FAST APPROACH NEEDED...

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FAST APPROACH NEEDED...

by Fab » Sun Nov 23, 2008 3:07 pm
Q31:
A solid yellow stripe is to be painted in the middle of a certain highway. If 1 gallon of paint covers an area of p square feet of highway, how many gallons of paint will be needed to paint a stripe t inches wide on a stretch of highway m miles long? (1 mile = 5,280 feet and 1 foot = 12 inches)

A. (5,280 mt) / 12p
B. (5,280 pt) / 12m
C. (5,280 pmt) / 12
D. (5,280)(12m) / pt
E. (5,280)(12p) / mt
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Q32:
Last year the price per share of Stock X increased by k percent and the earnings per share of Stock X increased by m percent, where k is greater than m. By what percent did the ratio of price per share to earnings per share increase, in terms of k and m?

A. k/m%
B. (k – m) %
C. [100(k – m)] / (100 + k) %
D. [100(k – m)] / (100 + m) %
E. [100(k – m)] / (100 + k + m) %
Source: — Problem Solving |

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by earth@work » Sun Nov 23, 2008 3:21 pm
Q31.....IMO A
length=5280*m ft; width=t/12ft ; area=5280*m*t/12 sq ft
p sq ft------1 gallon
5280*m*t/12 sq ft-------(1/p)*5280*m*t/12
=A. (5,280 mt) / 12p
this is the fastest i cud think of!

Q32.....IMO D
let P be original price & E original earning
inc. price= P(1+k/100); increased earning=E(1+m/100)
Ratio original price to earning=P/E
Ration of increased price to inc. earning =(P(1+k/100))/(E(1+m/100))
difference =(P(1+k/100))/(E(1+m/100))-P/K=(P/K)*((k-m)/100+m))
Percent increase = (P/K)*((k-m)/100+m))*100/(P/K)
D. [100(k – m)] / (100 + m) %
i think there shud be a more simpler/faster method for this