GMAT Problem Solving Discussions

MBA
4586
ps25 Says
Can someone please post tough absolute, inequality, Perms & Combs, and probability questions please?

I don't know whether the following question is exactly tough or not but I found it tricky.


A computer wholesaler sells 8 different computers each is priced differently.If the wholesaler chooses 3 computers for display at a trade show,what is the probability (all things being equal) that the 2 most expensive computers will be among the 3 chosen for display?

Answer :3/28
4587
I don't know whether the following question is exactly tough or not but I found it tricky.


A computer wholesaler sells 8 different computers each is priced differently.If the wholesaler chooses 3 computers for display at a trade show,what is the probability (all things being equal) that the 2 most expensive computers will be among the 3 chosen for display?

Answer :3/58


How many ways can 3 computers be chosen from set of 8 computers?(Denominator) -

8!/(3!*5!)

Suppose 2 most expensive computers are put on display, remaining 6 computers

How many ways can 1 computer(2 are chosen already!) from set of 6 computers(Numerator) -

6!/(1!*5!)

Probability = 6!/5! * (3!*5!)/8! = 3!/(8*7) = 3/28

Not sure how 3/58 came in answer. Can anyone confirm? (besides 58 = 29*2 and 29 is prime number and is much large than 8 :-()?
4588
How many ways can 3 computers be chosen from set of 8 computers?(Denominator) -

8!/(3!*5!)

Suppose 2 most expensive computers are put on display, remaining 6 computers

How many ways can 1 computer(2 are chosen already!) from set of 6 computers(Numerator) -

6!/(1!*5!)

Probability = 6!/5! * (3!*5!)/8! = 3!/(8*7) = 3/28

Not sure how 3/58 came in answer. Can anyone confirm? (besides 58 = 29*2 and 29 is prime number and is much large than 8 :-()?

hey!! your solution is perfect 😞 i posted the answer 3/58 by mistake
4589
protyusha Says
hey!! your solution is perfect 😞 i posted the answer 3/58 by mistake



thanks for correcting. inequality + absolute value questions in PS and DS would be very much appreciated, any challenging or tough to crack
4590

Hi,

Probability sum.

5 employees in a firm A,B,C,D,E

Only 3 can get leave at a given time. All apply together for same dates.

What is the probability that A and B will get leave at the same time??

I am not sure how to solve this just using math only combination and permutations formulae. Would be great if some one could help.

Thanks,
Murthy

4591
Hi,

Probability sum.

5 employees in a firm A,B,C,D,E

Only 3 can get leave at a given time. All apply together for same dates.

What is the probability that A and B will get leave at the same time??

I am not sure how to solve this just using math only combination and permutations formulae. Would be great if some one could help.

Thanks,
Murthy

Only 3 employees can get leave. So in how many ways can 3 people be selected from group of 5?

5!/(3!*2!) (Denominator)
= 10

A and B get leave on same day, in how many ways can 1 people (2 people got leave A and B) be selected from group of 3 (2 people got leave - A and B)?

3!/(2!*1!) (Numerator)
= 3

So probability is 3/10 or 30%

Please confirm with OA
4592

Solution Y is 30 percent liquid X and 70 percent water. If 2 kilograms of water evaporate from 8 kilograms of solution Y and 2 kilograms of solution Y are added to the remaining 6 kilograms of liquid, what percent of this new solution is liquid X?


(A) 30%
(B) 33 1/3 %
(C) 37 1/2 %
(D) 40%
(E) 50%

4593
Solution Y is 30 percent liquid X and 70 percent water. If 2 kilograms of water evaporate from 8 kilograms of solution Y and 2 kilograms of solution Y are added to the remaining 6 kilograms of liquid, what percent of this new solution is liquid X?


(A) 30%
(B) 33 1/3 %
(C) 37 1/2 %
(D) 40%
(E) 50%



Answer is C
4594
ps25 Says
Answer is C


COuld you give me an explanation?
4595
deepa7685 Says
COuld you give me an explanation?


Hi deepa,

Initial composition is as follows : -

8 Kg (Sol Y) = 2.4 Kg of Liquid X (30% Of 8 Kg) + 5.6 Kg of Water (70% of 8Kg)

Up on evaporation of 2 kg of water, composition is :-

8 - 2 = 6 kg (Sol y) = 2.4 Kg of Liquid X + 3.6 Kg of water

Additional 2 Kg of solution Y would have :-
2 Kg (Sol Y) = 0.6 Kg of Liquid X(30% of 2 Kg) + 1.4 Kg of Water (70% of 2Kg)

Added it to 6 Kg you would have

8 Kg (Sol Y) = 3 Kg of Liq X + 5 Kg of Water

3 Kg Liquid X / 8 Kg of Sol Y * 100 = 37.5 % And thats your answer
4596
Maya N Says
This is a great discussion thread. Along with these questions the GMAT Quantitative Review is also available which has over 300 actual questions from the past GMAT tests.



In all of your four posts so far, you have campaigned for Wiley publishers. Seems like this forum allows advertising
4597

Hello Puys, in case of GMAT, I'm just a starter. I have recently been placed and I was thinking about appearing for GMAT with some experience. Lt me put it this way. I have to begin from 0. I know nothing about the test except the 3 sections called Quants, VA, and DI. So if someone can put some inputs and show the directions how to begin the preparation, it would be helpful. What are the colleges accepting GMAT, their cut-offs, and the courses available through GMAT. Someone who knows can throw some light on this, including good, better and best colleges. Would really be helpful. 😁

4598

If M= (4)^1/2+(4)^1/3+(4)^1/4. What is the value of M?
Can anyone tell me a simple approach to this problem.....Explanations would be appreciated...

4599

For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n inclusive, if p is the smallest prime factor of h(100) +1, then p is

a) between 2 & 10
b) between 10 & 20
c) between 20 & 30
d) between 30 & 40
e) greater than 40.

Explanations will be appreciated.....

4600
If M= (4)^1/2+(4)^1/3+(4)^1/4. What is the value of M?
Can anyone tell me a simple approach to this problem.....Explanations would be appreciated...


Dont think it can be simplified. what are the options, or answer.
4601
For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n inclusive, if p is the smallest prime factor of h(100) +1, then p is

a) between 2 & 10
b) between 10 & 20
c) between 20 & 30
d) between 30 & 40
e) greater than 40.

Explanations will be appreciated.....



h(100)+1 = 2*50! + 1, so its going to be prime, so answer e.

Please confirm
4602
h(100)+1 = 2*50! + 1, so its going to be prime, so answer e.

Please confirm


Yes The right ans is E
4603
deepa7685 Says
Yes The right ans is E


can you tell what is solution to earlier problem 4^1/2 ....
4604
ps25 Says
Dont think it can be simplified. what are the options, or answer.


1) greater than 2
2)between 2 & 3
3) between 3 & 4
4)equal to 4
5) greater than 4

So obviously the ans is greater than 4 but I wondering id there was a simpler method
4605

If M= (4)^1/2+(4)^1/3+(4)^1/4. What is the value of M?

1) greater than 2
2)between 2 & 3
3) between 3 & 4
4)equal to 4
5) greater than 4

So obviously the ans is greater than 4 but I wondering id there was a simpler method




4^1/2 = 2
M = 2+(4)^1/3 + (4)^1/4

sqaure root of 4 is 2 and sqare root of 2 is 1.41 (that means 4^1/4 is 1.41)

4^1/3 > 4^1/4 so 4^1/3 is greater than 1.4

so M = 2+Greater than 1.4 + 1.4
= 3.4 + Grater than 1.4
= Greater than 4.8

So E

Your earlier question gave an impression that you want to 'simplify' which is not same as 'estimating' the value of expression.