In a certain brick wall, each row of bricks above the bottom row

STRATEGY: Upon reading any GMAT Problem Solving question, we should always ask, Can I use the answer choices to my advantage?
In this case, we can easily test the answer choices.
Now let's give ourselves up to 20 seconds to identify a faster approach.
In this case, we can also solve the question algebraically.
Since both options seem pretty fast, I'll go with testing answer choices

We'll begin with answer choice C, the middlemost answer choice, because we can often eliminated answer choices based on the results of testing the middlemost answer.
(C) 16. This answer choice suggests that the bottom row contains 16 bricks.
This means the row above it contains 15 bricks, the next row contains 14 bricks, the next row contains 13 bricks, and the top row contains 12 bricks.
Add all five rows to get: 16 + 15 + 13 + 12 + 11 = 67, which is no good because we want 75 bricks.
Eliminate choice C. Also, since we need MORE than 67 bricks, we can eliminate choices A and B, since they start off with even fewer bricks than choice C does.

At this point, we need only test one more answer choice. For example, if we test D and it works, we're done. If we test D and it doesn't work, we're still done because the correct answer must be E.

Let's best choice D.
(D) 17. This answer choice suggests that the bottom row contains 17 bricks.
This means the row above it contains 16 bricks, the next row contains 15 bricks, the next row contains 14 bricks, and the top row contains 13 bricks.
Add all five rows to get: 17 + 16 + 15 + 14 + 13 = 75. Perfect!!

Answer: D

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ALTERNATE APPROACH: Apply Algebraic Techniques
Let x = # of brick in bottom (1st) row
So x - 1 = # of brick in 2nd row
So x - 2 = # of brick in 3rd row
So x - 3 = # of brick in 4th row
So x - 4 = # of brick in 5th row

TOTAL # of bricks = 75

So, x + (x-1) + (x-2) + (x-3) + (x-4) = 75
Simplify: 5x - 10 = 75
5x =85
x = 17

Answer: D

Cheers,
Brent

Here it's given 75 total bricks in 5 rows.
Now 75/5 = 15 , so the first row will be greater than 15 for sure.
We can apply AP sum where sum = 75, rows= n = 5 and difference =1
75 = (5/2)[2*a+ (5-1)*1] where a = first number i.e. the last row of the wall
solving this we get a = 13 , so the first row or the last element of the A.P will be 13+(5-1) = 13+ 4 = 17

So, the answer is D.

let bottom row has x bricks
then,
x+x-1+x-2+x-3+x-4 = 75
5x-10=75
x= 17

Ans:D

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