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Quant Questions-please help me solve these

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Quant Questions-please help me solve these [#permalink] New post 18 Mar 2007, 18:04
Guys!

Please help me solve these Quant questions:

1) On a certain sight seeing tour, the ratio of number of women to the number of children was 5 to 2. What was the number of men on the sight seeing tour?
A) On the sight seeing tour, the ratio of the number if children to the number of men was 5 to 11.
B) The number of women on the sight seeing tiur was less than 30.

2) Is M+Z>0
A) M-3Z>0
B) 4Z-M>0

3) A certain law firm consists of 4 senior partners and 6 junior partners. How many different of 3 partners can be formed in which at least one member of the group is a senior partner?
A) 48
B)100
C) 120
D) 288
E) 600

4) At least 100 students at a certain high school study Japanese. If 4 percent of the students at the high school who study French also study Japanese, do more students at the school study French than Japanese?
A) 16 students at the school study both French and Japanese.
B) 10 percent of the students at the school who stdy Japanese also study French.

5) is [x-y]>[x]-[y]
A) y<x
B) xy<0

6)A certain city with a population of 132,000 is to be divided into 11 voting districts, and no district is to have a population that is more than 10% greater than the population of any other district. What is the minimum possible population that the least populated district could have?
A) 10,700
B) 10,800
C) 10,900
D) 11,000
E) 11,100
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 [#permalink] New post 18 Mar 2007, 18:18
1.

I am going to say C

WE know W/C=5/1

A) C/M = 5/11

Ratios cant give us the actual values

Insuff


B) W<30

Insuff doesn't give us anything on its own, we still dont know about the men.

Together:

We know that children must = 5 now, because if it were any other number, than it would make M be a non integer, or a number that would force W>30. So therefore W=25, C=5, and M=11.

This was tough for me, so I might be wrong, someone double check.

6. I believe is D

If took D and had 1 division of 11,000 and the other 10 being 12,100, that would make a total of 132,100 voters, which means you can make a combination that would only give 132,000 voters.

If you did this with C, then you would only end up with 131,890 voters, which isn't enough for 10,900 to be the minimum.

Last edited by Tuneman on 18 Mar 2007, 18:39, edited 4 times in total.
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Re: Quant Questions-please help me solve these [#permalink] New post 18 Mar 2007, 18:19
nitinneha wrote:

3) A certain law firm consists of 4 senior partners and 6 junior partners. How many different of 3 partners can be formed in which at least one member of the group is a senior partner?
A) 48
B)100
C) 120
D) 288
E) 600



(B) 100.

C(10,3) - C(6,3) = 120 - 20 = 100.

C(10,3) - total number of variations of 3 partners from 4 SP and 6 JP.

C(6,3) = 20. this case in only about selection of 3 partners from 6 JP. Here it means that there is no SP in the group. So we have to except the case.
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 [#permalink] New post 18 Mar 2007, 19:27
My answers are
1. C
2. C
3. B
4. C
5. B
6. D

Q1. I did this by normalising (Tuneman , you got it right)
From stmt 1
W:C=5:2
C:M=2:11
So
W:C:M=25:10:22 ( take LCM of 2 and 5)
This tells us # women has to be multiple of 25, but it does not give exact number so INSUFF
Stmt 2 W < 30 does not help INSUFF
Combning both we get W = 25

Q2.
From Stmt 1
We have 3Z < M
From Stmt 2
we have M < 4Z
Since we dont know if M and Z are +ve or -ve above are INSUFF individually

Combining
3Z < M < 4Z
Since 3Z < 4Z , Z has to be +ve
so M has to be +ve
and therefore M + Z has to be +ve

Q3.As explained by pi10t

Q4.
Stmt 1
4% of FRENCH = 16 therefore 100% of FRENCH = 25 * 16 = 400
Stmt 2
10% of JAPANESE = 16 therefore 100% of JAPANESE = 10 * 16 = 160

stmt 1 and 2 by itself is INSUFF, together SUFF

Q5.
Not sure about this one
This is what I did
Actually tried 4 cases
1. Both +ve : |X-Y| = |X| -|Y|
2. Both -ve : |X-Y| = |X| -|Y|
3. X -ve and Y +ve : |X-Y| > |X| -|Y|
4. X +ve and Y -ve : |X-Y| > |X| -|Y| Modified the equality sign here

So if we know the signs we can answer the question.

As you see
Stmt 1 does not tell us about sign
While
Stmt 2 tells us that atleast one of them has to be -ve; So case 3 and 4 above indicates LHS > RHS


Q6.
http://www.gmatclub.com/phpbb/viewtopic.php?t=39492

Last edited by kyatin on 20 Mar 2007, 18:16, edited 1 time in total.
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Q.4 [#permalink] New post 19 Mar 2007, 02:24
In Q the answer is B

Stmt 1
4% of FRENCH = 16 therefore 100% of FRENCH = 25 * 16 = 400
We dont know about the no. of Japanese.
SO INSUFF

Stmt 2 - 10 percent of the students at the school who study Japanese also study French.
The question stem tells us that 4 percent of the students at the high school who study French also study Japanese.

hence 10% of Japanese = 4% of French. This is possible only when the number of students who take French is more than those who take Japanese.

B is sufficient
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 [#permalink] New post 19 Mar 2007, 15:59
I agree it should be B for Q4. Like GMAThopeful pointed out.
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 [#permalink] New post 20 Mar 2007, 17:01
OA are:

1) C
2) C
3) B
4) B
5) B
6) D
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 [#permalink] New post 20 Mar 2007, 18:00
Could someone explain the answer to Q 5 again? According to Kyatin, enswer could not be B. Because B tells us that one of the x and y is -ve. But doesn't tell which is negative. Hence doesn't give a definitive answer to the question.
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 [#permalink] New post 20 Mar 2007, 18:13
nitinneha wrote:
Could someone explain the answer to Q 5 again? According to Kyatin, enswer could not be B. Because B tells us that one of the x and y is -ve. But doesn't tell which is negative. Hence doesn't give a definitive answer to the question.


Nitin
My answer is indeed B. I made one goof up in typing. Please see corrected original post.
  [#permalink] 20 Mar 2007, 18:13
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