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Abe, Beth, Carl and Duncan are four siblings, among which Abe and Carl

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Abe, Beth, Carl and Duncan are four siblings, among which Abe and Carl  [#permalink]

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Updated on: 13 Aug 2018, 02:11
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Abe, Beth, Carl and Duncan are four siblings, among which Abe and Carl are twins and Beth and Duncan are also twins. When the present ages of the four siblings are multiplied, the product is 900. If Beth is older than Abe, what is the age of Duncan? Assume the ages of all siblings to be integers.

(1) The difference between Beth’s age and Abe’s age is a prime number.

(2) If Carl had been born four years earlier, the difference between Duncan’s age and Carl’s age would have been a prime number.

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Originally posted by EgmatQuantExpert on 09 Apr 2015, 23:37.
Last edited by EgmatQuantExpert on 13 Aug 2018, 02:11, edited 6 times in total.
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Re: Abe, Beth, Carl and Duncan are four siblings, among which Abe and Carl  [#permalink]

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Updated on: 07 Aug 2018, 02:38
7
10
Detailed Solution

Step-I: Given Info

We are told about four siblings Abe, Beth, Carl and Duncan such that Abe and Carl are twins and Beth and Duncan are also twins. We are also given that the product of the present ages of the four siblings is 900. Further we are told that Beth is older than Abe and we are asked to find the age of Duncan

Step-II: Interpreting the Question Statement

Since Abe and Carl are twins, their ages would be same, let’s assume it to be $$x$$. Similarly, since Beth and Duncan are twins, their ages would be same, let’s assume it to be $$y$$.

We are told that Beth is older than Abe, i.e. $$y > x$$ and the product of the ages of the siblings is 900, so we can write $$x^ 2 * y^2 = 900$$.

We can observe here that 900 is written as product of two squares, since 900 can be prime factorized as $$900 = 2^2 * 3^2 * 5^2$$, the possible set of values of ($$x, y$$) can be:

• (1, 30) or
• (2, 15) or
• (3, 10) or
• (5,6)

Let’s proceed to the solutions to see if we can get a unique value of $$x$$ with this understanding.

Step-III: Statement I

Statement tells us that $$y$$ $$–$$ $$x$$ is a prime number. Let’s evaluate our possible cases to see if we can find a unique value for $$x$$:

• (1, 30) –>= 29-> Prime
• (2, 15) –>= 13 -> Prime
• (3, 10) – > = 7 -> Prime
• (5,6) – > = 1 -> Not Prime

We observe here that, there are three possible values for $$x$$, hence statement-I is not sufficient to arrive at the answer.

Step-IV: Statement II

Statement-II tells us that had Carl been born four years earlier, the difference between Duncan’s age and Carl’s age would have been a prime number. Since Carl’s present age is $$x$$, had he been born four years earlier, his present age would be ($$x +4$$).
The Statement tells us that $$y$$ $$–$$ $$(x +4)$$ is a prime number. Let’s evaluate our possible cases to see if we can find a unique value for $$x$$.

• (1, 30) –>= 25-> Not Prime
• (2, 15) –>= 9 -> Not Prime
• (3, 10) – > = 3 -> Prime
• (5,6) – > = 3 -> Prime

We observe here that there are two possible values for $$x$$, hence statement-II is not sufficient to arrive at the answer.

Step-V: Combining Statements I & II

Statement-I gives us the possible values of ($$x, y$$) as (1, 30), (2, 15) and (3, 10). Statement-II gives us the possible values of ($$x, y$$) as (3, 10) and (5,6).

Combining statement-I & II give us only possible option for values of ($$x, y$$) which is (3, 10).
Thus combination of St-I & II is sufficient to answer the question. Hence, the correct answer is Option C

Key Takeaways

1. Prime factorize a number to understand the ways in which a number can be represented

atom - you were right in describing why Statement-II alone is not sufficient but you did not consider the combinations of statement- I & II.

Regards
Harsh

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Originally posted by EgmatQuantExpert on 11 Apr 2015, 01:00.
Last edited by EgmatQuantExpert on 07 Aug 2018, 02:38, edited 2 times in total.
General Discussion
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Joined: 05 Apr 2014
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Re: Abe, Beth, Carl and Duncan are four siblings, among which Abe and Carl  [#permalink]

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10 Apr 2015, 08:38
3
1
Let the ages be A,B,C & D respectively.
Given A=C & B=D, A*B*C*D=900=>B^2*A^2=900=>B*A=30. Possible combinations are (30*1),(15*2),(10*3),(6*5)
Statement 1: B-A=P(prime) => (30*1),(15*2), & (10*3) satisfy the equation. Therefore statement 1 is insufficient.
Statement 2: D-(C+4)=P or B-(A+4)=P or B-A-4=P. Now only (10*3) satisfy. Therefore B=D=10 and statement 2 is sufficient.
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Re: Abe, Beth, Carl and Duncan are four siblings, among which Abe and Carl  [#permalink]

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10 Apr 2015, 09:56
Shuvabrata88,

I agree with the approach which you have taken but I think the answer should be E as statement 2 simply states that diff b/w D and C is prime so it could be either D-C-4 or C+4-D; in this case (5*6) combination also works out to be prime.

Initial ques stem mentions that B>A i.e D>C but if we add 4 to C then this statement may not hold true.

I may be over complicating the question, lets wait for OE
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Re: Abe, Beth, Carl and Duncan are four siblings, among which Abe and Carl  [#permalink]

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10 Apr 2015, 11:26
EgmatQuantExpert

Had a happy time solving, please provide the OA asap
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Re: Abe, Beth, Carl and Duncan are four siblings, among which Abe and Carl  [#permalink]

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11 Apr 2015, 11:40
Oh yes Harsh, my bad ; thanks for the detailed solution
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Re: Abe, Beth, Carl and Duncan are four siblings, among which Abe and Carl  [#permalink]

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12 Apr 2015, 08:37
2
e-gmat representative

i am sorry but i disagree with your evaluation of statement two. following are my points of dispute:
It is given that beth ( & hence duncan, since they are twins) is older than abe ( & correspondingly carl). That is a fact and cannot be challenged, right ? Also, fact 2: clearly states difference between Duncan's age and Carl's age..makes sense because we would want to end up with negetive number which would occur if the reverse ( C - D) were to happen..so far so good. now consider the possible age value sets (keeping in mind the logical validity of the statement )

" Since Carl’s present age is x, had he been born four years earlier, his present age would be (x +4).
The Statement tells us that y – (x +4) is a prime number. " this is your explanation..totally with you till this point. HOWEVER,

y – (x +4) when applied to the set (5,6) yields -3 (which is not positive first of all forget being prime) and not 3 as you said..absolutely illogical and hence not a valid set i feel. the other sets (resultant-prime or otherwise) at least yield a positive number since x and x+4 are both lesser than y.
And if we were to take this into account, the only valid set is (3,10) from stmnt 2..sufficient ans: B

WOULD REQUEST your thoughts on this and would kindly appreciate if A MATH EXPERT would care to enlighten or clarify.

regards
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Re: Abe, Beth, Carl and Duncan are four siblings, among which Abe and Carl  [#permalink]

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12 Apr 2015, 11:12
smashbiker84,

I agree that D>C is given in the question, but how would you know D>C+4?? we can have C+4>D right? Hence we must need to consider D-C-4 and C+4-D; in this case 5,6 gives you the prime output

Hope it helps

Cheers
AT
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Re: Abe, Beth, Carl and Duncan are four siblings, among which Abe and Carl  [#permalink]

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12 Apr 2015, 20:04
1
This is my question.
Well we can come up with the following pairs (A,B) or (D, C) = (15, 2), (6, 5), (10, 3), or (30, 1).
From statement 2 we get that had Carl been born four years earlier, the difference would have been a prime number. So from the given information we know that Carl's age is either 2,5,3,1. So if he was born 4 years earlier then his age could have been (-2, 1, -1, -3). Well obviously only 1 is possible, so the difference between 6 and 1 be a prime number.

But clearly that is not the way to approach this problem, nor is this approach taken by anyone. I just want to know what am I missing here.

Thanks!
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Re: Abe, Beth, Carl and Duncan are four siblings, among which Abe and Carl  [#permalink]

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Updated on: 07 Aug 2018, 02:39
Hi smashbiker84 We are not challenging any facts here. Statement-II presents a 'What If' scenario which gives another relation between the ages of the siblings. The question statement talks about the present situation, while statement-II talks about an alternate situation. For example, assume a situation where you are 3 years elder to your sister. If your sister were to be born 4 years earlier, she would become a year elder to you. This is not contradicting the present situation, but building upon the present situation.

To answer your 2nd question, please note that difference between two things is positive in real world scenarios. For example, difference between a son's age and his father's age is 30 years and the difference between the father’s and his son’s age is also 30 years. We do not say that the difference between son’s age and father’s age is -30 years. The "-" is only an indication that you have subtracted the larger item from the smaller item. Hence, when we are talking about difference between two things, it would mean the absolute difference or the magnitude of the difference.
.

Hi nphatak, if a person is born earlier, his age would increase, so instead of subtracting 4 from the present age, you would need to add 4 in the present age.
For ex: If a person is born in 2004, his present age would be 11, but if he is born four years earlier i.e. in 2000, his present age would be (11 + 4) = 15.
So, in this case Carl's age would be (6,9,7,5) instead of (-2, 1, -1, -3).

Hope its clear!

Regards
Harsh

Originally posted by EgmatQuantExpert on 13 Apr 2015, 02:47.
Last edited by EgmatQuantExpert on 07 Aug 2018, 02:39, edited 1 time in total.
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Re: Abe, Beth, Carl and Duncan are four siblings, among which Abe and Carl  [#permalink]

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20 Apr 2017, 02:25
Option C

In terms of age, A = C & B = D.
Also, A*B*C*D = 900 =A^2*B^2 = C^2*D^2

A*B = C*D = 30

Which means, (A,B) or (B,A) could possibly be (1,30) ; (2,15) ; (3,10) ; (5,6).

I: B - A = Prime number. Let us look at the possible options:
B - A = |30 -1| = 29 or |15 - 2| = 13 or |10 - 3| = 7 or |6 - 5| = 1.

Except for the last combination (6,5), rest satisfy the condition of being Prime number, so there is NO unique solution. Hence, I is Insufficient.

II: Carl's age today, had he born four years ago, would be C +4.
So, D - (C + 4) = B - (A+4) = Prime number. Now again looking at the possible combinations, we can see:

30 - (1+4) = 25 Not a Prime
15 - (2+4) = 9 Not a Prime
10 - (3+4) = 3 Prime

Again, no Unique solution. Insufficient.

Now, if we combine both I + II:
I:B-A = Prime Number
Also, II:B - (A+4) = Prime Number
From possible solutions, only one combination B = 10 & A = 3 satisfy both the conditions.
As, B-A = 10 -3 = 7 Prime
B - (A+4) = 10 - (3+4) = 3 Prime
Hence, Sufficient.
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Re: Abe, Beth, Carl and Duncan are four siblings, among which Abe and Carl  [#permalink]

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31 Oct 2017, 14:38
EgmatQuantExpert wrote:
Detailed Solution

Step-I: Given Info

We are told about four siblings Abe, Beth, Carl and Duncan such that Abe and Carl are twins and Beth and Duncan are also twins. We are also given that the product of the present ages of the four siblings is 900. Further we are told that Beth is older than Abe and we are asked to find the age of Duncan

Step-II: Interpreting the Question Statement

Since Abe and Carl are twins, their ages would be same, let’s assume it to be $$x$$. Similarly, since Beth and Duncan are twins, their ages would be same, let’s assume it to be $$y$$.

We are told that Beth is older than Abe, i.e. $$y > x$$ and the product of the ages of the siblings is 900, so we can write $$x^ 2 * y^2 = 900$$.

We can observe here that 900 is written as product of two squares, since 900 can be prime factorized as $$900 = 2^2 * 3^2 * 5^2$$, the possible set of values of ($$x, y$$) can be:

• (1, 30) or
• (2, 15) or
• (3, 10) or
• (5,6)

Let’s proceed to the solutions to see if we can get a unique value of $$x$$ with this understanding.

Step-III: Statement I

Statement tells us that $$y$$ $$–$$ $$x$$ is a prime number. Let’s evaluate our possible cases to see if we can find a unique value for $$x$$:

• (1, 30) –>= 29-> Prime
• (2, 15) –>= 13 -> Prime
• (3, 10) – > = 7 -> Prime
• (5,6) – > = 1 -> Not Prime

We observe here that, there are three possible values for $$x$$, hence statement-I is not sufficient to arrive at the answer.

Step-IV: Statement II

Statement-II tells us that had Carl been born four years earlier, the difference between Duncan’s age and Carl’s age would have been a prime number. Since Carl’s present age is $$x$$, had he been born four years earlier, his present age would be ($$x +4$$).
The Statement tells us that $$y$$ $$–$$ $$(x +4)$$ is a prime number. Let’s evaluate our possible cases to see if we can find a unique value for $$x$$.

• (1, 30) –>= 25-> Not Prime
• (2, 15) –>= 9 -> Not Prime
• (3, 10) – > = 3 -> Prime
• (5,6) – > = 3 -> Prime

We observe here that there are two possible values for $$x$$, hence statement-II is not sufficient to arrive at the answer.

Step-V: Combining Statements I & II

Statement-I gives us the possible values of ($$x, y$$) as (1, 30), (2, 15) and (3, 10). Statement-II gives us the possible values of ($$x, y$$) as (3, 10) and (5,6).

Combining statement-I & II give us only possible option for values of ($$x, y$$) which is (3, 10).
Thus combination of St-I & II is sufficient to answer the question. Hence, the correct answer is Option C

Key Takeaways

1. Prime factorize a number to understand the ways in which a number can be represented

Regards
Harsh

This interpretation of between of in the problem seems to go against what is used on GMAT--"difference between x and y" is interpreted as "x-y"--this is also shown in GMAT club math book; am I missing something here because the difference has a direction associated with it (+/-) so (5 and 6) would actually result in (6-(5+4)) [representing Duncan - Carl Born Four Years Ealier] is actually negative and thus cannot be a prime so statement II alone should be sufficient and the right answer should be B...
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Re: Abe, Beth, Carl and Duncan are four siblings, among which Abe and Carl  [#permalink]

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11 Aug 2018, 09:35
i too am getting statement B as sufficient.
Re: Abe, Beth, Carl and Duncan are four siblings, among which Abe and Carl   [#permalink] 11 Aug 2018, 09:35
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