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# Carla has 1/4 more sweaters than cardi

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Math Expert
Joined: 02 Sep 2009
Posts: 52295
Carla has 1/4 more sweaters than cardi  [#permalink]

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10 Dec 2018, 00:12
00:00

Difficulty:

85% (hard)

Question Stats:

46% (02:34) correct 54% (02:20) wrong based on 57 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

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Concentration: Finance, Marketing
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Re: Carla has 1/4 more sweaters than cardi  [#permalink]

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10 Dec 2018, 00:22
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.
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Intern
Joined: 19 Jan 2018
Posts: 2
GMAT 1: 610 Q44 V30
Carla has 1/4 more sweaters than cardi  [#permalink]

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05 Jan 2019, 06:40
Don't know why this one felt hard, from the text you get:

(1) $$S = \frac{5C}{4}$$
(2) $$C = \frac{3T}{5} => T = \frac{5C}{3}$$

$$T+S = \frac{5C}{3} + \frac{5C}{4} = \frac{35C}{12}$$

Since C>0, minimum T+S is when C = 12: ans 35 (D).
Carla has 1/4 more sweaters than cardi &nbs [#permalink] 05 Jan 2019, 06:40
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