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02 Jul 2016, 03:58
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
61% (01:31) correct
39%
(01:42)
wrong
based on 87
sessions
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Is the area of the rectangle more than the area of the square?
(1) The side of the square is half the sum of the length and the breadth of the rectangle.
(2) The length and breadth of the rectangle are 15 cm and 10 cm respectively.
(1) The side of the square is half the sum of the length and the breadth of the rectangle.
(2) The length and breadth of the rectangle are 15 cm and 10 cm respectively.
shashanksagar
GMAT 1: 720 Q50 V37
Posts: 62
Updated on: 02 Jul 2016, 05:39
Originally posted by shashanksagar on 02 Jul 2016, 05:01.
Last edited by shashanksagar on 02 Jul 2016, 05:39, edited 1 time in total.
Last edited by shashanksagar on 02 Jul 2016, 05:39, edited 1 time in total.
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IMO A..
a = (l+b)/2
Squaring both sides..
a^2 = ( l^2 + b^2 )/2 + lb
So area of square = some positive number + area of rectangle
So area of square > area of rectangle
P.s. : there was a mistake in my solution. . The denominator would be 4 not 2.
Sent from my SM-N910H using Tapatalk
a = (l+b)/2
Squaring both sides..
a^2 = ( l^2 + b^2 )/2 + lb
So area of square = some positive number + area of rectangle
So area of square > area of rectangle
P.s. : there was a mistake in my solution. . The denominator would be 4 not 2.
Sent from my SM-N910H using Tapatalk
02 Jul 2016, 05:02
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Is the area of the rectangle more than the area of the square?
Say the length of a side of the square is x and the lengths of the length and breadth of the rectangle are a and b. So, the question is whether ab > x^2.
(1) The side of the square is half the sum of the length and the breadth of the rectangle --> x = (a+b)/2. The question becomes whether ab > ((a+b)/2)^2 --> is 4ab > a^2 + 2ab + b^2 --> is 0 > (a - b)^2. Since the square of a number cannot be negative, then the answer to this question is NO. Sufficient.
(2) The length and breadth of the rectangle are 15 cm and 10 cm respectively. We don't know x. Not sufficient.
Answer: A.
Hope it's clear.
Say the length of a side of the square is x and the lengths of the length and breadth of the rectangle are a and b. So, the question is whether ab > x^2.
(1) The side of the square is half the sum of the length and the breadth of the rectangle --> x = (a+b)/2. The question becomes whether ab > ((a+b)/2)^2 --> is 4ab > a^2 + 2ab + b^2 --> is 0 > (a - b)^2. Since the square of a number cannot be negative, then the answer to this question is NO. Sufficient.
(2) The length and breadth of the rectangle are 15 cm and 10 cm respectively. We don't know x. Not sufficient.
Answer: A.
Hope it's clear.
