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Three consecutive even numbers are such that

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by AAPL » Thu Apr 12, 2018 4:58 am
Three consecutive even numbers are such that thrice the first number exceeds double the third by two, the third number is

A. 12
B. 6
C. 8
D. 10
E. 14

The OA is E.

3 consecutive even integers = n, n+2, n+4

The prompt can be translated to: 3n=2(n+4)+2, simplified we get: 3n=2n+10.

Scanning the answer choices, the simplest "n" value to test would be 10. Plug 10 in for all "n"s.

3(10)=2(10)+10, simplified 30=30. This is the only answer choice that will produce this result. "N" has to be 10. Plug 10 into the equation for the third value and the result is 14.

Has anyone another strategic approach to solve this PS question? Regards!
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by Vincen » Thu Apr 12, 2018 10:18 am
Hello AAPL.

I would solve it just like you did it.

Let's call x, x+2 and x+4 the three even consecutive numbers.

Then, we can set that $$3\cdot x\ =\ 2\cdot\left(x+4\right)+2$$ $$3x\ =\ 2x+10$$ $$3x-2x=10$$ $$x=10\ \ .$$ Since x is the first number, we have that x+4=10+4=14 is the third number.

Therefore, the correct answer is the option E.
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by Scott@TargetTestPrep » Mon Apr 16, 2018 3:39 pm
AAPL wrote:Three consecutive even numbers are such that thrice the first number exceeds double the third by two, the third number is

A. 12
B. 6
C. 8
D. 10
E. 14
We can let x = 1st even number, (x + 2) = 2nd even number and (x + 4) = 3rd even number. Thus we have:

3x = 2(x + 4) + 2

3x = 2x + 10

x = 10

Since x = 10, the third number is 10 + 4 = 14.

Answer: E

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