If the average (arithmetic mean) of a and b is 6, and a^2 - b^2 = 2

Question

If the average (arithmetic mean) of a and b is 6, and  a^2 - b^2 = 2 , what is the value of  a - b  ?

A)  \frac{1}{6}

B)  \frac{1}{3}

C)  \frac{1}{2}

D) 2

E) 3

pushpitkc · Answer

Formula used:    a^2 - b^2 = (a + b)(a - b)

The average of a and b is 6 translates to  \frac{a + b}{2} = 6  ->  a + b = 12

Since we have  a^2 - b^2 = 2 ,  (a + b)(a - b) = 12  ->  12 * (a - b) = 2

Therefore, the value of (a - b) is  \frac{2}{12} = \frac{1}{6}  (Option A)

gmatzpractice · Answer

A 
(a+b)/2 = 6 thus a+b = 12
(a+b)(a-b) = 2
a-b = 2/12

jackspire · Answer

EgmatQuantExpert · Answer

Solution

Given:  
 • The average of a and b is 6
 o Or,  \frac{a + b}{2}  = 6
o Or, a + b = 12 
• Also,  a^2 – b^2  = 2

To find:  
 • The value of a – b

Approach and Working:   
We know  a^2 – b^2  = (a + b) (a – b)

Replacing the known values, we can write
 • 12 (a – b) = 2
Or, a – b =  \frac{2}{12} = \frac{1}{6}

Hence, the correct answer is option A.

Answer: A

JeffTargetTestPrep · Answer

Since (a + b)/2 = 6, then (a + b) = 12. We recognize that a^2 - b^2 is a difference of squares, which factors to 
 
(a - b)(a + b) = 2; thus:

(a - b)(12) = 2

a - b = 2/12 = 1/6

Answer: A

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