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Mrs.Robbins started half an hour later than usual for market [#permalink]

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08 Nov 2010, 10:08

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Mrs. Robbins started half an hour later than usual for market. But by increasing her speed to 3/2 times her usual speed she reached 10 mins earlier than usual. What is her usual time for this journey?

Mrs.Robbins started half an hour later than usual for market. But by increasing her speed to 3/2 times her usual speed she reached 10 mins earlier than usual. What is her usual time for this journey?

A) 2 hrs B)1 hr C)43 MINS D)1,5 hrs

Let the usual speed be \(s\) and usual time \(t\) minutes, then as the distance covered is the same we will have: \(st=1.5s*(t-30-10)\) --> \(t=120\) minutes (2 hours).

Mrs.Robbins started half an hour later than usual for market. But by increasing her speed to 3/2 times her usual speed she reached 10 mins earlier than usual. What is her usual time for this journey?

A) 2 hrs B)1 hr C)43 MINS D)1,5 hrs

Since she covers the same distance, if she increases her speed by 3/2, the time taken by her becomes 2/3 of her usual time. (When distance is constant, speed will be inversely proportional to time.)

The 1/3rd difference in time taken in 30 mins + 10 mins = 40 mins. then her usual time must be 40 * 3 = 120 mins = 2 hrs.
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Re: Mrs.Robbins started half an hour later than usual for market [#permalink]

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04 Aug 2013, 11:28

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Mrs. Robbins started half an hour later than usual for market. But by increasing her speed to 3/2 times her usual speed she reached 10 mins earlier than usual. What is her usual time for this journey?

s=original speed t=original time

I always have had difficulty setting up these types of problems; rate, work, time, etc. This one is no different! We are looking to set the original time and speed to a new, lesser time and greater speed. In effect, the smaller value for the time multiplied by the greater speed balances it out, correct?

s*t = 1.5s*(t-30-10) When she starts off 30 minutes late and arrives 10 minutes early, she effectively spends 30 minutes less time on the road (because she set off late) and another 10 minutes less because she got there early. In effect, she shaves 40 minutes off her normal time spent driving. st = 1.5s*(t-40) st = 1.5st-60s t = 1.5t-60 -.5t=-60 t=120

Re: Mrs.Robbins started half an hour later than usual for market [#permalink]

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27 Aug 2014, 15:52

Hello from the GMAT Club BumpBot!

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Mrs.Robbins started half an hour later than usual for market [#permalink]

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28 Aug 2014, 01:25

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cleetus wrote:

Mrs. Robbins started half an hour later than usual for market. But by increasing her speed to 3/2 times her usual speed she reached 10 mins earlier than usual. What is her usual time for this journey?

A. 2 hrs B. 1 hr C. 43 MINS D. 1,5 hrs

This question does not have the 5th option?

Usual Speed = s

Usual Time = t

When Late

Time = t - 30 - 10 = t-40 (For 30 minutes, she was idle at home; for 10 Minutes, she was idle at station)

Mrs. Robbins started half an hour later than usual for market. But by increasing her speed to 3/2 times her usual speed she reached 10 mins earlier than usual. What is her usual time for this journey?

A. 2 hrs B. 1 hr C. 43 MINS D. 1,5 hrs

This question does not have the 5th option?

Usual Speed = s

Usual Time = t

When Late

Time = t - 30 - 10 = t-40 (For 30 minutes, she was idle at home; for 10 Minutes, she was idle at station)

Speed \(= \frac{3s}{2}\)

\(st = \frac{3s}{2} (t-40)\)

t = 120 Minutes

Answer = A

This is not a GMAT question so its format doesn't align, but it is based on the fundamentals tested in GMAT.
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Re: Mrs.Robbins started half an hour later than usual for market [#permalink]

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02 Sep 2015, 10:26

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Mrs.Robbins started half an hour later than usual for market [#permalink]

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06 Jan 2016, 18:27

i tried to plug in numbers, but forgot about the 30 minutes delay... if t=2, then D = 2r. then we have new rate 1.5r, and distance is 2r. so 2r/1.5r = 4/3, or 80 minutes. since we have 30 minutes delay, total - 110 minutes. 10 minutes earlier than regular. A works and don't even need to test other numbers.

Re: Mrs.Robbins started half an hour later than usual for market [#permalink]

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24 May 2016, 22:34

Mrs.Robbins started half an hour later than usual for market. But by increasing her speed to 3/2 times her usual speed she reached 10 mins earlier than usual. What is her usual time for this journey?

A) 2 hrs B)1 hr C)43 MINS D)1,5 hrs

Let distance covered by Mrs.Robbins is D,speed is S,usual time is T. D=ST----1 New speed is (3/2)S. New time is T-(30/60)-(10/60) as she starts 30 mins late and reaches 10 mins early. As distance covered is same so :D/((3/2)S)=T-(30/60)-(10/60)----2. Solving 2 and substituting 1 in 2 we get T=2hours or 120 mins. Ans:A

Mrs.Robbins started half an hour later than usual for market [#permalink]

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09 Aug 2017, 05:24

cleetus wrote:

Mrs. Robbins started half an hour later than usual for market. But by increasing her speed to 3/2 times her usual speed she reached 10 mins earlier than usual. What is her usual time for this journey?

A. 2 hrs B. 1 hr C. 43 MINS D. 1,5 hrs

Can anyone spot what's wrong with my reasoning? I couldn't

Time to walk D x km/h: T Time to walk D @ 3/2 x: 2/2 T She arrived 1/6 h earlier, thus total time was 5/6 T

(Time taken) = (Time stopped) + (Time to walk D) 5/6 T h = 0,5 h + 2/3 T T = 3 h

1) You assume that T=1h but that information is not provided. If we'd knew T we'd have the answer directly

Time to walk D x km/h: T Time to walk D @ 3/2 x: 2/23 T She arrived 1/6 h earlier, thus total time was 5/6 T->That means 10 minutes = 1/6T for you and subsequently 0,5h = 3/6T

(Time taken) = (Time stopped) + (Time to walk D) 5/6 T h = 0,5 h3/6T + 2/3 T// That would mean: 5/6 T=3/6 T+4/6 T = 7/6 T T = 3 h

2) The question asks for her usual travel time not the time she needed on this day

Therefore you are looking probably for something like this:

3/3T (her normal travel time) = 2/3T (time to walk D) + 40 (the time she needs less [leave 30 minutes later and arrive 10 minutes earlier]) => 1/3T = 40 => T= 120

1) You assume that T=1h but that information is not provided. If we'd knew T we'd have the answer directly

Time to walk D x km/h: T Time to walk D @ 3/2 x: 2/23 T She arrived 1/6 h earlier, thus total time was 5/6 T->That means 10 minutes = 1/6T for you and subsequently 0,5h = 3/6T

(Time taken) = (Time stopped) + (Time to walk D) 5/6 T h = 0,5 h3/6T + 2/3 T// That would mean: 5/6 T=3/6 T+4/6 T = 7/6 T T = 3 h

2) The question asks for her usual travel time not the time she needed on this day

Therefore you are looking probably for something like this:

3/3T (her normal travel time) = 2/3T (time to walk D) + 40 (the time she needs less [leave 30 minutes later and arrive 10 minutes earlier]) => 1/3T = 40 => T= 120

Re: Mrs.Robbins started half an hour later than usual for market [#permalink]

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10 Aug 2017, 07:53

Bunuel wrote:

cleetus wrote:

Mrs.Robbins started half an hour later than usual for market. But by increasing her speed to 3/2 times her usual speed she reached 10 mins earlier than usual. What is her usual time for this journey?

A) 2 hrs B)1 hr C)43 MINS D)1,5 hrs

Let the usual speed be \(s\) and usual time \(t\) minutes, then as the distance covered is the same we will have: \(st=1.5s*(t-30-10)\) --> \(t=120\) minutes (2 hours).

Answer: A.

Hope it helps.

Hello Bunnel,

I am just failing to understand this. I am sure this is a silly one.

When the question says "started half an hour later than usual" so she usually starts @9am now she will start @ 9.30. Shouldn't it be t+30 and then " she reached 10 mins earlier" so t+30-10 mins. or t+20 ?

Mrs. Robbins started half an hour later than usual for market. But by increasing her speed to 3/2 times her usual speed she reached 10 mins earlier than usual. What is her usual time for this journey?

A. 2 hrs B. 1 hr C. 43 MINS D. 1,5 hrs

We can let Mrs. Robbins’ regular speed = r and her distance = d. Thus, her increased rate is 3r/2. Her regular time is d/r, and so her new time will be d/(3r/2) = 2d/3r. Since she started ½ hour, or 30 minutes, late but arrived at the market 10 minutes early, her new time is 40 minutes shorter than her old time and we have:

2d/3r + 40 = d/r

Multiplying by 3r, we have:

2d + 120r = 3d

120r = d

120 = d/r

Since d/r = her old time, her old time must be 120 minutes, or 2 hours.

Answer: A
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