# Backsolving in GMAT

Backsolving is a technique of trying out the answer choices in GMAT questions. Each choice is selected one at a time and is assumed to be the right choice. Analyze the value or conditions by injecting the choices into the question. Backsolving is incredibly time saving in algebraic equations and number properties problems.

Case 1: To find the value of an equation
Example: Which number gives the lowest value for the equation?
1/X + (1/X) ^2 + 5X

Approach 1: Start from the middle value in the answer choice.
For example, if the five answer choices are 14,5,19,8,21

Plug 14 into the equation and check the value. If the value is too high, find out an answer choice with a much smaller value, like 5.

Approach 2:  If you inspect the equation, the first two terms will result in small numbers, smaller than 1, according to the given answer choices. However, the 3rd term of the equation is the leading one. A small value in 5X will make the equation’s net value smaller, and a big value in 5X will make the equation’s net value bigger. So, the answer is almost immediately x = 5.

Case 2: To satisfy a condition
Example: If y # 3 and 2x/y is a prime integer greater than 2, which of the following must be true?

I. x = y
II. y = 1
III. x and y are prime integers.

(A) None

(B) I only

(C) II only

(D) III only

(E) I and II

For 1: Let us plug-in x=y=1
2x/y = 2, which is not greater than 2

For 2: y=1
Therefore 2x/y = 2x
Plug-in value for x:
x = 1, 2x = 2 which is equal to 2 and not greater than 2

For 3: x and y are prime integers

x = 13 y= 2
2x/y = 13, which is a prime integer greater than 2
x=2 y=13
2x/y = 0.3, which is less than 2 and not a prime integer

Note: If a certain value satisfies the condition, always try with other values before confirming the answer.

For 3: x=13, y=2 satisfied the condition (Where x is a large prime number and y, a small one), but when we plugged in opposite numbers, x=2 and y =13 (Where x is a small prime number and y, a large prime number) the condition failed. Always test with opposite numbers to validate the condition.

Credits:
Leo John, Luis R. Villegas H

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