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If a^2b^2c^3 = 4500. Is b + c = 7?

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

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22 Jan 2011, 01:11
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If $$a^2*b^2*c^3 = 4500$$. Is $$b+c = 7$$ ?

(1) a, b and c are positive integers

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

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22 Jan 2011, 03:50
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5
MichelleSavina wrote:
Q) If a^2.b^2.c^3 = 4500. Is b+c = 7 ?
(1) a, b and c are positive integers
(2) a > b

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.

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

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22 Jan 2011, 01:34
1
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.
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Re: If a^2b^2c^3 = 4500. Is b + c = 7?  [#permalink]

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22 Jan 2011, 01:53
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
i need an explanation.......
if it is E, then y so....
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Re: If a^2b^2c^3 = 4500. Is b + c = 7?  [#permalink]

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22 Jan 2011, 02:12
4
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,
1. a = 3, b = 2
2. a = 6, b = 1
Thus (b + c) can be equal to either 6 or 7.

Not sufficient

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

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19 May 2012, 09:01
If a^2b^2c^3 = 4500. Is b+c = 7 ?
(1) a, b and c are positive integers
(2) a > b

2^23^25^3 = 4500 and from statement 1 and 2, it can be deduced that b=2, c=5 and b+c =7, why the answer is E, shouldn't it be C?
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Re: If a^2b^2c^3 = 4500. Is b + c = 7?  [#permalink]

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19 May 2012, 16:42
gmihir wrote:
If a^2b^2c^3 = 4500. Is b+c = 7 ?
(1) a, b and c are positive integers
(2) a > b

2^23^25^3 = 4500 and from statement 1 and 2, it can be deduced that b=2, c=5 and b+c =7, why the answer is E, shouldn't it be C?

a,b and c don't have to be prime.
a=6 and b=1 will also satisfy the 2 statements alone and b+c=1+5 = 6. insuff together. hence E
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Re: If a^2b^2c^3 = 4500. Is b + c = 7?  [#permalink]

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20 May 2012, 10:32
Good question....But is GMAC really expecting us to figure that out in a span of 2 min???
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Re: If a^2b^2c^3 = 4500. Is b + c = 7?  [#permalink]

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20 May 2012, 11:48
dchow23 wrote:
Good question....But is GMAC really expecting us to figure that out in a span of 2 min???

yes...such questions may very well appear in the actual GMAT
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Re: If a^2b^2c^3 = 4500. Is b + c = 7?  [#permalink]

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22 May 2012, 09:00
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.
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Re: If a^2b^2c^3 = 4500. Is b + c = 7?  [#permalink]

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22 May 2012, 09:41
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

fantastic problem, but i got stumped.. i selected C as the answer and now i understand why the answer is E
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Re: If a^2b^2c^3 = 4500. Is b + c = 7?  [#permalink]

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11 Feb 2014, 00:19
Great question! Fell in to the trap and picked C. E makes perfect sense.
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Re: If a^2b^2c^3 = 4500. Is b + c = 7?  [#permalink]

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23 Nov 2017, 15:56
1
Hi All,

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

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

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30 Jul 2018, 14:31
Explanation:

Given a2 b2 c3 = 4500
⇒ a2 b2 c3 = 22 32 53
⇒ a = ±2, b = ±3 and c = 5
or a = ±3, b = ±2 and c = 5
or a = ±1, b = ±6 and c = 5
or a = ±6, b = ±1 and c = 5
or a, b and c can be decimal numbers.

Considering statement 1:
a, b and c are positive integers
⇒ a = 2, b = 3 and c = 5
or a = 3, b = 2 and c = 5
⇒ b + c = 8 or 7 (depending whether a = 2 or 3, respectively)
Since we are not getting a definite answer from above statement , statement 1 itself is not sufficient to provide the answer.

Considering statement 2:
a > b
Let us consider 4 possible values of a (2, −2, 3 and −3):
If we consider a = 2, then b has to be −3 (since statement 2 says a > b). Also, c = 5 ⇒ b + c = 2.
If we consider a = −2, then b has to be −3 (since statement 2 says a > b). Also, c = 5 ⇒ b + c = 2.
If we consider a = 3, then b may be 2 or −2 (in both cases, a > b). Also, c = 5 ⇒ b + c = 7 or 3.
And if we consider a = −3, then no possible value of b satisfies a > b.
Hence, b + c may be either 2 or 7 or 3. ⇒ b + c may or may not be 7.
Further, if a, b and c are decimals then there are infinite combinations of b and c.
Since we are not getting a definite answer from above statement , statement 2 also itself is not sufficient to provide the answer.

Considering statement 1 and 2 both:
Statement 1 gives: b = 1, 2, 3 or 6
Statement 2 gives: b = 1 or 2 or some decimal number.

Hence, the value of b + c can be 6 or 7.

Since we are not getting a definite answer from combination of statements 1 and 2.

I think, I deserve a Kudus :D Lol
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Re: If a^2b^2c^3 = 4500. Is b + c = 7?  [#permalink]

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17 Aug 2018, 20:25
it's a hard question. easy to ommit the case of 1!
it didn't mention a,b,c are primes
Re: If a^2b^2c^3 = 4500. Is b + c = 7?   [#permalink] 17 Aug 2018, 20:25
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