GMAT Problem Solving Discussions

MBA
2116
Please solve this one....Please give the solution also...



Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?



1) 1/5
2) 1/4
3) 3/8
4) 2/5
5) 1/2



ans is 1/5?
2117
Please solve this one....Please give the solution also...



Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?



1) 1/5
2) 1/4
3) 3/8
4) 2/5
5) 1/2



Case 1 :
Consider 1st couple to sit together. (Hence they are considered as 1 unit) .
So, 1 unit and remaining 3 people i.e total 4 people can be arrange in 4! ways and the couple themselves can be arranged in 2! ways..
Total ways in which 1st couple sit together = 2!*4! ways..

Case 2 :
Consider 2nd couple to sit together:
Similar to above calculation,
Total ways in which 2nd couple sit together = 2!*4! ways..

Now, its like a venn diagram, where each of the above cases has 1 special case where both couples could be seated together and this special case has been included in each of the cases above , hence we need to subtract this special case once.

Special Case : (both couples together )
Now, when both couples are seated next to each other,
Total no of ways = 2!*2!*3! ( each couple considered as 1 unit , and 2! ways for each of the couple to be seated among themselves )

Total ways (atleast 1 couple seated next to each other) = case 1 + case 2 - special case

= 2!*4! + 2!*4! - 2!*2!*3!
=48+48-24
=72

and 5 people can be arranged in 5! Ways..

Hence , p(no couple together) = 1-p(atleast 1 couple together)
= 1 - 72/120
= 48/120
=2/5.....Ans
2118

Bhavin Thanks !!

Great solution..and awesome explanation....I was also doing in a similar manner but it was complex

1) Consider only 1st couple to sit together. (Hence they are considered as 1 unit) Second couple cannot sit together.There will be four cases.

Total ways in which only the first couple can sit together is = 24

2) Similar when only second couple sit together = 24

3) When both the couple sit together = 3! * 2! *2! = 24

Total = 24 + 24 +24 = 72
and 5 people can be arranged in 5! Ways..

Hence , p(no couple together) = 1-p(atleast 1 couple together)
= 1 - 72/120
= 48/120
=2/5

2119

Hi friends , i have commpressed the three of my questions and uploaded in Ps.zip

pls help m with the answers and the reasoning

i will post the OA's later

2120
Hi friends , i have commpressed the three of my questions and uploaded in Ps.zip

pls help m with the answers and the reasoning

i will post the OA's later


Answers below:-

Q29s30--B

Q34s30--A

Q8s30-- Dude i could not exactly get the meaning of 'dot' as given in the question.I have considered it as sq root(2).
ans will be--Perimeter is 12 units for each square i.e . option B.
2121

for Q29s30 although the the OA is A. Remaining two are correct

Below is the reasoning


slopes of parallel lines are equal

we have those points at (0,2) and (-3,0)
so the slope is 2/3

the only equation that gives this same slope is option A : 3y-2x =0

i.e y=2/3x

so both are parallel.

This is not original explanation.This is my version . so Pls correct me if i m wrong

2122
Answers below:-

.

Hey dude do look at my ds questuion too in the Ds thread
2123
Hi friends , i have commpressed the three of my questions and uploaded in Ps.zip

pls help m with the answers and the reasoning

i will post the OA's later

Ans for 29. is already posted

Ans for 34
year.....car...... ...truck .......total sales
1996 ... x ............. y ...........x+y
1997 ....0.89x ... 1.07y ....1.01(x+y)

0.89x + 1.07y = 1.01(x+y)

which come to x/y = 1/2
2124

A man and his wife appear in an interview for two vacancies in the same post. The probability of the husband's selection is (1/7) and the probability of wife's selection is (1/5). What is the probability that only one of them is selected?

A. 4/5
B. 2/7
C. 8/15
D. 4/7
E. 1/3

2125
A man and his wife appear in an interview for two vacancies in the same post. The probability of the husband's selection is (1/7) and the probability of wife's selection is (1/5). What is the probability that only one of them is selected?

A. 4/5
B. 2/7
C. 8/15
D. 4/7
E. 1/3


Answer: D

Probability of one getting selected = [ P(Husband getting selected) * P(Wife not getting selected) + P(Wife getting selected)P(Husband not getting selected)]* 2 (Vacancies)
2126
Answer: D

Probability of one getting selected = [ P(Husband getting selected) * P(Wife not getting selected) + P(Wife getting selected)P(Husband not getting selected)]* 2 (Vacancies)



The Answer Provided Is Different From You!
I will give the answers in a while, let few others try!
2127
Beths pizzeria offers x different toppings. What is the value of x?


(1) There are an equal number of different pizzas that can be made with
(x 2) toppings as there are different pizzas with just 2 toppings.



 (2) If the pizzeria were to add one additional topping, the number of different pizzas that could be made with 4 toppings would double.


Top of Form




Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.



Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.



BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.



EACH statement ALONE is sufficient.



Statements (1) and (2) TOGETHER are NOT sufficient.


Bottom of Form


can any one help me solving this problem


Neha
2128
A man and his wife appear in an interview for two vacancies in the same post. The probability of the husband's selection is (1/7) and the probability of wife's selection is (1/5). What is the probability that only one of them is selected?

A. 4/5
B. 2/7
C. 8/15
D. 4/7
E. 1/3

P(r)={P(Husband selected).P(Wife not selected)+P(Husband not selected).P(Wife selected)}*2
={1/7 *4/5+6/7*1/5}*2
=4/7
2129
A researcher plans to identify each participant in a certain medical experiment with a
code consisting of either a single letter or a pair of distinct letters written in alphabetical
order. What is the least number of letters that can be used if there are 12 participants, and
each participant is to receive a different code?
A. 4
B. 5
C. 6
D. 7
E. 8

pls provide the ans for that

Neha
2130
Beths pizzeria offers x different toppings. What is the value of x?


(1) There are an equal number of different pizzas that can be made with
(x 2) toppings as there are different pizzas with just 2 toppings.



 (2) If the pizzeria were to add one additional topping, the number of different pizzas that could be made with 4 toppings would double.


Top of Form



Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.



Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.



BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.



EACH statement ALONE is sufficient.



Statements (1) and (2) TOGETHER are NOT sufficient.


Bottom of Form


can any one help me solving this problem


Neha


Is answer D?
2131
A man and his wife appear in an interview for two vacancies in the same post. The probability of the husband's selection is (1/7) and the probability of wife's selection is (1/5). What is the probability that only one of them is selected?

A. 4/5
B. 2/7
C. 8/15
D. 4/7
E. 1/3

Answer: D

Probability of one getting selected = [ P(Husband getting selected) * P(Wife not getting selected) + P(Wife getting selected)P(Husband not getting selected)]* 2 (Vacancies)


I think there is no need to multiply with total no of vacancies...it does not matter if either of them occupy any vacancies as long as they are selected...

Hence, p= 1/7*4/5 + 6/7*1/5
= 10/35 = 2/7...Ans
2132

Yes, Thats correct my friend! :)
Good Job Bhavin!

2133
I think there is no need to multiply with total no of vacancies...it does not matter if either of them occupy any vacancies as long as they are selected...

Hence, p= 1/7*4/5 + 6/7*1/5
= 10/35 = 2/7...Ans



But probability of getting selected increases, doesn't it? I mean the more the vacancies, more are the chances you one of them gets the job.
2134
1. If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?


2. 2 couples and a single person are seated at random in a row of 5 chairs. What is the probability that neither of the couples sit together in adjacent chairs.

3. 9 people, including 3 couples, are to be seated in a row of 9 chairs.

What is the probability that


a. None of the Couples are sitting together


b. Only one couple is sitting together


c. All the couples are sitting together

Plz post explanations along with answers.
2135
vikas.mogle Says
But probability of getting selected increases, doesn't it? I mean the more the vacancies, more are the chances you one of them gets the job.


Hey vikas...

yes, agreed that probability of selection increases with more no of vacancies...But we are forgetting that the given prob of husband and wife of being selected is for a given no of vacancies. If vacancies increase, their individual prob would increase, but we should not multiply the given prob by no of vacancies..

For eg. For a given post, chances of me getting selected is say 0.2
so, if there are more than 5 vacancies i am more than 100 % certain of getting a job :grin:...My point is that prob of 0.2 is for a given no of vacancies in the co...any change in no of vacancies wud change the probability !!

hope this helps...