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If the length and width of rectangle R are each increased by 1, the ar

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If the length and width of rectangle R are each increased by 1, the ar  [#permalink]

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New post 07 Feb 2019, 07:50
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

81% (02:32) correct 19% (02:53) wrong based on 37 sessions

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Re: If the length and width of rectangle R are each increased by 1, the ar  [#permalink]

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New post 07 Feb 2019, 07:55
1
Area of R = (L+1)(W+1) = 72
WL+l+W = 71...........(1)
New area = (L-1)(W-1) = 35
WL-L-W = 34.............(2)
(1)-(2) = 2L+2W = 37 = Perimeter

B is the answer.
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Re: If the length and width of rectangle R are each increased by 1, the ar  [#permalink]

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New post 07 Feb 2019, 10:06
Bunuel wrote:
If the length and width of rectangle R are each increased by 1, the area of the new rectangle will be 72. If the length and width of rectangle R are each decreased by 1, the area of the new rectangle will be 35. What is the perimeter of rectangle R?

A. 34
B. 37
C. 48
D. 50
E. 51


(x+1)*(y+1)=72
(x-1)*(y-1)=35
solve and subtract eqn
-2(x+y)= 37
which is perimeter

IMO B
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Re: If the length and width of rectangle R are each increased by 1, the ar  [#permalink]

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New post 10 Feb 2019, 08:39
Bunuel wrote:
If the length and width of rectangle R are each increased by 1, the area of the new rectangle will be 72. If the length and width of rectangle R are each decreased by 1, the area of the new rectangle will be 35. What is the perimeter of rectangle R?

A. 34
B. 37
C. 48
D. 50
E. 51


We can create the equations:

(L + 1)(W + 1) = 72

LW + L + W + 1 = 72

LW + L + W = 71 → (Eq. 1)

and

(L - 1)(W - 1) = 35

LW - L - W + 1 = 35

LW - L - W = 34 → (Eq. 2)

Subtracting Eq. 2 from Eq. 1, we have:

2L + 2W = 37

Since 2L + 2W is the perimeter of rectangle R, the perimeter of rectangle R is 37.

Answer: B
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Re: If the length and width of rectangle R are each increased by 1, the ar   [#permalink] 10 Feb 2019, 08:39
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