BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Really challenging - Word problems

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 271
Joined: Tue May 22, 2012 3:22 am
Thanked: 7 times
Followed by:3 members

Really challenging - Word problems

by \'manpreet singh » Mon Jul 02, 2012 3:28 am
Q.1)A museum offers four video programs that run continuously throughout the day, each program starting anew as soon as it is finished. The first program runs every 15 minutes, the second every 30 minutes, the third every 45 minutes, and the fourth every 40 minutes; the first show of each program starts at 10:00 A.M. and the last showing of each program ends at 4:00 P.M. If a tour group can watch the programs in any order, but needs at least ten minutes between programs to regroup, what is the least amount of time the group can take to watch all four programs?

Q.2)Velma has exactly one week to learn all 71 Japanese hiragana characters. If she can learn at most a dozen of them on anyone day and will only have time to learn four of them on Friday, what is the least number of hiragana that Velma will have to learn on Saturday?

Q.3)Shaggy has to learn the same 71 hiragana characters, and also has one week to do so; unlike Velma, he can learn as many per day as he wants. However, Shaggy has decided to obey the advice of a study-skills professional, who has advised him that the number of characters he learns on anyone day should be within 4 of the number he learns on any other day.
a. What is the least number of hiragana that Shaggy could have to learn on Saturday?
b. What is the greatest number of hiragana that Shaggy could have tolearn on Saturday?
Top
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Mon Jul 02, 2012 3:38 am
'manpreet singh wrote:Q.1)A museum offers four video programs that run continuously throughout the day, each program starting anew as soon as it is finished. The first program runs every 15 minutes, the second every 30 minutes, the third every 45 minutes, and the fourth every 40 minutes; the first show of each program starts at 10:00 A.M. and the last showing of each program ends at 4:00 P.M. If a tour group can watch the programs in any order, but needs at least ten minutes between programs to regroup, what is the least amount of time the group can take to watch all four programs?
To minimize the total time, as many breaks as possible must last only 10 minutes each.
List the starting times of each show and create a schedule from the available times.

The 15-minute video begins at:
10:00, 10:15, 10:30, 10:45
11:00, 11:15, 11:30, 11:45
etc.

The 30-minute video begins at:
10:00, 10:30
11:00, 11:30
12:00, 12:30
etc.

The 40-minute video begins at:
10:00, 10:40, 11:20
12:00, 12:40, 1:20
etc.

The 45-minute video begins at:
10:00, 10:45, 11:30, 12:15
1, 1:45, 2:30, 3:15
etc.

Note that no video prior to 2pm starts 10 minutes after the 45-minute video.
Prior to 2pm, the 45-minute video must be followed by a break of at least 15 minutes.
(While the 1:45 showing of the 45-minute video ends at 2:30 and thus could be followed by the 2:40 showing of the 40-minute video, the resulting schedule would not minimize the total amount of time needed to view all 4 videos.)

Schedule:
10:00-10:30 = 30-minute video.
10:30-10:40 = 10-minute break.

10:40-11:20 = 40-minute video.
11:20-11:30 = 10-minute break.

11:30-12:15 = 45-minute video.
12:15-12:30 = 15-minute break.

12:30-12:45 = 15-minute video.

Total time = 2 hours, 45 minutes.

Since the number of 10-minute breaks has been maximized, there is no way to minimize the time further.
Last edited by GMATGuruNY on Mon Jul 02, 2012 9:54 am, edited 5 times in total.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
GMAT Score:770

by Anurag@Gurome » Mon Jul 02, 2012 6:51 am
'manpreet singh wrote:Q.1)A museum offers four video programs that run continuously throughout the day, each program starting anew as soon as it is finished. The first program runs every 15 minutes, the second every 30 minutes, the third every 45 minutes, and the fourth every 40 minutes; the first show of each program starts at 10:00 A.M. and the last showing of each program ends at 4:00 P.M. If a tour group can watch the programs in any order, but needs at least ten minutes between programs to regroup, what is the least amount of time the group can take to watch all four programs?
This problem can be solved by listing down all the program times and checking which order of watching programs minimizes the total time including the break times. But I'm going to show a methodical approach to tackle this kind of problems.

Note that the order of watching the programs is important.
Total time they should take = (A + 10) + (B + 10) + (C + 10) + D
Where A, B, C, and D are either 15 or 30 or 40 or 45.

Now to satisfy the conditions strictly,
  • 1. (A + 10) should be multiple of B
    2. (A + B + 20) should be a multiple of C
    3. (A + B + C + 30) should be a multiple of D
Hence, the possibilities are
  • 1. Only choice --> A = 30 and B = 40
    2. Already we have set A = 30, B = 40. Hence, C = 15 or 45
    3. We have to sub-cases,
    • a) For C = 15, (A + B + C + 30) = 115 --> Not multiple of 45
      To make it multiple of 45 we need to add another (135 - 115) = 20 minutes to the last break
      Making the total = (135 + 45) = 180 minutes
      b) For C = 45, (A + B + C + 30) = 145 --> Not multiple of 15
      To make it multiple of 15 we need to add another (150 - 145) = 5 minutes to the last break
      Making the total = (150 + 15) = 165 minutes
Hence, least amount of time the group can take to watch all four programs = 165 minutes = 2 hours and 45 minutes
GMATGuruNY wrote:12:00-12:45 = 45 minute show.
A 45 minute program cannot start at 12:00 PM.
It will run at 10:00 AM, 10:45 AM, 11:30 AM, 12:15 PM, ... so on.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
GMAT Score:770

by Anurag@Gurome » Mon Jul 02, 2012 7:09 am
'manpreet singh wrote:Q.2)Velma has exactly one week to learn all 71 Japanese hiragana characters. If she can learn at most a dozen of them on anyone day and will only have time to learn four of them on Friday, what is the least number of hiragana that Velma will have to learn on Saturday?
Velma has to learn (71 - 4) = 67 characters in 6 day.
To minimize the number of characters in any one day, we have to maximize the number of characters in other 5 days.

Hence, in 5 days except Friday and Saturday, Velma will learn 5*12 = 60 characters.

Hence, the least number of characters that Velma will have to learn on Saturday = (67 - 60) = 7
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
GMAT Score:770

by Anurag@Gurome » Mon Jul 02, 2012 7:36 am
'manpreet singh wrote:Q.3)Shaggy has to learn the same 71 hiragana characters, and also has one week to do so; unlike Velma, he can learn as many per day as he wants. However, Shaggy has decided to obey the advice of a study-skills professional, who has advised him that the number of characters he learns on anyone day should be within 4 of the number he learns on any other day.
a. What is the least number of hiragana that Shaggy could have to learn on Saturday?
b. What is the greatest number of hiragana that Shaggy could have tolearn on Saturday?
If the minimum number of characters Shaggy learns in one day is Min, then maximum number of characters he can learn in one day is (Min + 4)

So, Min + 6(Min + 4) ≥ 71
--> 7Min ≥ (71 - 24) = 47
--> Min ≥ 47/7 = 6.(Something)

Hence, least number of characters that Shaggy could have to learn on Saturday = 7

Similarly, if the maximum number of characters Shaggy learns in one day is Max, then minimum number of characters he can learn in one day is (Max - 4)

So, Max + 6(Max - 4) ≤ 71
--> 7Max ≤ (71 + 24) = 95
--> Max ≤ 95/7 = 13.(Something)

Hence, greatest number of characters that Shaggy could have to learn on Saturday = 13
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Master | Next Rank: 500 Posts
Posts: 271
Joined: Tue May 22, 2012 3:22 am
Thanked: 7 times
Followed by:3 members

by \'manpreet singh » Mon Jul 02, 2012 11:45 pm
Thanks Anurag and Mitch!!! Your methods are simple and easy to Grasp.
In the last question where i miss judged was the statement below...
the number of characters he learns on anyone day should be within 4 of the number he learns on any other day.
I thought each day was having a difference of 4 to the previous day....so the statement did me in here... :? :x
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 206
Joined: Sun Jun 24, 2012 5:44 pm
Thanked: 5 times
Followed by:3 members

by [email protected] » Mon Oct 29, 2012 11:45 pm
Anurag can u plz explain the relavance of signs greater than and less than in both the two scenarios. I fail to understand how can the characters exceed or be less than 71.

Min + 6(Min + 4) ≥ 71

Max + 6(Max - 4) ≤ 71

Anurag@Gurome wrote:
'manpreet singh wrote:Q.3)Shaggy has to learn the same 71 hiragana characters, and also has one week to do so; unlike Velma, he can learn as many per day as he wants. However, Shaggy has decided to obey the advice of a study-skills professional, who has advised him that the number of characters he learns on anyone day should be within 4 of the number he learns on any other day.
a. What is the least number of hiragana that Shaggy could have to learn on Saturday?
b. What is the greatest number of hiragana that Shaggy could have tolearn on Saturday?
If the minimum number of characters Shaggy learns in one day is Min, then maximum number of characters he can learn in one day is (Min + 4)

So, Min + 6(Min + 4) ≥ 71
--> 7Min ≥ (71 - 24) = 47
--> Min ≥ 47/7 = 6.(Something)

Hence, least number of characters that Shaggy could have to learn on Saturday = 7

Similarly, if the maximum number of characters Shaggy learns in one day is Max, then minimum number of characters he can learn in one day is (Max - 4)

So, Max + 6(Max - 4) ≤ 71
--> 7Max ≤ (71 + 24) = 95
--> Max ≤ 95/7 = 13.(Something)

Hence, greatest number of characters that Shaggy could have to learn on Saturday = 13
Top
Master | Next Rank: 500 Posts
Posts: 116
Joined: Wed Oct 03, 2012 10:44 pm
Thanked: 5 times
Followed by:1 members

by theachiever » Tue Oct 30, 2012 7:19 am
Hi Guys Need Help with this Problem

Of the Three digit integers greater than 700,how many have two digits that are equal to each other and the remaining digit different from the other two?

a)90
b)82
c)80
d)45
e)36
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 279
Joined: Mon Jun 25, 2012 10:56 pm
Thanked: 60 times
Followed by:10 members

by anuprajan5 » Tue Oct 30, 2012 8:07 am
The answer should be C



The way to solve this would be to reverse this around.

numbers between 700 and 999 - 299


a. Find all the numbers where the digits are different from each other.
Using slot method,
Hundreds digits - 3 ways (7,8,9)
Tens digit - 9 ways (0-9 except for the digit chosen for the hundreds digit)
Units digit - 8 ways (0-9 except for the hundreds and tens digit)

3*8*9 = 216


b. Find all the numbers where all the digits are equal - 777,888,999 (3 cases)

[spoiler]299-216-3 = 80 ways. Answer C[/spoiler]

Oh and it would be great if you put up future posts as new ones.
Regards
Anup

The only lines that matter - are the ones you make!

https://www.youtube.com/watch?v=kk4sZcG ... ata_player
Top
Master | Next Rank: 500 Posts
Posts: 116
Joined: Wed Oct 03, 2012 10:44 pm
Thanked: 5 times
Followed by:1 members

by theachiever » Tue Oct 30, 2012 8:57 pm
Thanks Anup. Will surely post future queries as a new post...... :)






anuprajan5 wrote:The answer should be C



The way to solve this would be to reverse this around.

numbers between 700 and 999 - 299


a. Find all the numbers where the digits are different from each other.
Using slot method,
Hundreds digits - 3 ways (7,8,9)
Tens digit - 9 ways (0-9 except for the digit chosen for the hundreds digit)
Units digit - 8 ways (0-9 except for the hundreds and tens digit)

3*8*9 = 216


b. Find all the numbers where all the digits are equal - 777,888,999 (3 cases)

[spoiler]299-216-3 = 80 ways. Answer C[/spoiler]

Oh and it would be great if you put up future posts as new ones.
Top
User avatar
Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Wed Oct 16, 2013 6:18 am

by adirising » Wed Oct 16, 2013 6:24 am
Anurag@Gurome wrote:
'manpreet singh wrote:Q.2)Velma has exactly one week to learn all 71 Japanese hiragana characters. If she can learn at most a dozen of them on anyone day and will only have time to learn four of them on Friday, what is the least number of hiragana that Velma will have to learn on Saturday?
Velma has to learn (71 - 4) = 67 characters in 6 day.
To minimize the number of characters in any one day, we have to maximize the number of characters in other 5 days.

Hence, in 5 days except Friday and Saturday, Velma will learn 5*12 = 60 characters.

Hence, the least number of characters that Velma will have to learn on Saturday = (67 - 60) = 7
Hi ,

How I interpret this question is that Velma can learn only 4 characters on Friday and can only learn at max 12 characters on any one day, this means say on Thursday she learns 12 characters. We are left with now 71-12-4=55 characters.

Therefore, she has to learn a minimum of 11 characters on the remaining five days as the threshold of learning max character in a day is 12.

Please correct if I am wrong
Top
Post new topic Post Reply
• Page 1 of 1