gmatophobia wrote:
Bob expected to spend a total of \($9.00\) to buy a given amount of pasta salad at a fixed price per pound. However, the price of the salad was \($0.20\) more per pound than Bob had expected. Consequently, he spent \($9.00\) and bought \(\frac{1}{2}\) pound less of the salad. How much did Bob spend per pound for the salad?
A. \($1.80 \)
B. \($1.85 \)
C. \($1.90 \)
D. \($1.95 \)
E. \($2.00 \)
The actual price per pound and the expected price per pound yield two weights are 1/2 pound apart.
Implication:
It is extremely likely that -- when divided into the paid amount of 900 cents -- one price-per-pound will yield a weight that is an INTEGER value, while the other yields a weight that is 0.5 MORE than an integer value.
We can PLUG IN THE ANSWERS, which represent the actual price per pound.
When divided into 900 cents, options B, C and D -- 185 cents, 190 cents, and 195 cents -- will yield neither an integer value nor 0.5 more than an integer value.
Option A implies an expected price of 160 cents per pound, 20 cents less than the actual price per pound (180 cents).
When divided into 900 cents, 160 will yield neither an integer value nor 0.5 more than an integer value.
The correct answer is almost certain to be E.
E: Actual price = 200 cents per pound, expected price = 180 cents per pound
Weight purchased at the actual price \(= \frac{900-cents}{200-cents-per-pound} = 4.5\) pounds
Weight that would have been purchased at the expeced price \(= \frac{900-cents}{180-cents-per-pound} = 5\) pounds
Success!
The actual weight is 0.5 pound less than the expected weight.