sandy wrote:
If a, b, c, and d are consecutive integers such that a<b<c<d, then in terms of a, the sum a + b + d =
(A) a + 4
(B) 2a + 3
(C) 3a + 2
(D) 3a + 3
(E) 3a +4
The key here is that a, b, c, and d are CONSECUTIVE integers. So, each integer is 1 GREATER than the one before it.
So, if the 1st integer = a
Then the 2nd integer must = a + 1
The 3rd integer = a + 2
The 4th integer = a + 3
In other words,
a = a
b = a + 1
c = a + 2
d = a + 3
In terms of a, the sum a + b + d =a + b + d = a + (a + 1) + (a + 3)
= 3a + 4
Answer: E
Cheers,
Brent