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Question

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

jaynayak · Answer

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.

jaynayak · Answer

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

jaynayak · Answer

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.

apollo168 · Answer

Hi Jay isnt 3C3 =1 Thanks for your help your really great bet you got an 800 in gmat

apollo168 · Answer

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?

jaynayak · Answer

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

jaynayak · Answer

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

rxs0005 · Answer

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

haas_mba07 · Answer

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

apollo168 · Answer

Thanks for the help guys

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