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Learn how Kamakshi achieved a GMAT 675 with an impressive 96th %ile in Data Insights. Discover the unique methods and exam strategies that helped her excel in DI along with other sections for a balanced and high score.
10 Nov 2022, 11:41
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
42% (02:17) correct
58%
(01:37)
wrong
based on 33
sessions
History
Date
Time
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Is the area of rectangle A greater than the area of rectangle B?
(1) The perimeter of rectangle A is greater than the perimeter of rectangle B.
(2) The diagonal of rectangle A is greater than the diagonal of rectangle B.
(1) The perimeter of rectangle A is greater than the perimeter of rectangle B.
(2) The diagonal of rectangle A is greater than the diagonal of rectangle B.
10 Nov 2022, 22:51
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- Length =\( l_1\)
- Breadth=\( b_1\)
Rectangle B
- Length =\( l_2\)
- Breadth=\( b_2\)
As the variables represent length, all the variables should be greater than 0.
Question
\(l_1*b_1 > l_2*b_2\)
Statement 1
\(2(l_1+b_1) > 2 (l_2+b_2)\)
Dividing both sides by 2, we have
\(l_1 + b_1 > l_2 + b_2\)
\(l_1 - l_2 > b_2 - b_1\)
Case 1:
\(l_1 = 3; l_2 = 1; b_2 = 3; b_1 = 2\)
\(l_1*b_1 > l_2*b_2\) ? -- Yes
Case 2:
\(l_1 = \frac{1}{2}; l_2 = 2; b_2 = 2; b_1 = 4\)
\(l_1*b_1 > l_2*b_2\) ? -- No
As we have both Yes and No. We can eliminate Statement 1.
Statement 2
\(\sqrt{(l_1^2+b_1^2)} > \sqrt{l_2^2+b_2^2}\)
Squaring both sides
\((l_1^2+b_1^2) > (l_2^2+b_2^2)\)
\((l_1^2 - l_2^2) > (b_2^2+b_1^2)\)
\((l_1 + l_2)(l_1 - l_2) > (b_2 + b_1)(b_2 - b_1)\)
Let's see if the same set of examples in 1, work in here.
Case 1:
\(l_1 = 3; l_2 = 1; b_2 = 3; b_1 = 2\)
\(l_1*b_1 > l_2*b_2\) ? -- Yes
Case 2:
\(l_1 = \frac{1}{2}; l_2 = 2; b_2 = 2; b_1 = 4\)
\(l_1*b_1 > l_2*b_2\) ? -- No
As we have both Yes and No. We can eliminate Statement 2.
Combined
Combined the statements won't help, as the same set of examples can be used.
Option E
10 Nov 2022, 22:57
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Bunuel
Same perimeter but with sides closer to each other or identical sides will have more area.
This is the reason why a square will have more area than a rectangle (with different values of length and breadth) with similar perimeter.
There are so many unknowns that the two sentences combined will also not give you an answer.
(1) The perimeter of rectangle A is greater than the perimeter of rectangle B.
Case I. Let the perimeter of A be 12 with sides 5 and 1. Area= 5 square units
Case II. a)Let the perimeter of B be 10 with sides 3 and 2. Area= 6 square units > 5
b) But sides 4 and 1 will give 4 square units <5
Insufficient
(2) The diagonal of rectangle A is greater than the diagonal of rectangle B.
We can take similar rectangles as in statement I
Case I. Diagonal of A = \(\sqrt{26}\)
Case II. a) Diagonal of B = \( \sqrt{13}\) b) \(\sqrt{17}\)
We know in above cases, we get different possibilities
Insufficient
Combined.
Above cases still remain
Insufficient
