GMAT Problem Solving Discussions

star_highway · May 21, 2009, 10:41am

The Total ways to arrange 5 people=!5
case1. When first couple is sitting together=!4*!2
cas2.When second couple sitting together=!4*!2
case3.when both couples sitting together=!3*!2*!2
so the number of ways when couples are sitting together=case1+case2-case3
=96-24=72
so probability=72/120=3/5

your thought process was right but you made a critical error in last step....

the question is to find the prob for no couple sitting together... so the answer should be 1 - 72/120 = 2/5

guy_with_guts · May 21, 2009, 10:52am

your thought process was right but you made a critical error in last step....

the question is to find the prob for no couple sitting together... so the answer should be 1 - 72/120 = 2/5

OOps!!!!!!!!!!!!!!!!!!!!!!!!!!
Blunder

amar_sinha_25505345 · May 21, 2009, 11:27am

i keep visiting this thread and share some of the good questions from my old GMAT notes. here is another one.

5 chairs, 2 couple and one single person.
What is the probability that no couple sit together.

total no. of ways of arranging 5 people = n=5!
no. of ways where first couple sit together is =n1=2*4!
no. of ways where second couple sit together is n2=2*4!
no. of ways where both couple sit together is n3=4*3!
no. of ways where none of the couples sit together is n-(n1+n2)+n3
= 5! - (2*4! + 2*4!) + 4*3!
= 5! - 3*4!
= 2*4!
So the probability = 2*4!/5! = 2/5

star_highway · May 21, 2009, 1:13pm

i keep visiting this thread and share some of the good questions from my old GMAT notes. here is another one.

5 chairs, 2 couple and one single person.
What is the probability that no couple sit together.

total no. of ways of arranging 5 people = n=5!
no. of ways where first couple sit together is =n1=2*4!
no. of ways where second couple sit together is n2=2*4!
no. of ways where both couple sit together is n3=4*3!
no. of ways where none of the couples sit together is n-(n1+n2)+n3
= 5! - (2*4! + 2*4!) + 4*3!
= 5! - 3*4!
= 2*4!
So the probability = 2*4!/5! = 2/5

Right !!!

Another way of doing it...

Let coulples are x1y1 and x2y2 and Single person is S

Total possible combination = 5! = 120

1st position is S
2 - 4 ways - any of the rest 4 can take it
3 - 2 ways - one member of the other coulple can take it
4 - 1 way - left member of couple on 2nd postion
5 - 1 way - left member of couple on 3rd position
Total number of ways = 8

2nd position is S
1 - 4 ways - any of the rest 4 can take it
3 - 2 ways - if x1 is on 1, y1 cannot be on 3 as 4,5 will have x2y2 together
4 - 1 way - left member of couple on 1st postion
5 - 1 way - left member of couple on 3rd position
Total number of ways = 8

3rd position is S
1 - 4 ways
2 - 2 ways
4 - 2 ways
5 - 1 way
(i think by now you can figure out, how I am counting)
Total number of ways = 16

4th position is S == same as S in 2nd position = 8
5th position is S == same as S in 1st position = 8

Total = 8+8+16+8+8 = 48

probability = 48/120 = 2/5

It looks like a longer process, because i had to type it. With pen and paper its quite fast...

nikhil2877 · May 22, 2009, 7:27am

Guys - Started off with my prep yesterday (Second attempt). Sat for Prep 1 and came out with dismal 630. Last year in July my GMAT score itself was 640... Looks like I have a journey to take. Wanted to discuss some of the problems in Prep1.

If X =/= 0 then (X^2)/X =
i.e Square root of X-sqaure divided by X =

A -1
B 0
C 1
D X
E X / X

Thanks
Nikhil

aniruddha.maiti · May 22, 2009, 7:42am

it should be E X|/X.. ..can be -x or +x.

shashankaggarwa · May 22, 2009, 7:47am

Guys - Started off with my prep yesterday (Second attempt). Sat for Prep 1 and came out with dismal 630. Last year in July my GMAT score itself was 640... Looks like I have a journey to take. Wanted to discuss some of the problems in Prep1.

If X =/= 0 then (X^2)/X =
i.e Square root of X-sqaure divided by X =
A -1
B 0
C 1
D X
E X / X
Thanks
Nikhil

i think its E.

wats the OA ??

nikhil2877 · May 22, 2009, 7:47am

Thanks Aniruddha... That's the right answer....

ramviswa · May 22, 2009, 10:45am

Guys - Started off with my prep yesterday (Second attempt). Sat for Prep 1 and came out with dismal 630. Last year in July my GMAT score itself was 640... Looks like I have a journey to take. Wanted to discuss some of the problems in Prep1.

If X =/= 0 then (X^2)/X =
i.e Square root of X-sqaure divided by X =

A -1
B 0
C 1
D X
E X / X

Thanks
Nikhil

E is my answer.

Square root of an integer would always be positive. Here square root of x^2 is equal to mod X

amar_sinha_25505345 · May 22, 2009, 4:33pm

the answer should be 1. if we take y = sqrt(x^2)/x
that means y = + sqrt(x^2)/x
= x/x = 1.
had it been y^2 = ax,
then y = +/- sqrt(ax)

amar_sinha_25505345 · May 22, 2009, 4:36pm

E is my answer.

Square root of an integer would always be positive. Here square root of x^2 is equal to mod X

I got my mistake.....you are right Viswa....E should be the answer.....

shashankaggarwa · May 23, 2009, 2:41pm

guys!!! lets have some new brainstorming question for PS !!!!

srworld · May 24, 2009, 12:32pm

Q1: When a tree was first planted, it was 4 feet tall, and the height of the tree increased by a constant amount each year for next 6 years. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increased each year?
A) 3/10 B) 2/5 C) 1/2 D) 2/3 E) 6/5

Q2: The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100%, which of the following is closest to the percentage change in the concentration of chemical A required to keep the reaction rate unchanged?
A) 100% decrease
B) 50% decrease
C) 40% decrease
D) 40% increase
E) 50% increase

jigar_p_civil · May 24, 2009, 12:46pm

Q1: When a tree was first planted, it was 4 feet tall, and the height of the tree increased by a constant amount each year for next 6 years. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increased each year?
A) 3/10 B) 2/5 C) 1/2 D) 2/3 E) 6/5

Q2: The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100%, which of the following is closest to the percentage change in the concentration of chemical A required to keep the reaction rate unchanged?
A) 100% decrease
B) 50% decrease
C) 40% decrease
D) 40% increase
E) 50% increase

Q1: When a tree was first planted, it was 4 feet tall, and the height of the tree increased by a constant amount each year for next 6 years. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increased each year?
A) 3/10 B) 2/5 C) 1/2 D) 2/3 E) 6/5

if the height of tree increase =X /year

6x+4=(4x+4) + 1/4(4x+4)

after solving x=2/3

The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100%, which of the following is closest to the percentage change in the concentration of chemical A required to keep the reaction rate unchanged?
A) 100% decrease
B) 50% decrease
C) 40% decrease
D) 40% increase
E) 50% increase

here

assum A=5 B=10

so ratio=2.5

to keep ratio 2.5 with B=20 ----> A=50 means 7.07

so 7.07-5/5 =41%

means 40% increase

srworld · May 24, 2009, 1:06pm

Q1: When a tree was first planted, it was 4 feet tall, and the height of the tree increased by a constant amount each year for next 6 years. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increased each year?
A) 3/10 B) 2/5 C) 1/2 D) 2/3 E) 6/5

if the height of tree increase =X /year

6x+4=(4x+4) + 1/4(4x+4)

after solving x=2/3

The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100%, which of the following is closest to the percentage change in the concentration of chemical A required to keep the reaction rate unchanged?
A) 100% decrease
B) 50% decrease
C) 40% decrease
D) 40% increase
E) 50% increase

here

assum A=5 B=10

so ratio=2.5

to keep ratio 2.5 with B=20 ----> A=50 means 7.07

so 7.07-5/5 =41%

means 40% increase

Ans1: D
Ans2: D

jigar_p_civil · May 24, 2009, 1:17pm

Ans1: D
Ans2: D

Hmmm correct both

ramviswa · May 27, 2009, 5:02pm

How to find the perimeter of the shaded area? (See attachment)

shashankaggarwa · May 27, 2009, 5:06pm

ramviswa Says
How to find the perimeter of the shaded area? (See attachment)

Nice question.

i m shaking my brains !!!

shakti_1982 · May 27, 2009, 5:16pm

ramviswa Says
How to find the perimeter of the shaded area? (See attachment)

Is cd//diameter
and whats the radius nothing given.
What is x

ramviswa · May 27, 2009, 5:17pm

Nice question.

i m shaking my brains !!!

This is a solid q and it took me some time to arrive at the answer.

In the figure to the right, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region?

(5/3)P + 5/3 (ROOT 3)

(5/3)P + 10/3 (ROOT 3)

(10/3)P + 5/3 (ROOT 3)

(10/3)P + 10/3 (ROOT 3)

(10/3)P + 20/3 (ROOT 3)

P - pie
Answer D