PR Math Workout Practice Set Q12

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PR Math Workout Practice Set Q12

by rbakerv » Fri Feb 20, 2015 1:29 pm
If y^c=y^d+1, what is the value of y?

(1) y<1
(2) d=c

PR said the correct answer is C but I think it's E. If you go through each statement independently you can cross off A and D. Then, you take both and c=d -> c=c+1 and y<1, 0 is basically the only value that works; however, it doesnt work when c=0 because then Y would be 1 and 0. Based on my thinking I thought neither was sufficient so I put E.

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by [email protected] » Fri Feb 20, 2015 4:39 pm
Hi rbakerv,

Is the information in the question meant to be formatted in the following way....?

Y^C = Y^(D+1)

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by Brent@GMATPrepNow » Fri Feb 20, 2015 5:48 pm
rbakerv wrote:If y^c = y^(d+1), what is the value of y?

(1) y < 1
(2) d = c
A few considerations:

case 1) If y = 1, then y^c = y^(d+1) for ALL values of c and d
case 2) If y = 0, then y^c = y^(d+1) for ALL values of c and d
case 3) If y = -1, then y^c = y^(d+1) if c and (d+1) are both ODD or both EVEN
case 4) If y equals any value OTHER THAN 1, 0 or -1, then y^c = y^(d+1) only if c and (d+1) are EQUAL


Target question: What is the value of y?

Given: y^c = y^(d+1)

Statement 1: y < 1
This rules out case 1 above.
However, this still leaves several possibilities that yield conflicting answers to the target question.
Here are two:
Case a: y = 0, c = 1 and d = 1. Notice that this satisfies the given information [y^c = y^(d+1)] because 0^1 = 0^(1+1). In this case y = 0
Case b: y = 3, c = 2 and d = 1. Notice that this satisfies the given information [y^c = y^(d+1)] because 3^2 = 3^(1+1). In this case y = 3
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: d = c
This rules out cases 3 and 4 above. Here's why:
case 3: if c and d are equal, then c and (d+1) CANNOT both be EVEN or ODD
case 4: if c and d are equal, then c and d+1 cannot be equal
However, this still leaves several possibilities that yield conflicting answers to the target question.
Here are two:
Case a: y = 0, c = 1 and d = 1. Notice that this satisfies the given information [y^c = y^(d+1)] because 0^1 = 0^(1+1). In this case y = 0
Case b: y = 1, c = 1 and d = 1. Notice that this satisfies the given information [y^c = y^(d+1)] because 1^1 = 1^(1+1). In this case y = 1
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 rules out case 1 above
Statement 2 rules out cases 3 and 4 above
This leaves only case 2, which means y must equal 0
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer = C

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by Brent@GMATPrepNow » Fri Feb 20, 2015 5:57 pm
rbakerv wrote:If y^c=y^d+1, what is the value of y?

(1) y<1
(2) d=c

PR said the correct answer is C but I think it's E. If you go through each statement independently you can cross off A and D. Then, you take both and c=d -> c=c+1 and y<1, 0 is basically the only value that works; however, it doesnt work when c=0 because then Y would be 1 and 0. Based on my thinking I thought neither was sufficient so I put E.
I have a feeling that you are concluding that, since the bases are the same in the equation y^c = y^(d+1), it must be the case that c = d+1
This is not correct.
For example, 0^1 = 0^2, but we can't conclude that 1=2

The ONLY time we can take y^c = y^(d+1) and conclude that c = d+1 is if y equals some value OTHER THAN -1, 0 or 1

Also, when you got to the point where you concluded that c=c+1, you should stop and check your work, because this equation is not possible for ANY value of c.

Cheers,
Brent
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by rbakerv » Sat Feb 21, 2015 11:00 am
Brent,
Thanks for your response. I think we're in agreement here but my confusion is when you try statement 1 and 2 together. What I meant to say was if c=d you could basically rewrite the question y^c=y^(c+1). I'm probably overthinking this but, using case 2 and both statements combined, if y has to be 0, the equation doesn't hold when c=0 because you get 0^0=1 and 0^(0+1)=0 on either side.

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Richard

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by Brent@GMATPrepNow » Sat Feb 21, 2015 11:09 am
rbakerv wrote:Brent,
Thanks for your response. I think we're in agreement here but my confusion is when you try statement 1 and 2 together. What I meant to say was if c=d you could basically rewrite the question y^c=y^(c+1). I'm probably overthinking this but, using case 2 and both statements combined, if y has to be 0, the equation doesn't hold when c=0 because you get 0^0=1 and 0^(0+1)=0 on either side.

Best,
Richard
Hi Richard,

The whole 0^0 issue is somewhat contentious (see https://www.beatthegmat.com/what-is-the- ... 27026.html). Some say 0^0 = 1, and others insist that 0^0 is undefined.
I agree with Ian (from the link); the GMAT will not test this concept.

Your point that the equation doesn't hold when c=0 may or may not be true (depending on who's evaluation of 0^0 you accept), HOWEVER, this doesn't apply here.

The question tells us that y^c = y^(d+1). It doesn't tell us that this equation holds for ALL values of c and d. It just tells us that the left side equals the right side.

Cheers,
Brent
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