# Quant Question of the Week

**QUESTION 1**

If the operation # is defined for all f and g by the equation f#g = (f^2 g)/2, then 3#(-1# - 2) =

A) - 9/2

B) -1

C) 4

D) 9/2

E) 81/4

SOLUTION

Parentheses work the same way with unfamiliar operators as they do with more familiar ones: first evaluate -1# - 2, then use the result as the second term with 3. So, first, say f = -1 and g = -2:

(f^2 g)/2= ((-1)^2 (-2))/2= (-2)/2= -1

Now, you're looking for the result of 3# - 1, so evaluate f = 3 and g = -1:

(f^2 g)/2= ((3)^2 (-1))/2= (-9)/2

The answer is A

**QUESTION 2**

If x > y, increasing the original price of an item by x% and then decreasing the new price by y% is equivalent to multiplying the original price by

A) 1/(100(x-y))

B) 1/100(x-y-xy/100)

C) (x-y-xy)/100

D)1+1/100 (x-y+xy/100)

E)1 – ((x-y-xy)/100)

SOLUTION

x% is equal to x/100 , so when you increase a price (call it p) by

x%, you're multiplying it by (1+x/100). So, after the first increase, the new price is p(1 +x/100) . Similarly, decreasing a price by y% is the same as multiplying it by (1 -y/100), so the final price is p(1 +x/100)(1 - y/100) . Thus, we're multiplying the original price by

(1 +x/100 )(1-y/100) x

1 +x/100 - y/100 - xy/10000

1 +1/100 (x – y - xy/100)

The answer is B