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Carla has 1/4 more sweaters than cardigans, and 2/5 fewer cardigans

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Math Expert
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Carla has 1/4 more sweaters than cardigans, and 2/5 fewer cardigans  [#permalink]

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New post 10 Dec 2018, 01:12
1
8
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

60% (02:40) correct 40% (02:24) wrong based on 102 sessions

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Re: Carla has 1/4 more sweaters than cardigans, and 2/5 fewer cardigans  [#permalink]

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New post 10 Dec 2018, 01:22
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.
Hence, D is the answer.
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Re: Carla has 1/4 more sweaters than cardigans, and 2/5 fewer cardigans  [#permalink]

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New post 05 Jan 2019, 07:40
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).
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Re: Carla has 1/4 more sweaters than cardigans, and 2/5 fewer cardigans  [#permalink]

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New post 08 Jun 2019, 02:47
Bunuel wrote:
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


Let those types be x and y and z, x+y+z>1, Integers and we have to find x+z

x= y(5/4)

y=z(3/5)

Now we have to make sure all x y and z are Integers and minimum value possible

let z=20, y=12
Then x= 15

x+z Minima= 35
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Re: Carla has 1/4 more sweaters than cardigans, and 2/5 fewer cardigans  [#permalink]

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New post 19 Aug 2019, 01:36
1
Bunuel wrote:
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


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
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Re: Carla has 1/4 more sweaters than cardigans, and 2/5 fewer cardigans   [#permalink] 19 Aug 2019, 01:36
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