speed

I believe that if this question is treated as two scenarios, the first when he is on time and the second when he arrives 16 minutes later. By using the formula rt=d for boh scenarios. For the normal rate I assigned an arbitrary number - let's say 12 (a good number because it is a multiple of 4). If 12 is the normal speed then three quarters of 12, or 9, is the rate when he is late. Thenormal time it takes him to get to work is assigned variable t, therefore the delayed time is t+16. So scenario one (normal rate and time) would equal 12t and the slower rate and time would equal 9(t+16). Set these two equal to each other, 12t = 9(t+16) and solve. The normal time it takes him to get to work is 48 minutes, answer b.

I forget to mention why they are set equal to each other - when plugged into the rate formula they both equal distance.

12t = d and 9(t+16) = d

Also, this works for other arbitrary numbers not just 12, make sure you take three quarters of which ever number you select to obtain he slower rate.

Hope this explanation helps.

Jonathan

rupsk wrote:Traveling at three-fourth of his normal speed John reaches his office late by 16 minutes. Find the time taken by John to reach his office if he travels at his normal speed.</p>

A. 45 min
B. 48 minutes
C. 40 minutes
D. 50 minutes
E. 55 minutes

I had solved by looking at the answer is there any other way?

Traveling at 3/4 the speed = traveling for 4/3 the time.
Since the normal time is increased by 1/3:
16 = (1/3)t
t = 48.

The correct answer is B.

We also could plug in the answers:

Answer choice B: 48 minutes.
Let r = 4 miles per minute.
Then d = r*t = 4*48 = 192 miles.
3/4 the rate = (3/4)*4 = 3 miles per minute.
Time at the slower rate = d/r = 192/3 = 64 minutes.
Difference in time = 64-48 = 16. Success!

rupsk wrote:Traveling at three-fourth of his normal speed John reaches his office late by 16 minutes. Find the time taken by John to reach his office if he travels at his normal speed.</p>

A. 45 min
B. 48 minutes
C. 40 minutes
D. 50 minutes
E. 55 minutes

I had solved by looking at the answer is there any other way?

Hi-- I always wonder, how can test makers come up with so many different question on Time-Speed-Distance when there is only 1 simple formula Speed = Distance/Time? Irony!

That's the beauty of Math.
Well, lets concentrate on this question.

From the formula of TSD, we know that Speed is inversely proportional to Time

This means that if Time increases then Speed Decreases and Vice Versa.

It is given that travelling at (3/4)T John reaches 16 mins late.
Reverse this number and we have (4/3)T i.e 1/3rd extra of original time..
Equate this to the extra time John took..

(1/3)T= 16
or T=48 mins..

The correct answer is Option B

(d/t)(3/4) = d/(t+16)...solve for t. ans = 48

rupsk wrote:Traveling at three-fourth of his normal speed John reaches his office late by 16 minutes. Find the time taken by John to reach his office if he travels at his normal speed.</p>

A. 45 min
B. 48 minutes
C. 40 minutes
D. 50 minutes
E. 55 minutes

I had solved by looking at the answer is there any other way?

There is a cool formula for solving such problem as follows :

If a person/body travels at a/b of his normal speed and the change in time taken to cover the same distance is t

Then the Change in time taken is

{( b/a) - 1 } * original time


Here a/b is 3/4 and change in time taken is 16

Now let's apply the formula

{ ( 4/3) - 1 } * original Time taken to cover the same distance = 16


Now ,

1/3 * Original Time = 16

Hence original time taken is 48....

thanks for all different solution, it gives me different direction to solve this kind of problem.