If Sandy cycles at 10 Km/hr, he reaches the school late by 4 minutes;

Since distance = speed * time, is a constant, \(s_1t_1 = s_2t_2\)

Let the correct time to reach be t hours

s1 = 10 km/hr and t1 = (t + 4/60) hours

s2 = 12 km/hr and t2 = (t - 2/60) hours

Therefore \(10 * (t + \frac{4}{60}) = 12 * (t - \frac{2}{60)}\)

\(10t + \frac{40}{60} = 12t - \frac{24}{60}\)

\(2t = \frac{64}{60}\)

\(t = \frac{32}{60}\)

Substituting t for the LHS gives the distance = \(10 * (\frac{32}{60} + \frac{4}{60})\) = 6 km


Option D

Arun Kumar
_________________

Start with Base Speed = 10 km/hr

And

Base Time = (T + 4) minutes


Given a Constant Distance, if we increase the Speed by (p / u)

———->the the Time taken to travel that Constant Distance will Decrease by: p / (u + p)


We increase the speed from 10 kmph to12 kmph——-> increase by (12 - 10) / 10 =
1/5

The Time needed will decrease by: 1 / (5 + 1) = 1/6


Let the Original Base Time of (T + 4) = X minutes

From being + 4 minute late to -2 minutes early is a Difference of 6 fewer minutes

X - 6 min = X - (1/6)X

The 6 minutes saved corresponds to a 1/6 decrease in the Base Time

6 = (1/6)X

X = 36 minutes


Thus at the Base Rate or 10 kmph it takes 36 minutes to get to school

S * T = D

(10) * (36/60) =

(10) * (6/10) =

6 km

Posted from my mobile device

Let the distance be D.

At 10 KM/h, time requires D/10 hours.

At 12 KM/h, time requires D/12 hours.

At 10 KM/h, Sandy is 4 minutes late and at 12 KM/h, he is 2 minutes early. So the time difference is 6 minutes or 6/60 = 1/10 hours.

So, (D/10) - (D/12) = 1/10

D = 6.

Option:D

The distance from Sandy’s home to his school is constant.

When distance is constant, Time and Speed are reciprocals of each other. This means,

\(\frac{S1 }{ S2}\) = \(\frac{T2 }{ T1}\)

In this question, S1 = 10 km/hr and S2 = 12 km/hr; therefore, \(\frac{S1 }{ S2}\) = \(\frac{5}{6}\). This means,
\(\frac{T2 }{ T1}\) = \(\frac{5}{6}\)

Therefore, T2 = (\(\frac{5}{6}\)) T1; this means that T2 is lesser than T1 by a magnitude of (\(\frac{1}{6}\)) T1; the amount by which T2 is lesser than T1 represents the savings in time in the second case compared to the first case.

T1 is the time taken by him when he cycles at 10km/hr, in this case, he reaches school late by 4 mins; T2 is the time taken by him when he cycles at 12 km/hr, in this case, he reaches school early by 2 mins.

Therefore, savings in time = 6 mins.
We have already established that savings in time = (\(\frac{1}{6}\)) T1. This means,
(\(\frac{1}{6}\)) T1 = 6 or T1 = 36 minutes = \(\frac{3}{5}\) hours.

Distance of his school from home = S1 * T1 = 10 *\( \frac{3}{5} \)= 6 km.

The correct answer option is D.
_________________

crack

Here is another same type of question. But upon solving, I am not getting the correct answer. Kindly help.
Q. If John travels to his school from his home at 2 km/hr he will be 15 minutes late. If he travels at 5 km/hr he will be 30 minutes early. Find the distance from his home to the school.
(a) 2.5 km
(b) 3 Km
(c) 4 Km
(d) 3.5 km
(e) 4.5 Km

Thank you.

Regarding the additional question: Here is another same type of question. But upon solving, I am not getting the correct answer. Kindly help.
Q. If John travels to his school from his home at 2 km/hr he will be 15 minutes late. If he travels at 5 km/hr he will be 30 minutes early. Find the distance from his home to the school.
(a) 2.5 km
(b) 3 Km
(c) 4 Km
(d) 3.5 km
(e) 4.5 Km

2(t+15/60)=5(t-30/60)
180/60=3t
t=60/60

Then plug 60/60 back into first equation:
2(60/60+15/60)=2(75/60)=150/60=2.5
Answer A