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10 Dec 2018, 01:12
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D
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Difficulty:
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70% (02:36) correct
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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
19 Aug 2019, 01:36
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Bunuel
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
Given:
1. Carla has \(\frac{1}{4}\) more sweaters than cardigans.
2. Carla has \(\frac{2}{5}\) fewer cardigans than turtle necks.
3. She has at least one of each item.
Asked: What is the minimum total number of turtlenecks plus sweaters that Carla could have?
Let number of sweaters, cardigans and turtle necks be s, c & t respectively
1. Carla has \(\frac{1}{4}\) more sweaters than cardigans.
s:c = 5:4
2. Carla has \(\frac{2}{5}\) fewer cardigans than turtle necks.
c:t = 3:5
s:c:t = 15:12:20
Minimum total number of turtlenecks plus sweaters that Carla could have = 15 + 20 = 35
IMO D
General Discussion
10 Dec 2018, 01:22
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Sweaters = Cardigans + 1/4*Cardigans
Sweaters = 4/5*Cardigans
Turtleneck = Cardigans - 2/5*Cardigans
Turtleneck = 3/5*Cardigans
Turtleneck + Sweaters = 7/5*Cardigans
Among the answer choices, Only 35 is a multiple of 7.
Hence, D is the answer.
Sweaters = 4/5*Cardigans
Turtleneck = Cardigans - 2/5*Cardigans
Turtleneck = 3/5*Cardigans
Turtleneck + Sweaters = 7/5*Cardigans
Among the answer choices, Only 35 is a multiple of 7.
Hence, D is the answer.
