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Two PS problems..

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by moneyman » Tue May 22, 2007 8:52 pm
Pls explain the answers for these two questions..

A ladder 25 feet long is leaning against a wall that is perpendicular to level ground. The bottom of the ladder is 7 feet from the base of the wall. If the top of the ladder slips down 4 feet, how many feet will the bottom of the ladder slip?
(A) 4
(B)5
(C)8
(D)9
(E)15



If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?
(A) 100 (B) 120 (C) 140
(D) 150 (E) 160



Thanks..
Maxx
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by Cybermusings » Tue May 22, 2007 9:50 pm
Visualise this as a right angled triangle with the leaning ladder being the hypotenuese and the bottom - base of the wall distance being the base of the triangle...So the height of "triangle" would be sqr rt. of (25^2 - 7^2) = (625 - 49) = sqr rt. of 576 = 24

If the top slips down 4 feet...the length of the ladder (hence the hypotenuse) will still stay 25 feet...the height of the triangle will become 24-4 = 20 feet...hence the base of the triangle would be sqr rt. (25^2 - 20^2) = (625 - 400 ) = 15...

Hence the bottom will slip by 15 - 7 = 8 feet
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by Cybermusings » Tue May 22, 2007 9:55 pm
Distance = Speed * Time
70 = (x+5) * 1 [if x is the actual speed]
x = 65 miles / hr

If he would have driven it for 2 hrs longer @ 75 miles/ hr he would have covered 75*2 = 150 additional miles
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Thanks

by moneyman » Wed May 23, 2007 5:27 am
Thank u very much..Now i understand my silly mistakes..
Maxx
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