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A square and an equilateral triangle have perimeters S and T respectiv

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Joined: 02 Sep 2009
Posts: 58402
A square and an equilateral triangle have perimeters S and T respectiv  [#permalink]

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27 Sep 2017, 04:51
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15% (low)

Question Stats:

88% (01:41) correct 12% (01:45) wrong based on 44 sessions

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A square and an equilateral triangle have perimeters S and T respectively. If s and t are the respective lengths of a side of the square and a side of the triangle, then, in terms of their perimeters, s – t =

(A) (S – T)/7
(B) (4T – 3S)/7
(C) (3S – 4T)/7
(D) (4T – 3S)/12
(E) (3S – 4T)/12

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Joined: 29 Aug 2017
Posts: 35
A square and an equilateral triangle have perimeters S and T respectiv  [#permalink]

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27 Sep 2017, 06:07
1
Bunuel wrote:
A square and an equilateral triangle have perimeters S and T respectively. If s and t are the respective lengths of a side of the square and a side of the triangle, then, in terms of their perimeters, s – t =

(A) (S – T)/7
(B) (4T – 3S)/7
(C) (3S – 4T)/7
(D) (4T – 3S)/12
(E) (3S – 4T)/12

Perimeter of square = S
Side of square = s = S/4
Perimeter of equilateral triangle = T
Side of equilateral triangle = t = T/3
s - t $$= \frac{S}{4}- \frac{T}{3} = \frac{(3S - 4T)}{12}$$

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Posts: 2816
Re: A square and an equilateral triangle have perimeters S and T respectiv  [#permalink]

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29 Sep 2017, 11:21
Bunuel wrote:
A square and an equilateral triangle have perimeters S and T respectively. If s and t are the respective lengths of a side of the square and a side of the triangle, then, in terms of their perimeters, s – t =

(A) (S – T)/7
(B) (4T – 3S)/7
(C) (3S – 4T)/7
(D) (4T – 3S)/12
(E) (3S – 4T)/12

Since the square has a perimeter of S, its side length s = S/4. Similarly, since the equilateral triangle has a perimeter of T, its side length t = T/3. Thus, s - t = S/4 - T/3 = 3S/12 - 4T/12 = (3S - 4T)/12.

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Re: A square and an equilateral triangle have perimeters S and T respectiv  [#permalink]

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15 Jan 2019, 08:54
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Re: A square and an equilateral triangle have perimeters S and T respectiv   [#permalink] 15 Jan 2019, 08:54
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