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# GMATPREP QUANT ?

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Intern
Joined: 25 Apr 2006
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27 Jun 2006, 03:35
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Guys, need your help on the following questions:

1) Six machines working together at constant rate can complete the job in 12 hours. How many machines are needed to complete the same job in 8 hours?

- 2, -3, - 4, -6, -8

OE is 3, could somebody explain this.

2) If 2^x-2^x-2=3(2^13) what is x

- 9, 11, 13, 15, 17

3) a wire of 40 meters is cut in two. one piece is used to form a circle with radius r, the other piece is used for a square. no wire is left. what is the total area in m2 of the circular and the square region in terms of r

A) pr2
b) pr2 +10
c) pr2 +0.25p2r2
d) pr2 +(40-2pr)^2
e) pr2 + (10-0.5pr)^2 (OE)

THANKS MUCH FOR CLARIFICATIONS
Manager
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27 Jun 2006, 04:03
1) Work = Rate * Time

Rate = #machines*rate of each

Let reat of each machine be X and Y be the # of machines required to do the job in 8 days.

Work = 6X * 12---->72X
Work = XY*8---->8XY

8XY = 72X
Y = 9

Ans 9-6 = 3

2) Total Area = Area of circle + Area of Square

= 2pir + (40 - 2pir)^2 ---> since 2pir out of 40 was used to make the circle, the square region has a side of 40 - 2pir
Manager
Joined: 12 Apr 2006
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27 Jun 2006, 05:29
Answer 1: Rate of each machine = 1/12 * 1/6 = 1/72

so machines required are 1/72 = x/8 => x = 9 means total of 3 extra.

2^x(3/4) = 3(2^13) => 2^x = 2^15 Hence the value of x is 15.

Answer 3: Length of wire used to make circle is 2Pr
Length of wire used to make square is 40-2Pr implies each side of square = 10-0.5Pr

Combined area = area of circle + area of square
Pr^2 + (10-0.5Pr)^2
Manager
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27 Jun 2006, 19:05
By the way, did the original formulation of the first question asked "How many ADDITIONAL (or MORE) machines are needed to complete the same job in 8 hours?" Because if it did not, the answer should be 9, not 3.
27 Jun 2006, 19:05
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