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A 25 feet long ladder is placed against the wall with its base 7 feet
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Updated on: 13 Dec 2018, 06:11
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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
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Originally posted by Rohit23 on 13 Dec 2018, 06:09.
Last edited by Bunuel on 13 Dec 2018, 06:11, edited 1 time in total.
Renamed the topic.



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A 25 feet long ladder is placed against the wall with its base 7 feet
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13 Dec 2018, 06:26
Rohit23 wrote: 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. The distance that the base is drawn out is in the range:
a) (2 , 7) b) (5 , 8) c) (9 , 10) d) (3 ,6) e) None of these
\(?\,\,:\,\,\,2x\,\,{\rm{in}}\,\,{\rm{which}}\,\,{\rm{range}}\) \(x + y = \sqrt {{{25}^2}  {7^2}} = \sqrt {\left( {25  7} \right)\left( {25 + 7} \right)} = 24\,\,\,\,\, \Rightarrow \,\,\,\,\,y = 24  x\) \({\left( {24  x} \right)^2} + {\left( {2x + 7} \right)^2} = {25^2}\,\,\,\,\, \Rightarrow \,\,\,\,\, \ldots \,\,\,\,\, \Rightarrow \,\,\,\,\,\,5x\left( {x  4} \right) = 0\,\,\,\,\,\mathop \Rightarrow \limits^{x\, > \,0} \,\,\,\,\,2x = 8\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{E}} \right)\) This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
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Re: A 25 feet long ladder is placed against the wall with its base 7 feet
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14 Dec 2018, 09:52
Rohit23 wrote: 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 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 \(= 24  x\) \((7 + 2x)^2 + (24x)^2 = 25^2\) \(4x^2 + 28x + 49 + x^2  48x + 576 = 625\) \(5x^2  20x + 625 = 625\) \(5x^2  20x = 625  625 => 5x^2 20x = 0\) \(x^2  4x = 0\) \(x(x4) = 0\) \(x = 0\) 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



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A 25 feet long ladder is placed against the wall with its base 7 feet
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18 Dec 2018, 04:55
Is the question asking for the drawn out distance which is 8, which is included in the range of answer b (5,8) or the new base distance which is 7+8? The question is not clear Bunuel



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Re: A 25 feet long ladder is placed against the wall with its base 7 feet
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18 Dec 2018, 06:18
Alamehchadi wrote: Is the question asking for the drawn out distance which is 8, which is included in the range of answer b (5,8) or the new base distance which is 7+8? The question is not clear Hi Alamehchadi , Please see my solution above, where the question stem was slightly modified to avoid the confusion to what has been asked. The solution (8) is NOT included in the range (5,8) , from the fact that (5,8) is the set of real numbers x such that 5 < x < 8. Regards, Fabio.
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Re: A 25 feet long ladder is placed against the wall with its base 7 feet
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18 Dec 2018, 06:26
fskilnik wrote: Alamehchadi wrote: Is the question asking for the drawn out distance which is 8, which is included in the range of answer b (5,8) or the new base distance which is 7+8? The question is
Please see my solution above, where the question stem was slightly modified to avoid the confusion to what has been asked.
The solution (8) is NOT included in the range (5,8) , from the fact that (5,8) is the set of real numbers x such that 5 < x < 8.
Regards, Fabio. Thanks for your reply fskilnikIf the range (5,8) is 5<x<8. Can you please tell me how the range including 5 and 8 is written? Is it [5,8] or something else? Posted from my mobile device



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Re: A 25 feet long ladder is placed against the wall with its base 7 feet
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18 Dec 2018, 06:30
Alamehchadi wrote: fskilnik wrote: Alamehchadi wrote: Is the question asking for the drawn out distance which is 8, which is included in the range of answer b (5,8) or the new base distance which is 7+8? The question is
Please see my solution above, where the question stem was slightly modified to avoid the confusion to what has been asked.
The solution (8) is NOT included in the range (5,8) , from the fact that (5,8) is the set of real numbers x such that 5 < x < 8.
Regards, Fabio. Thanks for your reply fskilnikIf the range (5,8) is 5<x<8. Can you please tell me how the range including 5 and 8 is written? Is it [5,8] or something else? Posted from my mobile deviceHi, Alamehchadi ! Thanks for the kudos and for the confidence in my answer. Yes, [5,8] is the usual notation for the set of real numbers x such that 5 <= x <= 8 . Another common notation for (5,8) is ]5,8[ , as you probably have already seen/used when younger. Regards and success in your studies, Fabio. P.S.: it is possible to have (say) (5,8] , meaning 5 < x < = 8 , etc.
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Re: A 25 feet long ladder is placed against the wall with its base 7 feet
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19 Dec 2018, 05:12
Using Pythagoras the height of the wall is (252 72)= 24ft. Now assume that the base is out by “x” ft. So, the height should come down by “x/2” ft. Using Pythagoras : 252 (24x/2)2= (7+x)2 Solving this, we will get x= 8ft. Hence, E.




Re: A 25 feet long ladder is placed against the wall with its base 7 feet
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