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Deb normally drives to work in 45 minutes at an average speed of 40 mi

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Deb normally drives to work in 45 minutes at an average speed of 40 mi [#permalink]

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Tough and Tricky questions: Distance/Rate .



Deb normally drives to work in 45 minutes at an average speed of 40 miles per hour. This week, however, she plans to bike to work along a route that decreases the total distance she usually travels when driving by 20% . If Deb averages between 12 and 16 miles per hour when biking, how many minutes earlier will she need to leave in the morning in order to ensure she arrives at work at the same time as when she drives?

A. 135
B. 105
C. 95
D. 75
E. 45
[Reveal] Spoiler: OA

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Re: Deb normally drives to work in 45 minutes at an average speed of 40 mi [#permalink]

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New post 06 Oct 2014, 09:37
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Deb travels 30 miles to work usually.

When she takes the different route she travels only 80% of that distance i.e. 24 miles.

She takes between 3/2 hours to 2 hours to do that [since her speed avg between 12 to 16 m/h]

Therefore the earliest time she has to start than her usual time to start is 75 min [ 120 - 45 ] and latest she would is 45 min than her usual time.

But 45 can't be answer for she would need more time than 45 min when she is travelling slower than 16 m/h .

Hence 75 min is the time she should start before her usual time [they haven't given...60 min, 65, 70 min etc....so take the max time].

Hence Ans D.

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Re: Deb normally drives to work in 45 minutes at an average speed of 40 mi [#permalink]

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QUESTION:
Deb normally drives to work in 45 minutes at an average speed of 40 miles per hour. This week, however, she plans to bike to work along a route that decreases the total distance she usually travels when driving by 20% . If Deb averages between 12 and 16 miles per hour when biking, how many minutes earlier will she need to leave in the morning in order to ensure she arrives at work at the same time as when she drives?

A. 135
B. 105
C. 95
D. 75
E. 45


SOLUTION:

Use formula D = RT

Car:
T1: 45 min
R1: 40 mph
D1: [(40*45)/60] = 30 miles

Bike:
T1:?
R2: 12 - 16 mph
D2: 08*D1 = 24 miles
T1: [(24*60)/12] = 120 min (Only 12 mph speed yields an answer given in the choices)

Therefore, Deb has to leave 120 min - 45 min = 75 min early

ANSWER: D

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Re: Deb normally drives to work in 45 minutes at an average speed of 40 mi [#permalink]

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Bunuel wrote:

Tough and Tricky questions: Distance/Rate .



Deb normally drives to work in 45 minutes at an average speed of 40 miles per hour. This week, however, she plans to bike to work along a route that decreases the total distance she usually travels when driving by 20% . If Deb averages between 12 and 16 miles per hour when biking, how many minutes earlier will she need to leave in the morning in order to ensure she arrives at work at the same time as when she drives?

A. 135
B. 105
C. 95
D. 75
E. 45


Long question stem - take one sentence at a time.

"Deb normally drives to work in 45 minutes at an average speed of 40 miles per hour."

Distance to work = Rate * Time = 40 * 45/60 = 30 miles

"This week, however, she plans to bike to work along a route that decreases the total distance she usually travels when driving by 20%."

Distance to work decreases = 30 * 80/100 = 24 miles

If Deb averages between 12 and 16 miles per hour when biking, how many minutes earlier will she need to leave in the morning in order to ensure she arrives at work at the same time as when she drives?

To ENSURE that she arrives BEFORE or at the same time as usual, she must consider her slowest average speed and calculate the maximum time she could take. She should start with that margin. If her speed is 12 mph (slowest), she will take 24/12 = 2 hours to bike to work. Normally she takes 45 mins so she must start 1 hour 15 mins before to ensure that she reaches on or before time (depending on her speed).

1 hr 15 mins = 60 + 15 = 75 mins

Answer (D)
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Re: Deb normally drives to work in 45 minutes at an average speed of 40 mi [#permalink]

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New post 10 Dec 2015, 09:51
VeritasPrepKarishma wrote:
Bunuel wrote:

Tough and Tricky questions: Distance/Rate .



Deb normally drives to work in 45 minutes at an average speed of 40 miles per hour. This week, however, she plans to bike to work along a route that decreases the total distance she usually travels when driving by 20% . If Deb averages between 12 and 16 miles per hour when biking, how many minutes earlier will she need to leave in the morning in order to ensure she arrives at work at the same time as when she drives?

A. 135
B. 105
C. 95
D. 75
E. 45


Long question stem - take one sentence at a time.

"Deb normally drives to work in 45 minutes at an average speed of 40 miles per hour."

Distance to work = Rate * Time = 40 * 45/60 = 30 miles

"This week, however, she plans to bike to work along a route that decreases the total distance she usually travels when driving by 20%."

Distance to work decreases = 30 * 80/100 = 24 miles

If Deb averages between 12 and 16 miles per hour when biking, how many minutes earlier will she need to leave in the morning in order to ensure she arrives at work at the same time as when she drives?

To ENSURE that she arrives BEFORE or at the same time as usual, she must consider her slowest average speed and calculate the maximum time she could take. She should start with that margin. If her speed is 12 mph (slowest), she will take 24/12 = 2 hours to bike to work. Normally she takes 45 mins so she must start 1 hour 15 mins before to ensure that she reaches on or before time (depending on her speed).

1 hr 15 mins = 60 + 15 = 75 mins

Answer (D)


why the speeds such as 15 ,16 not considered. please explain. thankyou

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Re: Deb normally drives to work in 45 minutes at an average speed of 40 mi [#permalink]

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Bunuel wrote:
[Deb normally drives to work in 45 minutes at an average speed of 40 miles per hour.


45 minutes = \(\frac{45}{60}\)=>3/4 hours

Distance = Speed x Time

Total distance from office to home is 40x\(\frac{3}{4}\)=>30 miles

Bunuel wrote:
This week, however, she plans to bike to work along a route that decreases the total distance she usually travels when driving by 20% .


Effective distance = 30x\(\frac{80}{100}\)=>24 miles.

Bunuel wrote:
If Deb averages between 12 and 16 miles per hour when biking, how many minutes earlier will she need to leave in the morning in order to ensure she arrives at work at the same time as when she drives?


Let speed be 12 miles/hour ( Minimum speed )

As we know , \(Time\) \(taken\)= \(\frac{Distance}{Speed}\)

Time required = 24/12=>2 hours ( Maximum Time )

Let speed be 16 miles/hour ( Maximum speed )

Time required = 24/16=>\(1\) \(\frac{1}{2}\) hours ( Minimum time)


Since the speed varies from 12 to 16 miles/hour and so does time from 2 hours to 1.5 hours , we must take maximum time taken for claculating the difference.

robu
robu wrote:
why the speeds such as 15 ,16 not considered. please explain. thankyou


We are not taking "15" because this problem boils down to a maximum / minimum missue....

The more your speed = The less time you take ; The less your speed = The more time you take

We can say Time taken \(α\) \(\frac{1}{speed}\) { Speed is inversely proportional to time }


I have taken 16 , just to show here that time taken decreases from 2 hours to 1.5 hours.

However our problem requires us t find out
Bunuel wrote:
how many minutes earlier will she need to leave in the morning in order to ensure she arrives at work at the same time as when she drives?


The issue is pretty similar to our daily chores in life

While going for office we always take the maximium time required and plan our journey accordingly to avoid any uncertain incident while reaching office.

So, we take the minimum speed, here and consider 12.


Rest I think is clear for you , plz feel free to revert in case of any doubt, I will be more than happy to help you. :-D
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Re: Deb normally drives to work in 45 minutes at an average speed of 40 mi [#permalink]

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Re: Deb normally drives to work in 45 minutes at an average speed of 40 mi   [#permalink] 24 Oct 2017, 02:15
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