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