Bunuel wrote:
Oscar buys 13 pencils and 3 erasers for 1.00. A pencil costs more than an eraser, and both items cost a whole number of cents. What is the total cost, in cents, of one pencil and one eraser?
(A) 10
(B) 12
(C) 15
(D) 18
(E) 20
Letting p = the number of pencils purchased and e = the number of erasers purchased, we can create the equation:
13p + 3e = 100
3e = 100 - 13p
e = (100 - 13p)/3
In order for (100 - 13p) to be a multiple of 3, we need to have p = 1, 4, 7, etc.
If p = 1, then e = 87/3 = 29 (but this is not possible since a pencil costs more than an eraser).
If p = 4, then e = 48/3 = 16 (again this is not possible).
If p = 7, then e = 9/3 = 3 (this is possible).
If p = 10, then e = -30/3 = -10 (this is not possible since e can’t be negative, so we can stop here).
Thus, the total cost of one pencil and one eraser is 7 + 3 = 10 cents.
Alternate Solution:
Letting p denote the cost of a pencil (in cents) and e denote the cost of an eraser (in cents), we can create the equation:
13p + 3e = 100
If we let n be the cost of one pencil and one eraser, then p + e = n or, equivalently, 3p + 3e = 3n. Let’s subtract this equation from the equation we formed earlier:
10 p = 100 - 3n
p = 10 - 3n/10
Since the cost of a pencil must be a whole number, in cents, and since 3 is not divisible by 10, we see that n (the cost of one pencil and one eraser) must be a multiple of 10. We can eliminate all answer choices except A and E.
If n = 20, then p = 10 - (3*20)/10 = 10 - 6 = 4 cents and an eraser is 20 - 4 = 16 cents, which is not possible because we are given that the cost of a pencil is greater than the cost of an eraser.
We know at this point that the answer is A, but let’s verify:
If n = 10, then p = 10 - (3*10)/10 = 10 - 3 = 7 and e = 10 - 7 = 3. So, the cost of one pencil is 7 cents, and the cost of one eraser is 3 cents, which agrees with the information given in the question.
Answer: A
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