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

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

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

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.

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.

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.

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

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!!

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:

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