Jeff drove to work from this home, averaging 40 miles per ho

LET DISTANCE = D

if T IS REQUIRED TIME TO REACH .

T = \(\frac{D}{40} - \frac{12}{60} = \frac{D}{48} - \frac{7}{60}\)
SOLVING D = 20

hence A

I didn't understand the equations but great method! Please elaborate..

\(\frac{12}{60}\) the reason you are subtracting is because its \(\frac{12 minutes}{60 minutes}\) late?

Let say
jeff need to reach in time T.
LET distance to office = d
now first day time taken by jeff = distance/speed = d/40
this time is 12 minutes more than T
therefore T = d/40 -12/60

SIMILARLY second day time taken by jeff = d/48
This is 7 minute more than T
Therefore T = d/40-7/60

NOW EQUATING both equation of T.
\(\frac{d}{40} -\frac{12}{60} = \frac{d}{40}-\frac{7}{60}\)

solving D = 20

hope it helps

You may equate distance also

i.e., 40*(t+12/60) = 48* (t+7/60)

You get t=18/60

d= 40*(18/60+12/60) = 20 miles
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still i didn't understood why 12/60 and 7/60.....no where they've mentioned 60min right...
please answer

The point is that the rates are given in miles per hour. So, we are converting 12 and 7 minutes into hours:

12 minutes = 12/60 hours.
7 minutes = 7/60 hours.

Does this make any sense?
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Thank you Bunuel.... concept is clear

I tried a different method here:
I used the following thought process...if the two day trips differed in time by 5 min (12 min late vs. 7 min late), then we can set up the two time equations and set them equal to 5 min or 1/12 hrs.

d/40 - d/48 = 1/12 (5 min); common denominator is 240, so we have (6d-5d) / 240 = 1/12 hr.

if we solve for d, we get: d / 240 = 1/12; 12d = 240 therefore, d = 20 miles.

Choose A.

I have a quick question.

For T = d/40 -12/60, why wouldn't we add 12/60 (i.e. 12 minutes) to the d/40? After all, doesn't his time increase with his slower speed of d/40?

Thanks!

Edit: I think I got it!

Let's pretend he needs to get home in 60 minutes. The time he takes on the first day works out to be 72 minutes. From this we subtract 12 minutes to even out the equation. "T" represents his time if he is on time. d/40 represents his time on the day in question (which will be greater than his regular time) and 12/60 or 7/60 represents the extra time he took, if subtracted from his slower time would represent his normal time.

Does that make sense?

First day: t = d/40 - 12/60
Second day: t = d/48 - 7/60

d/40 - 12/60 = d/48 - 7/60
d/40 = d/48 + 5/60

(LCM of 40 and 48 is 240)

d = 40*(d/48) + 40*(5/60)
d = 40d/48 + 200/60
d = 200d/240 + 800/240
d = (200d+800)/240
240d = 200d+800
40d = 800
d = 20

ANSWER: 20

I had the following:
d = (t + 12/60)*40
d = (t + 7/60)*48

However, I don't understand why the t's have to be same. Can someone explain?

Because t is the usual time, for which Jeff is on time at work.
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Let 't' is the usual time taken(without getting late)... Since the ration of speed in both case is 40:48 ie 5:6 , the ratio of time taken will be reverse ie 6/5. SO t+12 / t+ 7 = 6/5 giving t = 18 mins. substitute in any - D= 40*(18+12)/60 = 20 miles
divide by 60 to change into hour...

Since avg. speed is given in hour ; let t be be original time when Jeff covered the distance.

as we know speed=distance/time; 40(t+12/60)=x and from second 48(t+7/60)=x

equation 1 and 2 40(t+12/60)=48(t+7/60) will give t=.3

to know distance covered put t=.3 in 1st equation 40(.3+.2)=20 So (A)

Distance = Speed x Time

Equation (1)

\(d = 40 (t + \frac{12}{60})\)

Equation (2)

\(d = 48 (t+\frac{7}{60})\)

Equating (1) & (2)

\(40 (t + \frac{12}{60}) = 48 (t+\frac{7}{60})\)

\(t = \frac{18}{60}\)

Placing value of t in Equation (1)

\(d = 40 (\frac{18}{60} + \frac{12}{60})\)

\(= 40 * \frac{30}{60}\)

= 20

Answer = A

40(t+12/60) = 48(t+7/60)

solve for t,
=> t = 18/60 or 3/10

now distance = speed*time
i.e. d = 40(3/10 + 1/5) = 20.
Hence, A.

Bunuel pls what if "The next day he left home for work at 10 min late than yesterday"

For the original question, let v and t be the speed and time to reach the office without any delays.

Thus distance = vt

Case 1, 12 minutes late:

vt = 40(t+12/60) ...(1)

Case 2, 7 minutes late:

vt = 48(t+7/60) ...(2)

Thus 40(t+12/60) = 48(t+7/60) ---> t = 0.3 hour. Thus the distance = 40 (0.3+12/60) = 20 miles. A is the correct answer.

Now, coming back to your question, if the question said "The next day he left home for work at 10 min late than yesterday" ---> the time that Jim would be late will now be 10+7 = 17 minutes late (as he would be 7 minutes late when he left at the same time , so he would be 17 minutes late if left his home 10 minutes later that yesterday!) and the equations will become :

40(t+12/60) = 48(t+17/60) .
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Two equations can be created from the prompt using the D = RT equation:

(1) D = (40mi/hr)(t+ 1/5)
(2) D = (48mi/hr)(t+ 7/60)

Plug (1) into (2) --> 40t+8 = D --> 40t+8 = 48t+(28/5)

Simplify --> (40-28)/5 = 8t --> 12/5 = 8t --> 3/10 = t

Plug this back into either of the original equations (i.e. (1) or (2)) and you'll find distance to be 20 mi.

A.

40t=48[t-(1/12)]
t=1/2 hour
40*1/2=20 miles
A