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

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

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

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


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


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

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
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.

Since 45 minutes = ¾ hour, the distance between her house and work is 40 x ¾ = 30 miles. This is her driving route. Since her biking route is 20% shorter, her biking route is 0.8 x 30 = 24 miles.

To guarantee she will arrive at the same time as when she drives, we have to assume she bikes at her slow rate of 12 m/h. Thus it will take her 24/12 = 2 hours, or 120 minutes, to bike to work. So she has to leave 120 - 45 = 75 minutes earlier.

Answer: D

Question is confusing - should have stated " to ensure she arrives at work IN LESS THAN OR at the same time as when she drives".

Given:
1. Deb normally drives to work in 45 minutes at an average speed of 40 miles per hour.
2. This week, however, she plans to bike to work along a route that decreases the total distance she usually travels when driving by 20% .
Asked: 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?

1. Deb normally drives to work in 45 minutes at an average speed of 40 miles per hour.
Distance from home to work = 40 mph * 45/60 hours =30 miles

2. 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 for bike = 30 * (100% - 20%) = 24 miles
Largest Time taken by bike = 24/12 = 2 hours

Difference in times in driving and biking = 2*60 - 45 = 120 - 45 = 75 mins

IMO D

To ensure that Deb arrives on time, we must consider the worst possible case:
the time required if she bikes at the lowest possible speed (12 mph).

Rate and time have a RECIPROCAL relationship.
\(\frac{lowest-biking-speed}{driving-speed} = \frac{12}{40} = \frac{3}{10}\)
The biking rate is 3/10 the driving rate.
Thus -- if Deb were to bike the normal route -- she would require 10/3 of her normal time:
10/3 * 45 = 150 minutes

But the biking route is 4/5 of the driving route.
Thus -- to arrive at work on time -- Deb will need to bike for 4/5 of the blue time above:
4/5 * 150 minutes = 120 minutes

Since the time increases from 45 minutes to 120 minutes -- an increase of 75 minutes -- Deb must leave 75 minutes early.

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