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GRE Prep Club Team Member
Joined: 20 Feb 2017
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Carla has 1/4 more sweaters than cardigans, and 2/5 fewer cardigans
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06 May 2021, 00:19
06 May 2021, 00:19
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Question Stats:
40% (02:27) correct
60% (02:08) wrong based on 15 sessions
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Carla has \(\frac{1}{4}\) more sweaters than cardigans, and \(\frac{2}{5}\) fewer cardigans than turtle necks. If she has at least one of each item, what is the minimum total number of turtlenecks plus sweaters that Carla could have?
A. 10
B. 15
C. 20
D. 35
E. 45
A. 10
B. 15
C. 20
D. 35
E. 45
Re: Carla has 1/4 more sweaters than cardigans, and 2/5 fewer cardigans
[#permalink]
06 May 2021, 08:43
06 May 2021, 08:43
1
Let the turtlenecks = \(x\)
\(\frac{2}{5}\) fewer cardigans than turtle necks
cardigans = \(\frac{3x}{5}\)
\(\frac{1}{4}\) more sweaters than cardigans
sweaters = \(\frac{5}{4}\) * \(\frac{3x}{5}\) = \(\frac{3x}{4}\)
Now sweater + turtlenecks = \(x\) + \(\frac{3x}{4}\)
= \(\frac{7x}{4}\)
Now count can't be fraction so answer will be multiple of 7. Hence 35
Answer D
\(\frac{2}{5}\) fewer cardigans than turtle necks
cardigans = \(\frac{3x}{5}\)
\(\frac{1}{4}\) more sweaters than cardigans
sweaters = \(\frac{5}{4}\) * \(\frac{3x}{5}\) = \(\frac{3x}{4}\)
Now sweater + turtlenecks = \(x\) + \(\frac{3x}{4}\)
= \(\frac{7x}{4}\)
Now count can't be fraction so answer will be multiple of 7. Hence 35
Answer D
Re: Carla has 1/4 more sweaters than cardigans, and 2/5 fewer cardigans
[#permalink]
21 Sep 2022, 19:25
21 Sep 2022, 19:25
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