If x^3∗y^4=2000, what is y?
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- sachin_yadav
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Target question: What is the value of y?If x³y� = 2000, then what is the value of y?
1) x is an integer
2) y is an integer
Given: x³y� = 2000
This begs us to find the PRIME FACTORIZATION of 2000
2000 = (2)(2)(2)(2)(3)(3)(3)
= 3³2�
At this point, it might APPEAR that we can already determine the values of x and y.
It SEEMS that x = 3 and y = 2
However, it could also be the case that x = 3 and y = -2, since 2000 can also be written as (-2)(-2)(-2)(-2)(3)(3)(3)
NOTE: there are also other possible cases to consider if we don't restrict x and y to integer values.
Statement 1: x is an integer
Consider these two contradictory cases
Case a: x = 3 and y = 2, in which case y = 2
Case b: x = 3 and y = -2, in which case y = -2
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: y is an integer
There are several values of x and y that satisfy statement 2. Here are two:
Consider these two contradictory cases
Case a: x = 3 and y = 2, in which case y = 2
Case b: x = 3 and y = -2, in which case y = -2
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined
The same two contradictory cases apply:
Case a: x = 3 and y = 2, in which case y = 2
Case b: x = 3 and y = -2, in which case y = -2
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT
Answer = E
Cheers,
Brent
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- ceilidh.erickson
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Brent provided a great explanation, but I just want to point out a few big-picture takeaways that we can get from this problem:
- If the answer feels too obvious, there's probably something wrong. In this problem the obvious case that comes to mind after we prime-factor is x = 5 and y = 2
The GMAT likes to trick us! There will often be one obvious answer, and then another less obvious possibility: a negative, a non-integer, a square root, zero, etc. Try DS #113 in OG2016 for another similar example.
- Pay attention to even v. odd exponents. If you have both in a given problem, it's likely that they're testing positives v. negatives. If this question had asked for the value of x (the one with the odd exponent), then both statements together would have been sufficient. If x^3 = 5^3, then x must equal 5.
The GMAT won't test us on the easy one - they'll test us on the tricky one! If y^4 = 2^4, y could be 2 or -2.
Start to track these kinds of GMAT traps!
- If the answer feels too obvious, there's probably something wrong. In this problem the obvious case that comes to mind after we prime-factor is x = 5 and y = 2
The GMAT likes to trick us! There will often be one obvious answer, and then another less obvious possibility: a negative, a non-integer, a square root, zero, etc. Try DS #113 in OG2016 for another similar example.
- Pay attention to even v. odd exponents. If you have both in a given problem, it's likely that they're testing positives v. negatives. If this question had asked for the value of x (the one with the odd exponent), then both statements together would have been sufficient. If x^3 = 5^3, then x must equal 5.
The GMAT won't test us on the easy one - they'll test us on the tricky one! If y^4 = 2^4, y could be 2 or -2.
Start to track these kinds of GMAT traps!
Last edited by ceilidh.erickson on Fri Jan 22, 2016 4:09 pm, edited 1 time in total.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
- sachin_yadav
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It's probably easiest to factor first:
2000 = 2 * 2 * 2 * 2 * 5 * 5 * 5
Neither statement alone is sufficient. If x is an integer, then we could have x = 5 and y = 2, or we could have x = 10 and y = (fourth root of 2). Same idea with y: whatever y is, x can always be some bizarro irrational number.
Together, we have x³y� = 5³2�, with x and y integers. x³ = 5³ and y� = 2� is certainly one solution, but then y could equal 2 or -2! So we're left with a tricky E.
2000 = 2 * 2 * 2 * 2 * 5 * 5 * 5
Neither statement alone is sufficient. If x is an integer, then we could have x = 5 and y = 2, or we could have x = 10 and y = (fourth root of 2). Same idea with y: whatever y is, x can always be some bizarro irrational number.
Together, we have x³y� = 5³2�, with x and y integers. x³ = 5³ and y� = 2� is certainly one solution, but then y could equal 2 or -2! So we're left with a tricky E.