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avohden
Joined: 09 Jul 2013
Last visit: 14 Mar 2015
Posts: 403
Given Kudos: 630
Status:1,750 Q's attempted and counting
Affiliations: University of Florida
Location: United States (FL)
Schools: Warrington (A)
GMAT 1: 570 Q42 V28
GMAT 2: 610 Q44 V30
GMAT 3: 600 Q45 V29
GMAT 4: 590 Q35 V35
GPA: 3.45
WE:Accounting (Accounting)
24 Oct 2013, 09:57
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
75% (01:57) correct
25%
(01:59)
wrong
based on 388
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A certain store sold pens for $0.35 each and pencils for $0.25 each. If a customer purchased both pens and pencils from the store for a total of $2.50, what total number of pens and pencils did the customer purchase?
(A) 8
(B) 9
(C) 10
(D) 11
(E) 12
GH-04.05.13 - OE to follow
(A) 8
(B) 9
(C) 10
(D) 11
(E) 12
GH-04.05.13 - OE to follow
24 Oct 2013, 10:04
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avohden
\(0.35x+25y=2.5\);
\(35x+25y=250\);
\(7x+5y=50\).
Since x and y must be positive integers, then only one solution exists: x=5 and y=3 --> x+y=8.
Answer: A.
avohden
Joined: 09 Jul 2013
Last visit: 14 Mar 2015
Posts: 403
Given Kudos: 630
Status:1,750 Q's attempted and counting
Affiliations: University of Florida
Location: United States (FL)
Schools: Warrington (A)
GMAT 1: 570 Q42 V28
GMAT 2: 610 Q44 V30
GMAT 3: 600 Q45 V29
GMAT 4: 590 Q35 V35
GPA: 3.45
WE:Accounting (Accounting)
27 Oct 2013, 10:41
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Official Explanation
Answer: A Algebraically, the question looks like this: 2.5 = 0.35x + 0.25y
However, that's the only equation we get, so we can't solve for x and y this way. There would be an infinite number of possible values of x and y. Instead, we must use an intelligent version of guessing and checking. Since at least one pencil and one pen is purchased, the customer must spend at least $2.25 ($2.50 minus $0.25) on pens. There are six possible numbers of pens that fit that requirement:
1 pen: $0.35
2 pens: $0.70
3 pens: $1.05
4 pens: $1.40
5 pens: $1.75
6 pens: $2.10
Many of those aren't applicable here, though. For instance, if the customer bought 6 pens, he would need to spend an additional $0.40 on pencils. At $0.25 per pencil, that's impossible. Consider each of the other 5:
1 pen: $2.15 on pencils
2 pens: $1.80 on pencils
3 pens: $1.45 on pencils
4 pens: $1.10 on pencils
5 pens: $0.75 on pencils
Only the last of these works. $0.75 is 3 pencils for $0.25 each. That's 5 pens and 3 pencils for a total of 8 pens and pencils. Choice (A) is correct.
Answer: A Algebraically, the question looks like this: 2.5 = 0.35x + 0.25y
However, that's the only equation we get, so we can't solve for x and y this way. There would be an infinite number of possible values of x and y. Instead, we must use an intelligent version of guessing and checking. Since at least one pencil and one pen is purchased, the customer must spend at least $2.25 ($2.50 minus $0.25) on pens. There are six possible numbers of pens that fit that requirement:
1 pen: $0.35
2 pens: $0.70
3 pens: $1.05
4 pens: $1.40
5 pens: $1.75
6 pens: $2.10
Many of those aren't applicable here, though. For instance, if the customer bought 6 pens, he would need to spend an additional $0.40 on pencils. At $0.25 per pencil, that's impossible. Consider each of the other 5:
1 pen: $2.15 on pencils
2 pens: $1.80 on pencils
3 pens: $1.45 on pencils
4 pens: $1.10 on pencils
5 pens: $0.75 on pencils
Only the last of these works. $0.75 is 3 pencils for $0.25 each. That's 5 pens and 3 pencils for a total of 8 pens and pencils. Choice (A) is correct.
