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Need hel solving some problem from Gmatprep.....thanks:)

Problem Solving — algebra and arithmetic
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by billyr » Thu Feb 12, 2009 8:43 am
Q1: A company has two type of machine R and S. Operating at constant rate, R does the job in 36 hr, s does the same job in 18 hr. If the company use the same number of each type of machine to do job in 2 hr, how many machine type r used?
a)2
b)4
c)6---->answer
d)10
e)12


Q2: If 2^x - 2^x-2= 3(12^13) what is the value of x?
a)9
b)11
c)13
d)15------->answer
e)17

Q3: For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If P is the smallest prime factor of h(100)+1, then p is?

a)between 2 and 10
b)between 10 and 20
c)between 20 and 30
d)between 30 and 40
e)greater than 40--------> answer


Q4:If n is a positive integer and the product of all the integer from 1 to n inclusive is a multiple of 990, what is the least possible value for n?
a)10
b)11------>answer
c)12
d)13
e)15



Data sufficiency:
Q5: What is the tenth digit of positive integer r?

(1) the tens digit of r/10 is 3

(2)the hundred digit of 10r is 6


The answer is B (statement 2 alone sufficient) but i totally lost about that.


Q6:If x&y are positive integer, what value is xy?

(1)the greatest common factor x&y is 10

(2)the least common multiple of x& y is 180


the answer is C(Both statement is sufficient)----but i don't know how.





Thanks a lot
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Source: — Quantitative Reasoning |
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by BuckeyeT » Thu Feb 12, 2009 9:04 am
Just a reminder that you should try to post only 1 question in a post. It might increase the number of responses.

I can take a quick stab at #1, and I might be able to look at the others later.

1. This is a work problem. R = 1/36, S = 1/18.
R + S = hours to complete job inverted
1/36 + 1/18 = 3/36 = 1/12 or 12 hours to complete the job together.
If it takes them 12 hours together, and you need it done in 2 hours, then
12/2 = 6 times faster or 6 pairs of R and S.
So, the number of R machines = 6.

C.
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by BuckeyeT » Thu Feb 12, 2009 9:11 am
I'm assuming Q2 is written incorrectly...

Q2: If 2^x - 2^x-2= 3(2^13) what is the value of x?

Start factoring:
2^x - 2^(x-2) can be factored into...
2^(x-2) * (2^2 - 1) or...
2^(x-2) * (4-1) ...
2^(x-2) * 3

Look back at the equivalency...
2^(x-2) * 3 = 3(2^13) divide both sides by 3...
2^(x-2) = 2^13 ...
So, x-2 = 13...
x = 15

Answer D.
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by BuckeyeT » Thu Feb 12, 2009 11:56 am
Q4:If n is a positive integer and the product of all the integer from 1 to n inclusive is a multiple of 990, what is the least possible value for n?

To be a multiple of 990, you have at least the same factors are 990. So, let's look at the factors of 990.
990 = 2*495
=2*5*99
=2*5*11*9
=2*5*11*3*3

Since we have know that the greatest prime factor is 11, we know that n has to go through at LEAST n=11 to remain true. Therefore, 11 is the least possible value of n in order for 1*2*...*n to be a multiple of 990.

Q5: What is the tens digit of positive integer r?

(1) the tens digit of (r/10) is 3.
Let's assume, (r/10) = 30.
r = 10*30
r = 300
The tens place is 0. But, we need it to be consistent.
Let's assume, (r/10) = 88.
r = 10*88
r = 880
The tens place is 8. Therefore, we proved this statement insuffient because it's not the same for all r.

(2) The hundreds digit of 10r is 6.
Let's assume, 10r = 600.
r = 600/10
r = 60
The tens place is 6. But, we need it to be consistent.
Let's assume, 10r = 694.
r = 694/10
r = 69.4
The tens place is 6. And, you'll notice it will happen for all others. Therefore, this statement is sufficient.

Answer B.

The much easier way to look at this question is just to determine which way the decimal point will shift (and how that impacts the tens place). No calculation should be necessary.
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