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Just finish with the GMAT real questions help gona retake

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Just finish with the GMAT real questions help gona retake [#permalink] New post 04 Aug 2006, 02:20
1) Problem solving question

When x divided by 5 the remainder is 3. When x divided by 7 remainder is 4. When y divided by 5 remainder is 3. When y divided by 7 remainder is 4. Given x>y x-y could be?

2) 8 women 3 men. Need to form 3 person committee. How many can be formed with at least 1 man?

3) Data Suff question

What is the remainder when x is divided by something (sorry forgot)

1) When x is divided by 6 the remainder is something(forgot)

2) When x is divided by 5 the remainder is something again.

Just wana know how you go about solving this problem Thanks
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Re: Just finish with the GMAT real questions help gona retak [#permalink] New post 04 Aug 2006, 03:05
apollo168 wrote:
1) Problem solving question

When x divided by 5 the remainder is 3. When x divided by 7 remainder is 4. When y divided by 5 remainder is 3. When y divided by 7 remainder is 4. Given x>y x-y could be?



1) two ways
Method 1: subsitute numbers
x = 5a + 3 and 7b +4
y = 5c+3 and 7d+4
since x > y
X can be 53 and y can be 18
x-y = 35
Method two:
x-y = 5a+3 - 5c-3 = 5a+5c
Hence x-y is a multiple of 5
Also
x-y = 7b+4 - 7d-4 = 7b+7d
Hence x-y is a multiple of 7
LCM of 5 and 7 is 35
Hence x-y can be 35 or multiple of 35.
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Re: Just finish with the GMAT real questions help gona retak [#permalink] New post 04 Aug 2006, 03:11
apollo168 wrote:
1) Problem solving question

2) 8 women 3 men. Need to form 3 person committee. How many can be formed with at least 1 man?



At least 1 man.
hence commitee can have MMM, MMW and MWW

Choosing a commitee with all men = 3C3 = 3 ways
Chossing a commitee with two men and 1 women = 3C2 * 8C1 = 24
Chossing a commitee with one man and 2 women = 3C1 * 8C2 = 84
Total = 3+24+84 = 111
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Re: Just finish with the GMAT real questions help gona retak [#permalink] New post 04 Aug 2006, 03:14
apollo168 wrote:
1) Problem solving question

3) Data Suff question

What is the remainder when x is divided by something (sorry forgot)

1) When x is divided by 6 the remainder is something(forgot)

2) When x is divided by 5 the remainder is something again.

Just wana know how you go about solving this problem Thanks


When x is divided by 6 say the remainder is 2
x = 6q + 2
Putting values we get x = 8 or 14 etc
When x is divided by 5 ..... we get different remainders , hence not suff.

You can go like this if you want.
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 [#permalink] New post 04 Aug 2006, 03:16
Hi Jay isnt 3C3 =1 Thanks for your help your really great bet you got an 800 in gmat
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 [#permalink] New post 04 Aug 2006, 03:32
Hi Jay,

Let say these are the given

What is the remainder when x is divided by 7

1) When x is divided by 5 the remainder is 3
2) When x is divided by 6 the remainder is 1

I want to do it using equations since ive realize that during the gmat sometimes plugging in values can be time consuming

so from number 1 x=5a+3
from number 2 we have x= 6b+1

from there how do you arrive at a conclusion?
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 [#permalink] New post 04 Aug 2006, 03:44
apollo168 wrote:
Hi Jay isnt 3C3 =1 Thanks for your help your really great bet you got an 800 in gmat


Still to take GMAT .....

Sorry my mistake ... 3C3 is 1 .... and the answer is 109.... there is another way to do it too........

Find the number of commitees without any men. 8C3
Find the number of ways commitees can be formed with 11 persons = 11C3

At least one men = 11C3 - 8C3
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 [#permalink] New post 04 Aug 2006, 03:56
apollo168 wrote:
Hi Jay,

Let say these are the given

What is the remainder when x is divided by 7

1) When x is divided by 5 the remainder is 3
2) When x is divided by 6 the remainder is 1

I want to do it using equations since ive realize that during the gmat sometimes plugging in values can be time consuming

so from number 1 x=5a+3
from number 2 we have x= 6b+1

from there how do you arrive at a conclusion?


I think in some situations you have to plug in numbers.
Using your equations..... find two or three values of x and divide by 7.

Clearly both are Insuff.
1) x = 18 or 23
when 18 , 18/7 remainder = 4
When 23, 23/7 remainder = 3
2) x = 7 or 13
when 7, 7/7 remainder = 0
when 13, 13/7 remainder = 6

Together.
x = 13 (13/5 = 3 and 13/6 = 1)
When divided by 7 remainder = 6
x= 43 (43/5 =3 and 43/6 = 1)
When divided by 7 remainder = 1

Hence not suff. The answer would be E
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 [#permalink] New post 04 Aug 2006, 04:41
For the Combinatiosn Q i am gettgin the below :

3 members in a committee :

1 man min so can be chosen in 3 ways .

2 positions left with 10 people they can be chosen in 10C2 ways = 45

total ways = 45 * 3 = 135
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 [#permalink] New post 04 Aug 2006, 08:46
For 1... agree with 35 and multiples of 35...

For 2 committe...
Key is atleast 1 man (could be 1/2/3)
We can find answer by :
(Total # of ways of 3 members) - (Total # of ways of 3 member w/o any man)

11x10x9/3x2 - 8x7x6/3x2 = 165 - 56 = 109
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 [#permalink] New post 04 Aug 2006, 10:18
Thanks for the help guys
  [#permalink] 04 Aug 2006, 10:18
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