Ann and Bob drive separately to a meeting. Ann's average : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 21 Jan 2017, 16:18

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Ann and Bob drive separately to a meeting. Ann's average

Author Message
TAGS:

### Hide Tags

Intern
Joined: 14 May 2012
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 4

Ann and Bob drive separately to a meeting. Ann's average [#permalink]

### Show Tags

19 May 2012, 05:47
00:00

Difficulty:

15% (low)

Question Stats:

73% (02:22) correct 27% (01:12) wrong based on 138 sessions

### HideShow timer Statistics

Ann and Bob drive separately to a meeting. Ann's average driving speed is greater than Bob's avergae driving speed by one-third of Bob's average driving speed, and Ann drives twice as many miles as Bob. What is the ratio of the number of hours Ann spends driving to the meeting to the number of hours Bob spends driving to the meeting?

A. 8:3
B. 3:2
C. 4:3
D. 2:3
E. 3:8

My approach is given below
Rb = X mph
Db = 2 * Da miles = 2Y miles
Db = Rb * Tb
Tb = Db/Rb
= 2Y/X

Ra = 1/3 of Rb mph = X + X/3 = 4X/3 mph
Da = Y miles
Da = Ra * Ta
Ta = Da/Ra
= Y/(4X/3) = 3Y/4X

Ta/Tb = 3Y/4X * X/2Y
= 3/8
Ta:Tb = 3:8

[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 36590
Followers: 7092

Kudos [?]: 93362 [2] , given: 10557

Re: Ann and Bob drive separately to a meeting. Ann's average [#permalink]

### Show Tags

19 May 2012, 06:03
2
KUDOS
Expert's post
info2scs wrote:
Ann and Bob drive separately to a meeting. Ann's average driving speed is greater than Bob's avergae driving speed by one-third of Bob's average driving speed, and Ann drives twice as many miles as Bob. What is the ratio of the number of hours Ann spends driving to the meeting to the number of hours Bob spends driving to the meeting?

A. 8:3
B. 3:2
C. 4:3
D. 2:3
E. 3:8

My approach is given below
Rb = X mph
Db = 2 * Da miles = 2Y miles
Db = Rb * Tb
Tb = Db/Rb
= 2Y/X

Ra = 1/3 of Rb mph = X + X/3 = 4X/3 mph
Da = Y miles
Da = Ra * Ta
Ta = Da/Ra
= Y/(4X/3) = 3Y/4X

Ta/Tb = 3Y/4X * X/2Y
= 3/8
Ta:Tb = 3:8

The red part is not correct, since Ann drives twice as many miles as Bob it should be Db = Da/2 miles = Y/2 miles.

Also you used a lot of unnecessary variables to solve this question.

Say the rate of Bob is 3mph and he covers 6 miles then he needs 6/3=2 hours to do that.
Now, in this case the rate of Ann would be 3+3*1/3=4mph and the distance she covers would be 6*2=12 miles, so she needs 12/4=3 hours for that.

The ratio of Ann's time to Bob's time is 3:2.

Hope it helps.
_________________
Senior Manager
Joined: 13 May 2011
Posts: 317
WE 1: IT 1 Yr
WE 2: Supply Chain 5 Yrs
Followers: 21

Kudos [?]: 252 [0], given: 11

Re: Ann and Bob drive separately to a meeting. Ann's average [#permalink]

### Show Tags

19 May 2012, 06:43
Approach B: Pick Number
if Bob's Rate = 12, Ann's Rate= 16
if Bob's distance is= 10, Ann's distance= 20
Time Ann/ Time Bob= (20/16) /(10/12)= 3/2
Intern
Joined: 08 Nov 2012
Posts: 8
Followers: 0

Kudos [?]: 4 [0], given: 1

Re: Ann and Bob drive separately to a meeting. Ann's average [#permalink]

### Show Tags

06 Mar 2013, 06:06
Bunuel wrote:
info2scs wrote:
Ann and Bob drive separately to a meeting. Ann's average driving speed is greater than Bob's avergae driving speed by one-third of Bob's average driving speed, and Ann drives twice as many miles as Bob. What is the ratio of the number of hours Ann spends driving to the meeting to the number of hours Bob spends driving to the meeting?

A. 8:3
B. 3:2
C. 4:3
D. 2:3
E. 3:8

My approach is given below
Rb = X mph
Db = 2 * Da miles = 2Y miles
Db = Rb * Tb
Tb = Db/Rb
= 2Y/X

Ra = 1/3 of Rb mph = X + X/3 = 4X/3 mph
Da = Y miles
Da = Ra * Ta
Ta = Da/Ra
= Y/(4X/3) = 3Y/4X

Ta/Tb = 3Y/4X * X/2Y
= 3/8
Ta:Tb = 3:8

The red part is not correct, since Ann drives twice as many miles as Bob it should be Db = Da/2 miles = Y/2 miles.

Also you used a lot of unnecessary variables to solve this question.

Say the rate of Bob is 3mph and he covers 6 miles then he needs 6/3=2 hours to do that.
Now, in this case the rate of Ann would be 3+3*1/3=4mph and the distance she covers would be 6*2=12 miles, so she needs 12/4=3 hours for that.

The ratio of Ann's time to Bob's time is 3:2.

Hope it helps.

Dear Bunuel,

I always try to solve these problems with different methodologies and I was also thinking about this one:

SpeedA/SpeedB = 4/3 ---> hence times are in ratio TimeA/TimeB = 3/4 ---> but A covered twice as many miles of Bob (at constant speed), so it took the double amount of time:
TimeA/TimeB = 3*2/4 = 3/2

What do you think?

Thank you,

Patrizio
Math Expert
Joined: 02 Sep 2009
Posts: 36590
Followers: 7092

Kudos [?]: 93362 [0], given: 10557

Re: Ann and Bob drive separately to a meeting. Ann's average [#permalink]

### Show Tags

06 Mar 2013, 07:57
Patrizio wrote:
Bunuel wrote:
info2scs wrote:
Ann and Bob drive separately to a meeting. Ann's average driving speed is greater than Bob's avergae driving speed by one-third of Bob's average driving speed, and Ann drives twice as many miles as Bob. What is the ratio of the number of hours Ann spends driving to the meeting to the number of hours Bob spends driving to the meeting?

A. 8:3
B. 3:2
C. 4:3
D. 2:3
E. 3:8

My approach is given below
Rb = X mph
Db = 2 * Da miles = 2Y miles
Db = Rb * Tb
Tb = Db/Rb
= 2Y/X

Ra = 1/3 of Rb mph = X + X/3 = 4X/3 mph
Da = Y miles
Da = Ra * Ta
Ta = Da/Ra
= Y/(4X/3) = 3Y/4X

Ta/Tb = 3Y/4X * X/2Y
= 3/8
Ta:Tb = 3:8

The red part is not correct, since Ann drives twice as many miles as Bob it should be Db = Da/2 miles = Y/2 miles.

Also you used a lot of unnecessary variables to solve this question.

Say the rate of Bob is 3mph and he covers 6 miles then he needs 6/3=2 hours to do that.
Now, in this case the rate of Ann would be 3+3*1/3=4mph and the distance she covers would be 6*2=12 miles, so she needs 12/4=3 hours for that.

The ratio of Ann's time to Bob's time is 3:2.

Hope it helps.

Dear Bunuel,

I always try to solve these problems with different methodologies and I was also thinking about this one:

SpeedA/SpeedB = 4/3 ---> hence times are in ratio TimeA/TimeB = 3/4 ---> but A covered twice as many miles of Bob (at constant speed), so it took the double amount of time:
TimeA/TimeB = 3*2/4 = 3/2

What do you think?

Thank you,

Patrizio

You are writing two different things in red.
_________________
Intern
Joined: 08 Nov 2012
Posts: 8
Followers: 0

Kudos [?]: 4 [0], given: 1

Re: Ann and Bob drive separately to a meeting. Ann's average [#permalink]

### Show Tags

06 Mar 2013, 09:20
Bunuel wrote:
Patrizio wrote:
Bunuel wrote:
Ann and Bob drive separately to a meeting. Ann's average driving speed is greater than Bob's avergae driving speed by one-third of Bob's average driving speed, and Ann drives twice as many miles as Bob. What is the ratio of the number of hours Ann spends driving to the meeting to the number of hours Bob spends driving to the meeting?

A. 8:3
B. 3:2
C. 4:3
D. 2:3
E. 3:8

My approach is given below
Rb = X mph
Db = 2 * Da miles = 2Y miles
Db = Rb * Tb
Tb = Db/Rb
= 2Y/X

Ra = 1/3 of Rb mph = X + X/3 = 4X/3 mph
Da = Y miles
Da = Ra * Ta
Ta = Da/Ra
= Y/(4X/3) = 3Y/4X

Ta/Tb = 3Y/4X * X/2Y
= 3/8
Ta:Tb = 3:8

The red part is not correct, since Ann drives twice as many miles as Bob it should be Db = Da/2 miles = Y/2 miles.

Also you used a lot of unnecessary variables to solve this question.

Say the rate of Bob is 3mph and he covers 6 miles then he needs 6/3=2 hours to do that.
Now, in this case the rate of Ann would be 3+3*1/3=4mph and the distance she covers would be 6*2=12 miles, so she needs 12/4=3 hours for that.

The ratio of Ann's time to Bob's time is 3:2.

Hope it helps.

Dear Bunuel,

I always try to solve these problems with different methodologies and I was also thinking about this one:

SpeedA/SpeedB = 4/3 ---> hence times are in ratio TimeA/TimeB = 3/4 ---> but A covered twice as many miles of Bob (at constant speed), so it took the double amount of time:
TimeA/TimeB = 3*2/4 = 3/2

What do you think?

Thank you,

Patrizio

You are writing two different things in red.[/quote]

Ok, let's say it better. If distance for A and B was the same, the time A/B would be in ratio 3/4. Considering the fact that the distance of A is twice the distance covered by Bob the ratio becomes 3*2/4 (since distance and time have a positive proportionality).
Does it work now?

Thanks,

Patrizio
Re: Ann and Bob drive separately to a meeting. Ann's average   [#permalink] 06 Mar 2013, 09:20
Similar topics Replies Last post
Similar
Topics:
Ann is 5 years older than Sue, and Jill is twice as old as Ann. If S 3 07 Feb 2016, 09:59
3 Seven years ago Bob was k times as old as Ann. If Ann is now 11 years 5 19 Feb 2015, 08:48
9 In how many ways can Ann, Bob, Chuck, Don and Ed be seated 8 25 Nov 2010, 18:06
9 A sum of money is to be divided among Ann, Bob and Chloe. First, Ann r 12 10 Nov 2010, 09:14
8 A certain bank uses the formula above to approximate the ann 8 09 Oct 2007, 11:00
Display posts from previous: Sort by