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Two rectangles with integer sides, C and D, have the same area but dif

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Two rectangles with integer sides, C and D, have the same area but dif  [#permalink]

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New post 04 Sep 2018, 23:28
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Two rectangles with integer sides, C and D, have the same area but different perimeters. If the ratio of the perimeters of C and D is 8 to 7 and the area of both is 48, what is the perimeter of C?


A. 24
B. 28
C. 32
D. 44
E. 52

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Re: Two rectangles with integer sides, C and D, have the same area but dif  [#permalink]

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New post 05 Sep 2018, 00:44
Bunuel wrote:
Two rectangles with integer sides, C and D, have the same area but different perimeters. If the ratio of the perimeters of C and D is 8 to 7 and the area of both is 48, what is the perimeter of C?


A. 24
B. 28
C. 32
D. 44
E. 52



C(l,w): (4,12)
D(l,w): (6, 8)

Perimeter of rectangle C=2(4+12)=32

Ans. (C)
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Two rectangles with integer sides, C and D, have the same area but dif  [#permalink]

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New post Updated on: 06 Sep 2018, 11:49
Bunuel wrote:
Two rectangles with integer sides, C and D, have the same area but different perimeters. If the ratio of the perimeters of C and D is 8 to 7 and the area of both is 48, what is the perimeter of C?


A. 24
B. 28
C. 32
D. 44
E. 52


ab=cd=48
factors of 48 are
1*48
2*24
3*16
4*12
6*8
also,

2(a+b)/2(c+d) = 8/7
a+b/c+d = 8/7
a+b can be 8,16,24...
c+d can be 7,14,21...

from the list of factors
4*12 and 6*8 fit the bill for C and D respectively

perimeter of C = 2(16) = 32

Originally posted by CounterSniper on 05 Sep 2018, 00:46.
Last edited by CounterSniper on 06 Sep 2018, 11:49, edited 1 time in total.
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Re: Two rectangles with integer sides, C and D, have the same area but dif  [#permalink]

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New post 05 Sep 2018, 03:23
CounterSniper wrote:
Bunuel wrote:
Two rectangles with integer sides, C and D, have the same area but different perimeters. If the ratio of the perimeters of C and D is 8 to 7 and the area of both is 48, what is the perimeter of C?


A. 24
B. 28
C. 32
D. 44
E. 52


ab=cd=48
factors of 48 are
1*48
2*24
3*16
4*12
6*8
also,

2(a+b)/2(c+d) = 8/7
a+b/c+d = 8/7
a+b can be 8,16,24...
c+d can be 7,14,21...

from the list of factors
4*12 and 6*8 fits the bill for C and D respectively

perimeter of C = 2(16) = 32


Can you walk me through how you know that those factors 'fit the bill'?
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Two rectangles with integer sides, C and D, have the same area but dif  [#permalink]

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New post 05 Sep 2018, 03:39
1
memyselfi wrote:
CounterSniper wrote:
Bunuel wrote:
Two rectangles with integer sides, C and D, have the same area but different perimeters. If the ratio of the perimeters of C and D is 8 to 7 and the area of both is 48, what is the perimeter of C?


A. 24
B. 28
C. 32
D. 44
E. 52


ab=cd=48
factors of 48 are
1*48
2*24
3*16
4*12
6*8
also,

2(a+b)/2(c+d) = 8/7
a+b/c+d = 8/7
a+b can be 8,16,24...
c+d can be 7,14,21...

from the list of factors
4*12 and 6*8 fits the bill for C and D respectively

perimeter of C = 2(16) = 32


Can you walk me through how you know that those factors 'fit the bill'?


let a+b be 8x and c+d be 7x
when we substitute x= 1
we have a+b=8
can we find a particular value of a and of b where a*b=48 ?
for finding this value , we need list of factors of 48 (why ? because the sides are integers)
we can not find a set of values of a and b

similarly when x = 2
we have a+b=16
and a*b =4*12 "fits the bill"

perimeter = 2(4+12)=32

hope it makes sense !!
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Re: Two rectangles with integer sides, C and D, have the same area but dif  [#permalink]

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New post 05 Sep 2018, 05:38
IMO C.

The area of both C and D is 48. Possible configurations are 12*4 and 6*8. The ratio between perimeters of C and D is 8:7. Clearly C has a greater perimeter than D. And thus calculation perimeters for C and D from the possible configurations = 2(12+4) & 2(8+6) = 32:28 = 8:7.
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Re: Two rectangles with integer sides, C and D, have the same area but dif &nbs [#permalink] 05 Sep 2018, 05:38
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Two rectangles with integer sides, C and D, have the same area but dif

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