saxenarahul021 wrote:A certain box contains only red pencils and blue pencils. If 1 red pencil is removed, there will be the same number of red pencils as blue pencils in the box. If 1 blue pencil is removed instead, there will be twice as many red pencils as blue pencils in the box. How many pencils are in the box?

Algebraic Approach:
Say, number of red pencils in the box = R and number of blue pencils in the box = B

Now, (R - 1) = B
And, R = 2(B - 1)

Hence, B = 2(B - 1) - 1 = 2B - 3 ---> B = 3
Hence, R = (B + 1) = 4

The correct answer is D.

Tricky Approach:

Say, the total number of pencils in the box = N

Now after removing one red pencil, number of blue pencil = number of red
Hence, N must be odd.
Only possible options are A, B, and D.


And after removing one blue pencil, 2*(number of blue pencil) = number of red pencil.
Hence, (N - 1) must be divisible by 3.
Only possible options is D.