Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 22 May 2017, 22:35

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Just finish with the GMAT real questions help gona retake

Author Message
Senior Manager
Joined: 14 Jul 2006
Posts: 281
Followers: 1

Kudos [?]: 2 [0], given: 0

Just finish with the GMAT real questions help gona retake [#permalink]

### Show Tags

04 Aug 2006, 03:20
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
SVP
Joined: 30 Mar 2006
Posts: 1730
Followers: 1

Kudos [?]: 82 [0], given: 0

Re: Just finish with the GMAT real questions help gona retak [#permalink]

### Show Tags

04 Aug 2006, 04:05
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.
SVP
Joined: 30 Mar 2006
Posts: 1730
Followers: 1

Kudos [?]: 82 [0], given: 0

Re: Just finish with the GMAT real questions help gona retak [#permalink]

### Show Tags

04 Aug 2006, 04:11
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
SVP
Joined: 30 Mar 2006
Posts: 1730
Followers: 1

Kudos [?]: 82 [0], given: 0

Re: Just finish with the GMAT real questions help gona retak [#permalink]

### Show Tags

04 Aug 2006, 04:14
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.
Senior Manager
Joined: 14 Jul 2006
Posts: 281
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

04 Aug 2006, 04:16
Senior Manager
Joined: 14 Jul 2006
Posts: 281
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

04 Aug 2006, 04:32
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?
SVP
Joined: 30 Mar 2006
Posts: 1730
Followers: 1

Kudos [?]: 82 [0], given: 0

### Show Tags

04 Aug 2006, 04:44
apollo168 wrote:

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
SVP
Joined: 30 Mar 2006
Posts: 1730
Followers: 1

Kudos [?]: 82 [0], given: 0

### Show Tags

04 Aug 2006, 04:56
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
Director
Joined: 07 Jun 2004
Posts: 612
Location: PA
Followers: 7

Kudos [?]: 792 [0], given: 22

### Show Tags

04 Aug 2006, 05:41
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
VP
Joined: 02 Jun 2006
Posts: 1261
Followers: 2

Kudos [?]: 87 [0], given: 0

### Show Tags

04 Aug 2006, 09:46
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
Senior Manager
Joined: 14 Jul 2006
Posts: 281
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

04 Aug 2006, 11:18
Thanks for the help guys
04 Aug 2006, 11:18
Display posts from previous: Sort by