GMAT Problem Solving Discussions

bhavin422 · April 22, 2009, 12:17pm

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?

Total no of committees = 10*8*6/3! = 80

There are a couple of more ways to approach this one...
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.

i guess this has been posted in earlier posts..

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, 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)

will post the 3rd problem in some time...at work now 😞

d2rockstar · April 22, 2009, 2:11pm

bhavin422 Says
will post the 3rd problem in some time...at work now :(

for first question

I guess it should be 10*8*6=480

Can you explain why do u divide it by 3!?

tryitnow · April 22, 2009, 2:51pm

1. A student got 7, 8 and 5 in 3 consecutive exams (out of 10). The weightage assigned to each is 60%, 50% and 40% respectively. What is the average score?

Shall I divide by 3 or by w1 + w2 + w3? I bet the answer is 3.4 but one of my friend thinks it should be divided by w1 + w2 + w3.

2. Out of 4 integers the avg. of the first 3 is 16 and the last 3 is 15 the last number being 18. Whats the first number?

This problem is in Arun Sharma Avg. LODI Q20.. They have 21 as the correct answer but I am getting 20. Am I missing something or is it a misprint?

Thanks in advance!

bhavin422 · April 22, 2009, 3:24pm

for first question

I guess it should be 10*8*6=480

Can you explain why do u divide it by 3!?

this is a problem of combinations and not permutations i.e the order of selection of 3 people in the committee does not matter, all that matters is that who r d ppl on committee ...

So, u divide by 3 !, which is the total no of arrangements within them ....

I.e If suppossing A, B, C are d selected 3 ppl out of 10 on committee..

ABC, ACB, BAC, BCA, CAB, CBA are nt different...they r one n the same thing ..

bhavin422 · April 22, 2009, 3:33pm

1. A student got 7, 8 and 5 in 3 consecutive exams (out of 10). The weightage assigned to each is 60%, 50% and 40% respectively. What is the average score?

Shall I divide by 3 or by w1 + w2 + w3? I bet the answer is 3.4 but one of my friend thinks it should be divided by w1 + w2 + w3.

2. Out of 4 integers the avg. of the first 3 is 16 and the last 3 is 15 the last number being 18. Whats the first number?

This problem is in Arun Sharma Avg. LODI Q20.. They have 21 as the correct answer but I am getting 20. Am I missing something or is it a misprint?

Thanks in advance!

Sum 1 :

u divide by sum of weights and not 3 ...
avg = (0.6*7+0.5*8+0.4*5)/(0.6+0.5+0.4) = 6.8

reverse logic also confirms that average should always be within the extreme values ..3.4 is not correct

Sum 2 :

a+b+c = 48
b+c+d= 45

Hence, a-d = 3
i.e a-18=3

Hence first term is 21

d2rockstar · April 22, 2009, 3:58pm

this is a problem of combinations and not permutations i.e the order of selection of 3 people in the committee does not matter, all that matters is that who r d ppl on committee ...

So, u divide by 3 !, which is the total no of arrangements within them ....

I.e If suppossing A, B, C are d selected 3 ppl out of 10 on committee..

ABC, ACB, BAC, BCA, CAB, CBA are nt different...they r one n the same thing ..

sorry boss...but as far as I know we r jus selecting ppl...we never arranged them...we selected one out of 10, den 1 out of 8 and den 1 out of 6..its all selection how did arrangement creeped in????

bhavin422 · April 23, 2009, 8:19am

d2rockstar Says
sorry boss...but as far as I know we r jus selecting ppl...we never arranged them...we selected one out of 10, den 1 out of 8 and den 1 out of 6..its all selection how did arrangement creeped in????

Ok..let us try to analyse from a different angle...

Selection for 1st position = 10C1= 10 ways ...say person A
Selection for second position = 8C1 = 8 ways( since u cannot select d spouse) ....say person B
Selection for 3rd position = 6C1 = 6 ways ...say person C...

my point was ...how would it matter if person A was selected for second or third position and same is applicable for B and C as well ...it is multiple counting ...hence u need to divide by total no of arrangements within them...

If u r not convinced this way....

Another approach :

we need to select at the max only 1 person from every couple..
3 couples out of 5 can be selected in 5C3 ways ..
and for each of the 3 couples there are 2 ways for selecting either a husband or wife ..

hence total selections possible = 5C3*2*2*2 = 10*2^3 = 80 ...

Approach 3:

selection of 3 ppl out of 10 without a couple = total no of ways of selecting of 3 ppl out of 10 - no of ways of selecting 3 out of 10 including a couple

= 10C3 - 5C1*8 (1 out of 5 couples in 5C1 ways * 1 out of rem 8 )
= 120 - 40
= 80 ...Ans

Hope this helps 😃

vikas130678 · April 23, 2009, 9:39am

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.

Hi , I am trying to post answers here based on the above two questions explanations-
C.=>4!*3!*2!=288
B.=>8!*2!-288
A.=>Total arrangements-only one couple sitting
=9!-B

Plz evaluate the procedure.

vikas130678 · April 23, 2009, 9:42am

1.The telephone company wants to add an area code composed of 2 letters to every phone number. In order to do so, the company chose a special sign language containing 124 different signs. If the company used 122 of the signs fully and two remained unused, how many additional area codes can be created if the company uses all 124 signs?

2.In how many ways can you sit 7 people on a bench if Suzan wont sit on the middle seat or on either end?

msalot · April 23, 2009, 6:40pm

Mary typed a six-digit number, but the two 4s she typed didn't show. What appeared was 2002. How many diďŹ€erent six-digit numbers could she have typed?

A.4

B.8

C.10

D.15

E.20[LEFT]

Help Guys

[/LEFT]

srijack · April 23, 2009, 7:11pm

1.The telephone company wants to add an area code composed of 2 letters to every phone number. In order to do so, the company chose a special sign language containing 124 different signs. If the company used 122 of the signs fully and two remained unused, how many additional area codes can be created if the company uses all 124 signs?
2.In how many ways can you sit 7 people on a bench if Suzan wont sit on the middle seat or on either end?

Hi
1.There are certain number of codes formed using 122 signs. Now using 2 new signs along with 122 one can create 244+244+2=490 additional codes ( two distinct signs can form two codes, 1 new sign along with 122 signs can give 244 codes ( AB,BA) ), 2 new signs form two more codes.

2.lets make suzan sit in the middle
1,2,3,4,5,6,7
1- in 6 ways first position can be filled
2- 5 ways
3- 4 ways
4- 1 way (suzan)
5- 3 ways
6- 2 ways
7- 1 way
total = 6*5*4*3*2*1
or suzan sitting at left end 6!
or suzan sitting at rt end 6!
in total 3*6! ways suzan is sitting in the middle,or either ends whcih should be subtracted from total no of ways.
7!
so the no of ways = 7!- (3*6!)
or the ans is 4*6!

rds-
sri.

star_highway · April 24, 2009, 5:19am

Mary typed a six-digit number, but the two 4s she typed didnt show. What appeared was 2002. How many diďŹ€erent six-digit numbers could she have typed?

A.4

B.8

C.10

D.15

E.20[LEFT]

Help Guys

[/LEFT]

_ 2 _ 0 _ 0 _ 2 _

there are 5 possible locations for first 4.

if the first 4 goes at location
1, the second 4 can go at 5 different locations, number of ways = 5
2, the second 4 can go at 4 different locations, number of ways = 4
3, the second 4 can go at 3 different locations, number of ways = 3
4, the second 4 can go at 2 different locations, number of ways = 2
5, the second 4 can go at 1 different location, number of ways = 1

Total number of different 6 digit numbers = 5+4+3+2+1 = 15

Option D

Good question BTW .....

bhavin422 · April 24, 2009, 6:07am

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.

excellent sum ...havent seen a similar kind for GMAT

this is like a venn diagram involving intersection of 3 groups ..

Grp 1 = 1st couple together, likewise Grp 2 and Grp 3
Grp 12 = couples 1 n 2 together, likewise Grp23 n Grp 13
Grp 123 = all couples together

Grp 1 = Grp 2 = Grp3 = 8! * 2! (couple treated as a unit)
Grp12 = Grp13 = Grp23 = 7! * 2! * 2!
Grp 123 = 6! * 2! * 2! *2! ( 3 couples as 3 units, and 3 stags, so 6 units )

Now, fr 3 way intersection , union is given by

Union(atleast 1 couple together) = Grp1 + Grp2 + Grp3 - Grp12 - Grp13 - Grp23 + Grp123

= 3*8!*2!-3*7!*2!*2!+6!*2!*2!*2! ...(1)

Ans A:
no couple together = total poss - atleast 1 couple together
= 9! -

Ans B:
only grp1 = Grp1 - Grp13 - Grp12 + Grp123
= 8!*2! - 7!*2!*2! - 7!*2! *2! + 6! *2! *2!*2! ....eqn (2)

Likewise for Only grp2 and Grp 3 as well

Hence,

Only one grp together=3 *

Ans C :

All couples together = 6!*2!*2!*2!

Pls correct for approach ...am sure there has to be a shorter way ...without calci , nt sure hv to get these sum within 2-3 mins on exam

mr_derivatives · April 24, 2009, 6:44am

Two players P & Q each have a well shuffled standard pack of cards (with no jokers). The players deal their cards 1 at a time, from the top of the deck, checking for an exact match. Player P wins if, once the packs are fully dealt, no matches are found. And player Q wins if at least one match occurs. What is the probability that player P wins?

Answer is 1 1/1! + 1/2! 1/3! + ... + 1/52!

Pls reply if this answer is correct and also explanation.
Thanks

bhavin422 · April 24, 2009, 1:38pm

Two players P & Q each have a well shuffled standard pack of cards (with no jokers). The players deal their cards 1 at a time, from the top of the deck, checking for an exact match. Player P wins if, once the packs are fully dealt, no matches are found. And player Q wins if at least one match occurs. What is the probability that player P wins?

Answer is 1 1/1! + 1/2! 1/3! + ... + 1/52!

Pls reply if this answer is correct and also explanation.
Thanks

yes...this is a problem of dearrangement...
i guess this is the best way to solve this problem..

srworld · April 25, 2009, 8:04pm

Answer is 5C2 which is 5!/(5-2)! 2! = 5!/3! 2! = 10.

_ 2 _ 0 _ 0 _ 2 _

1st 4 can be placed at any of the 5 positions. Once 1st 4 is placed, the 2nd 4 can be placed at any of the remaining 4 positions. So total number of ways 2 4's can be placed here is 5*4. Since both the number are 4 so positions have duplicate and the correct answer is 5*4/2 = 20/2 = 10.

ramviswa · April 26, 2009, 6:05am

A student worked without interruption to complete a two part test in 1 hour. If he completed the first part in 1/2 the time it took him to complete the second part, how many minutes did it take him to complete the second part?

a. 24
b. 36
c. 36
d. 38
e. 40

Option e is my answer but the correct answer is b. I disagree. If anyone get the answer as b) please post your explanation.

ramviswa · April 26, 2009, 6:16am

One more question

A person takes a spherical ball and three nails are fixed on it. Then he joins the three nails by a tight string and each segment of the string is 5 cm. What will be the area of the closed figure so formed in square units?

a) 25(root 3)/4
b) less than 25(root 3)/4
c) less than or equal to 25(root 3)/4
d) greater than 25(root 3)/4
e) greater than or equal to 25(root 3)/4

jigar_p_civil · April 26, 2009, 7:11am

A student worked without interruption to complete a two part test in 1 hour. If he completed the first part in 1/2 the time it took him to complete the second part, how many minutes did it take him to complete the second part?

a. 24
b. 36
c. 36
d. 38
e. 40

Option e is my answer but the correct answer is b. I disagree. If anyone get the answer as b) please post your explanation.

yes friend i agree with u ans is 40

jigar_p_civil · April 26, 2009, 7:19am

One more question

A person takes a spherical ball and three nails are fixed on it. Then he joins the three nails by a tight string and each segment of the string is 5 cm. What will be the area of the closed figure so formed in square units?

a) 25(root 3)/4
b) less than 25(root 3)/4
c) less than or equal to 25(root 3)/4
d) greater than 25(root 3)/4
e) greater than or equal to 25(root 3)/4

here with the three points its became equilateral triangle

so ans is 25sqrt3/4

but spherical ball so i think ans is less than 25sqrt3/4 (i am not sure")