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A stationery shop sold pencils for $0.70 each and erasers for $0.50 ea
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28 Jan 2024, 23:32
28 Jan 2024, 23:32
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Question Stats:
80% (02:11) correct
20% (04:57) wrong based on 5 sessions
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A stationery shop sold pencils for $0.70 each and erasers for $0.50 each. If a girl purchased both pencils and erasers from the shop for a total of $6.30, what is the total number of pencils and erasers the girl purchased? The girl purchased at least one of both the items.
(A) 9
(B) 10
(C) 11
(D) 12
(E) 13
(A) 9
(B) 10
(C) 11
(D) 12
(E) 13
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
Re: A stationery shop sold pencils for $0.70 each and erasers for $0.50 ea
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31 Jan 2024, 09:27
31 Jan 2024, 09:27
1
The answer is C.
Let the number of pencils = p
Let the number of erasers = e
--> 0.7p + 0.5e = 6.3
Let's test values for p:
When p = 1 --> 0.7*1 + 0.5e = 6.3 --> 0.5e = 5.6 (not possible, we cannot buy a part of an eraser)
When p = 2 --> 0.7*2 + 0.5e = 6.3 --> 0.5e = 4.9 (not possible)
When p = 3 --> 0.7*3 + 0.5e = 6.3 --> 0.5e = 4.2 (not possible)
When p = 4 --> 0.7*4 + 0.5e = 6.3 --> 0.5e = 3.5 (possible) --> e = 7
Hence, p + e = 4 + 7 = 11
Let the number of pencils = p
Let the number of erasers = e
--> 0.7p + 0.5e = 6.3
Let's test values for p:
When p = 1 --> 0.7*1 + 0.5e = 6.3 --> 0.5e = 5.6 (not possible, we cannot buy a part of an eraser)
When p = 2 --> 0.7*2 + 0.5e = 6.3 --> 0.5e = 4.9 (not possible)
When p = 3 --> 0.7*3 + 0.5e = 6.3 --> 0.5e = 4.2 (not possible)
When p = 4 --> 0.7*4 + 0.5e = 6.3 --> 0.5e = 3.5 (possible) --> e = 7
Hence, p + e = 4 + 7 = 11
gmatclubot
Re: A stationery shop sold pencils for $0.70 each and erasers for $0.50 ea [#permalink]
31 Jan 2024, 09:27
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