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# Oscar buys 13 pencils and 3 erasers for 1.00. A pencil costs more than

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Joined: 02 Sep 2009
Posts: 58396
Oscar buys 13 pencils and 3 erasers for 1.00. A pencil costs more than  [#permalink]

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25 Mar 2019, 01:09
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Difficulty:

25% (medium)

Question Stats:

76% (01:57) correct 24% (02:37) wrong based on 29 sessions

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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

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Re: Oscar buys 13 pencils and 3 erasers for 1.00. A pencil costs more than  [#permalink]

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25 Mar 2019, 01:26
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

given
13x+3y=100
and x>y
max value x = 7 so y = 3
x+y= 10
IMO A
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Re: Oscar buys 13 pencils and 3 erasers for 1.00. A pencil costs more than  [#permalink]

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27 Mar 2019, 18:53
1
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.

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Re: Oscar buys 13 pencils and 3 erasers for 1.00. A pencil costs more than  [#permalink]

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28 Mar 2019, 07:42
13x + 3y = 100
x>y
x+y=int

x+y=int => Var * int = 100 (Var int as well)
Var * 10 = 100; var = 10 - ok
var * 12 = 100; var = not int - not ok
var * 15 = 100; var = not int - not ok
var * 18 = 100; var = not int - not ok
var * 20 = 100; var = 5 - ok

So answer is A or E
Lets check E
If x>y than
x=3 and y = 2
or
X=4 and y=2
13*3+3*2 = 100 ? not
13*4 + 4*2 = 100 ? not

Re: Oscar buys 13 pencils and 3 erasers for 1.00. A pencil costs more than   [#permalink] 28 Mar 2019, 07:42
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