One day, Rocky walked from his home to his office at three fourths of

Question

One day, Rocky walked from his home to his office at three fourths of his usual speed. When he reached midway, he realised that he was 10 minutes late at that point. He, then, increased his speed by 25% and completed the remaining journey. What is the time (in minutes) taken by Rocky to reach his office that day?

(A) 48
(B) 60
(C) 64
(D) 72
(E) 80

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Official Answer

D

(A) 48
(B) 60
(C) 64
(D) 72
(E) 80

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Jagmeister · Answer

My answer: (D) 72 mins

My Approach:
let us assume that the distance from his home to his office is 'd'
let us assume that on a normal day, Rocky travels at speed 's'
on that particular day, he travelled at a speed of  \frac{3s}{4}

now it is given that after travelling half the distance i.e  \frac{d}{2} , he realizes that he is 10 mins late. Let us put this in an equation

using the formula time =  \frac{distance}{speed}

time taken on a normal day + 10 = time taken on that day

\frac{(d}{2)/(s)} + 10 = \frac{(d}{2)/(3s/4)}

solving the above we get d = 60s

now for the remaining half of the distance he increased his speed by 25%. Hence, his new speed is 125% of  \frac{3s}{4} = \frac{15s}{8}

Total time taken to cover the distance on that day =

time taken for first half + time taken for second half

\frac{(d}{2)/(3s/4)} + \frac{(d}{2)/(15s/8)}

substitute d = 60s in the above equation and solve to get   72 mins

ddtDDT · Answer

usual speed: v
usual time: t

(3/4)*v*(t+10) = v*t
=> t = 30 (mins)

1st half speed: v1
1st half time: t1

2nd half speed: v2
2nd half time: t2

t1= t+10 = 30 + 10 =40

v1*t1=v2*t2 => t2=t1*v1/v2 = t1*(1/1.25) = 40*(4/5) = 32

=> total time = t1+t2 = 72

IanStewart · Answer

I think this phrase can legitimately be interpreted in two different ways. If I was heading to the dentist for a 2pm appointment, and when I was halfway there I realized I was 10 minutes late, I'd think the time was 2:10pm. But if that's the intended meaning of this question, it involves time travel (the times required to travel the distance become negative).

The question means to say that midway, he has taken 10 minutes longer than normal. Traveling at 3/4 of his normal speed, he'll take 4/3 as long to travel the same distance, or 1/3 more time, so if he takes 10 minutes longer than normal, and that represents 1/3 of his normal time, he usually takes 30 minutes. At his slow speed, he took 40. He then travels the rest of the distance 25% more quickly, so at 5/4 of 3/4 of his usual speed, or 15/16 of his usual speed. So that will take him 16/15 as long as normal, and since it normally takes him 30 minutes, at this speed it will take (16/15)(30) = 32 minutes. So in total, his slower-than-normal trip takes 72 minutes to complete.

$!vakumar.m · Answer

For first part, distance = D/2, speed is 3S/4, time (T1) = 10 + D/2S (as with s speed, for travelling D/2 distance time taken in D/2s)

substituting the above in, Dist = Speed * Time and solving we get D=60S , and T1 = 10 + 30 = 40 min

For second part, distance = D/2, speed in 25 % more i.e 1.25 of 3S/4 = 15S/16 
and from the formula, T2 = (D/2) / (15S/16) = 32 min

therefore, to travel total distance, time taken = T1 + T2 = 40 + 32 = 72 min
So Ans is D

samarpan.g28 · Answer

Please correct me if I am wrong. In the second half when the speed is increased by 25% then the new speed will be 15s/16, not 15s/8. The answer will be 72 mins.

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