Timing Troubles

Assume that the time(minute) Lewis used to the station is $a$ , and the delay time of the clock(how long the power cut last night last) is $x$ .

$8am$ (clock) ------ $a$ min.------> $9:30am$ (actual time)

When Lewis returns home, the speed used is $2/3$ of the speed in the morning. Since the distance of the station and Lewis's house is constant, so the time used when returning to home is $3/2$ of the time used in the morning.

$7:30pm$ (actual time) ------ $(3/2)a$ min.------> $7pm$ (clock)

We can make the following equations:

$9 hrs 30 min.-a-x=8 hrs$ ----(1)

$7 hrs 30 min.+\frac{3}{2}a-x=7 hrs$ ----(2)

(1)-(2) $=2 hrs-\frac{5}{2}a=1 hr$

$a=\frac{2}{5} hrs=\frac{2}{5}\times60= 24$ (min.)

Subsitute $a$ into (1),

$9 hrs 6 min.-x=8 hrs$

$x=9 hrs 6 min.-8 hrs=1 hrs 6 min.=(60+6) min.=\boxed{66}min.$

Therefore, the answer is $\boxed{66}$ .

Let the actual time at which I left the house be '8am + x minutes'. Hence the time taken to walk to the station is (90-x) minutes since I arrived at the station at exactly 9.30 am. Similarly, the actual time at which I returned home can be represented as '7 pm + x minutes'. The time taken to return home is therefore (x-30) minutes. Now, (x-30)/(90-x) = 1/(2/3) = 3/2. We have 2(x-30)= 3(90-x) or 5x=330. Hence x=66.

Say the power cut lasted for "x" minutes. Say the distance between his house and station is "d" And his speed in the morning when going from house to station is "y". Therefore the duration he worked is from 9:30 to 7:30 which is 10 hours. But the total duration he was out of his house is from 8:00 am to 7:00 am.[Since there was no power cut during the day] which is equal to 11 hours . Therefore the total time taken by him to go from house to station and to come from station to house is 1 hour. Say he took "t" minutes time to go from house to station at a speed of y. therefore he took (60-t) minutes to from station to house at a speed of 2y/3. So t=d/y and (60-t)=3d/2y from the two equations - 60-(d/y)=3d/2y solving this we will get d=24y. So he will took 24 minutes to go from house to station and 36 minutes to go from station to house. Now from this we can say that when he left for his house at 8:00 am after traveling for 24 minutes reached the station where he saw the actual time to be 9:30 am. So when it was 8:24 am in his electric clock the actual time was 9:30 am . So the power cut lasted for the duration of time between 9:30 am to 8:24 am which is equal to 1 hour 6 minutes=66 minutes.

The key to this simple question is the fact that a) the total time elapsed is 11 hours no matter which clock you refer to and b) time from home to station (morning) is 1.5 times the time from station to home, since the speed. This gives us the equations $2.5\times x = 60$ , which gives us 24 minutes to station from home, which in turn means that the clock is 66 min late. Thanks

Let, the power cut lasted x minutes. So, the clock is delayed by x minutes. When I wake up, it's actually 8hr + x min. Time taken for travel = [9hr + 30min] - [8 hr + x min] = 1 hr + (30-x)min Time taken for work = 19hr + 30 min - 9hr - 30 min = 10 hr. When it was 9:30, my clock showed 9 hr + (30-x) min Time shown by my clock when work was finished = 10 hr + [9hr + (30-x)min] = 19hr + (30-x) min. Now, note that when I reached home, the clock showed 7 pm. So, definitely, x>30 and hence, we write 19hr + (30-x)min as 18 hr + (90-x) min. Similarly, time taken for travel = 1 hr + (30-x)min is written as (90-x) min. Time shown when he returns home = 18hr + (90-x) min + 1.5(90-x) min = 19 hr This only means the total minutes should be 60. Hence, 2.5(90-x) = 60 which gives x = 66

Let $x$ be the difference between real time and clock time. Let $y$ be the time of the journey.

From the information given, we know that:

$x + y = 90$ minutes

And

$x \times \frac{3}{2} - y = 1.5 x - y = -30$ minutes

If we sum the two equations, we get:

$2.5x = 60$ $x = 24$

Using substitution, we get the equation:

$24 + y = 90$

which gives the result of:

$y= \boxed{66}$

Let the time delay be $t$ and the speed be $s$ . Also, $\text{distance} = \text{speed} \times \text{time}$

Thus, the distance traveled in the morning was $(90 - t)s$ . The observed time it took to reach the office was $90$ minutes but infact it was lesser because the clock was ahead of $8$ $\text{am}$ when the person left the house. So to correct for the time difference, we subtract $t$ .

Now, while coming back the observed time taken was $-30$ . The correct time difference was $-30 + t$ to correct for the clock. The speed was $\dfrac{2}{3} s$ .

Thus, we can equate the two as the distance traveled was the same both times.

$(90 - t) s = \dfrac{2}{3}s(-30 + t)$

or

$(90 - t) = \dfrac{2}{3}(t - 30)$ (Cancelling the $s$ and rearranging $t - 30$ .

Solving this, we get $\boxed{t = 66}$ .

T = clock lag [minutes]

w = actual walking time for morning speed [minutes]

3w/2 = actual walking time in evening at 2/3 speed [minutes]

'Apparent' walking times -

  in the morning: 9:30 - 8:00 = 90 minutes = w + T

  in the evening: 7:00 - 7:30 = -30 minutes = 3/2w - T

Solve to get T = 66 minutes and w = 24 minutes

Let the man takes time t to go from his house to office & the clock in his house runs x mins late. Let the distance between his house & office is d, his speed be v. So, d = vt or t = d/v -------(1) Now, in the 1st journey, t = 90 - x ----------------(2) (Assuming the house clock is right, he takes 90 mins to complete his journey. But as the clock is running x mins late, the actual time of travel is 90 - x which is equal to t.) Now, for the return journey, let the time be t{1} d = (2/3)v t{1} or t{1} = 3d/2v From equation (1), t{1} = (3/2) t So, now, (3/2) t = -30 + x ---------------(3) (Assuming the house clock is right, he takes -30 mins to complete his journey. But as the clock is running x mins late, the actual time of travel is -30 + x which is equal to t{1}.) Now solving equations (1) & (2), we get, x = 66 mins So the electric clock at his house runs 66 mins late.

Lewis left his house at what the delayed clock said was 8:00 am, or 480 minutes after 12:00 am, then arrived at the station at 9:30 am, or 570 minutes after 12:00 am, meaning that $480+\gamma+x=570, \gamma+x=90,$ where $x$ is the clock's delay and $\gamma$ is the time it takes him to get to the station. He then leaves the station after work at 7:30 pm, or 1170 minutes after 12:00 am, and traveled back home at $\frac {2}{3}$ the speed he traveled in the morning, arriving back home at 7:00 pm, or 1140 minutes after 12:00 am. This situation yields the equation $1170+1.5\gamma-x=1140, 1.5\gamma-x=-30$ We can then simply isolate $x$ and get our answer $-1.5\gamma-1.5x=-135$ $1.5\gamma-x=-30$ $-2.5x=-165$ $x=\boxed{66}$