- Staff Selection Commission Mathematics 1999 to 2017 - TIME AND DISTANCE Part 3

Staff Selection Commission Mathematics - TIME AND DISTANCE TYPE-III



Ans .

(2) 18 km./hr.


    Explanation :





Ans .

(2) 5000.


    Explanation :

    Speed of person = 3 kmph
    =\( \frac{3000}{60}\)m./min
    = 50 m./min.
    Length of the diagonal of square field
    = 50 × 2 = 100 metre
    Required area = \( \frac{1}{2}\)*( 100 ) 2
    = 5000 sq. metre



TEST YOURSELF



Ans .

(1) 10 m/sec.


    Explanation :

    36 km/hr.
    36*\( \frac{5}{18}\)
    = 10 m/sec.





Ans .

(2) 216 kmph


    Explanation :

    60 metres per sec.
    60* \( \frac{18}{5}\)
    = 216 km per hr.





Ans .

(3) 55 km.


    Explanation :

    Distance = 20 kms
    Time = 2 hours
    Speed = \( \frac{Distance}{Time}\)
    \( \frac{20}{2}\)=10 km per hr..
    Now, we have, Speed = 10 km per hr
    Time=\( \frac{11}{2}\) hour
    Distance = Speed × Time
    10*\( \frac{11}{2}\)=55 km





Ans .

(4) 5 km


    Explanation :

    Man’s speed =\( \frac{1}{3}\)of the speed of car=\( \frac{1}{3}\)* 60= 20 km per hr..
    Time taken to reach office= 15 minutes =\( \frac{15}{60}\)=\( \frac{1}{4}\) hour
    Distance between his house and office = Speed × Time
    20*\( \frac{1}{4}\)= 5 km.





Ans .

(2) 37.5 hour


    Explanation :

    Speed = 6 km/hr
    Time taken = 5 hours
    \ Distance covered
    = 6 × 5 = 30 kms
    \ Time required to cover 30 kms
    at the speed of 8 km/hr.
    \( \frac{Distance}{Speed}\)=\( \frac{30}{8}\)=3\( \frac{3}{4}\)=37.5 hour





Ans .

(2) 5.16 kmph


    Explanation :

    Case I.
    Distance = 10 kms
    Speed = 4 km/hr.
    Time taken (t 1 ) =\( \frac{10}{4}\)=\( \frac{5}{2}\)hr.
    Case II.
    Distance = 21 kms
    Speed = 6 km/hr
    . Time taken (t 2 ) =\( \frac{21}{6}\)=\( \frac{7}{2}\)hr.
    Total time taken =\( \frac{5}{2}\)+\( \frac{7}{2}\)= 6 hrs.
    Total distance covered
    = 10 + 21 = 31 kms
    \ Average Speed
    \( \frac{Total distance}{Total time}\)=\( \frac{31}{6}\)=5\( \frac{1}{6}\)=5.16 km per hr..





Ans .

(2) 13.4 kmph.


    Explanation :

    Let the speed between P and Q be x km.
    Then time taken to cover x km.
    P to Q =\( \frac{x}{20}\)
    Time taken to cover x km from
    Q to P at 10 km per hr. P to Q=\( \frac{x}{10}\)
    Total distance covered
    = x + x = 2x km.
    Time taken to cover 2x km
    \( \frac{x}{20}\) + \( \frac{x}{10}\)=\( \frac{3x}{20}\)
    Average Speed
    \( \frac{2x}{\frac{3x}{20}}\)=\( \frac{2x*20}{3x}\)=\( \frac{40}{3}\)=13.4 km per hr..





Ans .

(2) 8.1 kmph


    Explanation :

    Here, the man covers equal distance at different speeds. Us- ing the formula, the Average Speed is given by
    \( \frac{3}{\frac{1}{5}+\frac{1}{10}+\frac{1}{15}}\)=\( \frac{90}{11}\)= 8.1 km per hour..





Ans .

(1) 640 kmph.


    Explanation :

    As distance is covered along four sides (equal) of a square at different speeds, the average speed of the aeroplane
    \( \frac{4}{\frac{1}{400}+\frac{1}{600}+\frac{1}{800}+\frac{1}{1200}}\) \( \frac{48000}{75}\)= 640 km per hr..





Ans .

(2) 38.57 kmph.


    Explanation :

    Length of journey = 150 kms
    \( \frac{1}{3}\)rd of journey =\( \frac{150}{3}\)= 50 kms
    Remaining \( \frac{2}{3}\) journey = 150 – 50 = 100 kms
    Time taken in \( \frac{1}{3}\)rd journey at 30 km per hr. t1=\( \frac{5}{3}\) hr
    Time taken in \( \frac{2}{3}\) rd journey at 45 km per hr. t2=\( \frac{20}{9}\) hr
    Total time taken in whole jour- ney = t 1 + t 2
    \( \frac{5}{3}\)+\( \frac{20}{9}\)=\( \frac{35}{9}\)hr
    Average Speed \( \frac{150}{\frac{35}{9}}\)=\( \frac{270}{7}\)=38.57 km per hr..





Ans .

(3) 150 km..


    Explanation :

    Let time taken to reach of fice at 50 kmph be x hrs
    Then time taken to reach office at 60 kmph =x+ \( \frac{30}{60}\)hrs
    As, distance covered is same, x* 50 =60 {x+\( \frac{30}{60}\)}
    50x = 60x + 30
    x = 3 hrs
    Hence, distance = 3 × 50
    = 150 km





Ans .

(4) 5 km.


    Explanation :

    Let time taken to reach school at 4 kmph be x hrs.
    Then time taken to reach school at 5 kmph =x+ \( \frac{15}{60}\)hr
    Since, distance is equal.
    4x= 5{x+\( \frac{15}{60}\)}
    x=\( \frac{5}{4}\)hr.
    Hence, distance between school & house =4*\( \frac{5}{4}\)km = 5 km





Ans .

(1) 25 kmph.


    Explanation :

    Let the original speed of the car = x km per hr.
    When it is increased by 5 km per hr, the speed = x + 5 km per hr.
    As per the given information in the question,
    \( \frac{300}{x}\)-\( \frac{300}{x+5}\)=2
    \( \frac{1500}{x+5x}\)=2
    x 2 + 5x = 750
    x 2 + 5x – 750 = 0
    x 2 + 30x – 25x – 750 = 0
    x (x + 30) – 25 (x + 30) = 0
    (x + 30) (x – 25) = 0
    x = – 30 or 25
    The negative value of speed is inadmissible.
    Hence, the required speed = 25 km per hr





Ans .

(2) 12 kmph.


    Explanation :

    Time = 10 hours,
    Speed = 48 km per hr.
    Distance = Speed × Time
    = 48 × 10 = 480 km
    Now, this distance of 480 kms is to be covered in 8 hours. Hence, the required Speed
    \( \frac{Distance}{New time}\)=\( \frac{480}{8}\)
    = 60 km per hr.Increase in speed = 60 – 48 = 12 km per hr.





Ans .

(3) 4 km


    Explanation :

    Let the distance be x kms.
    Time taken at 4 km per hr. t 1=\( \frac{x}{4}\)hr
    Time taken at 3 km per hr. t 2=\( \frac{x}{3}\)hr
    Difference in timings = 10 + 10 = 20 minutes
    or \( \frac{20}{60}\)=\( \frac{1}{3}\)hour
    \( \frac{x}{3}\)-\( \frac{x}{4}\)=\( \frac{1}{3}\)
    \( \frac{x}{12}\)=\( \frac{1}{3}\)
    x = 4 km.
    Hence the required distance = 4 kms.





Ans .

(4) 8 kmph


    Explanation :

    Let the speed of Rickshaw be 'x ' .
    Then, time taken to cover 16 km on foot and 24 km on Rikshaw =\( \frac{16}{4}\)+\( \frac{24}{x}\)hr
    and time taken to travel 24 km
    on foot & 16 km on Rikshaw=\( \frac{16}{x}\)+\( \frac{24}{4}\)hr
    According to question,
    \( \frac{16}{4}\)+\( \frac{24}{x}\)+1=\( \frac{16}{x}\)+\( \frac{24}{4}\)
    \( \frac{24-16}{x}\)=1
    x = 8 km/hr





Ans .

(1) 30 minutes


    Explanation :

    Since I walk at \( \frac{3}{4}\)of my usual speed the time taken is \( \frac{4}{3}\) of my usual time.
    \( \frac{4}{3}\)of usual time
    = Usual time + Time I reach late
    \( \frac{1}{3}\)of usual time
    = 10 minutes
    Usual time = 10 × 3 = 30 minutes.





Ans .

(2) 50 minutes


    Explanation :

    \( \frac{5}{3}\)of usual speed means\( \frac{3}{5}\) of usual time as he reaches earlier.
    \( \frac{3}{5}\)usual time + 20 minutes=Usual time
    20 minutes= 1-\( \frac{3}{5}\)usual time
    =\( \frac{2}{5}\) usual time
    Usual time \( \frac{20*5}{2}\)=50 minutes





Ans .

(2) 7.5 hours


    Explanation :

    New speed is\( \frac{3}{4}\)of the usual speed
    New time taken =\( \frac{4}{3}\)of the usual time
    \( \frac{4}{3}\)of the usual time – Usual time =\( \frac{5}{2}\)
    \( \frac{1}{3}\)of the usual time =\( \frac{5}{2}\)
    Usual time =\( \frac{5}{2}\)*3
    =\( \frac{15}{2}\)hours or 7.5 hrs





Ans .

(3) 35 km.


    Explanation :

    When B meets A at R,
    by then B has walked a distance (XY + YR) and A,the distance XR.
    That is both of them have togeth- er walked twice the distance from X to Y, i.e., 42 kms.





Ans .

(4) 18 km


    Explanation :





Ans .

(1) 36 km


    Explanation :

    Let the total distance travelled be x kms.
    Case I : Speed for the first one-third distance \( \frac{x}{3}\)kms =10 km per hr.
    Time taken =\( \frac{x}{30}\)hours
    Similarly, time taken for the next one-third distance=\( \frac{x}{27}\)hour
    and time taken for the last one third distance =\( \frac{x}{24}\)hour
    Total time taken to cover x kms \( \frac{x}{30}\)+\( \frac{x}{27}\)+\( \frac{x}{24}\)hour Case II :
    Time taken for one-half distance at the speed of 10 km per hr. \( \frac{x}{20}\)hr
    and time taken for remaining \( \frac{1}{2}\) of distance \( \frac{x}{16}\)hrs. at 8 km per hr.
    Total time taken \( \frac{x}{20}\)+\( \frac{x}{16}\)hr
    Time taken in (Case II – Case I)
    1 minute=\( \frac{1}{60}\)hr
    According to the question
    \( \frac{x}{20}\)+\( \frac{x}{16}\)-\( \frac{x}{30}\)+\( \frac{x}{27}\)+\( \frac{x}{24}\)hour=\( \frac{1}{60}\)
    \( \frac{x}{2160}\)=\( \frac{1}{60}\)
    x=36 km
    Hence the required distance = 36 km.





Ans .

(3) 8 hours


    Explanation :

    hours at 4 km per hr. and y hours at 5 km per hr. and cov- ers a distance of 35 kms.
    Distance = 4x + 5y = 35 ...(i)
    Now, he walks at 5 km per hr.
    for x hours and at 4 km per hr.
    for y hours and covers a distance (35 + 2) = 37 kms
    Distance = 5x + 4y = 37...(ii)
    By 5 × (i) – 4 × (ii) we have
    20x + 25y = 175
    20x + 16y = 148
    By solving these equations, y=3
    Putting the value of (y) in equa- tion (i), we have
    4x + 5 × 3 = 35
    4x = 35 – 15 = 20
    x = 5
    Total time taken
    = x + y = 5 + 3 = 8 hours.





Ans .

(4) 100 km.


    Explanation :

    Obviously,\( \frac{4}{5}\)of total time in train = 2 hour
    Total time in train=\( \frac{5}{4}\)*2=\( \frac{5}{2}\)hr
    Total time to cover 400 km is 4 hours
    \ Time spent in travelling by air= 4-\( \frac{5}{2}\)=\( \frac{3}{2}\)hr
    If 400 kms is travelled by air, then time taken = 2 hours
    \ In 2 hours, distance covered by air = 400 kms
    In \( \frac{3}{2}\)hr distance covered \( \frac{400}{2}\)*\( \frac{3}{2}\) = 300 kms
    Distance covered by the train = 400 – 300 = 100 kms.





Ans .

(1) 40 kmph


    Explanation :

    Let the original speed be x km/hr then, increased speed
    = (x + 10) km/hr
    According to question,
    \( \frac{100}{x}\)-\( \frac{100}{x+10}\)=\( \frac{30}{60}\)
    100[\( \frac{1}{x}\)-\( \frac{1}{x+10}\)]=\( \frac{1}{2}\)
    10 × 200 = x (x + 10)
    x 2 + 10x – 2000 = 0
    x 2 + 50x – 40x – 2000 = 0
    x (x + 50) – 40 (x + 50) = 0
    x = – 50, 40
    Speed can’t be negative.
    Hence, Original speed = 40 kmph





Ans .

(2) 100 days.


    Explanation :

    Working hours per day= 24 – 9 = 15 hrs.
    Total working hours for 40 days = 15 × 40 = 600 hrs.
    On doubling the distance, the time required becomes twice but on walking twice as fast, the time required gets halved. Therefore, the two together cancel each other with respect to time re- quired. Increasing rest to twice reduces walking hours per day to 24 – (2 × 9) = 6 hrs.
    \ Total number of days required to cover twice the distance, at twice speed with twice the rest. \( \frac{600}{6}\)=100 days





Ans .

(3)16.58 minutes.


    Explanation :

    In 1 minute the monkey climbs 12 metres but then he takes 1 minute to slip down 5 metres. So, at the end of 2 min- utes the net ascending of the monkey is 12 – 5 = 7 metres.
    So, to cover 63 metres the above process is repeated \( \frac{63}{7}\)=9 times. Obviously, in 9 such hap- penings the monkey will slip 8 times, because on 9th time, it will climb to the top.
    Thus, in climbing 8 times and slipping 8 times, he covers 8 × 7 = 56 metres.
    Time taken to cover 56 metres \( \frac{56*2}{7}\)= 16 minutes = 16 minutes
    \( \frac{7}{12}\)minutes
    Total time taken = 16 + \( \frac{7}{12}\) =16.58 minutes





Ans .

(4) 1100 metres


    Explanation :





Ans .

(1) 210 leaps


    Explanation :

    Grey hound and hare make 3 leaps and 4 leaps respective- ly.
    This happens at the same time. The hare goes 1.75 metres in 1 leap.
    Distance covered by hare in 4 leaps = 4 × 1.75 = 7 metres The grey hound goes 2.75 metres in one leap
    Distance covered by it in 3 leaps = 3 × 2.75 = 8.25 metres
    Distance gained by grey hound in 3 leaps=(825-7) = 1.25 metres
    Distance covered by hare in 50 leaps = 50 × 1.75 metres = 87.5 metres
    Now, 1.25 metres is gained by grey hound in 3 leaps 87.5 metres is gained in \( \frac{3}{1.25}\)*87.5
    = 210 leaps.





Ans .

(2) 1 hour.


    Explanation :

    Let the original speed be x kmph then,
    new speed = (x – 200) kmph
    According to question,
    Time taken with new speed – time taken with original speed =30 min. i.e \( \frac{1}{2}\)
    \( \frac{600}{x-200}\)-\( \frac{600}{x}\)=\( \frac{1}{2}\)
    \( \frac{x-x+200}{x(x-200)}\)=\( \frac{1}{1200}\)
    24000 = x (x – 200)
    x 2 – 200x – 24000 = 0
    x 2 – 600x + 400x – 24000 = 0
    x (x – 600) + 400 (x – 600) = 0
    (x – 600) (x + 400) = 0
    x = 600, – 400
    Speed cannot be negative Hence, original speed = 600 kmph and duration of flight =1 hour





Ans .

(2) 15 kmph.


    Explanation :

    Let the speed of the second train be x km per hr. Then the speed of the first train is x + 5 km per hr.

    Let O be the position of the rail- way station from which the two trains leave. Distance travelled by the first train in 2 hours = OA = 2 (x + 5) km.
    Distance travelled by the 2nd train in 2 hours= OB = 2x km.
    By Pythagoras theorem, AB 2 = OA2+ OB 2
    50 2 = [2 (x + 5)] 2 + [2x] 2
    2500 = 4 (x + 5) 2 + 4x 2
    2500 = 4 (x 2 + 10x + 25) + 4x 2

    8x 2 + 40x – 2400 = 0
    x 2 + 5x – 300 = 0
    x 2 + 20x – 15x – 300 = 0
    x (x + 20) – 15 (x + 20) = 0
    (x – 15) (x + 20) = 0
    x = 15, – 20
    But x cannot be negative
    x = 15
    The speed of the second train is 15 km per hr. and the speed of the first train is 20 km per hr.





Ans .

(4) 9 kmph


    Explanation :

    The distance covered by man in 4 minutes \( \frac{6*1000*4}{60}\)= 400 metres
    The distance covered by carriage in 4 minutes = 200 + 400 = 600 metres
    Speed of carriage Speed of carriage=\( \frac{600}{4}\)*\( \frac{60}{1000}\) = 9 km per hr.





Ans .

(1) 34 kmph.


    Explanation :

    If the car were not moving, the person would have heard the two sounds at an interval of 12 minutes.
    Therefore, the distance travelled by car in 11 minutes 40 seconds is equal to the dis- tance that could have been cov- ered by sound in 12 min – 11 min. 40 seconds = 20 seconds. Distance covered by sound in 20 seconds
    = 330 × 20 = 6600 m
    In 11 min 40 seconds or 700 seconds the car travels 6600 m.
    In 1 second the car will travel
    \( \frac{6600}{700}\)=\( \frac{66}{7}\)metre
    Speed of the car = \( \frac{66}{7}\)metre per second
    \( \frac{66}{7}\)*\( \frac{18}{5}\)
    =34 kmph





Ans .

(2) 40 km.


    Explanation :

    When A and B cross each other at M for the first time, they have together covered the whole dis- tance PQ = 180 km.
    When they meet again at N, they have together covered total dis- tance equal to 3 times of PQ = 3 × 180 = 540 km.
    PM=\( \frac{5}{5+4}\)*180= 100 km
    QP + PN =\( \frac{4}{5+4}\)*540
    = 240 km
    or PN = 240 – QP = 240 – 180 = 60 km.
    Then, MN = PM – PN = 100 – 60 = 40 km.





Ans .

(2) 6.6 kmph


    Explanation :

    Distance covered by man in 3 minutes
    [\( \frac{4*1000}{60}\)]\( \frac{m}{minutes}\)*3 minutes =200 metres
    Total distance covered by the car in 3 min.
    = (200 + 130) m = 330 metres
    Speed of the car \( \frac{330}{3}\) = 110 m per minutes
    \( \frac{\frac{110}{1000}}{\frac{1}{60}}\)=6.6 kmph





Ans .

(3) 25 kmph


    Explanation :

    Suppose that Ram and Mohan meet at A. Let Ram’s speed be x km per hr.
    and Mohan’s speed be y km per hr. Then AP=\( \frac{25}{4}\)x
    km and AB = 4y km.
    Now, time taken by Ram in going from B to A =\( \frac{4y}{x}\)
    and the time taken by Mohan in going from P to A =\( \frac{25x}{4y}\)
    Obviously time taken is equal
    \( \frac{4y}{x}\)=\( \frac{25x}{4y}\)
    16y 2 = 25x 2
    \( \frac{y}{x}\)=\( \frac{5}{4}\)
    y=\( \frac{5}{4}\)x
    Here, x = 20 km per hr.
    y = Mohan’s speed
    \( \frac{5}{4}\)*20=25 km per hr..





Ans .

(4) 120 km ; 30 kmph


    Explanation :

    Let the original speed be x and distance be y
    Case I.
    Time taken by train to travel 30 km= \( \frac{30}{x}\)
    Time taken by train after acci- dent=\( \frac{y-30}{\frac{4}{5}x}\)
    Total time taken =\( \frac{30}{x}\)+\( \frac{y-30}{\frac{4}{5}x}\)
    Case II :
    Time taken by train to travel 48 km =\( \frac{48}{x}\)
    Time taken by train after accident =\( \frac{y-48}{\frac{4}{5}x}\)
    Total time taken =\( \frac{48}{x}\)+\( \frac{y-48}{\frac{4}{5}x}\)
    According to question,
    \( \frac{30}{x}\)+\( \frac{y-30}{\frac{4}{5}x}\) - \( \frac{48}{x}\)+\( \frac{y-48}{\frac{4}{5}x}\)=\( \frac{9}{60}\)
    \( \frac{90-72}{4x}\)=\( \frac{9}{60}\)
    x=30
    Hence, original speed = 30 kmph
    Also \( \frac{30}{x}\)+\( \frac{y-30}{\frac{4}{5}x}\) =\( \frac{y}{x}\)=\( \frac{45}{60}\)
    3x – y = –30
    3(30) – y = –30
    y = 120 km
    i.e. Distance = 120 km





Ans .

(2) 1200 km ; 100 kmph


    Explanation :

    Let A be the starting point, B the terminus. C and D are points where accidents take place.

    0.75=\( \frac{3}{4}\)
    By travelling at \( \frac{3}{4}\) of its original speed,
    the train would take \( \frac{4}{3}\)of its usual time i.e.,\( \frac{1}{3}\)more of the usual time.
    \( \frac{1}{3}\)of the usual time taken to travel the distance CB. = 4 – 1 = 3 hrs
    and \( \frac{1}{3}\)of the usual time taken to travel the distance
    DB=3\( \frac{1}{2}\)-1=2\( \frac{1}{2}\)
    Subtracting equation (ii) from (i)
    we can write,\( \frac{1}{3}\)of the usual time taken to travel the distance
    CD=3-2\( \frac{1}{2}\)=\( \frac{1}{2}\)hr
    Usual time taken to travel CD=\( \frac{\frac{1}{2}}{\frac{1}{2}}\)=\( \frac{3}{2}\)
    Usual sp eed of the train =\( \frac{150}{\frac{3}{2}}\)= 100 km per hr.
    Usual time taken to travel CB =\( \frac{3}{\frac{1}{3}}\)= 9 hrs.
    Total time = 3 + 9 = 12 hrs.
    Length of the trip = 12 × 100 = 1200 km.





Ans .

(3) 80 kmph; 460 km


    Explanation :

    Let P be the starting point, Q the terminus, M and N the places where accidents occur.
    At \( \frac{3}{4}\)th of the original speed,
    the train will take \( \frac{4}{3}\)of its usual time to cover the same distance
    i.e.,\( \frac{1}{3}\)rd more than the usual time.
    \( \frac{1}{3}\)rd of the usual time to travel a distance of 60 kms between MN = 15 min.
    Usual time to travel 60 kms
    = 15 × 3 = 45 min. =\( \frac{3}{4}\)hr
    Usual speed of the train per hour = 60 *\( \frac{4}{3}\)= 80 km per hr..

    Usual time taken to travel MQ=90 × 3
    = 270 minor \( \frac{9}{2}\)
    The distance MQ 80*\( \frac{9}{2}\)= 360 km.
    Therefore, the total distance PQ
    = PM + MQ
    = 100 + 360 = 460 kms.





Ans .

(4) 10 : 00 a.m.


    Explanation :

    Let they meet x hrs after 7 am.
    Di stance covered by A i n x hours = 20x km
    Distance covered by B in (x –1) hr.
    = 25 (x – 1) km
    20x + 25 (x – 1) = 110
    20x + 25x – 25 = 110
    45x = 110 + 25 = 135
    x = 3
    Trains meet at 10 a.m.





Ans .

(1) 467 th line.


    Explanation :

    Writing ratio = 200 : 150 = 4 : 3 In a given time first boy will be writing the line number \( \frac{4}{7}\)*817 \( \frac{3268}{7}\)th line
    =466\( \frac{6}{7}\)
    Hence, both of them shall meet on 467th line





Ans .

(2) 9 hours


    Explanation :

    Let the two men meet after t hours.

    Distance covered by the first man starting from A = 4 t km.
    Distance covered by the second man starting from B
    = 2 + 2 . 5 + 3 + ..... +[2+\( \frac{t-1}{2}\)]
    This is an arithmetic series of t terms with \( \frac{1}{2}\)as common difference
    By applying formula S=\( \frac{n}{2}\)[2 a + (n – 1) d]
    Where, n = no. of terms
    a = first term
    d = common difference
    We have its sum =\( \frac{t}{2}\)[2*2+(t-1)*\( \frac{1}{2}\)]
    2t+\( \frac{t2-t}{4}\)
    Total distance covered by two men =4 t + 2 t +\( \frac{t2-t}{4}\) =72
    or 24t+t2-t=288
    or t 2 – 9t + 32t – 288 = 0
    or t (t–9) + 32 (t – 9) = 0
    or (t – 9) (t + 32) = 0
    Either t – 9 = 0
    t = 9,-32
    Time cannot be negative. Hence, the two men will meet after 9 hrs.





Ans .

(3) 40 m ; 20 m/sec


    Explanation :

    Let the length of the train be x metres
    Then, the time taken by the train to cover (x + 50) metres is 4\( \frac{1}{2}\)sec
    Speed of the train \( \frac{x+50}{\frac{9}{2}}\)m/s
    Again, the time taken by the train to cover x metres in 2 seconds.
    Speed of the train = \( \frac{x}{2}\)metre per second ..(ii)
    From equations (i) and (ii), we have
    \( \frac{}{}\)=\( \frac{x}{2}\)
    4x + 200 = 9x
    5x = 200
    x = 40
    Length of the train
    = 40 metre
    Speed of the train=\( \frac{40}{2}\)=20 m/sec





Ans .

(4) 17.5 kmph


    Explanation :

    Bo th trains meet after 6 hours.
    The relative speed of two trains =\( \frac{162}{6}\)== 27 km per hr..
    The speed of the slower train starting from B =\( \frac{19}{2}\) km per hr..
    The speed of the faster train =w\( \frac{35}{2}\)= 17.5 km per hr..





Ans .

(1) 25 metres


    Explanation :

    Let the length of train be x metres and the length of platform be y metres.
    Speed of the train 25*\( \frac{5}{18}\)=\( \frac{125}{18}\)
    Time taken by train to pass the platform
    x+y *\( \frac{18}{125}\) or, x + y = 125 ...(i)
    Speed of train relative to man = (25 + 5) km per hr.
    30*\( \frac{5}{18}\)=\( \frac{25}{3}\)
    Time taken by the train to pass the man
    x*\( \frac{3}{25}\)
    x=100 metres
    length of the platform = 25 metres





Ans .

(2) 108 kmph ; 72 kmph


    Explanation :

    Let the speed of the train be x metre per sec. and y metre per sec. respectively.
    Sum of the length of the trains = 200 + 175 = 375 metres
    Case : I
    When the trains are moving in opposite directions Relative speed = (x + y) m per sec.
    In this case the time taken by the trains to cross each other =\( \frac{375}{x+y}\)
    \( \frac{375}{x+y}\)=\( \frac{15}{2}\)
    x + y = 50
    Case : II
    When the trains are moving in the same direction.
    Relative speed = (x – y) m per sec.
    In this case, the time taken by the trains to cross each other \( \frac{375}{x-y}\)=\( \frac{75}{2}\)
    x – y = 10
    Now, x + y = 50
    x – y = 10
    x=30
    Putting this value in equation (i), we have
    y = 50 – 30 = 20
    \ Speed of trains = 30 m per sec.
    =30*\( \frac{18}{5}\) = 108 km per hr.
    and 20 m per sec. = 20 * \( \frac{18}{5}\)
    = 72 km per hr.





Ans .

(3) 4.5 km.


    Explanation :

    Trains are running in oppo- site direction Relative speed of the two trains = 90 + 60 = 150 km per hr.
    Distance travelled in 4\( \frac{1}{2}\)seconds onds with speed of 150 km per hr
    =150*\( \frac{5}{18}\)=150*\( \frac{5}{18}\)*( \frac{9}{2}\)=\( \frac{375}{2}\)
    Let the length of the first train be x metres.
    Then the length of the second train be \( \frac{x}{2}\)
    \( \frac{3x}{2}\)=\( \frac{375}{2}\)
    3x = 375
    x = 125 metres
    Hence, the length of the first train = 125 metres
    Speed of the first train = 60 km per hr.
    60*\( \frac{5}{18}\)=\( \frac{50}{3}\)
    Time taken by the first train to cross the tunnel = 4 minutes and 37\( \frac{1}{2}\)sec
    240+\( \frac{75}{2}\)sec=\( \frac{555}{2}\)sec
    Speed of first train =\( \frac{50}{3}\)
    Distance covered by it in \( \frac{555}{2}\) sec
    =\( \frac{50}{3}\)*\( \frac{555}{2}\)
    = 4625 metres
    Hence, length of tunnel = 4625 – 125 = 4500 metres
    = 4.5 km





Ans .

(4) 50 metres


    Explanation :

    Let the length of the train be x km and its speed y km per hr.
    Case I : When it passes the man walking at 2 km per hr. in the same direction
    Relative speed of train = (y – 2) km per hr.
    \( \frac{x}{y-2}\)=9 sec
    =\( \frac{1}{400}\)hr
    Case II : When the train crosses the man walking at 4 km per hr. in the same direction.
    Relative speed of train= (y – 4) km per hr.
    \( \frac{x}{y-4}\)=10 sec
    \( \frac{x}{y-4}\)=\( \frac{1}{360}\)hr
    On dividing equation (i) by (ii), we have
    \( \frac{y-4}{y-2}\)=\( \frac{\frac{1}{400}}{\frac{1}{360}}\) =\( \frac{360}{400}\) =\( \frac{9}{10}\)
    10y – 40 = 9y – 18
    10y – 9y = 40 – 18
    y = 22 km per hr.
    \ From equaton (i), we have
    \( \frac{x}{22-2}\)=\( \frac{1}{400}\)
    x=50 metres





Ans .

(1) 50.4 kmph


    Explanation :

    Let the length of the train be x metres
    Then, in 18 sec. the train trav- els (x + 162) metres ...(i)
    and in 15 sec. the train travels (x + 120) metres
    In (18 – 15) = 3 sec. the train travels (x + 162) – (x + 120) = 42m.
    In 1 sec the train travels =14 metres
    In 18 sec. the train travels
    = 14 × 18 = 252 metres ...(iii)
    From equations (i) and (iii)
    \ x + 162 = 252
    Þ x = 252 – 162 = 90
    \ Length of the train = 90 metres
    Also, from equation (ii) we see that in 1hr. the train travels = 14 × 60 × 60 metres
    \( \frac{14*60*60}{1000}\)=50.4
    The speed of the train = 50.4 km per hr.





Ans .

(2) 20 m/sec.


    Explanation :

    Let the length of trains be x m and (x + 50)m and the speed of other train be y m per sec.
    The speed of the first train = 90 km per hr.
    90*\( \frac{5}{18}\)= 25 m per sec.
    Case I : Opposite direction, Their relative speed = (y + 25)m per sec.
    Distance covered = x + x + 50
    = 2x + 50 metres
    Time taken= \( \frac{2 x + 50}{y + 25}\)=10
    2x + 50 = 10y + 250 ...(i)
    Case II. Direction is Same Their relative speed = (25 – y) m per sec.
    Distance covered = x + x + 50
    = 2x + 50m
    Time taken= \( \frac{2 x + 50}{25-y}\)=90
    2x + 50 = 90 (25 – y)
    From equations (i) and (ii)
    10y + 250 = 2250 – 90y
    10y + 90y = 2250 – 250
    y=20
    Putting y = 20 in equation (i), we have 2x + 50= 10 × 20 + 250 = 450
    x=200
    x + 50 = 200 + 50
    = 250 metres.
    Hence, The length of the 1st train = 200 metres.
    The length of the 2nd train = 250 metres.
    The speed of the 2nd train = 20 m per sec





Ans .

(1)12.59 m/sec.


    Explanation :

    Let the length of the train be x m and its speed y m/sec. Distance covered in crossing the platform = 170 + x metres
    and time taken = 21 seconds Speed y= \( \frac{170+x}{21}\)
    Distance covered to cross the man = x metres and time taken = \( \frac{15}{2}\)sec
    Speed y=\( \frac{2x}{15}\)
    From equations (i) and (ii),
    \( \frac{170+x}{21}\)=\( \frac{2x}{15}\)
    2550 + 15x = 42x
    Þ 42x – 15x = 2550
    Þ 27x = 2550
    x=94.44m/sec
    and y=\( \frac{340}{27}\)
    y=12.59m/sec





Ans .

(2) 4 hours 21.6 sec..


    Explanation :

    The goods train leaves Delhi at 6 am and mail train at 12 noon, hence after 6 hours
    The distance covered by the goods train in 6 hours at 32 km per hr. = 32 * 6 = 192 kms
    The relative velocity of mail train with respect to goods train = 80 – 32 = 48 km per hr.
    To completely cross the goods train, the mail train will have to cover a distance
    = 192 km + 158m + 130m
    = 192km + 0.158 km + 0.130 km
    = 192.288 km more
    Since, the mail train goes 48 kms more in 1 hour.
    \ The mail train goes 192.288 kms more in \( \frac{192288}{1000}\)*\( \frac{1}{48}\)=\( \frac{2003}{500}\) = 4 hours 21.6 sec.





Ans .

(3) 9 kmph .


    Explanation :

    Let the speed of the motor- boat in still water be Z km per hr.
    Downstream speed= (Z + 3) km per hr. Upstream speed = (Z – 3) km per hr.
    Total journey time
    = 30 minutes =\( \frac{1}{2}\)hr
    We can write,
    \( \frac{2}{z-3}\)+\( \frac{2}{z+3}\)=\( \frac{1}{2}\)
    Z 2– 9 = 8Z
    Z 2 – 8Z – 9 = 0
    Z 2 + Z – 9Z – 9 = 0
    Z(Z + 1) – 9 (Z + 1) = 0
    (Z + 1) (Z – 9) = 0
    Z = – 1 or 9.
    Since speed can’t be negative Therefore, the speed of the mo- tor-boat in still water = 9 km per hr.





Ans .

(4) 2 kmph


    Explanation :

    Let the upstream speed be x km per hr. and downstream speed be y km per hr.
    Then, we can write,
    \( \frac{32}{x}\)+\( \frac{60}{y}\)=9and
    \( \frac{40}{x}\)+\( \frac{84}{y}\)=12
    Let \( \frac{1}{x}\)=m and \( \frac{1}{y}\)=n
    The above two equations can now be written as
    32 m + 60 n = 9 ...(i)
    and, 40 m + 84 n = 12 ...(ii)
    7 × (i) – 5 × (ii) gives 24 m = 3
    or x=8 4 × (ii) – 5 × (i) gives 36 n = 3
    y=12km per hr..
    Rate of current \( \frac{y-x}{2}\)
    = 2 km. per hr..





Ans .

(1) 2 kmph


    Explanation :

    Let the speed of boat and riv- er be x km per hr. and y km per hr. respectively. Then,
    The speed of boatman down- stream = (x + y) km per hr.
    and the speed of boatman up- stream = (x – y) km per hr.
    Time taken by boatman in going 21 km downstream=\( \frac{21}{x+y}\)hours
    Time taken by boatman in going 21 km upstream =\( \frac{21}{x-y}\)hours
    According to the question,
    \( \frac{21}{x+y}\)+\( \frac{21}{x-y}\)=10
    Now, time taken for 7 kms downstream =\( \frac{7}{x+y}\)
    and time taken for 3 kms upstream =\( \frac{3}{x-y}\)
    \( \frac{7}{x+y}\)-\( \frac{3}{x-y}\)=0
    Therefore,x + y = 7 and x – y = 3 On adding (iii) and (iv), we have
    2x = 10
    Þ x = 5
    \ y = 7 – x = 7 – 5 = 2
    \ Speed of river = 2 km per hr.





Ans .

(2) 12 kmph


    Explanation :

    Let the speed of the cyclist be x km per hr.
    Speed of the motorist= (x + 15) km per hr.
    Time taken by the motorist to cover half of the distance=\( \frac{9}{x+15}\)hr
    After covering 9 kms, the speed of motorist gets reduced by 20%
    New speed = x + 15*\( \frac{80}{100}\)=\( \frac{4(x+15)}{5}\)
    Time taken by the motorist to cover the remaining half distance=\( \frac{45}{4(x+15)}\)
    Total time taken by the motorist=\( \frac{9}{x+15}\)+\( \frac{1}{2}\)+\( \frac{45}{4*x+15}\)
    Total time taken by the cyclist=\( \frac{18}{x}\)
    Motorist reaches 15 minutes, i.e.,\( \frac{1}{4}\)
    \( \frac{18}{x}\)-\( \frac{9}{x+15}\)-\( \frac{1}{2}\)-\( \frac{45}{4*x+15}\)=\( \frac{1}{4}\)
    72x + 1080 – 36x – 2x 2 –
    30x – 45x = x 2 + 15x
    Þ 3x 2 + 54x – 1080 = 0
    Þ x 2 + 18x – 360 = 0
    Þ x 2 + 30x – 12x – 360 = 0
    Þ x (x + 30) – 12 (x + 30) = 0
    Þ (x + 30) (x – 12) = 0
    Þ x = – 30, 12
    The speed cannot be negative.
    \ The speed of the cyclist = 12 km per hr.





Ans .

(3) 1520 km.


    Explanation :

    Total distance travelled = 3990 km
    Distance = Time × Speed
    Ratio of time spent = 1 : 16 : 2
    Ratio of speed = 20 : 1 : 3
    Ratio of time × speed
    = 20 × 1 : 16 × 1 : 2 × 3
    = 20 : 16 : 6
    Sum of the ratios
    = 20 + 16 + 6 = 42
    Distance covered by sea=\( \frac{3990}{42}\)*16=1520 kms





Ans .

(4) 28 km.


    Explanation :

    Relative speed of insect = 30 + 42 = 72 km per hr.
    Distance between railway engine and insect = 20 km.
    Engine and insect will meet for the first time after =\( \frac{20}{72}\)hr.
    Distance covered in this period \( \frac{20}{72}\)* 42 =\( \frac{35}{3}\) km returning to A.
    The distance covered by engine in this period=\( \frac{20}{72}\)* 30 =\( \frac{25}{3}\)
    Remaining distance between
    A and engine 20- \( \frac{25}{3}\)+\( \frac{25}{3}\)=\( \frac{10}{3}\)
    Again, engine and insect will meet after =\( \frac{5}{108}\)hr
    The distance covered by the in- sect in this period \( \frac{5}{108}\)*42 =\( \frac{35}{18}\)
    and again the insect will cover \( \frac{35}{18}\) km in returning.
    Total distance covered by the insect =\( \frac{70}{3}\)+\( \frac{70}{18}\) +....
    [\( \frac{35}{3}\)+\frac{35}{3}\)=\frac{70}{3}\)and \frac{35}{18}\)+\frac{35}{18}\)=\frac{70}{18}\)....]
    =\( \frac{70}{3}\)[1+\( \frac{1}{6}\)........]
    It is a Geometric Progression to infinity with common ratio \( \frac{1}{6}\)
    =\( \frac{70}{3}\) *\( \frac{1}{\frac{5}{6}}\)=28 km





Ans .

(1) 12:48 pm..


    Explanation :


    Let P be at equal distance from Q and R after t hours.
    (87.5 – 33) + 5t
    t= 1 hr 18 minutes
    11.30 am + 1 hr. 18 min. = 12.48 pm
    At 12.48 pm, P would have cov- ered a distance
    = (12.48 pm – 8 am) × 25
    = 120 km
    Therefore, P will be at equal dis- tance from Q and R at 12.48 pm





Ans .

(2) 40 km.


    Explanation :

    Let the original speed of the person be x km/hr. and the dis- tance be y km.
    Case 1:
    \( \frac{y}{x}\)-\( \frac{y}{x+3}\)=40 minutes
    or \( \frac{3y}{x(x+3)}\)=\( \frac{2}{3}\)
    or, 2 x (x + 3) = 9y...(i)
    Case II :
    \( \frac{2y}{x(x-2)}\)=\( \frac{2}{3}\) or, x (x – 2) = 3y... (ii)
    On dividing equation (i) by (ii) we have,
    \( \frac{2(x+3)}{x-2}\)=\( \frac{2}{3}\) or, x = 12 km/hr.
    Original speed of the person = 12 km/hr.
    Putting the value of x in equa- tion (ii)
    12 (12 – 2) = 3y
    or, 3y = 12 × 10
    y=40
    The required distance =40 km





Ans .

(3) 80 km.


    Explanation :

    Let the speed of steamer in still water = x kmph \ Rate downstream = (x + 2) kmph
    Rate upstream = (x – 2) kmph Obviously, distance covered downstream and upstream are equal
    Þ 4 (x + 2) = 5 (x – 2)
    4x + 8 = 5x – 10
    Þ 5x – 4x = 10 + 8 Þ x = 18
    \ Rate downstream
    = 18 + 2 = 20 kmph
    Therefore, the required distance
    = Speed downstream × Time
    = 20 × 4 = 80 km.





Ans .

(2) 145 metre.


    Explanation :

    According to the question, when A covers the distance of 200 metres, B covers only 200– 20 = 180 metres
    Again, in 100 metre race, B beats C by 5 metres.
    Hence, if B runs 100 metres, C runs 100–5 = 95 metres Q If B runs 100 m, C runs = 95 m
    If B runs 180 m, C runs \( \frac{95*180}{100}\)= 171 m
    A : B : C = 200 : 180 : 171
    Hence, A will beat C by = 200–171 = 29 m in 200 m race.
    i.e., 29 × 5 = 145 m in 1 km race.





Ans .

(3) 35 kmph


    Explanation :

    Case I : When the cars are moving in the same direction.
    Let A and B be two places and C be the place of meeting.
    Let the speed of car starting from A be x kmph, and that of car starting from B be y kmph.
    Relative speed = (x – y) kmph
    According to the question.
    (x – y) × 8 = 80
    x – y = 10
    Case II : When the cars are mov- ing in the opposite directions and they meet at point C.
    Relative speed = (x + y) kmph
    Time taken = 1 hour 20 minutes
    1+\( \frac{1}{3}\)=\( \frac{4}{3}\) hr
    x+y *\( \frac{4}{3}\)=80
    x + y = 60
    Adding equations (i) and (ii),
    2x = 70
    x = 35 Þ From equation (ii),
    x + y = 60
    35 + y = 60
    Þ y = 60 – 35 = 25
    \ Speed of the faster car = 35 kmph





Ans .

(4) 8.3 m/sec.


    Explanation :

    Let B take x seconds to run 1000 m.
    \ Time taken by C
    = (x + 15) seconds
    \( \frac{x}{x+15}\)=\( \frac{9}{10}\)
    10x = 9x + 135
    Þ x = 135 seconds
    Now in a one kilometre race, A beats B by 15 seconds.
    It means A covers 1000 m in 135 – 15 = 120 seconds
    Speed of A =\( \frac{1000}{120}\)=8.3 m/sec.





Ans .

(1) 3510 metre.


    Explanation :

    Trains are running in oppo- site directions.
    Relative speed = 72 + 90 = 162 kmph
    =162*\( \frac{5}{18}\)=45
    Let the length of the first train be = x metre.
    \ Length of the second train \( \frac{3}{4}\)x distance travelled in 3 \( \frac{1}{2}\)onds at 45 m/sec =\( \frac{315}{2}\)
    This distance is equal to sum of the lengths of trains.
    x+\( \frac{3x}{4}\)=\( \frac{315}{2}\)
    x=90
    Hence, the length of the first train = 90 metre.
    Speed of first train = 72 kmph
    72* \( \frac{5}{18}\)= 20 m/sec
    Time taken by the first train to cross the tunnel
    = 3 minutes = 180 seconds
    \ Distance covered by it in 180 seconds
    = 180 × 20 = 3600 metre
    \ Length of (first train + tunnel) = 3600 metre
    \ Length of tunnel = 3600 – 90 = 3510 metre