GMATPrep Questions- clueless!

[nomadmba]

Here are three questions that have left me scratching my head. I have the answers, but not the methodology:

1. For which of the following functions is f(a+b)= f(a) + f(b) for all positive numbers a and b?

a. f(x) = x*x
b. f(x) = x + 1
c. f(x) = sq. root of x
d. f(x) = 2/x
e. f(x) = -3x


2. In the XY plane, at what two points does the graph of y= (x+a)(x+b) intersect the X axis?
1) a + b = 1
2) the graph intersects Y axis at (0, -6)

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?

I know the last should be fairly simple but my mind just doesn't seem to process permutation and combination questions...

Thanks.

[ameyp12]

Originally Posted by nomadmba

Here are three questions that have left me scratching my head. I have the answers, but not the methodology:

1. For which of the following functions is f(a+b)= f(a) + f(b) for all positive numbers a and b?

a. f(x) = x*x
b. f(x) = x + 1
c. f(x) = sq. root of x
d. f(x) = 2/x
e. f(x) = -3x


Thanks.

Its E

Method is replace x by a, b & (a+b)

1. x*x = x²
a² + b² = (a+b) ² - 2ab
not equal to (a+b)²

2. a+1 + b + 1= (a+b) + 2
not equal to (a+b) + 1

3. root a + root b not equal to root(a+b)

4. 2 / a + 2/b not equal to 2/(a+b)

5. -3a + (- 3b) = -3 (a+b) -- > Correct

[ameyp12]

Originally Posted by nomadmba

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?

I know the last should be fairly simple but my mind just doesn't seem to process permutation and combination questions...

Thanks.

Total 12 temp --> 5 Women So 7 men
To select 3 W & 1 M --> Selection ... not arrangement ... So its Combination.
To select 3 from 5 women we have 5C3 ways
To select 1 from 7 men we have 7C1 ways
Total= 5C3 * 7C1
I would suggest just go over basics of Permutation / Combination
Here is some stuff to go over

[nomadmba]

Thank you so much ameyp12. Unfortunately i can't download the zip file because i'm new to the forum I'm gonna look over perm and comb stuff in the books that i have.

I'm a little peeved because i actually took a class-room based princeton course and there my instructor basically told us to skim through probability and perm-comb stuff because there might only be 1 question on the test. Now i'm getting anxious because if that 1 or even 2 questions come towards the begining of my test, i'm pretty screwed.

Thanks for your help though.

[finsisher]

Originally Posted by nomadmba

Here are three questions that have left me scratching my head. I have the answers, but not the methodology:

1. For which of the following functions is f(a+b)= f(a) + f(b) for all positive numbers a and b?

a. f(x) = x*x
b. f(x) = x + 1
c. f(x) = sq. root of x
d. f(x) = 2/x
e. f(x) = -3x


2. In the XY plane, at what two points does the graph of y= (x+a)(x+b) intersect the X axis?
1) a + b = 1
2) the graph intersects Y axis at (0, -6)

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?

I know the last should be fairly simple but my mind just doesn't seem to process permutation and combination questions...

Thanks.

solun for first question, substitute a= b=1 so f(2) = f(1)+f(1)=2f(1) only E satisfies it hence E.

solun for question 2 , clearly the graph intersects x axis at x= -a and x = -b when y = 0
from stm1 1 we dont know what is a and b individually hence we cant find the points on which graph intersect X axis,
from stmt 2 , we know a=0 b =-6 or b=0 or a = -6 so answer should be B

[himanshusinghs]

2. In the XY plane, at what two points does the graph of y= (x+a)(x+b) intersect the X axis?
1) a + b = 1
2) the graph intersects Y axis at (0, -6)

answer.- y=x^2+x(a+b)+ab
graph intersects x-axis when y=0
therefore,x^2+x(a+b)+ab=0
=>x^2+x-6=0 (from 1 & 2)
therefore, x=-3,2
answer-C

[abhishek_mumbai]

1. For which of the following functions is f(a+b)= f(a) + f(b) for all positive numbers a and b?
a. f(x) = x*x
b. f(x) = x + 1
c. f(x) = sq. root of x
d. f(x) = 2/x
e. f(x) = -3x

replace a, b and a+b in these equations.. you can do it mentally.

the answer is E..

a. (a+b)*(a+b) = a*a + b*b (false)
b. a + b + 1 = a + 1 + b + 1 (false)
c. sqrt(a+b) = sqrt(a) + sqrt(b) [false]
d. 2/(a+b) = 2/a + 2/b (false)
e. -3(a+b) = -3a - 3b (TRUE)

IMO E

[abhishek_mumbai]

2. In the XY plane, at what two points does the graph of y= (x+a)(x+b) intersect the X axis?
1) a + b = 1
2) the graph intersects Y axis at (0, -6)
on X axis, y = 0 so
(x+a)(x+b) = 0
x^2 + x(a+b) + ab = 0

stmt 1 still have ab unresolved.
stmt 2 will solve ab but not a+b..
C can make this complete equation.. x^2 + x - 6 = 0

IMO C

[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

[abhishek_mumbai]

Originally Posted by nomadmba

Thank you so much ameyp12. Unfortunately i can't download the zip file because i'm new to the forum I'm gonna look over perm and comb stuff in the books that i have.

I'm a little peeved because i actually took a class-room based princeton course and there my instructor basically told us to skim through probability and perm-comb stuff because there might only be 1 question on the test. Now i'm getting anxious because if that 1 or even 2 questions come towards the begining of my test, i'm pretty screwed.

Thanks for your help though.

Yes..you are right.. if your target is 700 or above, you can;t afford to leave any loose ends.. It's better to learn these concept rather than giving the test again ..