Tricky Time and distance problem

Expert replies

guerrero: Master | Next Rank: 500 Posts; Posts: 227; Joined: Sun Apr 08, 2012 4:53 amFollowed by: 3 members; Thanked: 12 times

by guerrero » Fri Jan 11, 2013 2:57 pm

Q1) Mike and Darren run a race between points A and B, 5 miles apart. Mike starts at 9 am from A at a speed of 5 mh/h, reaches B, and returns at the same speed. Darren starts at 9:45 am from A at a speed of 10 mh/h, reaches B and comes back to A at the same speed.
1)At what time do Mike and Darren first meet each other?
a)10 am b) 10:10 am c) 10:20 am d) 10:30 am

2)At what time does Darren overtake Mike? a) 10:20 am b) 10:30 am c) 10:40 am d) 10:50 am

Q2)If a person increases his usual speed by 20%, he reaches his office 15 minutes early. By how many minutes will he be late to his office, if he reduces his usual speed by 20%:

(a) 10 (b) 20 (c) 15.5 (d) 22.5

Please provide with the approach and explanation as well. Help appreciated. Regards

Like this post

Source: Beat The GMAT — Problem Solving | Fri Jan 11, 2013 2:57 pm

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Fri Jan 11, 2013 6:54 pm

guerrero wrote:If a person increases his usual speed by 20%, he reaches his office 15 minutes early. By how many minutes will he be late to his office, if he reduces his usual speed by 20%:

(a) 10 (b) 20 (c) 15.5 (d) 22.5

Let r = his usual rate and t = his usual time.

20% faster:
In other words, he travels at 120% of his usual rate.
Since rate and time are RECIPROCALS:
(120/100)r = (100/120)t = (5/6)t.
Since he takes 5/6 of his usual time, the 15 minutes saved at the faster speed is equal to (1/6)t:
15 = (1/6)t
t = 90 minutes.

20% slower:
In other words, he travels at 80% of his usual rate.
Since rate and time are RECIPROCALS:
(80/100)r = (100/80)t = (5/4)t.
Since he takes 5/4 of his usual time, he will be late by 1/4 of his usual time:
(1/4)90 = 22.5.

The correct answer is D.

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

Like this post

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Jan 12, 2013 4:36 am

guerrero wrote:Q1) Mike and Darren run a race between points A and B, 5 miles apart. Mike starts at 9 am from A at a speed of 5 mh/h, reaches B, and returns at the same speed. Darren starts at 9:45 am from A at a speed of 10 mh/h, reaches B and comes back to A at the same speed.
1)At what time do Mike and Darren first meet each other?
a)10 am b) 10:10 am c) 10:20 am d) 10:30 am

For Mike and Darren to meet, the following must be traveled:

A -------------------Mike------------------------> B
A --------Darren---------><---------Mike---------- B

The figure above shows that the total distance traveled -- Mike's distance + Darren's distance -- must be equal to the distance from A to B and back (10 miles).

From 9am-9:45am, the distance traveled by Mike = r*t = 5(3/4) = 15/4 miles.
Remaining distance = 10 - 15/4 = 25/4 miles.
From 9:45am on, Mike and Darren WORK TOGETHER to cover the remaining distance.
Combined rate for Mike and Darren = 5+10 = 15 miles per hour.
Time for Mike and Darren to travel the remaining distance = d/r = (25/4)/15 = 5/12 hours = 25 minutes.
9:45am + 25 minutes = 10:10am.

The correct answer is B.

2)At what time does Darren overtake Mike? a) 10:20 am b) 10:30 am c) 10:40 am d) 10:50 am

From 9am-9:45am, the distance traveled by Mike alone = r*t = 5(3/4) = 15/4 miles.
To overtake Mike, Darren must CATCH-UP by 15/4 miles..
The CATCH-UP rate is the DIFFERENCE between Darren's rate and Mike's rate:
10-5 = 5 miles per hour.
(Since every hour Darren travels 10 miles while Mike travels only 5 miles, Darren CATCHES UP by 5 miles every hour.)
Time for Darren to overtake Mike = d/r = (15/4)/5 = 3/4 hours.
Since Darren starts at 9:45am, the time at which he overtakes Mike = 9:45 + 3/4 hours = 10:30am.

The correct answer is B.

Like this post

subhakam: Senior | Next Rank: 100 Posts; Posts: 38; Joined: Mon Jun 21, 2010 12:37 pm

by subhakam » Mon Jan 14, 2013 8:11 pm

GMATGuruNY wrote:
guerrero wrote:If a person increases his usual speed by 20%, he reaches his office 15 minutes early. By how many minutes will he be late to his office, if he reduces his usual speed by 20%:

(a) 10 (b) 20 (c) 15.5 (d) 22.5
Let r = his usual rate and t = his usual time.

20% faster:
In other words, he travels at 120% of his usual rate.
Since rate and time are RECIPROCALS:
(120/100)r = (100/120)t = (5/6)t.
Since he takes 5/6 of his usual time, the 15 minutes saved at the faster speed is equal to (1/6)t:
15 = (1/6)t
t = 90 minutes.

20% slower:
In other words, he travels at 80% of his usual rate.
Since rate and time are RECIPROCALS:
(80/100)r = (100/80)t = (5/4)t.
Since he takes 5/4 of his usual time, he will be late by 1/4 of his usual time:
(1/4)90 = 22.5.

The correct answer is D.

How can we solve this using algebra? Can you post that method as well please?

Like this post

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon Jan 14, 2013 9:40 pm

subhakam wrote:
GMATGuruNY wrote:
guerrero wrote:If a person increases his usual speed by 20%, he reaches his office 15 minutes early. By how many minutes will he be late to his office, if he reduces his usual speed by 20%:

(a) 10 (b) 20 (c) 15.5 (d) 22.5
How can we solve this using algebra? Can you post that method as well please?
Let r = his usual rate and t = his usual time.
Distance = r*t.

When he travels at 120% of his usual rate, the time is 15 minutes less than his usual time:
Distance = (1.2r)(t-15).

Since the distance is the same in each case, we get:
r*t = (1.2r)(t-15)
t = 1.2t - 18
.2t = 18
t = 90.

Thus, the distance = r*t = r*90 = 90r.

When he travels at 80% of his usual rate, the time is x minutes longer than his usual time of 90 minutes:
Distance = (.8r)(x + 90).

Since the distance is unchanged, we get:
90r = (.8r)(x + 90)
90 = .8x + 72
18 = .8x
x = 22.5.

Like this post

vishugogo: Master | Next Rank: 500 Posts; Posts: 131; Joined: Tue Aug 30, 2011 4:50 amFollowed by: 6 members; Location: India; Thanked: 28 times

by vishugogo » Wed Jan 16, 2013 1:26 am

GMATGuruNY....in the second question how is this statement coming -- To overtake Mike, Darren must CATCH-UP by 15/4 miles..

GMATGuruNY wrote:
guerrero wrote:Q1) Mike and Darren run a race between points A and B, 5 miles apart. Mike starts at 9 am from A at a speed of 5 mh/h, reaches B, and returns at the same speed. Darren starts at 9:45 am from A at a speed of 10 mh/h, reaches B and comes back to A at the same speed.
1)At what time do Mike and Darren first meet each other?
a)10 am b) 10:10 am c) 10:20 am d) 10:30 am
For Mike and Darren to meet, the following must be traveled:

A -------------------Mike------------------------> B
A --------Darren---------><---------Mike---------- B

The figure above shows that the total distance traveled -- Mike's distance + Darren's distance -- must be equal to the distance from A to B and back (10 miles).

From 9am-9:45am, the distance traveled by Mike = r*t = 5(3/4) = 15/4 miles.
Remaining distance = 10 - 15/4 = 25/4 miles.
From 9:45am on, Mike and Darren WORK TOGETHER to cover the remaining distance.
Combined rate for Mike and Darren = 5+10 = 15 miles per hour.
Time for Mike and Darren to travel the remaining distance = d/r = (25/4)/15 = 5/12 hours = 25 minutes.
9:45am + 25 minutes = 10:10am.

The correct answer is B.

2)At what time does Darren overtake Mike? a) 10:20 am b) 10:30 am c) 10:40 am d) 10:50 am
From 9am-9:45am, the distance traveled by Mike alone = r*t = 5(3/4) = 15/4 miles.
To overtake Mike, Darren must CATCH-UP by 15/4 miles..
The CATCH-UP rate is the DIFFERENCE between Darren's rate and Mike's rate:
10-5 = 5 miles per hour.
(Since every hour Darren travels 10 miles while Mike travels only 5 miles, Darren CATCHES UP by 5 miles every hour.)
Time for Darren to overtake Mike = d/r = (15/4)/5 = 3/4 hours.
Since Darren starts at 9:45am, the time at which he overtakes Mike = 9:45 + 3/4 hours = 10:30am.

The correct answer is B.

Like this post

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Wed Jan 16, 2013 5:19 am

vishugogo wrote:GMATGuruNY....in the second question how is this statement coming -- To overtake Mike, Darren must CATCH-UP by 15/4 miles..
guerrero wrote:Q1) Mike and Darren run a race between points A and B, 5 miles apart. Mike starts at 9 am from A at a speed of 5 mh/h, reaches B, and returns at the same speed. Darren starts at 9:45 am from A at a speed of 10 mh/h, reaches B and comes back to A at the same speed.

At what time does Darren overtake Mike?

a) 10:20 am b) 10:30 am c) 10:40 am d) 10:50 am

From 9am - 9:45am, the distance traveled by Mike alone = r*t = 5(3/4) = 15/4 miles.
Thus, when Darren starts to travel at 9:45am, Mike is AHEAD of Darren by 15/4 miles.
In order to OVERTAKE Mike -- in other words, in order to PASS Mike -- Darren must CATCH UP by 15/4 miles.
To CATCH UP means to reach the person AHEAD of you.

Put another way:
Darren will overtake Mike when the two of them have each traveled the SAME DISTANCE.
At 10:30am:
The distance traveled by Mike in the 1.5 hours from 9-10:30am = r*t = 5(1.5) = 7.5 miles.
The distance traveled by Darren in the 3/4 hours from 9:45-10:30am = r*t = 10(.75) = 7.5 miles.
Since Mike and Darren will have traveled the same distance -- 7.5 miles -- Darren will OVERTAKE Mike at 10:30am.

Other CATCH-UP problems:

https://www.beatthegmat.com/tom-reads-at ... 51462.html
https://www.beatthegmat.com/rates-work-t119850.html
https://www.beatthegmat.com/p-and-q-circ ... 22287.html
https://www.beatthegmat.com/interesting- ... 16571.html
https://www.beatthegmat.com/rate-problem-t111575.html

Like this post

subhakam: Senior | Next Rank: 100 Posts; Posts: 38; Joined: Mon Jun 21, 2010 12:37 pm

by subhakam » Wed Jan 16, 2013 4:03 pm

GMATGuruNY wrote:
subhakam wrote:
GMATGuruNY wrote:
guerrero wrote:If a person increases his usual speed by 20%, he reaches his office 15 minutes early. By how many minutes will he be late to his office, if he reduces his usual speed by 20%:

(a) 10 (b) 20 (c) 15.5 (d) 22.5
How can we solve this using algebra? Can you post that method as well please?
Let r = his usual rate and t = his usual time.
Distance = r*t.

When he travels at 120% of his usual rate, the time is 15 minutes less than his usual time:
Distance = (1.2r)(t-15).

Since the distance is the same in each case, we get:
r*t = (1.2r)(t-15)
t = 1.2t - 18
.2t = 18
t = 90.

Thus, the distance = r*t = r*90 = 90r.

When he travels at 80% of his usual rate, the time is x minutes longer than his usual time of 90 minutes:
Distance = (.8r)(x + 90).

Since the distance is unchanged, we get:
90r = (.8r)(x + 90)
90 = .8x + 72
18 = .8x
x = 22.5.
Thank you very much GMATGuru!!

Like this post

vishugogo: Master | Next Rank: 500 Posts; Posts: 131; Joined: Tue Aug 30, 2011 4:50 amFollowed by: 6 members; Location: India; Thanked: 28 times

by vishugogo » Thu Jan 17, 2013 12:58 am

thank you very much...

GMATGuruNY wrote:
vishugogo wrote:GMATGuruNY....in the second question how is this statement coming -- To overtake Mike, Darren must CATCH-UP by 15/4 miles..
guerrero wrote:Q1) Mike and Darren run a race between points A and B, 5 miles apart. Mike starts at 9 am from A at a speed of 5 mh/h, reaches B, and returns at the same speed. Darren starts at 9:45 am from A at a speed of 10 mh/h, reaches B and comes back to A at the same speed.

At what time does Darren overtake Mike?

a) 10:20 am b) 10:30 am c) 10:40 am d) 10:50 am

From 9am - 9:45am, the distance traveled by Mike alone = r*t = 5(3/4) = 15/4 miles.
Thus, when Darren starts to travel at 9:45am, Mike is AHEAD of Darren by 15/4 miles.
In order to OVERTAKE Mike -- in other words, in order to PASS Mike -- Darren must CATCH UP by 15/4 miles.
To CATCH UP means to reach the person AHEAD of you.

Put another way:
Darren will overtake Mike when the two of them have each traveled the SAME DISTANCE.
At 10:30am:
The distance traveled by Mike in the 1.5 hours from 9-10:30am = r*t = 5(1.5) = 7.5 miles.
The distance traveled by Darren in the 3/4 hours from 9:45-10:30am = r*t = 10(.75) = 7.5 miles.
Since Mike and Darren will have traveled the same distance -- 7.5 miles -- Darren will OVERTAKE Mike at 10:30am.

Other CATCH-UP problems:

https://www.beatthegmat.com/tom-reads-at ... 51462.html
https://www.beatthegmat.com/rates-work-t119850.html
https://www.beatthegmat.com/p-and-q-circ ... 22287.html
https://www.beatthegmat.com/interesting- ... 16571.html
https://www.beatthegmat.com/rate-problem-t111575.html

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

559 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review