Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Good set of PS 2 [#permalink]
01 Jul 2010, 13:23

------------------------------------------------ 2. What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits? (A) 271/900 (B) 27/100 (C) 7/25 (D) 1/9 (E) 1/10

Total 3 digit numbers 900, 3 digit number with no 7 =8*9*9=648, P(at least one 7)=1-P(no 7)=1-648/900=252/900=7/25

Answer: C. ------------------------------------------ How do you derive 3 digit number with no '7 =8*9*9' ???

Re: Good set of PS 2 [#permalink]
13 Aug 2010, 11:07

Hi Bunuel

for the time and work question in this problem set can you pls explain this step in a bit more detail:

"Then 6 more people was hired --> speed of construction increased by 1.6, days needed to finish 55/1.6=34.375"

How did you figure out the speed increased by 1.6?

For such problems I tend to reduce the question to how much work did the workers do in 1 day and then to how much work did each worker do in one day and then multiply that by 6 to get what 6 workers would have done in a day; add that to what 10 workers would have done in a day; take reciprocal of the fraction to see how much time 16 workers would take for that work---

your method seems much better.... can you explain that step of figuring out the increased speed of 1.6.... thanks.

Re: Good set of PS 2 [#permalink]
13 Aug 2010, 12:50

1

This post received KUDOS

Expert's post

gmat1011 wrote:

Hi Bunuel

for the time and work question in this problem set can you pls explain this step in a bit more detail:

"Then 6 more people was hired --> speed of construction increased by 1.6, days needed to finish 55/1.6=34.375"

How did you figure out the speed increased by 1.6?

For such problems I tend to reduce the question to how much work did the workers do in 1 day and then to how much work did each worker do in one day and then multiply that by 6 to get what 6 workers would have done in a day; add that to what 10 workers would have done in a day; take reciprocal of the fraction to see how much time 16 workers would take for that work---

your method seems much better.... can you explain that step of figuring out the increased speed of 1.6.... thanks.

Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance.

We know that 10 people need 55 days to complete the job --> let the combined rate of 10 people be x --> \(Time*Rate=x*55=Job \ done\); Now, if combined rate of 10 people is \(x\), then combined rate of 16 man will be \(1.6x\) (1.6 times more than x as 16 is 1.6 times more than 10), so \(new \ time*new \ rate=same \ job\) --> \(t_2*1.6x=x*55\) --> \(t_2=\frac{55}{1.6}\approx{34.8}\) (as rate increased 1.6 times then time needed to do the same job will decrease 1.6 times).

Re: Good set of PS 2 [#permalink]
23 Aug 2010, 12:09

I found question 4 interesting, and I usually attempt it by plugging in. A little more time consuming but I find it helpful. Lets assume the job is to make 1100 widgets, that means that each man is expected to make 1 widget per day, and a ten man crew can make 10 widgets. So after 55 days, 550 widgets have been made, thus 550 are left. Since they hire 6 more guys, it will take a little over 34 days to compete the job at a rate of 16 widgets per day. Since they do not work on any day that it rains this means that they worked for 35 out of the 40 days it took to finish the job. 40-35=5.

Re: Good set of PS 2 [#permalink]
30 Sep 2010, 13:55

I don't know if it's the broken English, but question 9 doesn't make any sense. 15 people applied to both college X and college Y. Those 15 people are composed of 20% of the people who applied to college X and 25% of the people who applied to college Y. So .2x + .25y = 15.

What if 50 people applied to college X and 20 people applied to college Y? Then you have (.2)(50) + (.25)(20) = 10 + 5 = 15. But 70 isn't an answer choice.

Re: Good set of PS 2 [#permalink]
30 Sep 2010, 14:59

Expert's post

TehJay wrote:

I don't know if it's the broken English, but question 9 doesn't make any sense. 15 people applied to both college X and college Y. Those 15 people are composed of 20% of the people who applied to college X and 25% of the people who applied to college Y. So .2x + .25y = 15.

What if 50 people applied to college X and 20 people applied to college Y? Then you have (.2)(50) + (.25)(20) = 10 + 5 = 15. But 70 isn't an answer choice.

9. Of the applicants passes a certain test, 15 applied to both college X and Y. If 20 % of the applicants who applied college X and 25% of the applicants who applied college Y applied both college X and Y, how many applicants applied only college X or college Y? (A) 135 (B) 120 (C) 115 (D) 105 (E) 90

Question means the following: 15 people applied to both college X and Y, these 15 people represent 20% of all applicants who applied to X and 25% of all applicants who applied to Y.

\(0.2*x=15\) --> \(x=75\) --> # of people who applied only to X is: \(75-15=60\); \(0.25*y=15\) --> \(y=60\) --> # of people who applied only to Y is: \(60-15=45\);

# of people who applied only to X OR Y is: \(60+45=105\).

Re: Good set of PS 2 [#permalink]
02 Oct 2010, 02:03

Bunuel wrote:

ANSWERS (OA):

1. A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there? (A) 28 (B) 32 (C) 48 (D) 60 (E) 120

As most of the combination problems this one can be solved in more than 1 way:

Sisters sit separately: 1. one of them is on the front seat (2 ways). Others (including second sister) can be arranged in: 2 (drivers seat)*3! (arrangements of three on the back seat)=12 ways. Total for this case: 2*12=24 Or 2. both by the window on the back seat (2 ways). Others can be arranged in: 2 (drivers seat)*2 (front seat)*1(one left to sit between the sisters on the back seat)=4 ways. Total for this case=8. Total=24+8=32.

"Another way: Total number of arrangements-arrangements with sisters sitting together=2*4*3!-2*2(sisters together)*2*2*1(arrangement of others)=48-16=32"

Answer: B.

Bunuel,

For Q1, Can u please elaborate more on this >> Another way: Total number of arrangements-arrangements with sisters sitting together=2*4*3!-2*2(sisters together)*2*2*1(arrangement of others)=48-16=32

I was able to get answer by first method as u have mentioned. But, I am failing to get the computation mentioned in another method.

Re: Good set of PS 2 [#permalink]
02 Oct 2010, 04:27

Expert's post

samark wrote:

Bunuel wrote:

ANSWERS (OA):

1. A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there? (A) 28 (B) 32 (C) 48 (D) 60 (E) 120

As most of the combination problems this one can be solved in more than 1 way:

Sisters sit separately: 1. one of them is on the front seat (2 ways). Others (including second sister) can be arranged in: 2 (drivers seat)*3! (arrangements of three on the back seat)=12 ways. Total for this case: 2*12=24 Or 2. both by the window on the back seat (2 ways). Others can be arranged in: 2 (drivers seat)*2 (front seat)*1(one left to sit between the sisters on the back seat)=4 ways. Total for this case=8. Total=24+8=32.

"Another way: Total number of arrangements-arrangements with sisters sitting together=2*4*3!-2*2(sisters together)*2*2*1(arrangement of others)=48-16=32"

Answer: B.

Bunuel,

For Q1, Can u please elaborate more on this >> Another way: Total number of arrangements-arrangements with sisters sitting together=2*4*3!-2*2(sisters together)*2*2*1(arrangement of others)=48-16=32

I was able to get answer by first method as u have mentioned. But, I am failing to get the computation mentioned in another method.

Thanks!

Total # of arrangements: Drivers seat: 2 (either mother or father); Front seat: 4 (any of 4 family members left); Back seat: 3! (arranging other 3 family members on the back seat); So. total # of arrangements is 2*4*3!=48.

# of arrangements with sisters sitting together: Sisters can sit together only on the back seat either by the left window or by the right window - 2, and either {S1,S2} or {S2,S1} - 2 --> 2*2=4; Drivers seat: 2 (either mother or father); Front seat: 2 (5 - 2 sisters on back seat - 1 driver = 2); Back seat with sisters: 1 (the last family member left); So, # of arrangements with sisters sitting together is 4*2*2*1=16.

Re: Good set of PS 2 [#permalink]
03 Oct 2010, 09:09

my approach was like this -

Case 1a- Father Drives - Co-driver is Daughter1/Daughter2 So it is possible in 3!*2 ways. Case 1b -Father Drives Co-driver is occupied by Son or Mother + both daughters are at the end of back seat - this is possible in 4 ways...So totat to case 1 is 16 that is going to be symetrical when mother drives.Hence, total ways = 32

Re: Good set of PS 2 [#permalink]
27 Oct 2010, 22:41

was just wondering if these questions can be combined together into one document... or maybe in an online test format. Is it possible ? ( same comment as in the previous section) _________________

Regards, If you like my post, consider giving me akudo. THANKS!

Re: Good set of PS 2 [#permalink]
30 Oct 2010, 16:45

Perfect Answer

Economist wrote:

Bunuel wrote:

Please find below new set of PS problems:

4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8

Nobody attempted 4, let me give a try. total work = 110*10 = 1100 man days now, from day 1 to day 55, 10 men worked = 550 man days of work was done. from day 61 to day 100, 16 men worked = 640 man days of work was done.

so total work done should be 1190, the 90 days offset is due to rain on few days between 61 to 100th day...so the number of rainy days should be 90/16=5.625 ~ 6. C.

Re: Good set of PS 2 [#permalink]
09 Jan 2011, 11:22

Expert's post

144144 wrote:

can someone plz try and show me question 2 in the C way?

1 mean 1C9. thanks.

2. What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits? (A) 271/900 (B) 27/100 (C) 7/25 (D) 1/9 (E) 1/10

There are total 900 3 digit numbers;

There are 8*9*9=648 (you can write this as 8C1*9C1*9C1 if you like) 3-digit numbers without 7 in its digits (first digit can take 8 values from 1 to 9 excluding 7; second and third digits can take 9 values from 0 to 9 excluding 7),

P(at least one 7)=1-P(no 7)=1-648/900=252/900=7/25.

Re: Good set of PS 2 [#permalink]
09 Jan 2011, 15:10

pesfunk wrote:

Perfect Answer

Economist wrote:

Bunuel wrote:

Please find below new set of PS problems:

4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8

Nobody attempted 4, let me give a try. total work = 110*10 = 1100 man days now, from day 1 to day 55, 10 men worked = 550 man days of work was done. from day 61 to day 100, 16 men worked = 640 man days of work was done.

so total work done should be 1190, the 90 days offset is due to rain on few days between 61 to 100th day...so the number of rainy days should be 90/16=5.625 ~ 6. C.

I found one more solution to this problem by Bunnel but the answer is 5... that is B.. But here its mentioned C...

4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8

This one was solved incorrectly: Days to finish the job for 10 people 110 days. On the 61-st day, after 5 days of rain --> 5 days was rain, 55 days they worked, thus completed 1/2 of the job, 1/2 is left (55 days of work for 10 people). Then 6 more people was hired --> speed of construction increased by 1.6, days needed to finish 55/1.6=34.375, BUT after they were hired job was done in 100-60=40 days --> so 5 days rained. They needed MORE than 34 days to finish the job, so if it rained for 6 days they wouldn't be able to finish the job in 100(40) days. Answer: B.

Re: Good set of PS 2 [#permalink]
09 Jan 2011, 15:17

Expert's post

jullysabat wrote:

Economist wrote:

Bunuel wrote:

Please find below new set of PS problems:

4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8

Nobody attempted 4, let me give a try. total work = 110*10 = 1100 man days now, from day 1 to day 55, 10 men worked = 550 man days of work was done. from day 61 to day 100, 16 men worked = 640 man days of work was done.

so total work done should be 1190, the 90 days offset is due to rain on few days between 61 to 100th day...so the number of rainy days should be 90/16=5.625 ~ 6. C.

I found one more solution to this problem by Bunnel but the answer is 5... that is B.. But here its mentioned C...

4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8

This one was solved incorrectly: Days to finish the job for 10 people 110 days. On the 61-st day, after 5 days of rain --> 5 days was rain, 55 days they worked, thus completed 1/2 of the job, 1/2 is left (55 days of work for 10 people). Then 6 more people was hired --> speed of construction increased by 1.6, days needed to finish 55/1.6=34.375, BUT after they were hired job was done in 100-60=40 days --> so 5 days rained. They needed MORE than 34 days to finish the job, so if it rained for 6 days they wouldn't be able to finish the job in 100(40) days. Answer: B.

Solution provided by Economist is not correct (the one with answer C). OA's and solutions are given in my post on the 2nd page. OA for the quoted question is indeed B:

Given: 10-man crew needs 110 days to complete the construction.

"On the 61-st day, after 5 days of rain ..." --> as it was raining for 5 days then they must have bee working for 55 days thus completed 1/2 of the job, 1/2 is left (55 days of work for 10 men).

Then contractor "hired 6 more people" --> speed of construction increased 1.6 times, so the new 16-man crew needed 55/1.6=~34.4 days to complete the construction, but after they were hired job was done in 100-60=40 days --> so 5 days rained. (They needed MORE than 34 days to finish the job, so if it rained for 6 days they wouldn't be able to finish the job in 100(40) days.)

Re: Good set of PS 2 [#permalink]
09 Jan 2011, 16:14

4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8

This one was solved incorrectly: Days to finish the job for 10 people 110 days. On the 61-st day, after 5 days of rain --> 5 days was rain, 55 days they worked, thus completed 1/2 of the job, 1/2 is left (55 days of work for 10 people). Then 6 more people was hired --> speed of construction increased by 1.6, days needed to finish 55/1.6=34.375, BUT after they were hired job was done in 100-60=40 days --> so 5 days rained. They needed MORE than 34 days to finish the job, so if it rained for 6 days they wouldn't be able to finish the job in 100(40) days. Answer: B.[/quote]

Solution provided by Economist is not correct (the one with answer C). OA's and solutions are given in my post on the 2nd page. OA for the quoted question is indeed B:

Given: 10-man crew needs 110 days to complete the construction.

"On the 61-st day, after 5 days of rain ..." --> as it was raining for 5 days then they must have bee working for 55 days thus completed 1/2 of the job, 1/2 is left (55 days of work for 10 men).

Then contractor "hired 6 more people" --> speed of construction increased 1.6 times, so the new 16-man crew needed 55/1.6=~34.4 days to complete the construction, but after they were hired job was done in 100-60=40 days --> so 5 days rained. (They needed MORE than 34 days to finish the job, so if it rained for 6 days they wouldn't be able to finish the job in 100(40) days.)

Re: Good set of PS 2 [#permalink]
16 Jan 2011, 01:24

9. Of the applicants passes a certain test, 15 applied to both college X and Y. If 20 % of the applicants who applied college X and 25% of the applicants who applied college Y applied both college X and Y, how many applicants applied only college X or college Y? (A) 135 (B) 120 (C) 115 (D) 105 (E) 90

I am getting a single equation like .2X + 0.25Y = 15. I feel like missing something. Can someone explain what it could be?

gmatclubot

Re: Good set of PS 2
[#permalink]
16 Jan 2011, 01:24

Hello everyone! Researching, networking, and understanding the “feel” for a school are all part of the essential journey to a top MBA. Wouldn’t it be great... ...

Hey Everyone, I am launching a new venture focused on helping others get into the business school of their dreams. If you are planning to or have recently applied...