must be the case that 2x + 2y = 30
Divide both sides by 2 to get x + y = 15

We can now REPHRASE our target question...
REPHRASED target question: Does x + y = 15?

Statement 1: x - y = 9
There are several values of x and y that satisfy this condition. Here are two:
Case a: x = 12 and y = 3, in which case x + y = 15
Case b: x = 10 and y = 1, in which case x + y ≠ 15
Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: y/x = 1/4
Cross multiply to get: x = 4y
There are several values of x and y that satisfy this condition. Here are two:
Case a: x = 12 and y = 3, in which case x + y = 15
Case b: x = 8 and y = 2, in which case x + y ≠ 15
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that x - y = 9
Statement 2 tells us that x = 4y
We now have a system of TWO linear equations with TWO variables, so we COULD easily solve this system for x and y, which means we COULD determine whether x + y = 15.
Since we can answer the REPHRASED target question with certainty, the combined statements are SUFFICIENT

Answer = C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Brent@GMATPrepNow » Sat Sep 12, 2015 3:56 pm
infiniti007 wrote:Is 4^(x+y) = 8^(10)?

1.) x - y = 9
2.) y/x = 1/4

I've seen the solution for the above problem, but I'm confused why I could not solve for x and y using each statement separately?

Once you get down to the point that you're solving:
Does x + y = 15?

At this point, why could I not use Statement 1 as follows:
x - y = 9
So, x = 9 - y

Therefore, 9 - y + y = 15
9 = 15 False.

Statement 2:
y = 1/4x
So, x = 4y

Therefore, 4y + y = 15
5y = 15
y = 3

Plug back into Statement 2:
x = 4(3) = 12

Does x + y = 15? 12 + 3 = 15 True

I feel like I'm missing something very basic here.
You are trying to determine (with absolute certainty) whether or not x + y = 15
This is the QUESTION.

You are using this (x + y = 15) as a given fact, when our goal is to determine whether or not it is true.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Max@Math Revolution » Tue Sep 15, 2015 7:53 am
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem.
Remember equal number of variables and independent equations ensures a solution.

Is 4^(x+y) = 8^(10)?

1.) x - y = 9
2.) y/x = 1/4

Transforming the original condition and the question, 2^2(x+y)=2^3(10)=2^30? gives us 2(x+y)=30?, x+y=15? and therefore we have 2 variables. We need 2 equations to match the number of variables and equations, and since there is 1 each in 1) and 2), there is high probability that C is the answer and it turns out that C actually is the answer.

Normally for cases where we need 2 more equations, such as original conditions with 2 variable, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using ) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

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