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

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BTGmoderatorLU: Moderator; Posts: 2544; Joined: Sun Oct 15, 2017 1:50 pmFollowed by: 6 members

by BTGmoderatorLU » Mon Apr 15, 2019 1:44 pm

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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.

The OA is C

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Source: Beat The GMAT — Data Sufficiency | Mon Apr 15, 2019 1:44 pm

deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pmFollowed by: 5 members

by deloitte247 » Sun Apr 28, 2019 9:43 am

Let Abe age = A, Beth = B, Carl = C, and Duncan = D.
Given that A = C and B = D
$$y>x\ and\ x^2\cdot y^2=900$$ $$y>x\ and\ x^2\cdot y^2=900$$
What is the age of Duncan?
Statement 1 --> The difference between Beth's age and Abe's age is a prime number.
$$900\ as\ a\ product\ of\ the\ squares\ can\ be\ factorized\ as\ 2^2\cdot3^2\cdot5^2=900$$
$$Possible\ values\ for\ x^2\cdot y^2=900\ are\ \left(1\ and\ 30\right),\ \left(2\ and\ 15\right),\ \left(3\ and\ 10\right),\ or\ \left(5\ and\ 6\right)$$
y-x is a prime number. Possible cases are;
30 - 1 = 29 -> prime
15 - 2 = 13 -> prime
10 - 3 = 7 -> prime
6 - 5 = 1 -> Not a prime as there are more than one possible values for x.
Therefore, STATEMENT 1 IS NOT SUFFICIENT

Statement 2 --> If Carl had been born 4 years earlier, the difference between Duncan and Carl's age would have been a prime number.
$$y-\left(x+4\right)=prime\ number;$$
Now, factorizing 900 as product of two-square possible values of (x and y) = (1 and 30), (2 and 15), (3 and 10), or (5 and 6) for [y - (x + 4)] = prime.
So for (1 and 30), [30 - (1+4)] = 25 --> Not prime.
For (2 and 15), [15 - (2+4)] = 9 --> Not prime
For (3 and 10), [10 - (3+4)] = 3 --> Prime
For (5 and 6), [6 - (5+4)] = -3 --> Prime

As there are more possible value for x, therefore, STATEMENT 2 IS NOT SUFFICIENT.

Now, let's combine statement 1 and 2 together
The possible values for statement 1 are (1 and 30), (2 and 15) and (3 and 10). For statement 2, the possible values are (3 and 10) and (5 and 6).
Hence, the point of intersection of statement 1 and 2 is (3 and 10) where x = 3 and y = 10;
Thus, Duncan's age = 10.

Conclusively, both statement combined together are SUFFICIENT.

OPTION C IS THE CORRECT ANSWER

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