# GMAT Problem Solving: Absolute Value

Q-51 Practice / Problem Solving / Algebra / Absolute value / Q10
This question tests your understanding of absolute values of numbers. It also tests your ability to find values for which the expression will be positive and for those for which it will be negative and judge for what values will the expression be maximum and when it will be minimum.

• PS
• If a, b, and c are not equal to zero, what is the difference between the maximum and minimum value of S? $$S = 1+\frac { \left| a \right| }{ a } +\frac {2 \left| b \right| }{ b } +\frac {3 \left| ab \right| }{ ab } -\frac {4 \left| c \right| }{ c } \\$ A. 12 B. 14 C. 22 D. 20 E. 18 • Correct Answer Choice E. The difference between maximum and minimum value of S is 18. ## Explanatory Answer ## Detailed Solution Understanding Absolute Value $\frac { \left| a \right| }{ a } \\$ = 1 when a is positive. $\frac { \left| a \right| }{ a } \\$ = -1 when a is negative. ## When will the value of the expression be maximum? $\ S = 1+\frac { \left| a \right| }{ a } +\frac {2 \left| b \right| }{ b } +\frac {3 \left| ab \right| }{ ab } -\frac {4 \left| c \right| }{ c } \\$ The value of the expression will be maximum when all of the terms become positive. i.e., a, b, and ab should be positive and c should be negative. When a is positive, $\frac { \left| a \right| }{ a } \\$ = 1. When b is positive, $\frac {2 \left| b \right| }{ b } \\$ = 2. When a and b are positive, as required in the previous two steps, ab will be positive and the expression, $\frac {3 \left| ab \right| }{ ab } \\$ = 3. When c is negative, $\frac {4 \left| c \right| }{ c } \\$ = -4. Therefore, the maximum value = 1 + 1 + 2 + 3 -$-4) = 11.

Compute the Minimum Value

The value of the expression will be minimum if we make as many terms negative as possible. Higher the magnitude of the terms made negative, lower the value of the expression.

$\ S = 1+\frac { \left| a \right| }{ a } +\frac {2 \left| b \right| }{ b } +\frac {3 \left| ab \right| }{ ab } -\frac {4 \left| c \right| }{ c } \\$

c has to be positive for S to be minimum. The last term will then be -4.

If ab is negative, then $\frac {3 \left| ab \right| }{ ab } \\$ = -3.
If ab has to be negative, one of a or b has to be positive and the other has to be negative.

Possibility 1: If a > 0 and b < 0, $\frac { \left| a \right| }{ a } \\$ = 1 and $\frac {2 \left| b \right| }{ b } \\$ = -2.

The value of the expression is 1 + 1 - 2 - 3 - 4 = -7.

Possibility 2: If a < 0 and b > 0, $\frac { \left| a \right| }{ a } \\$ = -1 and $\frac {2 \left| b \right| }{ b } \\$ = 2.

The value of the expression is 1 - 1 + 2 - 3 - 4 = -5.

Therefore, the minimum value of the expression is -7.

The difference between Maximum and Minimum

Maximum value = 11.

Minimum value = -7

The difference is 18.

Choice E is the correct answer.