Quiz 6

Here, again the distance is constant. However, if we write the equation as we did in the previous question, we get:
s * t = (s + 5) * (t - 1) = 100

In this case, we can identify another constant as the time which Alok took less to cover the distance i.e. 1 hour. So, we can write:
100s100s - 100s+5100s+5 = 1 (Difference of time take in both the cases is 1 hour)

=> 100s + 500 - 100s = s(s+5)
=> s(s+5) = 500 (One should practice solving this kind of equations directly by factorizing 500 into 20 X 25, instead of solving the entire quadratic. However, if you are more comfortable with quadratic, keep using it. More important thing is to avoid silly mistakes once you have got the concept right)
Solving we get, s = 20 Km/hr
Therefore, Increased speed = s + 5 = 25 km/hr

We could have alternatively identified increase in speed by 5 Km/hr as a constant. Therefore, we could write the equations as:
100t−1100t−1 - 100t100t = 5
Solving, we get t = 5 hrs, t - 1 = 4hrs. Therefore, increased speed = 100t−1100t−1 = 25 Km/hr.

The question is What is the changed speed of Alok?

Hence, the answer is 25 Km/hr.

You may solve for D and S by writing the equations as:
Ds−15Ds−15 − DsDs = 45
DsDs - Ds+10Ds+10 = 20
You can solve these two equations to get D = 9750 Km which is a perfectly fine approach. However, if you are comfortable with alligation, you may save time on such TSD question as:

Both these equations can be solved quickly to get s = 65 Km/hr, t = 150 hr. Therefore, D = 9750 Km. I want to reiterate that using the first approach is perfectly fine. That\'s how I would do it. It is better to spend 20 seconds extra and get the right answer instead of getting it wrong with something you are not comfortable with.

The question is Find the distance he covers.

Hence, he covers 9750 km.

Scenario 1:

5D14s5D14s − D1sD1s = 1 + 24602460 (always write minutes in this form if the unit is Km/hr)
=> D1sD1s = 7575 * 4 = 285285 hours

Scenario 2:

5D24s5D24s − D2sD2s = 1
=> D2sD2s = 4 hours
i) Therefore, in his normal speed, he can travel 40 Km in 285285 - 4 = 8585 hours so,
s = 408/5408/5 = 25 Km/hr.
ii) D1sD1s = 285285 => D1 = 140 Km. Therefore, total distance D = 140 + 80 = 220 Km.

Hence, the answer is 25 Km/hr and 220 Km.

Total distance travelled by both of them for 22nd meeting = 1400 + (21 * 2 * 1400) = 43 * 1400
Distance travelled by each will be in proportion of their speed :
Therefore, distance travelled by A = 50/(50 + 20) * 43 * 1400 = 43000 (Note - Always do complicated calculations at last because things cancel out generally)
Now, for every odd multiple of 1400, A will be at Q and for every even multiple of 1400 A will be at P. So, at 42000 Km (1400 x 30, even multiple) A will beat P. So at their 22 meeting, A will be 1000 Km from P, therefore, 400 Km from Q.

The question is How far is A from Q when he meets B for the 22nd time?

Hence, the answer is 400 km

For 22nd meeting, total distance travelled = 22 * 2 * 1400 Km

Distance travelled by A = 5757* 44 * 1400 = 44000 Km (1400 * 31 + 600).

Therefore, A would be 600 Km from Q.

The question is " How far is A from Q when he meets B for the 22nd time? "

Hence, the answer is 600 Km

Speed of A = 54 * 100060∗60100060∗60 = 15 m/s
Speed of B = 36 * 100060∗60100060∗60 = 10 m/s

Relative speed = S1 + S2 = 15 + 10 m/s = 25 m/s
The length that needs to be crossed = length of train B = 150 m.
Therefore time taken = 1502515025 = 6s
What is the time taken for trains to completely cross each other? The length that needs to be crossed = 100 + 150 = 250 m. Time taken = 2502525025 = 10 s

The question is to Calculate the time taken by Arun to completely cross Train B.

Hence, the answer is 6 s.

the trains cross each other.

Total distance travelled by both the trains before meeting = D. This distance will be covered in the proportion of their speeds. Clarify with the diagram.

3 hours after meeting distance travelled by A = S_A * 3 and by B = S_B * 3
Therefore, 3 (S_A + S_B) = 675 => S_A + S_B = 225.

Now the remaining distance to be covered by first train is DS_B/(S_A + S_B)
Therefore, time taken = DSB(SA+SB)SADSB(SA+SB)SA = 16 --------- (1)
Similarly, DSA(SA+SB)SADSA(SA+SB)SA = 25 --------- (2)

Dividing equation 1 by 2 -

S2AS2BSA2SB2 = 25162516 => SASBSASB = 5454.
Therefore, S_A + 4545 * S_A = 225
=> S_A = 125 Km/hr and S_B = 100 Km/hr
From equation 2, DSADSA = 16 * 225100225100 = 36h,
which is the time taken for the first train to complete the journey.

The question is How long did it take for the first train to make the whole trip?

Hence, it took 36h for the first train to make the whole trip.

how they travel.

Distance between the two at any time t, D = √((200−40t)²+(20t)²)
If X is minimum, then √will also be minimum and vice versa.
So, we need to minimize (200−40t)²+(20t)²
Or, we need to minimize 400 (20 – 2t)² + 400t²
Or, we need to minimize (20 – 2t)² + t²
Or, we need to minimize 400 + 4t² – 80t + t²
Or, we need to minimize 5t² – 80t + 400
Or, we need to minimize t² – 16t + 80
Or, we need to minimize t² – 16t + 64 + 16
Or, we need to minimize (t – 4)² + 16
Or, t should be 4.
Or, the distance will be minimum in 4 hours’ time.
WE could do this using differentiation also, but we do not need to know differentiation to answer this question

The question is Find the time in which Arjun & Rakesh will be closer to each other?

Hence, the answer is 4 hr.