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Rohit23
Joined: 13 Dec 2018
Last visit: 12 Apr 2019
Posts: 36
Given Kudos: 1
Location: India
Schools: Stanford '21 HBS '21 Melbourne '21
GPA: 3.94
E
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Difficulty:
95%
(hard)
Question Stats:
32% (02:24) correct
68%
(02:55)
wrong
based on 74
sessions
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A 25 feet long ladder is placed against the wall with its base 7 feet from the wall. The base of the ladder is drawn out so that the top comes down by half the distance that the base is drawn out. This distance is in the range:
a) (2 , 7)
b) (5 , 8)
c) (9 , 10)
d) (3 ,6)
e) None of these
a) (2 , 7)
b) (5 , 8)
c) (9 , 10)
d) (3 ,6)
e) None of these
13 Dec 2018, 06:26
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Rohit23
\(?\,\,:\,\,\,2x\,\,{\rm{in}}\,\,{\rm{which}}\,\,{\rm{range}}\)
\(x + y = \sqrt {{{25}^2} - {7^2}} = \sqrt {\left( {25 - 7} \right)\left( {25 + 7} \right)} = 24\,\,\,\,\, \Rightarrow \,\,\,\,\,y = 24 - x\)
\({\left( {24 - x} \right)^2} + {\left( {2x + 7} \right)^2} = {25^2}\,\,\,\,\, \Rightarrow \,\,\,\,\, \ldots \,\,\,\,\, \Rightarrow \,\,\,\,\,\,5x\left( {x - 4} \right) = 0\,\,\,\,\,\mathop \Rightarrow \limits^{x\, > \,0} \,\,\,\,\,2x = 8\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{E}} \right)\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
sashiim20
Joined: 04 Dec 2015
Last visit: 05 Jun 2024
Posts: 608
Given Kudos: 276
Location: India
Concentration: Technology, Strategy
Schools: ISB '19 IIMB IIMA XLRI HEC Sept19 intake Rotman '21 NUS '21 NTU '20 Trinity MBA '20 Bocconi '21 Smurfit "21
WE:Information Technology (Consulting)
Schools: ISB '19 IIMB IIMA XLRI HEC Sept19 intake Rotman '21 NUS '21 NTU '20 Trinity MBA '20 Bocconi '21 Smurfit "21
Posts: 608
14 Dec 2018, 09:52
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Rohit23
Given base of triangle is \(7\) and hypotenuse \(25\).
We can find the third side of the triangle. Let the third side be \(a\).
\(25^2 = a^2 + 7 ^2\)
\(625 = a^2 + 49\)
\(a^2 = 625 - 49 = 576\)
\(a = 24\)
Given the base of the ladder is drawn out so that the top comes down by half the distance that the base is drawn out.
Let the top distance drawn out be \(x\) ; then the base distance drawn out is \(2x\).
Hypotenuse is the same \(= 25\)
New base \(= 7 + 2x\)
New value of other side \(= 24 - x\)
\((7 + 2x)^2 + (24-x)^2 = 25^2\)
\(4x^2 + 28x + 49 + x^2 - 48x + 576 = 625\)
\(5x^2 - 20x + 625 = 625\)
\(5x^2 - 20x = 625 - 625 => 5x^2- 20x = 0\)
\(x^2 - 4x = 0\)
\(x(x-4) = 0\)
\(x = 0\) or \(x = 4\)
\(x\) cannot be \(0\) because its given the ladder is drawn out hence has moved some distance. Therefore \(x = 4\)
\(x = 4 ; 2x = 8\)
Sides of new triangle will be;
\(7 + 2x = 7 + 8 = 15\)
\(24 - x = 24 - 4 = 20\)
Answer E
