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The width of rectangle T is 25 percent greater than the width of recta
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Bunuel
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The width of rectangle T is 25 percent greater than the width of recta [#permalink] New post  23 Jun 2022, 01:30
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A
B
C
D
E

Difficulty:

25% (medium)

Question Stats:

83% (01:16) correct 17% (02:15) wrong based on 12 sessions

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The width of rectangle T is 25 percent greater than the width of rectangle R, and the length of T is 10 percent greater than the length of R. The area of T is what percent greater than the area of R ?

A. 2.5%
B. 15%
C. 27.3%
D. 35%
E. 37.5%
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Kinshook
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Re: The width of rectangle T is 25 percent greater than the width of recta [#permalink] New post  23 Jun 2022, 02:00
Given: The width of rectangle T is 25 percent greater than the width of rectangle R, and the length of T is 10 percent greater than the length of R.

Asked: The area of T is what percent greater than the area of R ?

W'= 1.25W
L'= 1.1L

Area of rectangle T = W'L'= 1.25W*1.1L = 1.375* Area of rectangle R = (100%+37.5%) * Area of rectangel R

IMO E
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Re: The width of rectangle T is 25 percent greater than the width of recta [#permalink]
23 Jun 2022, 02:00
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