GMATPrep Questions- clueless!

[doctortt]

Hi just fyi - I thought it's combination when the order is not matter.

[givinggmat]

Originally Posted by abhishek_mumbai

3. There are 12 temp emloyees in a company. 4 of them will be hired at permanent staff. If 5 of 12 of the temp staff are women, how many of the possible groups of 4 temp employees can consist of 3 women and 1 man?

The order does not matter so it is a Permutation question

for women, 3 selected out of 5 => 5P3
for men, 1 selected out of 7 => 7P1

total = 5P3 * 7P1 = 420

Lets name 5 woman emps as - A B C D E
Let the one man be named as - M

Now, the grp - A B C M is same as B C A M; so the answer is -->

women selection = 5!/3!*2! = 10
sel of man = 7!/6!*1! = 7

total = 10*7 = 70

[ships12]

1)Answer is E
2) Answer is C, since we need both the statements to calculate the value of a and b.
3)Since it is a combination question it is 5C3.7C1 = 70

[saravanants]

1) Answer E.
2) 1 => a+b = 1 and 2 => ab = -6. Combining both can answer the question C.
3) This is a combination problem and not a permutation problem as we cannot select the same group again but in reversed order For example women#1,women#2,women#3 if formed as one group, then we cannot count another group as women#3, women#2, women#1. So right answer should be 5c3*7C1 = 10*7 = 70.

[lecast]

Originally Posted by abhishek_mumbai

3. There are 12 temp emloyees in a company. 4 of them will be hired at permanent staff. If 5 of 12 of the temp staff are women, how many of the possible groups of 4 temp employees can consist of 3 women and 1 man?

The order does not matter so it is a Permutation question

for women, 3 selected out of 5 => 5P3
for men, 1 selected out of 7 => 7P1

total = 5P3 * 7P1 = 420

If the order does not matter then it is a combination question.

[abhishek_mumbai]

Originally Posted by lecast

If the order does not matter then it is a combination question.

Yes.. you are right.. my mistake.. it is a combination question..

5C3 * 7C1 = 5C2 * 7C1 = 10 * 7 = 70..

[MBA2010HereWeGo]

Agree that answer to 10 is E. I approached this question by subsituting numbers into the function. And choice E was the correct one. This might be a longer process but using smaller number makes the calculation takes less time. I used a=1 and b=2.

Question 2, IMO C. Because we need to figure out what is the value of either a or b.

y=(x+a)(X+b) or y=(x)^2+x(a+b) + ab

stmt 1 a+b=1 wouldn't help alone (we need 2 unique equations)
stmt 2 y intercept is (0,6) wouldn't help on its own either

but combined we know

0=(x)^2+x(1)+6 or x=2 or -3 thus intersect at (2, 0) or (-3, 0).

Question 3

since order doesn't matter thus 5C3*7C1