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Updated on: 05 Jul 2019, 04:58
Originally posted by Bunuel on 17 Jun 2017, 06:46.
Last edited by Sajjad1994 on 05 Jul 2019, 04:58, edited 1 time in total.
Last edited by Sajjad1994 on 05 Jul 2019, 04:58, edited 1 time in total.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
85% (00:55) correct
15%
(01:03)
wrong
based on 48
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In the figure to the right, the length of QS is
(A) \(\sqrt{51}\)
(B) \(\sqrt{61}\)
(C) \(\sqrt{69}\)
(D) \(\sqrt{77}\)
(E) \(\sqrt{89}\)
2017-06-17_1744.png [ 10.38 KiB | Viewed 3845 times ]
(A) \(\sqrt{51}\)
(B) \(\sqrt{61}\)
(C) \(\sqrt{69}\)
(D) \(\sqrt{77}\)
(E) \(\sqrt{89}\)
Attachment:
2017-06-17_1744.png [ 10.38 KiB | Viewed 3845 times ]
Source: Nova GMAT
Difficulty Level: 550
Difficulty Level: 550
17 Jun 2017, 06:54
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IMO answer is B.
QS= root ( 6^2 + 5^2)= root ( 61)
QS= root ( 6^2 + 5^2)= root ( 61)
17 Jun 2017, 06:59
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Since PR = PQ + QR
PR = 5+3 = 8
Since the hypotenuse of the right angled triangle is 10.
The other two sides are 6 and 8(6,8,10 is a Pythagorean triplet)
Therefore, PS = 6
Since PS = 6 and PQ = 5
Using pythogoras theorem, \(QS^2 = PS^2 + PQ^2\)
\(QS^2 = 25 + 36\)
QS = \(\sqrt{61}\) (Option B)
PR = 5+3 = 8
Since the hypotenuse of the right angled triangle is 10.
The other two sides are 6 and 8(6,8,10 is a Pythagorean triplet)
Therefore, PS = 6
Since PS = 6 and PQ = 5
Using pythogoras theorem, \(QS^2 = PS^2 + PQ^2\)
\(QS^2 = 25 + 36\)
QS = \(\sqrt{61}\) (Option B)
