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More Exercises from GMAT official PREP- Help please!

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More Exercises from GMAT official PREP- Help please!

by bobdylan » Wed Apr 11, 2012 6:04 am
1) 2^5+2^5+ 3^5 + 3^5 +3^5=

2)2^x-2^x-2 = 3( 2^13), what's the value of x?
a)9
b)11
c)13
d)15
e)17

3) A 4-person task force is to be formed from the 4 men and 3 women who work in company G's human resources dept.If there are to be 2 men 2 women on this task force, how many different task forces can be formed?
a)14
b)18
c)35
d)56
e)144
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by sam2304 » Wed Apr 11, 2012 6:20 am
Please stick to one question per post.
1) 2^5+2^5+ 3^5 + 3^5 +3^5=
2^5(1 + 1) + 3^5(1 + 1 + 1) = 2^5*2 + 3^5 * 3 = 2^6 + 3^6
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by sam2304 » Wed Apr 11, 2012 6:21 am
2)2^x-2^x-2 = 3( 2^13), what's the value of x?
a)9
b)11
c)13
d)15
e)17
2^x-2^x-2 = 2^x - 2^x/2^2
=> 2^x - 2^x/4
=> 2^x(1 - 1/4)
=> 2^x(3/4)
2^x(3/4) = 3(2^13)
2^x = (4/3) * 3(2^13)
2^x = 4 * 2^13
2^x = 2^15
equating the powers we get x = 15
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by sam2304 » Wed Apr 11, 2012 6:21 am
3) A 4-person task force is to be formed from the 4 men and 3 women who work in company G's human resources dept.If there are to be 2 men 2 women on this task force, how many different task forces can be formed?
a)14
b)18
c)35
d)56
e)144
We are supposed to choose 2 men from 4 men and 2 women from 3 women.
2 men from 4 men = 4C2 = 6
2 women from 3 women = 3C2 = 3 ways
Total = 6 * 3 = 18 ways.
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by Birottam Dutta » Wed Apr 11, 2012 6:31 am
For the second question, put the various options into the equation to see which one matches.

a) x=9 => 2^9- 2^7= 2^7(2^2-1)= 2^7(4-1)=3(2^7)

In this way, as we go through the options, we see that x=15 satisfies the equation, hence D.

Note: As we want 2^13 and the second term is 2^x-2, we should scan through the options and check x=15 first because only in this case is x-2=13.


For third question,

2 men can be chosen out of 4 in 4C2 ways or 6 ways. 2 women out of 3 can be chosen in 3C2 ways or 3 ways.

So, total ways of choosing the team = 6*3= 18 ways. Hence B.

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