Quant Boosters - Rajesh Balasubramanian - Set 4

  • Here, the distance travelled by Sourav is constant in both the cases. So,
    20 * t = 25 * (t-1) = D
    20t = 25t - 25
    5t = 25
    t = 5 hrs
    So, Distance travelled = 20 * 5 = 100 Km.

  • Q21) Distance between the office and the home of Alok is 100 Km. One day, he was late by an hour than the normal time to leave for the office, so he increased his speed by 5 Km/hr and reached office at the normal time. What is the changed speed of Alok?
    a) 25 Km/hr
    b) 20 Km/hr
    c) 16 Km/hr
    d) 50 Km/hr

  • Here, again the distance is constant. However if we write the equation as we did in previous question, we get:
    s * t = (s + 5) * (t - 1) = 100
    In this case, we can identify other constant as the time which Alok took less to cover the distance i.e. 1 hour. So, we can write:
    100/s−100/(s+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 to solve these 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:
    100/(t−1) − 100/t = 5
    Solving, we get t = 5 hrs, t-1 = 4hrs. Therefore, increased speed = 100/(t−1) = 25 Km/hr.

  • Q22) Akash when going slower by by 15 Km/hr, reaches late by 45 hours. If he goes faster by 10 Km/hr from his original speed, he reaches early by by 20 hours than the original time. Find the distance he covers.
    a) 8750 Km
    b) 9750 Km
    c) 1000 Km
    d) 3750 Km

  • You may solve for D and S by writing the equations as:
    D/(s−15) − D/s = 45
    D/s − D/(s+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.

    Correct Answer: 9750 Km

  • Q23) Raj was travelling to his hometown from Mumbai. He met with a small accident 80 Km away from Mumbai and continued the remaining journey at 4/5 of his original speed and reached his hometown 1 hour and 24 minutes late. If he had met with the accident 40 Km further, he would have been an hour late.
    What is Raj's normal speed?
    a) 20 Km/hr
    b) 15 Km/hr
    c) 30 Km/hr
    d) 25 Km/hr

  • Scenario 1


    5D1/4s − D1/s = 1 + 24/60 (always write minutes in this form if the unit is Km/hr)

    D1/s = 7/5 ∗ 4 = 28/5 hours

    Scenario 2


    5D2/4s − D2/s = 1
    =) D2/s = 4 hours

    i) Therefore, in his normal speed, he can travel 40 Km in 28/5 − 4 = 8/5 hours
    so, s = 40/(8/5) = 25 Km/hr. Answer d)

    ii) D1/s = 28/5 =) D1 = 140 Km. Therefore, total distance D = 140 + 80 = 220 Km

  • Q24) Two persons A and B start moving at each other from point P and Q respectively which are 1400 Km apart. Speed of A is 50 Km/hr and that of B is 20 Km/hr. How far is A from Q when he meets B for the 22nd time?
    a) 1000 Km
    b) 400 Km
    c) 800 Km
    d) 1400 Km

  • Total distance travelled by both of them for 22nd meeting = 1400 + 21 x 2 x 1400 = 43 x 1400

    Distance travelled by each will be in proportion of their speed:-

    Therefore, distance travelled by A = 50/(50 + 20) x 43 x 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

  • Q25) What would happen in the previous question if both A and B had started at point P.
    a) 800 Km
    b) 600 Km
    c) 1000 Km
    d) 350 Km

  • For 22nd meeting, total distance travelled = 22 x 2 x 1400 Km
    Distance travelled by A = 5/7 ∗ 44 ∗ 1400 = 44000Km (1400 x 31 + 600).
    Therefore, A would be 600 Km from Q

  • Q26) Two trains A and B are 100 m and 150 m long and are moving at one another at 54 Km/hr and 36 Km/hr respectively. Arun is sitting on coach B1 of train A. Calculate the time taken by Arun to completely cross Train B.
    a) 10 s
    b) 6 s
    c) 4 s
    d) 8 s

  • Speed of A = 54 ∗ 1000/(60∗60) = 15 m/s
    Speed of B = 36 ∗ 1000/(60∗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 = 150/25 = 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 = 250/25 = 10 s.

  • Q27) Two trains start together from a Station A in the same direction. The second train can cover 1.25 times the distance of first train in the same time. Half an hour later, a third train starts from same station and in the same direction. It overtakes the second train exactly 90 minutes after it overtakes the first train. What is the speed of third train, if the speed of the first train is 40 Km/hr?
    a) 20 Km/hr
    b) 50 Km/hr
    c) 60 Km/hr
    d) 80 Km/hr

  • Speed of Train B = Sb = 1.25 x Sa = 1.25 x 40 = 50 Km/hr

    In half an hour, Train A would have mover = 40∗1/2 = 20 Km away from train C .

    Therefore, train C will have to cover 20 Km in relative speed to cross train A. So, time taken = 20/(Sc−40) (where Sc - 40 Km/hr is the relative speed of train C w.r.t A)

    Similarly, train C will have to cover 50∗1/2 = 25 Km in relative speed to cross train B. So, time taken = 25/(Sc−50).

    Given the difference between these two times is 90 minutes. Therefore, 25/(Sc−50)−20/(Sc−40) = 90/60 = 3/2.

    Now, instead of solving this equation, one can quickly put in the options in this equation to find Sc = 60 Km/hr.

    Note - You should be able to rule out A and B beforehand because speed of train Chas to be greater than Train A and B to overtake them.

  • Q28) Two trains left from two stations P and Q towards station Q and station P respectively. 3 hours after they met, they were 675 Km apart. First train arrived at its destination 16 hours after their meeting and the second train arrived at its destination 25 hours after their meeting. How long did it take the first train to make the whole trip?
    a) 18h
    b) 36h
    c) 25h
    d) 48h

  • 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 = SA x 3 and by B = SB x 3

    Therefore, 3 (SA + SB) = 675 =) SA + SB = 225.

    Now the remaining distance to be covered by first train is DSB/(SA + SB).

    Therefore, time taken = DSB/(SA+SB)SA = 16 ---------- (1)

    Similarly, DSA/(SA+SB)SB = 25 ---------- (2)

    Dividing equation 1 by 2

    SA^2/SB^2 = 25/16 => SA/SB = 5/4.
    Therefore, SA + 4/5 ∗ SA = 225 => SA = 125 Km/hr and SB = 100 Km/hr

    From equation 2, D/SA = 16 ∗ (SA + SB)/SB = 16 ∗ 225/100 = 36 h which is the time taken for the first train to complete the journey.

  • Q29) Arjun travels from A to B, a distance of 200 Km at the speed of 40 Km/hr. At the same time, Rakesh starts from point C at a speed of 20 Km/hr along a roadwhich is perpendicular to AB. Find the time in which Arjun & Rakesh will be closer to each other?
    a) 1.5 h
    b) 3.33 h
    c) 5 h
    d) 4 h

  • 0_1516623859456_f2d28826-9ddf-4bc1-b052-3391e0e60372-image.png

    Distance between the two at any time t, D = sqrt((200−40t)^2 + (20t)^2)
    One way to calculate is to find the minima using d/dt (D) = 0
    =) 1/(2 sqrt(Doesn't matter)) x 2 x (200 - 40t) x -40 + (2 x 20t) x 20) = 0
    =) 2 x (200 - 40t) x -40 = -40t x 20
    =) 400 - 80t = 20t
    =) t = 4 hours.

    Alternatively, you can find the right answer by putting the options in the equation.

  • Q30) Station X of length 900 meters has two station masters A and B. But as the station is not a busy one, they are mostly jobless and decide to conduct an experiment. They stand at either end of the station and decide to note the exact time when trains cross the stationmasters. They synchronize their watches and proceed to either end of the station. Two trains P and Q go past the station (neither train stops here), and after having taken down their readings, the station masters sit down to have a chat

    A: Train P entered the station at exactly 8:00:00
    B: Train Q entered the station at exactly 8:00:10 (10 seconds past 8)
    A: The last carriage of train P crossed me by at 8:00:20, and precisely two seconds after this, the engines of the two trains went past each other. (Engines are at the front of the train)
    B: The last carriage of train Q crossed me 22 seconds after the engine of P went past me.
    A: After the last carriage of train P crossed by me, it took 35 seconds for the engine of train Q to cross me.
    B: I got bored and I came here.

    Q1. What is the length of train Q?

    Q2. At what time do the rear ends of the two trains cross each other?

    Q3. How far from station master A do the rear ends of the two trains cross each other?

    Q4. You are told that the two trains enter the station at the same times mentioned and the length of the two trains are unchanged. Furthermore, train P continues to travel at the same speed (as computed above). At what minimum speed should train Q travel such that the rear ends of the two trains cross each other at a point within the length of the platform?

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