Problem Solving

If x, y, and z are consecutive even positive integers, which of the following could be equal to x + y + z ?

A. 141
B. 200
C. 318
D. 391
E. 412

I took a long *** method doing this question.. it was easily over 2 mins...
is there a shorter way to do this??

OA is C

x

y=x+2

z=x+4

x+y+z=3x+6=3(x+2)

Test out the answer choices:

A.) 3(x+2)=141

x+2=47

x=45

45 isn't an even integer.

B.) 3(x+2)=200

200 isn't divisible by 3.

C.) 3(x+2)=318

x=104.

This is our answer.

fruti_yum wrote:If x, y, and z are consecutive even positive integers, which of the following could be equal to x + y + z ?

A. 141
B. 200
C. 318
D. 391
E. 412

I took a long *** method doing this question.. it was easily over 2 mins...
is there a shorter way to do this??

OA is C

This question aims at 2 properties.
1) Sum of even numbers is always even. So this eliminates choices A and D.

2) The average of 3 consecutive even numbers will be the middle number (Say Y is the middle number). Also, the sum of the 3 consecutive numbers 3 * average = 3Y. So this means that the sum MUST be divisible by 3.

Simple check on choices B, C and E will lead us to C as the other two are not divisible by 3.

I dont know any quicker way to it...

hope this helps!

fruti_yum wrote:If x, y, and z are consecutive even positive integers, which of the following could be equal to x + y + z ?

A. 141
B. 200
C. 318
D. 391
E. 412

I took a long *** method doing this question.. it was easily over 2 mins...
is there a shorter way to do this??

OA is C

fruiti,

first off you can eliminate answer choices a and d, because they are odd. we are looking for the sum of consecutive even numbers. even+even+even=even.

second, since the problem tells us that x, y and z are even numbers, then the sequence formula is the addition of n, n+2, and n+4 which is 3n+6

third use that formula to check your remaining choices.

3n+6=318
3n=312
n=104

no need to go further...the rest of the options won't work!

you got this!!