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

Staff Selection Commission Mathematics - TIME AND DISTANCE TYPE-III



Ans .

(3) 12 km.


    Explanation :

    Let the distance be x km.
    \( \frac{x}{10}\)-\( \frac{x}{12}\)=\( \frac{12}{60}\)
    x=\( \frac{1}{5}\)*60
    = 12 km.





Ans .

(1) 60 km


    Explanation :

    Let the distance between stations be x km, then speed of train
    =\( \frac{x}{\frac{45}{60}}\)=\( \frac{4x}{3}\)
    \( \frac{3x}{4x-15}\)=\( \frac{4}{5}\)
    16x – 60 = 15x
    x = 60 km





Ans .

(1) 13.33 minutes


    Explanation :

    Speed of train =\( \frac{Distance}{Time}\)=\( \frac{10}{\frac{12}{60}}\)
    = 50 kmph
    New speed = 45 kmph
    Required time =\( \frac{10}{45}\)
    \( \frac{2}{9}\)*60 minutes
    =\( \frac{40}{3}\)=13.33 minutes.





Ans .

(1) 4 km.


    Explanation :

    Let the distance of the office
    be x km, then \( \frac{x}{5}\)-\( \frac{x}{6}\)=\( \frac{8}{60}\)
    x = 2 × 2 = 4 km





Ans .

(2) 4 km.


    Explanation :

    Let the distance of school be x km,
    then \( \frac{x}{3}\)-\( \frac{x}{4}\)=\( \frac{20}{60}\)
    \( \frac{x}{12}\)=\( \frac{1}{3}\)
    x= 4 km





Ans .

(3) 20 minutes


    Explanation :

    Distance between stations X and
    Y = Speed × Time
    = 55 × 4 = 220 km.
    New speed = 55 + 5 = 60 kmph
    Required time =\( \frac{220}{60}\)=\( \frac{11}{3}\)
    = 3 hours 40 minutes.
    Required answer
    = 4 hours – 3 hours 40 minutes
    = 20 minutes





Ans .

(3) 1 hour


    Explanation :

    Distance of journey = x km
    Difference of time = 12 – 3 = 9 minutes
    \( \frac{9}{60}\) hr=\( \frac{3}{20}\)hr
    \( \frac{x}{70}\)-\( \frac{x}{80}\)=\( \frac{3}{20}\)
    \( \frac{x}{56}\)=\( \frac{3}{2}\)
    x=84 km
    Required correct time \( \frac{84}{70}\)hr-12 minutes
    =72 – 12 = 60 minutes
    = 1 hour



TYPE-10



Ans .

(4) 45 km/hr


    Explanation :

    Let the length of train be x me- tres
    \ According to question
    Speed of the train =\( \frac{x}{10}\)m / sec
    Also, the speed of the train \( \frac{x+50}{14}\)m / sec. It passes the platform in 14 seconds]
    Both the speeds should be equal, i.e.,
    \( \frac{x}{10}\)=\( \frac{x+50}{14}\)
    or 14x = 10x + 500
    or 14x – 10x = 500
    or 4x = 500
    \ x = 125 metres
    Hence, Speed =\( \frac{125}{10}\)= 12 . 5 m / sec
    \( \frac{12.5*18}{5}\)km / hr .
    = 45 km/hr.





Ans .

(2) 176


    Explanation :

    Let length of train be x m
    Speed of train \( \frac{x+264}{20}\)
    Also, speed of train =\( \frac{x}{8}\)
    \( \frac{x}{8}\)=\( \frac{x+264}{20}\)
    5x = 2x + 528
    5x – 2x = 528
    x = 528 ÷ 3 = 176 m





Ans .

(4) 79.2 km/hr


    Explanation :

    Let the length of train be x me- tres.
    Then, speed of train when it passes a telegraph post = \( \frac{x}{8}\)m/sec and speed of train, when it passes the bridge =\( \frac{x+264}{20}\)
    Clearly,
    \( \frac{x}{8}\)=\( \frac{x+264}{20}\)
    5x = 2x + 528
    3x = 528
    x=176 m
    Speed of train
    \( \frac{176}{8}\)= 22 m/sec
    22*\( \frac{18}{5}\)kmph
    = 79.2 kmph





Ans .

(1) 25.2 km/hour.


    Explanation :

    Let the length of train be x metres.
    When the train crosses the standing man, its speed = \( \frac{x}{9}\)
    When the train crosses the plat- form of length 84 m, its speed \( \frac{x+84}{21}\)
    Obviously,\( \frac{x+84}{21}\)=\( \frac{x}{9}\)
    21x – 9x = 9 × 84
    12x = 9 × 84
    x=63 m
    Required speed =\( \frac{63}{9}\)=\( \frac{63}{9}\)*\( \frac{18}{5}\)= 25.2 kmph





Ans .

(4)45 km/hr.


    Explanation :

    Suppose length of train be x
    According to question
    \( \frac{x+50}{14}\) = \( \frac{x}{10}\)14x = 10x + 500
    4x = 500
    x=125 m
    Therefore, speed \( \frac{125}{10}\)*\( \frac{18}{5}\)= 45 kmph





Ans .

(4) 21.6 kmph.


    Explanation :

    Let the length of the train be x
    According to the question,
    Speed of the train \( \frac{x+90}{30}\)=\( \frac{x}{15}\)
    x + 90 = 2x
    x = 90 m
    Speed of train =\( \frac{90}{15}\)
    6 m/s =6*\( \frac{18}{5}\)= 21.6 kmph





Ans .

(3) 20 seconds.


    Explanation :

    Let the length of the train be x metre
    Speed of train when it crosses man=\( \frac{x}{10}\)
    Speed of train when it crosses platform =\( \frac{x+300}{25}\)
    According to the question,
    Speed of train=\( \frac{x}{10}\)=\( \frac{x+300}{25}\)
    25x = 10x + 3000
    15x = 3000
    x=200 m
    Length of train = 200 metre
    Speed of train= \( \frac{x}{10}\)=\( \frac{200}{10}\)= 20 m/sce
    Time taken in crossing a 200m long platform = \( \frac{200+200}{20}\)= 20 seconds





Ans .

(4) 330 m


    Explanation :

    Let the length of the train be x metres.
    Speed of train in crossing boy = \( \frac{x}{30}\)
    Speed of train in crossing platform \( \frac{x+110}{40}\)
    According to the question,
    \( \frac{x}{30}\)=\( \frac{x+110}{40}\)
    4x = 3x + 330
    x = 330 metres





Ans .

(3) 150


    Explanation :

    Let the length of train be x metre
    \( \frac{x}{15}\)=\( \frac{x+100}{25}\)
    5x = 3x + 300
    2x = 300
    x=150 metres





Ans .

(2) 40, 30


    Explanation :





Ans .

(2) 52 km/hr, 26 km/hr


    Explanation :

    Let the speed of trains be x and y metre/sec respectively,
    \( \frac{100+95}{x-y}\)=27
    x-y=\( \frac{65}{9}\)..(1)
    \( \frac{195}{x+y}\)=9
    x+y=\( \frac{195}{9}\)...(2)
    By equation (i) + (ii)
    2x= \( \frac{65}{9}\)+\( \frac{195}{9}\)=\( \frac{260}{9}\)
    x = \( \frac{130}{9}\)m/sec.
    \( \frac{130}{9}\)*\( \frac{18}{5}\)kmph = 52 kmph
    From equation (ii),
    y = \( \frac{65}{9}\)m/sec
    \( \frac{65}{9}\)*\( \frac{18}{5}\)
    = 26 kmph





Ans .

(2) 130.


    Explanation :

    Let the length of train be x me- tre, then
    \ Speed of train
    \( \frac{x}{7}\)=\( \frac{x+390}{28}\)
    x=\( \frac{390}{3}\)
    = 130 metres





Ans .

(3) 15.5 seconds.


    Explanation :

    Speed of train = 36 kmph
    36 * \( \frac{5}{18}\)= 10 m/sec
    Length of train = 10 × 10
    = 100 metres
    Required time= \( \frac{100+55}{10}\)
    = 15.5 seconds





Ans .

(2) 300.


    Explanation :

    Speed of train = 60 kmph
    60*\( \frac{5}{18}\)m/sec. =\( \frac{50}{3}\)m/sec.
    If the length of platform be
    = x metre, then
    Speed of train=\( \frac{Length of (train + platform)}{Time taken in crossing}\)
    50 × 10 = 200 + x
    x = 500 – 200 = 300 metre





Ans .

(4) 9:24 am


    Explanation :

    Let both trains meet after t hours since 7 a.m.
    Distance between stations A and B = x Km.
    \( \frac{x}{4}\)*t+\( \frac{x}{\frac{7}{2}}\)*(t-1)=x
    Speed=\( \frac{Distance}{Time}\)
    \( \frac{7t+8t-8}{28}\)=1
    15 t – 8 = 28
    15 t = 28 + 8 = 36
    t=2 hours 24 minutes
    Required time = 9 :24 a.m.





Ans .

(2) 12.1 seconds.


    Explanation :

    Speed of train = 72 kmph.
    \( \frac{72*5}{18}\)m/sec
    = 20 m./sec.
    Required time
    \( \frac{Length of train and bridge}{Speed of train}\)
    \( \frac{242}{20}\)
    = 12.1 seconds





Ans .

(2) 6 seconds.


    Explanation :

    Relative speed of train
    = (60 + 6) kmph.
    \( \frac{66*5}{18}\)m/sec
    =\( \frac{55}{3}\)m/sec.
    Length of train = 110 metre
    Required time =\( \frac{110}{\frac{55}{3}}\)
    = 6 seconds



TYPE-11



Ans .

(2) 20 m.


    Explanation :

    Let the time taken to complete the race by A,B, and C be x min- utes
    Speed of A =\( \frac{1000}{x}\)
    B =\( \frac{1000-50}{x}\) =\( \frac{950}{x}\)
    C =\( \frac{1000-69}{x}\)= =\( \frac{931}{x}\)
    Now, time taken to complete the race by
    B=\( \frac{1000}{\frac{950}{x}}\)=\( \frac{1000*x}{950}\)
    and distance travelled by C in
    \( \frac{1000x}{950}\)min
    \( \frac{1000x}{950}\)*\( \frac{931}{x}\)= 980 km.
    B can allow C
    = 1000 – 980 = 20 m





Ans .

(4) 12 minutes


    Explanation :

    Ratio of the speed of A, B and
    C = 6 : 3 : 1
    Ratio of the time taken
    =\( \frac{1}{6}\):\( \frac{1}{3}\):1=1:2:6
    Time taken by A
    \( \frac{72}{6}\)= 12 minutes





Ans .

(1) 17.24 seconds.


    Explanation :

    Let A take x seconds in covering
    1000m and b takes y seconds According to the question,
    x+20=\( \frac{900}{1000}y\)
    x+20=\( \frac{9}{10}y\)...(1)
    \( \frac{950}{1000}\)x + 25 = y....(2)
    From equation (i),
    \( \frac{10x}{9}\)+\( \frac{200}{9}\)=y
    \( \frac{10x}{9}\)+\( \frac{200}{9}\)=\( \frac{950}{1000}\)x + 25
    \( \frac{200x-171x}{180}\)=\( \frac{225-200}{9}\)
    \( \frac{29x}{180}\)=\( \frac{25}{9}\)
    x=\( \frac{500}{29}\)=17.24





Ans .

(3) 14.4 kmph


    Explanation :

    Time taken by Kamal
    \( \frac{100}{18*\frac{5}{18}}\)
    = 20 seconds
    Time taken by Bimal
    = 20 + 5 = 25 seconds
    Bimal’s speed = \( \frac{100}{25}\)=4 m/sec
    =\( \frac{4*18}{5}\)=14.4 kmph.





Ans .

(1) 95 m..


    Explanation :

    When A runs 1000m, B runs 900m.
    \ When A runs 500m, B runs 450 m.
    Again, when B runs 400m, C runs 360 m.
    \ When B runs 450m, C runs
    \( \frac{360}{400}\)*450 = 405 metres
    Required distance = 500 – 405
    = 95 metres





Ans .

(1) 11.9 metre


    Explanation :

    According to the question,
    \ When A runs 800 metres, B runs 760 metres
    \ When A runs 200 metres, B runs= \( \frac{760}{800}\)*200= 190 metres
    Again, when B runs 500 metres, C runs 495 metres.
    \ When B runs 190 metres, C runs =\( \frac{495}{500}\)*190= 188.1 metres
    Hence, A will beat C by 200 – 188.1 = 11.9 metres in a race of 200 metres.





Ans .

(3) 29 metres..


    Explanation :

    According to the question,
    Q When B runs 200 m metres, A runs 190 metres
    \ When B runs 180 metres, A runs=\( \frac{190}{200}\)*180= 171 metres When C runs 200m, B runs 180 metres.
    Hence, C will give a start to A by = 200 – 171 = 29 metres





Ans .

(1) 31.25 metre


    Explanation :

    According to the question,
    When A covers 1000m, B covers = 1000 – 40 = 960 m
    and C covers =1000 – 70 = 930 m
    When B covers 960m, C covers 930 m.
    \ When B covers 1000m, C covers=\( \frac{930}{960}\)*1000 = 968.75 metre
    Hence, B gives C a start of = 1000 – 968.75 = 31.25 metre





Ans .

(2) 20 min.


    Explanation :

    Relative speed
    = 95 – 75 = 15 kmph
    Required Time=\( \frac{Distance}{Relative speed}\)
    \( \frac{5}{15}\)*60
    =20 minutes





Ans .

(1) 15 minutes


    Explanation :

    Time taken by C = t hours
    Time taken by B =\( \frac{t}{3}\)hours
    Time taken by A =\( \frac{t}{6}\)hours
    Here,t=\( \frac{3}{2}\)hours
    Required time taken by A
    \( \frac{3}{\frac{2}{6}}\)=\( \frac{1}{4}\)
    \( \frac{1}{4}\)*60= 15 minutes





Ans .

(2) 40 min


    Explanation :

    2 hours 45 minutes
    2 + \( \frac{45}{60}\)hours=\( \frac{11}{4}\)hours
    Distance = Speed × Time
    4 * \( \frac{11}{4}\)= 11 km.
    Time taken in covering 11 km at 16.5 kmph
    =\( \frac{11}{16.5}\)
    =40 minutes.





Ans .

(2) 35 km.


    Explanation :

    Let the total distance be x km.
    Time =\( \frac{Distance}{Speed}\)
    According to the question,
    \( \frac{10}{6}\) +\( \frac{20}{16}\) +\( \frac{x-30}{3}\)=4 \( \frac{35}{60}\)
    =4\( \frac{7}{12}\)
    \( \frac{5}{3}\) +\( \frac{5}{4}\) +\( \frac{x}{3}\)-10 =\( \frac{55}{12}\)
    x=\( \frac{140}{12}\)*3=35 km





Ans .

(1) 1 hour.


    Explanation :

    Usual time = x minutes
    New time =\( \frac{4x}{3}\)
    Speed ∝ \( \frac{1}{Time}\)
    According to the question,
    \( \frac{4x}{3}\)– x = 20
    x= 1hour





Ans .

(2) 4 km/hr.


    Explanation :

    Let, A’s speed = x kmph.
    \ B’s speed = (7 – x) kmph
    Time=\( \frac{Distance}{Speed}\)
    According to the question,
    \( \frac{24}{x}\)+\( \frac{24}{7-x}\)=14
    \( \frac{24*7}{x(7-x)}\)=14
    x (7 – x) = 12 = 4 × 3 or 3 × 4
    Þ x (7 – x) = 4 (7 – 4) or 3 (7 – 3)
    Þ x = 4 or 3
    \ A’s speed = 4 kmph.





Ans .

(3) 2 hours


    Explanation :

    Relative speed
    = 12 + 10 = 22 kmph Distance covered
    = 55 – 11 = 44 km
    \ Required time
    \( \frac{44}{22}\)
    = 2 hours





Ans .

(2) 200.


    Explanation :

    Required time = LCM of 40
    and 50 seconds
    = 200 seconds





Ans .

(1) 2 km.


    Explanation :

    Distance between starting
    point and multiplex = x metre
    Time =\( \frac{Distance}{Speed}\)
    According to the question,
    \( \frac{x}{3}\) -\( \frac{x}{4}\)= \( \frac{5+5}{60}\)
    \( \frac{x}{12}\) =\( \frac{1}{6}\)
    x=2 km



TYPE-12



Ans .

(2) 19 minutes


    Explanation :

    Two ways walking time = 55 min...(i)
    One way walking + One way riding time = 37 min.....(ii)
    By 2 × (ii) – (i),
    2 ways riding time = 2×37–55 = 19 minutes.





Ans .

(3) 50 km.


    Explanation :

    Let the distance be x km
    Time taken by A =\( \frac{x}{40}\)hrs
    Time taken by B =\( \frac{x}{50}\)hrs
    \( \frac{x}{40}\)-\( \frac{x}{50}\)=\( \frac{15}{60}\)
    \( \frac{5x-4x}{200}\)=\( \frac{15}{60}\)
    x=50 km





Ans .

(1) 5 km/hr.


    Explanation :

    Let the speed of man be x kmph
    30 x – 30{x-\( \frac{x}{15}\)} = 10
    30{x-x+\( \frac{x}{15}\)}=10
    \( \frac{x}{15}\)=\( \frac{10}{30}\)
    x= 5kmph





Ans .

(1) 46 minutes 12 seconds


    Explanation :

    Required time = LCM of 252,
    308 and 198 seconds.
    Now, 252 = 2 × 2 × 3 × 3 × 7
    308 = 2 × 2 × 7 × 11
    198 = 2 × 3 × 3 × 11
    \ LCM = 2 × 2 × 3 × 3 × 7 × 11
    = 36 × 77 seconds
    \( \frac{36*77}{60}\)minutes
    = 46 minutes 12 seconds





Ans .

(4) 6 hours


    Explanation :

    Suppose, time taken while walking be x hours And, time taken on riding be y hours
    \ According to question
    x+y=4\( \frac{1}{2}\)hr
    Then, 2y = 3 hours
    y=1\( \frac{1}{2}\)hr
    x=4\( \frac{1}{2}\) - 1\( \frac{1}{2}\)=3 hr
    Time required to walk both ways = 6 hours





Ans .

(4) 9 km


    Explanation :

    Let the required distance be x km
    \( \frac{x}{\frac{9}{2}}\)+\( \frac{x}{3}\)=5
    x{\( \frac{2+3}{9}\)=5
    x=9 km





Ans .

(4) 26.7 km.


    Explanation :

    Distance covered by A in 4
    hours = 4 × 4 = 16 km
    Relative speed of B with respect
    to A = 10 – 4 = 6 km/hr
    Time taken to catch A
    \( \frac{16}{6}\)=\( \frac{8}{3}\)hr
    Required distance
    =\( \frac{8}{3}*10\)
    = 26.67 km.= 26.7 km





Ans .

(2) 480 km.


    Explanation :

    Suppose distance be x km
    \( \frac{x}{2*40}\)+\( \frac{x}{2*60}\)=10
    \( \frac{3x+2x}{240}\)=10
    x=480 km





Ans .

(1) 2 min 25 sec.


    Explanation :

    If A covers the distance of 1 km in x seconds, B covers the distance of 1 km in (x + 25) sec- onds. If A covers the distance of 1 km, then in the same time C covers only 725 metres.
    If B covers 1 km in (x + 25) sec- onds, then C covers 1 km in (x + 55) seconds.
    Thus in x seconds, C covers the distance of 725 m.
    \( \frac{x}{725}\)*1000=x+55
    x = 145
    A co vers the di stance of 1 km in 2 minutes 25 seconds.





Ans .

(4) 250 m


    Explanation :





Ans .

(2) 205 seconds


    Explanation :

    A beats B by 30 seconds and B beats C by 15 seconds. Clearly, A beats C by 45 seconds. Also, A beats C by 180 metres. Hence, C covers 180 metres in 45 seconds
    Speed of C =\( \frac{180}{45}\)= 4 m/sec
    Time taken by C to cover 1000 m =\( \frac{1000}{4}\)=250 sec
    Time taken by A to cover 1000 m = 250–45 = 205 sec





Ans .

(2) 27.


    Explanation :

    Difference of time
    = 6 min. – 5 min. 52 sec.
    = 8 seconds
    Distance covered by man in 5 min. 52 seconds
    = Distance covered by sound in 8 seconds
    = 330 × 8 = 2640 m.
    \ Speed of man
    \( \frac{2640 m}{5 min. 52 sec.}\)=\( \frac{2640}{352}\)
    \( \frac{2640}{352}\)*\( \frac{18}{5}\)kmph
    = 27 kmph





Ans .

(1) 70 km


    Explanation :

    Let the required distance be x km.
    Difference of time
    = 15 + 5 = 20 minutes
    =\( \frac{1}{3}\)hr
    According to the question,
    \( \frac{x}{35}\)-\( \frac{x}{42}\)=\( \frac{1}{3}\)
    \( \frac{x}{210}\)=\( \frac{1}{3}\)
    x=70 km





Ans .

(3) 7 hours 30 minutes


    Explanation :

    1-\( \frac{5}{6}\) of time taken by B
    =1 hour 15 minutes
    \ Time taken by B
    = 1 hour 15 minutes × 6
    =7 hours 30 minutes





Ans .

(1) 5


    Explanation :

    Abhay’s speed = x kmph
    Sameer’s speed = y kmph
    \( \frac{30}{x}\)-\( \frac{30}{y}\)=12
    \( \frac{30}{y}\)-\( \frac{30}{2x}\)=1
    On adding,
    \( \frac{30}{y}\)-\( \frac{30}{2x}\)=3
    \( \frac{30}{2x}\)=3
    x = 5 kmph





Ans .

(3) 4 hours 45 minutes


    Explanation :

    Time taken in walking both ways = 7 hours 45 minutes ....(i)
    Time taken in walking one way and riding back = 6 hours 15 minutes ....(ii)
    By equation (ii) × 2 – (i), we have
    Time taken by the man to ride both ways
    = 12 hours 30 minutes – 7 hours
    45 minutes
    = 4 hours 45 minutes





Ans .

(1) 25 km/hr.


    Explanation :

    Let the total distance be 100 km.
    Average speed
    \( \frac{Total distance covered}{Time taken}\)
    \( \frac{100}{ \frac{30}{20}+ \frac{60}{40}+ \frac{10}{10}}\)
    \( \frac{100*2}{8}\)=25 kmph





Ans .

(2) 6 km/hr


    Explanation :

    Let the speed of A = x kmph and that of B = y kmph
    According to the question,
    x × 6 + y × 6 = 60
    x + y = 10
    and \( \frac{2}{3}\)x* 5 + 2 y * 5 = 60
    10x + 30y = 180
    x + 3y = 18 ...(ii)
    From equations (i) × (3) – (ii)
    3x + 3y – x – 3y = 30 – 18
    12x = 12
    x = 6 kmph.





Ans .

(2) 81 km


    Explanation :

    Let the trains meet after t hours, then
    24t – 18t = 27
    6t = 27
    t=\( \frac{9}{2}\)hours
    QR = 18t = 18 * \( \frac{9}{2}\)= 81 km





Ans .

(3) 8 km/hr


    Explanation :

    Let the speed of Ravi be x kmph
    then, Ajay’s speed = (x + 4) kmph
    Distance covered by Ajay
    = 60 + 12 = 72 km
    Distance covered by Ravi
    = 60 – 12 = 48 km.
    According to the question
    \( \frac{72}{x+4}\)=\( \frac{48}{8}\)
    \( \frac{3}{x+4}\)=\( \frac{2}{x}\)
    3x = 2x + 8
    x = 8 kmph





Ans .

(2) 16 km


    Explanation :

    Let man walked for t hours.
    then, t × 4 + (9 – t) × 9 = 61
    4t + 81 – 9t = 61
    81 – 5t = 61
    5t = 20
    t = 4
    Distance travelled on foot
    = 4 × 4 = 16 km.





Ans .

(1) 6


    Explanation :

    Let the required distance be x km, then
    \( \frac{x}{5}\)-\( \frac{x}{6}\)=\( \frac{1}{5}\)
    \( \frac{x}{30}\)=\( \frac{1}{5}\) x=6 km





Ans .

(4) 6 km.


    Explanation :

    Let the required distance be x km.
    \( \frac{x}{3}\)-\( \frac{x}{4}\)=\( \frac{30}{60}\)
    \( \frac{x}{12}\)=\( \frac{1}{2}\)
    x= 6km





Ans .

(2) 60 km/hr


    Explanation :

    Let the speed of train be x kmph and that of car be y kmph, then
    \( \frac{60}{x}\)+\( \frac{240}{y}\)=4
    and ,\( \frac{100}{x}\)+\( \frac{200}{y}\)=\( \frac{25}{6}\)
    \( \frac{4}{x}\)+\( \frac{8}{y}\)=\( \frac{1}{6}\)
    By equation (i ) – equation (ii) × 30
    \( \frac{60}{x}\)+\( \frac{240}{y}\)-\( \frac{120}{x}\)-\( \frac{240}{y}\)=4-5
    -\( \frac{60}{x}\)=-1
    x=60 kmph





Ans .

(2) 38 minutes.


    Explanation :

    Ratio of the speed of A and B
    = A : B = 2 : 1 = 6 : 3
    B : C = 3 : 1
    A : B : C = 6 : 3 : 1
    Ratio of their time taken
    \( \frac{1}{6}\):\( \frac{1}{3}\):1=1:2:6
    Time taken by B
    \( \frac{2}{6}\)*114 minutes
    = 38 minutes





Ans .

(3) 54 km/hr


    Explanation :





Ans .

(2) 800


    Explanation :

    Total distance of trip
    =\( \frac{1200*5}{2}\)= 3000 km
    Part of journey covered by train
    1-\( \frac{2}{5}\)-\( \frac{1}{3}\)=\( \frac{4}{15}\)
    Distance covered by train
    3000*\( \frac{4}{15}\)= 800 km





Ans .

(1) 1.6 minutes


    Explanation :

    A’s speed =\( \frac{1000}{5}\) = 200 m/minute
    B’s speed =\( \frac{1000}{8}\) = 125 m/minute
    C’s speed =\( \frac{1000}{10}\) = 100 m/minute
    Distance covered by C in 2 min- utes = 200 metre
    Distance covered by B in 1 minute = 125 metre
    Relative speed of A with respect to C = 100 metre
    Time=\( \frac{200}{100}\)= 2 minutes
    Relative speed of A with respect
    ot B = 75 metre
    Time=\( \frac{125}{75}\)= \( \frac{5}{3}\) minutes =1.6minutes





Ans .

(2) v 1 : v 2 = 4 : 5.


    Explanation :





Ans .

(1) 1.20 am..


    Explanation :

    Time taken in covering 999km
    \( \frac{999}{55.5}\)= 18 hours
    Required time = 18 hours + 1
    hour 20 minutes
    = 19 hours 20 minutes
    i.e. 1 : 20 am





Ans .

(1) 12.5 metre/seccond.


    Explanation :

    Speed = 45 kmph
    \( \frac{45*1000}{60*60}\)metre/second
    \( \frac{45*5}{18}\)metre/second
    = 12.5 metre/second





Ans .

(1) 610 m..


    Explanation :

    Distance covered in 2nd
    minute = 90 – 50 = 40 metre
    Distance covered in 3rd minute
    = 130 – 90 = 40 metre
    \ Required distance
    = 50 + 40 × 14
    = 50 + 560 = 610 metre





Ans .

(3) 7: 56 AM


    Explanation :

    Here distance is constant.
    Speed ∝ \( \frac{1}{Time}\)
    Ratio of the speeds of A and B
    \( \frac{\frac{7}{2}}{4}\)=7:8
    A’s speed = 7x kmph (let)
    B’s speed = 8x kmph
    AB = 7x × 4 = 28x km.
    Let both trains cross each other after t hours from 7 a.m. According to the question,
    7x (t + 2) + 8x × t = 28x
    7t + 14 + 8t = 28
    15t = 28 – 14 = 14
    t=\( \frac{14}{15}\)hours
    \( \frac{14}{15}\)*60
    =56 minutes.
    Required time = 7 : 56 A.M.





Ans .

(4) 9 hours


    Explanation :

    Speed of plane = \( \frac{Distance}{Time}\)
    =\( \frac{6000}{8}\)= 750 kmph
    New speed = (750 + 250) kmph
    = 1000 kmph
    Required time =\( \frac{9000}{1000}\)
    = 9 hours





Ans .

(1) 45


    Explanation :

    Let speed of train be x kmph.
    Speed of car = y kmph.
    Case I,
    Time = \( \frac{Distance}{Speed}\)
    \( \frac{240}{x}\) + \( \frac{210}{y}\)=8\( \frac{2}{3}\)
    \( \frac{240}{x}\) + \( \frac{210}{y}\)=8\( \frac{26}{3}\)...(1)
    Case II,
    \( \frac{180}{x}\)+\( \frac{270}{y}\)=9....(2)
    By equation (i) × 3 – (ii) × 4,
    \( \frac{720}{x}\) +\( \frac{630}{y}\) -\( \frac{720}{x}\) -\( \frac{1080}{y}\)
    = 26 – 36
    =\( \frac{-450}{y}\)= –10
    = –10
    y = 45 kmph





Ans .

(3) 81 km/h.


    Explanation :

    Difference of time = 11 min- utes 45 seconds – 11 minutes = 45 seconds
    Distance covered by sound in 45 seconds = Distance covered by train in 11 minutes
    330 × 45 = 11 × 60 × Speed of train
    Speed of train
    \( \frac{330*45}{11*60}\)m/sec.
    \( \frac{45}{2}\) \( \frac{18}{5}\)
    =81 kmph





Ans .

(2) 45


    Explanation :

    Distance covered in 3 hours
    36 minutes i.e. 3\( \frac{36}{60}\)hours
    =5\( \frac{18}{5}\)= 18 km.
    Time taken at 24 kmph.
    \( \frac{18}{24}\)hours
    \( \frac{18}{24}\)*minutes
    =45 minutes





Ans .

(3) 2 hours


    Explanation :

    Let the origi nal speed of aeroplane be x kmph.
    According to the question,
    \( \frac{1200}{x-300}\)-\( \frac{1200}{x}\)=2
    1200{\( \frac{x-x+300}{x(x-300)}\)}=2
    x (x – 300) =\( \frac{1200*300}{2}\)
    x (x – 300) = 600 × 300
    x (x – 300) = 600 (600 – 300)
    x = 600 kmph.
    Scheduled duration of flight =\( \frac{1200}{600}\)= 2 hours





Ans .

(4) 240.


    Explanation :

    Consumption of petrol in cov- ering 540 km=\( \frac{540}{45}\)= 12 litres
    Required expenses
    = Rs. (12 × 20)
    = Rs. 240





Ans .

(2) 6 cm


    Explanation :

    18 km o 1.5 cm
    1 km= \( \frac{1.5}{18}\)cm
    72=\( \frac{1.5*72}{18}\)cm = 6 cm





Ans .

(2) 16 km.


    Explanation :

    Length of journey on foot = x km. (let).
    \ Length of journey on cycle = (61 – x ) km.
    According to the question,
    Time = \( \frac{Distance}{Speed}\)
    \( \frac{x}{4}\)+\( \frac{61-x}{9}\)=9
    \( \frac{9x+244-4x}{36}\)=9
    5x + 244 = 36 × 9 = 324
    5x = 324 – 244 = 80
    x=16 km





Ans .

(1) 16 km.


    Explanation :

    Let the distance covered on foot be x km.
    Distance covered on cycle = (61 – x) km.
    Time=\( \frac{Distance}{Speed}\)
    \( \frac{x}{4}\)+\( \frac{61-x}{9}\)=9
    \( \frac{x}{4}\)-\( \frac{x}{9}\)=9-\( \frac{61}{9}\)
    \( \frac{5x}{36}\)=\( \frac{20}{9}\)
    x=16 km





Ans .

(4) 3300 metre


    Explanation :

    Distance = Speed × Time
    = 330 × 10 = 3300 metre





Ans .

(2) 800 km.


    Explanation :

    Let total distance covered be 2x km.
    Total time = 14 hours 40 min- utes
    =14\( \frac{40}{60}\)hours=\( \frac{44}{3}\)hours
    Time = \( \frac{Distance}{Speed}\)
    According to the question,
    \( \frac{x}{60}\)\)+\( \frac{x}{50}\)=\( \frac{44}{3}\)
    \( \frac{11x}{300}\)=\( \frac{44}{3}\)
    x=400
    Total distance = 2x = 2 × 400 = 800 km





Ans .

(2) 80


    Explanation :

    Distance between both don- keys = 400 metre.
    Relative speed = (3 + 2) m./sec.
    = 5 m./sec.
    Required time
    =\( \frac{Distance}{Relative Speed}\)
    =\( \frac{400}{5}\)= 80 seconds