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

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Intern
Joined: 14 May 2012
Posts: 3

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

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

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19 May 2012, 06:47
00:00

Difficulty:

25% (medium)

Question Stats:

73% (01:22) correct 27% (01:12) wrong based on 147 sessions

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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

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

Math Expert
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129078 [2], given: 12194

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

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19 May 2012, 07:03
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.
_________________

Kudos [?]: 129078 [2], given: 12194

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Joined: 13 May 2011
Posts: 298

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

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19 May 2012, 07:43
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

Kudos [?]: 289 [0], given: 11

Intern
Joined: 08 Nov 2012
Posts: 8

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

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

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06 Mar 2013, 07:06
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

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

Math Expert
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129078 [0], given: 12194

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

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06 Mar 2013, 08:57
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.
_________________

Kudos [?]: 129078 [0], given: 12194

Intern
Joined: 08 Nov 2012
Posts: 8

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

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

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06 Mar 2013, 10:20
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

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

Re: Ann and Bob drive separately to a meeting. Ann's average   [#permalink] 06 Mar 2013, 10:20
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