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Re: If a^2b^2c^3 = 4500. Is b + c = 7? [#permalink]

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22 Jan 2011, 00:34

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Hi, The answer to this question should be C. 4500=2^2*3^2*5^3

This will yield C=5. Now possible values of 'a' are 2,-2,3,-3 and possible values of b are 3,-3,2,-2

Statement 1 tells a,b,c are positive integers. Hence possible values of 'b' reduces to 2,3. Not sufficient to answer the value of b+c. Statement 2 tells a>b. Hence possible values of b are -3, when a=2 and 2,-2 when a=3. Hence not sufficient. Now statement 1 and 2 together will yield 'b' as 2 Hence sufficient.Hope this helps . Thank You.

Re: If a^2b^2c^3 = 4500. Is b + c = 7? [#permalink]

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22 Jan 2011, 00:53

sashikanth wrote:

Hi, The answer to this question should be C. 4500=2^2*3^2*5^3

This will yield C=5. Now possible values of 'a' are 2,-2,3,-3 and possible values of b are 3,-3,2,-2

Statement 1 tells a,b,c are positive integers. Hence possible values of 'b' reduces to 2,3. Not sufficient to answer the value of b+c. Statement 2 tells a>b. Hence possible values of b are -3, when a=2 and 2,-2 when a=3. Hence not sufficient. Now statement 1 and 2 together will yield 'b' as 2 Hence sufficient.Hope this helps . Thank You.

hey... thanks this is what i answered... but the correct answer is E... i need an explanation....... if it is E, then y so....

Re: If a^2b^2c^3 = 4500. Is b + c = 7? [#permalink]

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22 Jan 2011, 01:12

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MichelleSavina wrote:

If a^2.b^2.c^3 = 4500. Is b+c = 7 ? (1) a, b and c are positive integers (2) a > b

4500 = (2²)(3²)(5³) = (a²)(b²)(c³) Comparing the terms we see that, c = 5 But a and b can have different sets of values.

1. a = ±2, b = ±3 2. a = ±3, b = ±2 3. a = ±1, b = ±6 4. a = ±6, b = ±1

Statement 1: a, b and c are positive integers Hence, we should discard the negative values of a and b. Still we have four possible sets of values for a and b,

1. a = 2, b = 3 2. a = 3, b = 2 3. a = 1, b = 6 4. a = 6, b = 1

Thus (b + c) can have different possible values.

Not sufficient

Statement 1: a > b Hence, we should discard the values of a and b such that a < b. Still we have different sets of values for a and b,

1. a = ±2, b = -3 2. a = 3, b = ±2 3. a = ±1, b = -6 4. a = 6, b = ±1

Thus (b + c) can have different possible values.

Not sufficient

1 & 2 Together : Still we can have two different sets of values for a and b,

Anurag Mairal, Ph.D., MBA GMAT Expert, Admissions and Career Guidance Gurome, Inc. 1-800-566-4043 (USA) +91-99201 32411 (India) http://www.facebook.com/Gurome

Q) If a^2.b^2.c^3 = 4500. Is b+c = 7 ? (1) a, b and c are positive integers (2) a > b

please explain...

Given: \(a^2*b^2*c^3 = 4500=2^2*3^2*5^3\). Question: is \(b+c=7\)

(1) a, b and c are positive integers --> sure it's possible that \(a=3\), \(b=2\) and \(c=5\) and in this case the answer will be YES but it's also possible that \(a=2\), \(b=3\) and \(c=5\) and in this case the answer will be NO. Not sufficient.

(2) a > b. Clearly insufficient.

(1)+(2) Note that we are not told that a, b and c are prime numbers, so again it's possible that \(a=3>2=b\) and \(c=5\) and in this case the answer will be YES but for example it's also possible that \(a=2^2*3^2=6^2\), \(b=1\) and \(c=5\) (\(a^2*b^2*c^3 =6^2*1^2*5^3=4500\)) and in this case the answer will be NO. Not sufficient.

Re: If a^2b^2c^3 = 4500. Is b + c = 7? [#permalink]

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22 May 2012, 08:00

Seriouslly very good question. Even after looking at the anwer I was not able to figure out how this can be possible. I thought about this and realised how our mind works in one direction. As it has been so well trained to do the prime factors as soon as you see any problem like this. It can not think any other situation.

Thanks for posting such question which can break the mindset.

We're told that (A^2)(B^2)(C^3) = 4500. We're asked if B+C = 7. This is a YES/NO question. This question really tests the 'thoroughness' of your thinking - and it's likely that many Test Takers would get this question wrong because they wouldn't consider all of the possible options.

To start, it would help to 'break down' 4500 into pieces... 4500 = (9)(500) = (3^2)(5)(100) = (3^2)(5)(2^2)(5^2) = (2^2)(3^2)(5^3)

Based on the starting equation, we know that C MUST be +5. However, there are more than just these two possible values for A and B. Since we're SQUARING terms, A and B could be (2 or -2) and (3 or -3). In addition, this question is NOT forcing us to deal with Prime Numbers, so there is another option besides "2 and 3" to consider.... "1 and 6" is another possible pairing of values for A and B.... (1^2)(6^2)(5^3) ...meaning that we would have to also consider that the values could be (1 or -1) and (6 or -6):

(1) A, B and C are POSITIVE integers

IF.... A=2, B=3, C=5 then the answer to the question is NO A=3, B=2, C=5 then the answer to the question is YES Fact 1 is INSUFFICIENT

2) A > B A=3, B=2, C=5 then the answer to the question is YES A=6, B=1, C=5 then the answer to the question is NO Fact 2 is INSUFFICIENT

Combined, we already have two groups of values that fit both Facts and produce different answers to the given question: A=3, B=2, C=5 then the answer to the question is YES A=6, B=1, C=5 then the answer to the question is NO Combined, INSUFFICIENT