2IIM Quant Notes - Time, Speed and Distance - Part (2/2)


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012 and 2014.


    Ramesh takes 6.5 hours to go from city A to city B at 3 different speeds 30 kmph, 45 kmph, and 60 kmph covering the same distance with each speed. The respective mileages per liter of fuel are 11 km, 14 km and 18 km for the above speeds. Ramesh's friend Arun is an efficient driver and wants to minimise his friend's car's fuel consumption. So he decides to drive Ramesh's car one day from city A to city B. How much fuel will he be able to save?
    4.2 liters
    4.5 liters
    0.7 liters
    0.3 liters

    Total time = 6.5 hrs
    Let the total distance be x
    Ramesh's speeds = 30, 45, 60 kmph
    6.5 = (x/3)/30 + (x/3)/45 + (x/3)/60
    6.5 = (2x+4x/3+x)/180
    6.5 = (6x+4x+3x)/540
    or x = 270 km
    With each speed distance covered = 270/3 = 90 km
    Fuel consumed when driving at 30 = 90/11 = 8.1 l
    Fuel consumed when driving at 45 = 90/14 = 6.4 l
    Fuel consumed when driving at 60 = 90/18 = 5
    Total fuel = 8.1 + 6.4 + 5 = 19.5 liters

    When Arun drives the car, he should drive at 60 kmph throughout so that he gets the maximum mileage!
    Fuel consumption when driving at 60 for entire journey = 270/18 = 15 liters

    Difference = 19.5 – 15 = 4.5

    Amar, Akbar and Antony decide to have a 'x' m race. Antony completes the race 14 m ahead of Amar. Akbar finishes 20 m ahead of Antony and 32 m ahead of Amar. What is Amar’s speed?
    9/10 th of Antony's speed
    5/8 th of Akbar's speed
    14/15 th of Antony's speed
    10/7 th of Akbar's speed

    When Antony completes, Amar would have run (x – 14) m; Time taken is the same
    Ratio of Amar’s speed : Antony’s speed = (x–14)/t1 : x/t1
    Or (x–14)/x ---- (1)

    When Akbar finishes, Amar would have run (x – 32) m; Antony would have run (x – 20) m; Time taken is the same
    At this point, the ratio of Amar’s speed : Antony’s speed = (x–32)/t2 : (x−20)/t2
    Or (x–32)/(x−20) ---- (2)

    Equating (1) and (2), we get
    (x–14)/x = (x–32)/(x−20)
    x^2 – 34x + 280 = x^2 – 32x
    2x = 280
    Or x = 140 m

    Ratio of Amar’s speed : Antony’s speed = (140–14)/140 = 126/140 = 9/10

    Tom, Jerry and Bill start from point A at the same time in their cars to go to B. Tom reaches point B first and turns back and meets Jerry at a distance of 9 miles from B. When Jerry reaches B, he too turns back and meets Bill at a distance of 7 miles from B. If 3 times the speed with which Tom drives his car is equal to 5 times Bill’s speed, what could be the distance between the points A and B
    40 miles
    24 miles
    31 miles
    63 miles

    Let the distance between A and B be ‘x’ miles.
    Let their speeds be T, J and B in miles/hour

    When Tom meets Jerry:
    Distance travelled by Tom = x + 9
    Distance travelled by Jerry = x – 9
    Time taken will be the same => (x+9)/T= (x–9)/J
    Or T/J = (x+9)(x−9) -- (1)

    When Jerry meets Bill:
    Distance travelled by Jerry = x + 7
    Distance travelled by Bill = x – 7
    Time taken will be the same => (x+7)/J= (x–7)/B
    Or J/B = (x+7)(x−7) -- (2)

    Given 3T = 5B
    or T/B = 5/3
    From (1) and (2)
    T/B = T/J * J/B
    5/3 = (x+9)/(x−9) * (x+7)/(x−7)
    5(x - 9)(x - 7) = 3 (x + 9) (x + 7)
    5x^2 – 80x + 315 = 3x^2 + 48x + 189
    2x^2 – 128x + 126 = 0
    x^2 – 64x + 63 = 0
    x = 63 or x = 1
    x has to be 63

    Kumar started from Chennai at x hrs y minutes and travelled to Vellore. He reached Vellore at y hrs z minutes. If the total travel time was z hrs and x minutes, his starting time in Chennai could have been ______ (Assume clock format to be 0 to 24 hrs).
    02:08 hrs
    13:03 hrs
    00:02 hrs
    12:01 hrs

    x, y or z cannot be greater than 24
    Also x + y + z cannot be greater than 24

    Given x hrs + z hrs = y hrs
    Also y min + x min = z min

    The only value of x that will satisfy the above 2 equations is 0.
    Hence, from among the choices given, the time could be 00:02 hrs

    When Sourav increases his speed from 20 Km/hr to 25 Km/hr, he takes one hour less than the usual time to cover a certain distance. What is the distance usually covered by him?
    125 Km
    100 Km
    80 Km
    120 Km

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

    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?
    25 Km/hr
    20 Km/hr
    16 Km/hr
    50 Km/hr

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

    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.
    8750 Km
    9750 Km
    1000 Km
    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:

    0_1486457727288_speed-22-1.png

    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.

    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.
    i) What is Raj's normal speed?
    a) 20 Km/hr b) 15 Km/hr c) 30 Km/hr d) 25 Km/hr
    ii) What is the distance between Mumbai and Raj's hometown?
    a) 140 Km b) 200 Km c) 220 Km d) 250 Km

    Scenario 1:

    0_1486458593097_speed-23-11.png

    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:

    0_1486458604813_speed-23-22.png

    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.

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

    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?
    1000 Km
    400 Km
    800 Km
    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.

    What would happen in the previous question if both A and B had started at point P.
    800 Km
    600 Km
    1000 Km
    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.

    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.
    10 s
    6 s
    4 s
    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.

    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?
    20 Km/hr
    50 Km/hr
    60 Km/hr
    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) (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 C has to be greater than Train A and B to overtake them.

    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?
    18h
    36h
    25h
    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.

    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 road which is perpendicular to AB. Find the time in which Arjun & Rakesh will be closer to each other?
    a) 5 h
    b) 3.33 h
    c) 5 h
    d) 4 h

    0_1486459473376_speed-29_1.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.


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