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26 Jul 2016, 09:51
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Difficulty:
15%
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Question Stats:
85% (01:21) correct
15%
(01:16)
wrong
based on 89
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A rectangle is equal in area to a square with sides of length 12. Is the diagonal of the rectangle greater in length than 20?
(1) The rectangle has a length of 16.
(2) The rectangle has a width of 9.
(1) The rectangle has a length of 16.
(2) The rectangle has a width of 9.
26 Jul 2016, 10:02
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D
Each Statement itself is sufficient.
1) lb=144
Using this we know that b= 144/16=9
So we can find length of diagonal ----> Sufficient to tell if is greater than 20
2)Similarly lb=144
Using this we know that l=144/9=16
So we can find length of diagonal ----> Sufficient to tell if is greater than 20
Each Statement itself is sufficient.
1) lb=144
Using this we know that b= 144/16=9
So we can find length of diagonal ----> Sufficient to tell if is greater than 20
2)Similarly lb=144
Using this we know that l=144/9=16
So we can find length of diagonal ----> Sufficient to tell if is greater than 20
DensetsuNo
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Updated on: 26 Jul 2016, 11:24
Originally posted by DensetsuNo on 26 Jul 2016, 10:15.
Last edited by DensetsuNo on 26 Jul 2016, 11:24, edited 3 times in total.
Last edited by DensetsuNo on 26 Jul 2016, 11:24, edited 3 times in total.
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We have a rectangle whose area has to be 144 and we are asked whether it's possible for the diagonal to be > 20.
\(a*b=144\)
and \(\sqrt{a^{2\; }+\; b^{2}}\; =\; d\)
The only thing we have to do is prove its limit.
1) If a=16 b=9, and viceversa in statement 2. *
but \(\sqrt{9^{2}\; +\; 16^{2}}\; =\; \sqrt{337}\; \; ->\; 17<\sqrt{337}<18\)
so there's no way for d to be ≥ 20.
D it is.
Thanks for the questions Bunuel. \(:)\)
*
\(\frac{144}{16}\; =\; \frac{12\cdot 12}{4\cdot 4}\; =\; \frac{4\cdot 4\cdot 3\cdot 3}{4\cdot 4}=\; 9\; <->\; \frac{144}{9}\; =\; \frac{12\cdot 12}{3\cdot 3}\; =\; \frac{4\cdot 4\cdot 3\cdot 3}{3\cdot 3}\; =\; 16\)
\(a*b=144\)
and \(\sqrt{a^{2\; }+\; b^{2}}\; =\; d\)
The only thing we have to do is prove its limit.
1) If a=16 b=9, and viceversa in statement 2. *
but \(\sqrt{9^{2}\; +\; 16^{2}}\; =\; \sqrt{337}\; \; ->\; 17<\sqrt{337}<18\)
so there's no way for d to be ≥ 20.
D it is.
Thanks for the questions Bunuel. \(:)\)
*
\(\frac{144}{16}\; =\; \frac{12\cdot 12}{4\cdot 4}\; =\; \frac{4\cdot 4\cdot 3\cdot 3}{4\cdot 4}=\; 9\; <->\; \frac{144}{9}\; =\; \frac{12\cdot 12}{3\cdot 3}\; =\; \frac{4\cdot 4\cdot 3\cdot 3}{3\cdot 3}\; =\; 16\)
