- Staff Selection Commission Mathematics 1999 to 2017 - TIME AND DISTANCE Part 2

Staff Selection Commission Mathematics - TIME AND DISTANCE TYPE-II



Ans .

(4) 400 m


    Explanation :

    Let the length of the train travelling at 48 kmph be x metres.
    Let the length of the platform be y metres.
    Relative speed of train = (48 + 42) kmph
    \( \frac{90*5}{18}\)m/sec
    = 25 m./sec. and 48 kmph
    \( \frac{48*5}{18}\)=\( \frac{40}{3}\)m./sec.
    According to the question,
    \( \frac{x+\frac{x}{2}}{25}\)=12
    \( \frac{3x}{2x*25}\)=12
    3x = 2 × 12 × 25 = 600
    x = 200 m.
    Also,\( \frac{200+y}{\frac{40}{3}}\)=45
    600 + 3y = 40 × 45
    3y = 1800 – 600 = 1200
    y=\( \frac{1200}{3}\)= 400 m.





Ans .

(2) 12 Noon


    Explanation :

    Let two trains meet after t hours when the train from town A leaves at 8 AM.
    Distance covered in t hours at 70 kmph + Distance covered in (t – 2) hours at 110 kmph = 500km
    70t + 110 (t – 2) = 500
    70t + 110t – 220 = 500
    180 t = 500 + 220 = 720
    t=\( \frac{720}{180}\)=4 hours
    Hence, the trains will meet at 12 noon.





Ans .

(3) 5 sec


    Explanation :

    Relative speed
    = (68 + 40) kmph = 108 kmph
    =\( \frac{108*5}{18}\) m/s or 30 m/s
    Required time
    \( \frac{Sum of the lengths of both trains}{Relative speed}\)
    \( \frac{70+80}{30}\)second = 5 seconds





Ans .

(3) 12


    Explanation :

    When a train crosses a telegraph post, it covers its own length.
    Speed of first train=\( \frac{120}{10}\)= 12 m/sec.
    Speed of second train=\( \frac{120}{15}\)= 8 m/sec.
    Relative speed = 12 + 8
    = 20 m/sec.
    Required time
    \( \frac{Total length of trains}{Relative speed}\)
    =\( \frac{2*120}{20}\)= 12 seconds





Ans .

(3) 12 sec.


    Explanation :

    Relative speed = 42 + 48 = 90 kmph
    \( \frac{90*5}{18}\)m/s = 25 m/s
    Sum of the length of both trains = 137 + 163 = 300 metres
    Required time =\( \frac{300}{25}\)= 12 seconds





Ans .

(1) 54 km/hr


    Explanation :

    Speed of second train = 43.2 kmph
    =\( \frac{43.2*5}{18}\) m/sec
    Let the speed of first train be x m per second, then \( \frac{150 + 120}{x + 12}\)=10
    27 = x + 12
    x = 15 m/s=15*\( \frac{18}{5}\)kmph = 54 kmph





Ans .

(1) 444


    Explanation :

    Let the trains meet after t hours
    Then, 21t – 16t = 60
    5t = 60 Þ t = 12 hours
    Distance between A and B = (16 + 21) × 12
    = 37 × 12 = 444 miles





Ans .

(3) 8 sec


    Explanation :

    Relative speed = 45 + 54
    = 99 kmph
    99*\( \frac{5}{18}\)m/sec
    Required time =\( \frac{108 + 112}{\frac{55}{2}}\)
    =\( \frac{220*2}{55}\)= 8 seconds





Ans .

(3) 3.42 sec


    Explanation :

    Let the length of each train be x metres
    Then, Speed of first train = \( \frac{x}{3}\)m/sec.
    Speed of second train =\( \frac{x}{4}\)m/sec.
    They are moving in opposite di- rections Relaive speed = \( \frac{x}{3}\)+ \( \frac{x}{4}\)
    =\( \frac{4x+3x}{12}\)=\( \frac{7x}{12}\)m/sec
    Total length = x + x = 2 x m.
    Time taken = \( \frac{2x}{\frac{7x}{12}}\)=\( \frac{24}{7}\)=3.42sec.





Ans .

(2) 85 km/hour


    Explanation :

    To tal length of both trains = 250 metres
    Let speed of second train = x kmph
    Relative speed = (65 + x) kmph
    =(65 + x )*\( \frac{5}{18}\)m/sec
    Time=\( \frac{Sum of length of trains}{Relative speed}\)
    6=\( \frac{250}{65 + x * \frac{5}{18}}\)
    =6*\( \frac{5}{18}\)*(65 + x ) = 250
    65+x=\( \frac{250*3}{5}\)
    65 + x = 150
    x = 150 – 65 = 85 kmph





Ans .

(3) 100.


    Explanation :

    Relative speed = (84 + 6)
    = 90 kmph
    =90*\( \frac{5}{18}\) m/sec.
    = 25 m/sec.
    Length of train = Relative speed × Time
    = 25 × 4 = 100 metre





Ans .

(3) 54 km/hr


    Explanation :

    =\( \frac{Speed of X}{Speed of Y}\)
    =\( \frac{Time taken by Y}{Time taken by X} ^\frac{1}{2} \)
    =\( \frac{45}{y}\)=\( \frac{3 hours 20 min}{4 hours 48 min.} ^\frac{1}{2} \)
    =\( \frac{45}{y}\)=\( \frac{200 minutes}{288 minutes.} ^\frac{1}{2} \)
    =\( \frac{10}{12}\)
    10y = 12 × 45
    y=54 kmph.





Ans .

(3) 180


    Explanation :

    Let P and Q meet after t hours.
    Distance = speed × time
    According to the question,
    30t – 20t = 36
    10t = 36
    t=3.6 hours.
    Distance between P and Q
    = 30t + 20t
    = 50t = (50 × 3.6) km.
    = 180 km





Ans .

(3) 9.5 kmph


    Explanation :

    Speed of train starting from Q = x kmph
    Speed of train starting from P = (x + 8) kmph
    According to the question, PR + RQ = PQ
    (x + 8) × 6 + x × 6 = 162
    [Distance = Speed × Time]
    6x + 48 + 6x = 162
    12x = 162 – 48 = 114
    x=\( \frac{114}{12}\)
    =9.5 kmph.





Ans .

(1) 875 km.


    Explanation :

    Let the trains meet after t hours.
    Distance = Speed × Time
    According to the question,
    75t – 50t = 175
    25t = 175
    t=7hours.
    Distance between A and B
    = 75t + 50t = 125t
    = 125 × 7 = 875 km.





Ans .

(4) 11 seconds


    Explanation :

    Relative speed = (50 + 58) kmph
    =108*\( \frac{5}{18}\) m/sec
    = 30 m/sec
    Required time
    =\( \frac{Total length of trains}{Relative speed}\)
    =\( \frac{150+180}{30}\) sec
    =\( \frac{330}{30}\)=11 sec.





Ans .

(1) 350 km.


    Explanation :

    Let the trains meet each other after t hours.
    Distance = Speed × Time
    According to the question,
    21t – 14t = 70
    7t = 70
    t=10
    Required distance
    = 21t + 14t = 35t
    = 35 × 10 = 350 km.



TYPE-5



Ans .

(3) 8 hours


    Explanation :

    Since the train runs at \( \frac{7}{11}\)of its own speed, the time it takes is \( \frac{11}{7}\)of its usual speed.Let the usual time taken be t hours.
    Then we can write,\( \frac{11}{7}\)t = 22
    t=14 hours
    Hence, time saved
    = 22 – 14 = 8 hours





Ans .

(1) 3.75 hours


    Explanation :

    \( \frac{3}{5}\)of usual speed will take \( \frac{5}{3}\)of usual time. time & speed are inversely
    proportional \( \frac{5}{3}\) of usual time
    = usual time + \( \frac{5}{2}\)
    =\( \frac{2}{3}\) of usual time = \( \frac{5}{2}\)
    usual time
    \( \frac{5}{2}\) * \( \frac{3}{2}\)=\( \frac{15}{4}\)=3.75 hours.





Ans .

(1) 35 kmph


    Explanation :

    1 hr 40 min 48 sec
    1 hr 40 + \( \frac{48}{60}\)
    1 hr 40 + \( \frac{4}{5}\)
    1 hr \( \frac{204}{5}\)
    1 + \( \frac{204}{300}\)hr=\( \frac{504}{300}\)hr
    Speed = \( \frac{42}{\frac{504}{300}}\)= 25 kmph
    Now \( \frac{5}{7}\)*usual speed = 25
    Usual speed = \( \frac{25*7}{5}\)= 35 kmph





Ans .

(3) 6 hours


    Explanation :

    \( \frac{4}{3}\)× usual time – usual time = 2
    \( \frac{1}{3}\)usual time = 2
    Usual time = 2 × 3 = 6 hours





Ans .

(2) 60 minutes


    Explanation :

    \( \frac{4}{3}\)of usual time
    = Usual time + 20 minutes
    \( \frac{1}{3}\)of usual time = 20 minutes
    Usual time = 20 × 3
    = 60 minutes





Ans .

(2) 4.5 hours


    Explanation :

    Time and speed are inversely proportional.
    \( \frac{4}{3}\)of usual time –usual time
    =\( \frac{3}{2}\)
    \( \frac{1}{3}\) * usual time= \( \frac{3}{2}\)
    Usual time=\( \frac{3*3}{2}\)=\( \frac{9}{2}\)=4.5 hours





Ans .

(1) 2 hours 30 minutes


    Explanation :

    Time and speed are inversely proportional.
    \( \frac{7}{6}\)* Usual time – Usual time
    = 25 minutes
    Usual time \( \frac{7}{6}\)-1
    = 25 minutes
    Usual time × \( \frac{1}{6}\)
    = 25 minutes
    Usual time = 25 × 6
    = 150 minutes
    = 2 hours 30 minutes





Ans .

(2) 1 hour 12 minutes


    Explanation :

    Time and speed are inversely proportional.
    Usual time * \( \frac{7}{6}\)– usual time
    = 12 minutes
    Usual time * \( \frac{1}{6}\)= 12 minutes
    Usual time = 72 minutes
    = 1 hour 12 minutes





Ans .

(2) 420 km


    Explanation :

    Fixed distance = x km and certain speed = y kmph (let).
    Case I,
    \( \frac{x}{y+10}\)=\( \frac{x}{y}\) - 1
    =\( \frac{x}{y+10}\) + 1=\( \frac{x}{y}\) .....(1)
    Case II,
    \( \frac{x}{y+20}\) = \( \frac{x}{y}\) -1 -\( \frac{3}{4}\)
    =\( \frac{x}{y}\)- \( \frac{4+3}{4}\)
    \( \frac{x}{y+20}\)+\( \frac{7}{4}\)=\( \frac{x}{y}\).....(2)
    From equations (i) and (ii),
    \( \frac{x}{y+10}\)+1=\( \frac{x}{y+20}\)+\( \frac{7}{4}\)
    \( \frac{x}{y+10}\)-\( \frac{x}{y+20}\)=\( \frac{7}{4}\)-1
    x*( \frac{y + 20 - y - 10}{y + 10 )( y + 20 )} )
    \( \frac{7-4}{4}\)=\( \frac{3}{4}\)
    \( \frac{x *10}{( y + 10 )( y + 20 )}\)=\( \frac{3}{4}\)
    3 (y + 10) (y + 20) = 40 x
    \( \frac{3 ( y + 10 )( y + 20 )}{40}\)=x...(3)
    From equation (i),
    \( \frac{3 ( y + 10 )( y + 20 )}{40(y+10)}\) + 1
    \( \frac{3 ( y + 10 )( y + 20 )}{40y}\)
    3 (y +20) + 40
    \( \frac{3 ( y + 10 )( y + 20 )}{y}\)
    3y 2 + 60y + 40 y = 3(y 2 + 30y + 200)
    3y 2 + 100y = 3y 2 + 90y + 600
    10y = 600 Þ y = 60
    Again from equation (i),
    \( \frac{x}{y+10}\)+1=\( \frac{x}{y}\)
    \( \frac{x}{60+10}\)+1=\( \frac{x}{60}\)
    \( \frac{x}{70}\)+1=\( \frac{x}{60}\)
    \( \frac{x+70}{70}\)+1=\( \frac{x}{60}\)
    6x + 420 = 7x
    7x – 6x = 420
    x = 420 km.





Ans .

(2) 20 km/hour


    Explanation :

    Total distance
    = 7 × 4 = 28 km.
    Total time
    \( \frac{7}{10}\)+\( \frac{7}{20}\)+\( \frac{7}{30}\)+\( \frac{7}{60}\) hours
    \( \frac{42 + 21 + 14 + 7}{60}\)hours
    =\( \frac{84}{60}\)hours=\( \frac{7}{5}\)hours
    Average speed
    =\( \frac{Total distance}{Total time}\)=\( \frac{28}{\frac{7}{5}}\)kmph
    =20 kmph





Ans .

(2) 72


    Explanation :

    1 m/sec = \( \frac{18}{5}\) kmph
    20 m/sec =\( \frac{20*18}{5}\)
    = 72 kmph





Ans .

(1) 15 m/sec


    Explanation :

    1 kmph =\( \frac{5}{18}\)m/sec
    54 kmph =\( \frac{5}{18}\)*54
    = 15 m/sec.





Ans .

(3) 40 km./hr.


    Explanation :

    Speed of car = x kmph.
    Distance = Speed × Time
    = 25x km.
    Case II,
    Speed of car =\( \frac{4x}{5}\)kmph
    Distance covered =\( \frac{4x}{5}\)*25
    = 20x km.
    According to the question,
    25x – 20x = 200
    5x = 200
    x=40 kmph.





Ans .

(4) 6.6 km. per hour


    Explanation :

    Speed of car = x kmph.
    Relative speed = (x – 4) kmph
    Time = 3 minutes =\( \frac{3}{60}\)hour=\( \frac{1}{20}\)hour
    Distance = 130 metre
    \( \frac{130}{1000}\)km=\( \frac{13}{100}\)km
    Relative speed =\( \frac{Distance}{Time}\)
    5x – 20 = 13
    5x = 20 + 13 = 33
    x=6.6 kmph.



TYPE-6



Ans .

(2) 10.8 km/hr


    Explanation :

    Total distance = 10 + 12
    = 22 km
    Total time = \( \frac{10}{12}\)+ \( \frac{12}{10}\)=\( \frac{244}{120}\)hours
    Required average speed
    \( \frac{Total Distance}{Total Time}\)=\( \frac{22}{\frac{244}{120}}\)=\( \frac{22}{244}\)*120
    = 10.8 km/hr.





Ans .

(1) 65.04 km/hr


    Explanation :

    Total distance = 10 + 12
    = 22 km
    Total time
    \( \frac{600}{80}\)+\( \frac{800}{40}\) +\( \frac{500}{400}\) +\( \frac{100}{50}\)
    \( \frac{246}{8}\)hr
    Average speed
    \( \frac{600 + 800 + 500 + 100}{\frac{246}{8}}\)
    \( \frac{2000*8}{246}\)
    =65.04 km/hr.





Ans .

(2) 36 kmph


    Explanation :

    Average speed
    =\( \frac{Total distance}{Time taken}\)
    =\( \frac{30* \frac{12}{60}+45*\frac{8}{60}}{\frac{12}{60}+\frac{8}{60}}\)
    = 12 × 3 = 36 kmph





Ans .

(3) 4 km/hr


    Explanation :

    If the same distance are covered at different speed of x kmph and y kmph, the average speed of the whole journey is given by =\( \frac{2xy}{x+y}\)kmph
    Required average speed =\( \frac{36}{9}\)=4 kmph





Ans .

(3) 6


    Explanation :

    If two equal distances are cov- ered at two unequal speed of x
    kmph and y kmph, then average =\( \frac{2xy}{x+y}\)kmph
    =\( \frac{96}{16}\)= 6 kmph





Ans .

(1) 3 km/hour more


    Explanation :

    Remaining distance
    = (3584 – 1440 – 1608) km
    = 536 km.
    This distance is covered at the rate of \( \frac{536}{8}\)= 67 kmph.
    Average speed of whole journey =\( \frac{3584}{56}\)=64 kmph
    Required difference in speed = (67 – 64) kmph i.e. = 3 kmph more





Ans .

(1) 8


    Explanation :

    Total distance
    = 24 + 24 + 24 = 72 km.
    Total time
    =\( \frac{24}{6}\)+\( \frac{24}{8}\)+\( \frac{24}{12}\)
    = (4 + 3 + 2) hours = 9 hours
    \ Required average speed
    =\( \frac{Total distance}{Total time}\)=8 kmph





Ans .

(4) 88.89 km/hr


    Explanation :

    If same distance are covered at two different speed of x and y kmph, the average speed of journey =\( \frac{2xy}{x+y}\)
    =\( \frac{2*100*80}{100+80}\)
    = 88.89 kmph





Ans .

(2) \( \frac{2xy}{x+y}\)


    Explanation :

    Required average speed \( \frac{2xy}{x+y}\)
    Since, can be given as corollary If the distance between A and B be z units, then
    Average speed =\( \frac{Total speed}{Time taken}\)
    \( \frac{z+z}{\frac{z}{x}+\frac{z}{y}}\)
    =\( \frac{2xy}{x+y}\)





Ans .

(1) 48 km/hr


    Explanation :

    Average speed
    \( \frac{2xy}{x+y}\)
    \( \frac{2*40*60}{40+60}\)
    = 48 kmph





Ans .

(1) 14*\( \frac{2}{5}\)km/hr


    Explanation :

    Average speed
    \( \frac{2xy}{x+y}\)
    \( \frac{2*12*18}{12+18}\)
    =14*\( \frac{2}{5}\)





Ans .

(2) 33*\( \frac{1}{3}\) km/hr


    Explanation :

    Let the total distance be x km
    Total time =\( \frac{\frac{x}{3}}{25}\)+\( \frac{\frac{x}{4}}{30}\)+\( \frac{\frac{5x}{12}}{50}\)
    =\( \frac{4x+5x}{300}\)
    =\( \frac{3x}{100}\)
    Average speed=\( \frac{Distance}{Time}\)
    =\(\frac{x}{\frac{3x}{100}}\)
    =33*\( \frac{1}{3}\) km/hr





Ans .

(1) 7 km/hr


    Explanation :

    Time taken to cover 30km at 6 kmph=\( \frac{30}{6}\)= 5 hour
    Time taken to cover 40 km = 5 hours
    \ Average speed=\( \frac{Total Distance}{Time}\)
    \( \frac{30+40}{10}\)
    =7 km/hr





Ans .

(1) 40 km/hr


    Explanation :

    Here same distances are covered at different speeds.
    \ Average speed
    \( \frac{2xy}{x+y}\)
    =\( \frac{2*36*45}{36+45}\)
    =40 kmph





Ans .

(1) 120 kmph


    Explanation :

    Here, the distances are equal.
    \ Average speed=\( \frac{2*100*150}{100+150}\)
    =120 kmph





Ans .

(2) 5*\( \frac{1}{3}\)


    Explanation :

    Total distance = 5 × 6 + 3 × 6
    = 30 + 18 = 48 km
    Total time = 9 hours
    \ Average speed
    \( \frac{48}{9}\)
    =5*\( \frac{1}{3}\)





Ans .

(3) 70 km


    Explanation :

    Let the length of journey be x km, then \( \frac{x}{35}\)-\( \frac{x}{40}\)=\( \frac{15}{60}\)=\( \frac{1}{4}\) x= 70 km





Ans .

(3) 20 km/hr


    Explanation :

    Average speed =\( \frac{Distance}{Time}\) =\( \frac{12}{\frac{3}{10}+\frac{3}{20}+\frac{3}{30}+\frac{3}{60}}\) \( \frac{12*60}{3*12}\) =20 km/hr





Ans .

(1) 30 km/hr


    Explanation :

    Distance covered 35*\( \frac{10}{60}\)+20*\( \frac{5}{60}\)
    =\( \frac{45}{6}\)km
    Total time = 15 minutes=\( \frac{1}{4}\)hr
    Required average speed =\( \frac{Distance}{Time}\)
    =30 kmph





Ans .

(2) 40


    Explanation :

    Total distance = 100 km.
    Total time \( \frac{50}{50}\)+\( \frac{40}{40}\)+\( \frac{10}{20}\)
    =\( \frac{5}{2}\)hr
    Average speed =\( \frac{100*2}{5}\)
    = 40





Ans .

(4) 24 km/hr


    Explanation :

    Required average speed \( \frac{2*30*20}{30+20}\)
    = 24 km/hr





Ans .

(3) 9.00 a.m.


    Explanation :

    If A and B meet after t hours, then
    4 t + 6 t = 20
    10 t = 20
    t = 2 hr
    Hence, both will meet at 9 a.m.





Ans .

(3) 24


    Explanation :

    Average speed =\( \frac{2*20*30}{20+30}\)
    = 24





Ans .

(1) 37.5


    Explanation :

    Average speed of whole journey \( \frac{2*50*30}{50+30}\)
    = 37.5 kmph





Ans .

(2) 4 km


    Explanation :

    Required distance of office from house = x km. (let)
    Time =\( \frac{Distance}{Speed}\)
    According to the question,
    \( \frac{x}{5}\)-\( \frac{x}{6}\)=\( \frac{2}{15}\)
    \( \frac{x}{30}\)=\( \frac{2}{15}\)
    x= 4 km





Ans .

(4) 14 hrs


    Explanation :

    Time =\( \frac{Distance}{Speed}\)=\( \frac{1050}{75}\)
    = 14 hrs





Ans .

(2) 45 km/hr


    Explanation :

    Total distance covered by train in 5 minutes
    = (500 + 625 + 750 + 875 + 1000)
    metre = 3750 metre
    = 3.75 km.
    Time = 5 minutes =\( \frac{5}{60}\)=\( \frac{1}{12}\)hr
    Speed of train=\( \frac{3.75}{\frac{1}{12}}\)
    = (3.75 × 12) kmph
    = 45 kmph





Ans .

(1) 12\( \frac{1}{2}\) km/hr


    Explanation :

    Distance covered in first 2 hours = 2 × 20 = 40 km.
    Remaining distance = 100 – 40 = 60 km.
    Time taken in covering 60 km at
    10 kmph \( \frac{60}{10}\)=6 hr
    Required average speed=\( \frac{Distance}{Time}\)
    \( \frac{100}{2+6}\)
    =12\( \frac{1}{2}\) km/hr





Ans .

(1) 68 kmph


    Explanation :

    Difference of time = 5 + 3 = 8 minutes
    \( \frac{8}{60}\)=\( \frac{2}{15}\)hr
    If the speed of motorbike be x kmph, then
    \( \frac{25}{50}\)-\( \frac{25}{x}\)=\( \frac{2}{15}\)
    11x = 25 × 30
    x=68.18 kmph
    x= 68 kmph





Ans .

(4) 4


    Explanation :

    Let the speed of cyclist while returning be x kmph.
    \ Average speed \( \frac{2*16*x}{16+x}\)
    6.4 × 16 + 6.4x = 32x
    32x – 6.4x = 6.4 × 16
    25.6x = 6.4 × 16
    x= 4 kmph





Ans .

(3) 40 km/hr


    Explanation :

    Total distance covered = 400 km. Total time =\( \frac{25}{2 }\)hr \( \frac{3}{4}\)of total journey \( \frac{3}{4}\) * 400 = 300 km. Time=\( \frac{Distance}{Speed}\) \( \frac{300}{30}\)=10 Remaining time =\( \frac{25}{2}\)-10 =\( \frac{5}{2}\) Remaining distance = 100 km. \ Required speed of car \( \frac{100}{\frac{5}{2}}\) =40 km/hr





Ans .

(3) 160 minutes


    Explanation :

    Durga’s average speed
    \( \frac{2*5*15}{5+15}\)
    =\( \frac{15}{2}\) kmph
    Distance of School = 5 km.
    Smriti’s speed =\( \frac{15}{4}\)
    Required time =2*\( \frac{5}{\frac{15}{4}}\)
    =\( \frac{8}{3}\)hr
    \( \frac{8}{3}\)*60 =160 minutes





Ans .

(4) 35.55 kmph


    Explanation :

    Here, distances are equal.
    \ Average speed
    \( \frac{2*32*40}{32+40}\)
    \( \frac{320}{9}\)
    = 35.55 kmph





Ans .

(1) 48 km/h


    Explanation :

    Here, distance is same.
    Average speed=\( \frac{2xy}{x+y}\)
    =\( \frac{2*40*60}{40+60}\)
    =48 km/h





Ans .

(1) 54 km/hr


    Explanation :

    Total distance covered by the bus = 150 km. + 2 × 60 km.
    = (150 + 120) km.
    = 270 km.
    \ Average speed=\( \frac{Distance}{Time}\)
    \( \frac{270}{5}\)
    = 54 km/hr





Ans .

(3) 10.9 kmph


    Explanation :

    Here distances are same
    =\( \frac{2*12*10}{12+10}\)
    =\( \frac{240}{22}\)
    = 10.9 kmph





Ans .

(1) 18 kmph.


    Explanation :

    Total distance covered
    = (50 + 40 + 90) km
    = 180 km
    Time = \( \frac{Distance}{Speed}\)
    Total time taken
    \( \frac{50}{25}\)+\( \frac{40}{20}\) +\( \frac{90}{15}\)hours
    = (2 + 2 + 6) hours
    = 10 hours
    Average speed
    =\( \frac{Total distance}{Total time taken}\)
    =\( \frac{180}{10}\)
    =18 kmph





Ans .

(3) 560 m.


    Explanation :

    Distance = Speed × Time
    = (80 × 7) km.
    = 560 km.





Ans .

(4) 18.75 metre/second


    Explanation :

    Required speed of car=\( \frac{Distance}{Time}\)
    \( \frac{216}{3.2}\)kmph
    \( \frac{216}{3.2}\)*\( \frac{5}{18}\)m/sec
    = 18.75 m./sec.



TYPE-7



Ans .

(1) 2 hours


    Explanation :

    Let the distance of destination be D km
    Let the speed of A = 3x km/hr
    then speed of B = 4x km/hr
    \ According to question,
    \( \frac{D}{3x}\)-\( \frac{D}{4x}\)=30 min
    =\( \frac{1}{2}\)hour
    \( \frac{D}{12x}\)=\( \frac{1}{2}\)
    \( \frac{D}{3x}\)=\( \frac{4}{2}\)= 2 hours
    Hence, time taken by A to reach destination = 2hrs.





Ans .

(1) 1.33 hour..


    Explanation :

    Ratio of speed = 3 : 4
    Ratio of time taken = 4 : 3
    Let the time taken by A and B be 4x hours and 3 x hours respec- tively.
    Then, 4x–3x =\( \frac{20}{60}\)
    x=\( \frac{1}{3}\)
    Time taken by A = 4x hours
    4*\( \frac{1}{3}\)
    =1.33 hour





Ans .

(3) 25 : 18.


    Explanation :

    Required ratio
    \( \frac{5}{6}\):\( \frac{3}{5}\)
    \( \frac{5*30}{6}\):\( \frac{30*3}{5}\)
    = 25 : 18





Ans .

(2) 3 : 2.


    Explanation :

    Required ratio of the speed of two trains
    = \( \frac{√9}{√4}\)
    3 : 2





Ans .

(3) 78 km/hr


    Explanation :

    Speed of second train
    \( \frac{364}{4}\)
    = 91 kmph
    7x = 91
    6x=\( \frac{91}{7x}\)*6x
    =78 kmph





Ans .

(3) 3 : 4.


    Explanation :

    Speed of truck
    = 550m/minute
    Speed of bus =\( \frac{33000}{45}\)
    Required ratio = 550 :\( \frac{2200}{3}\)
    =3:4





Ans .

(2) 1 : 3 : 9


    Explanation :

    Required ratio =\( \frac{1}{3}\):\( \frac{2}{2}\):\( \frac{3}{1}\)
    =\( \frac{1}{3}\):1:3
    =\( \frac{1}{3}\)*3:1*3:3*3
    = 1 : 3 : 9





Ans .

(3) 3


    Explanation :

    The winner will pass the other, one time in covering 1600m. Hence, the winner will pass the other 3 times in completing 5km race





Ans .

(3) 3 : 4


    Explanation :

    Distance covered on the first day
    \( \frac{4}{5}\)*70= 56 km
    Required ratio = 42 : 56
    = 3 : 4





Ans .

(1) 1 : 4


    Explanation :

    Let speed of cyclist = x kmph
    & Time = t hours
    Distance= \( \frac{xt}{2}\)while time = 2t
    Required ratio =\( \frac{xt}{2*2t}\):x
    = 1 : 4





Ans .

(3) 3 : 4


    Explanation :

    Speed of train = x kmph
    Speed of car = y kmph
    Case 1:
    \( \frac{120}{x}\)+\( \frac{600-120}{y}\)=8
    \( \frac{15}{x}\)+\( \frac{60}{y}\)=1...(1)
    Case 2
    \( \frac{200}{x}\)+\( \frac{400}{y}\)= 8 hours 20 min
    \( \frac{24}{x}\)+\( \frac{48}{y}\)=1...(2)
    \( \frac{15}{x}\)+\( \frac{60}{y}\)=\( \frac{24}{x}\)+\( \frac{48}{y}\)
    \( \frac{9}{x}\)=\( \frac{12}{y}\)
    \( \frac{x}{y}\)=\( \frac{9}{12}\)
    =3:4





Ans .

(2) 3 : 4


    Explanation :

    Let the speed of train be x kmph. and the speed of car be y kmph
    Time=\( \frac{Distance}{Speed}\)
    \( \frac{120}{x}\)+\( \frac{480}{y}\)=8
    \( \frac{15}{x}\)+\( \frac{60}{y}\)=1.....(1)
    \( \frac{200}{x}\)+\( \frac{400}{y}\)=\( \frac{25}{3}\)
    \( \frac{24}{x}\)+\( \frac{48}{y}\)=1....(2)
    From equations (i) and (ii),
    \( \frac{15}{x}\)+\( \frac{60}{y}\)=\( \frac{24}{x}\)+\( \frac{48}{y}\)
    \( \frac{x}{y}\)=\( \frac{9}{12}\)
    =3:4





Ans .

(3) 3 : 4.


    Explanation :

    Speed of truck =\( \frac{550 metre}{60 second}\)
    \( \frac{55}{6}\)m/sec
    Speed of bus =\( \frac{33 * 1000 metre}{\frac{3}{4}*60*60sec}\)
    =\( \frac{440}{36}\)
    Required ratio =\( \frac{55}{6}\):\( \frac{440}{36}\)
    = 55 × 6 : 440
    = 3 : 4





Ans .

(1) 2 : 3


    Explanation :

    Speed =\( \frac{Distance}{Time}\)
    Speed of car : Speed of train
    =\( \frac{80}{2}\):\( \frac{180}{3}\)
    = 40 : 60 = 2 : 3





Ans .

(3) 15 : 5 : 3


    Explanation :

    Speed ∝ \( \frac{1}{Time}\)
    Required ratio of time
    1:\( \frac{1}{3}\):\( \frac{1}{5}\)
    =15:\( \frac{1}{3}\)*15:\( \frac{1}{5}\)*15
    = 15 : 5 : 3



TYPE-8



Ans .

(1) 100 m.


    Explanation :

    Relative speed of police
    = 11 – 10 = 1 kmph
    =\( \frac{5}{18}\)
    Distance decreased in 6 min- =\( \frac{5}{18}\) × 6×60 = 100 m
    Distance remained between them = 200–100 = 100 m





Ans .

(1) 85 km/hr


    Explanation :

    Suppose the speed of first train be x kmph
    Speed of second train
    = 30 kmph
    \( \frac{30*1000}{60}\)= 500 m per min.
    According to question
    \( \frac{Total distance}{Relative speed}\)
    \( \frac{(66 + 88 )}{x-500}\)=0.168
    0.168x – 84 = 154
    0.168x = 238
    x=\( \frac{238}{0.168}\)
    \( \frac{238*1000}{168}\)*\( \frac{3}{50}\)
    = 85 kmph





Ans .

(1) 19 minutes.


    Explanation :

    The gap of 114 metre will be filled at relative speed. Required time
    \( \frac{114}{21-15}\)
    =19 minutes





Ans .

(4) 25 seconds.


    Explanation :

    Both trains are moving in the same direction.
    \ Their relative speed
    = (68 – 50) kmph = 18 kmph
    =18*\( \frac{5}{8}\)= 5 m/sec
    Total length = 50 + 75 = 125 m
    \ Required time =\( \frac{Total length}{Relative speed}\)=\( \frac{125}{5}\)
    =25 seconds.





Ans .

(2) 12 minutes


    Explanation :

    The constable and thief are running in the same direction
    \ Their relative speed
    = 8 – 7 = 1km.
    1*\( \frac{5}{18}\)
    Required time =\( \frac{200}{\frac{5}{18}}\)
    =720 sec
    =\( \frac{720}{60}\)
    =12 minutes





Ans .

(4) 140


    Explanation :

    Relative speed
    = (58 – 30) km/hr
    28*\( \frac{5}{18}\)
    \( \frac{70}{9}\)m/sec.
    Length of train =\( \frac{70}{9}\)*18
    = 140 metres





Ans .

(3) 75.


    Explanation :

    Relative speed
    = 56 – 29 = 27 kmph
    27*\( \frac{5}{18}\)
    \( \frac{15}{2}\)
    Distance covered in 10 sec- onds \( \frac{15}{2}\)*10
    = 75 m





Ans .

(1) 27 km/hr


    Explanation :

    Let the speed of the truck be x kmph
    Relative speed of the bus
    = 45 - x kmph
    Time=\( \frac{Distance}{Relative speed}\)
    \( \frac{30}{60*60}\)=\( \frac{\frac{150}{1000}}{45-x}\)
    (45 – x ) = 18
    x=27 kmph





Ans .

(2) 50 m.


    Explanation :

    Let the length of each train be x metre.
    Relative speed
    = 46 – 36 = 10 kmph
    =\( \frac{25}{9}\)
    =\( \frac{2x}{\frac{25}{9}}\)=36
    x = 50 metre





Ans .

(3) 3 km 750 m


    Explanation :

    Relative speed
    = 45– 40 = 5 kmph
    Required distance
    5*\( \frac{45}{60}\)
    \( \frac{15}{4}\)km
    = 3 km 750





Ans .

(3) 18.6


    Explanation :

    Let the speed of Scooter be x
    Distance covered by cycling in 3\( \frac{1}{2}\)hours = Distance covered
    by scooter in 2\( \frac{1}{4}\) hours
    12*\( \frac{7}{2}\)=x*\( \frac{9}{4}\)
    x=\( \frac{56}{3}\)
    = 18.6 kmph





Ans .

(2) 400 m


    Explanation :

    Relative speed
    \( \frac{1000}{8}\)-\( \frac{1000}{10}\)=\( \frac{1000}{40}\)
    Required time = 4 m/minute
    Distance covered by the thief =\( \frac{1000}{10}\)*4
    = 400 metres





Ans .

(1) 27.7 m


    Explanation :

    Relative speed = 40 – 20
    = 20 km/hour
    =\( \frac{20*5}{18}\)
    Length of the faster train =\( \frac{250}{9}\)= 27.7 metres





Ans .

(4) 90 km/h


    Explanation :

    Distance = Speed × Time
    = 80 × 4.5 = 360 km
    Required speed = \( \frac{360}{4}\)
    = 90 kmph.





Ans .

(2) 9


    Explanation :

    Required time =\( \frac{Sum of the lengths of trains}{Relative speed}\)
    Relative speed = 65 + 55
    = 120 kmph
    \( \frac{120*5}{18}\)
    Required time = \( \frac{180+120}{\frac{120*5}{18}}\)
    = 9 seconds





Ans .

(1) 125


    Explanation :

    When two trains cross each other, they cover distance equal to the sum of their length with relative speed.
    Let length of each train = x metre
    Relative speed = 90 – 60
    = 30 kmph
    \( \frac{30*5}{18}\)
    =\( \frac{25}{3}\)m/sec
    \( \frac{2x}{\frac{25}{3}}\)=30
    2x = 250
    x = 125 metres





Ans .

(4) 72.


    Explanation :

    Relative speed = 35 – 25
    = 10 kmph
    =\( \frac{10*5}{18}\)m/sec
    Total length = 80 + 120
    = 200 metres
    Required time =\( \frac{Sum of the length of trains}{Relative speed}\)
    =\( \frac{200*18}{10*5}\)
    = 72 seconds





Ans .

(1) 24


    Explanation :

    Distance covered by the first goods train in 8 hours = Distance covered by the second goods train in 6 hours.
    18 × 8 = 6 * x
    x=\( \frac{18*8}{6}\)
    = 24 kmph





Ans .

(3) 12.


    Explanation :

    Relative speed
    = (33 + 39) kmph
    = 72 kmph
    \( \frac{72*5}{18}\)m/sec
    = 20 m/sec.
    \ Time taken in crossing
    =\( \frac{Length of both trains}{Relative speed}\)
    =\( \frac{240}{20}\)
    =12 seconds





Ans .

(2) 4 p.m..


    Explanation :

    Distance covered by the thief in half an hour =\( \frac{1}{2}\)*40 =20 km
    Relative speed of car owner
    = 50 – 40 = 10 km
    \ Required time
    =\( \frac{Difference of distance}{Relative speed}\)
    \( \frac{20}{10}\)
    = 2 hours
    i.e. at 4 p.m.





Ans .

(1) 50 m


    Explanation :

    Length of each train = x metre

    Relative speed = 46 – 36
    = 10 kmph
    =\( \frac{10*5}{18}\)
    =\( \frac{25}{9}\)m/sec
    Time taken in crossing
    \( \frac{Length of both trains}{Relative speed}\)
    36=\( \frac{2x}{\frac{25}{9}}\)
    x = 50 metre





Ans .

(3) 1320 km


    Explanation :

    Let both trains meet after t hours.
    \ Distance = speed × time
    60t – 50t = 120
    10t = 120 t = 12 hours
    Required distance
    = 60t + 50t
    = 110t = 110 × 12
    = 1320 km





Ans .

(3) 6.


    Explanation :

    Let both cars meet at C after t hours.
    Distance covered by car A = AC = 35t km
    Distance covered by car B
    = BC = 25t km
    AC – BC = AB = 60 km.
    35t – 25t = 60
    10t = 60
    t = 6 hours





Ans .

(2) 88.


    Explanation :

    Let the speed of train C be x kmph.
    Relative speed of B
    = (100 – x ) kmph.
    Time taken in crossing
    \( \frac{Length of both trains}{Relative speed}\)
    \( \frac{2}{60}\)=\( \frac{\frac{150+250}{1000}}{100-x}\)
    100 – x = 12
    x = 100 – 12 = 88 kmph.





Ans .

(1) 32 kmph


    Explanation :

    Let the speed of goods train be x kmph.
    Distance covered by goods train in 10 hour = distance cov- ered by passenger train in 4 hours
    10x = 80 × 4
    x = 32 kmph.





Ans .

(4) 3.75 km..


    Explanation :

    Relative speed = 45 – 40
    = 5 kmph.
    Gap between trains after 45 minutes = 5*\( \frac{45}{60}\)
    = 3.75 km.





Ans .

(3) 500 metre


    Explanation :

    Distance between thief and policeman = 400 metre
    Relative speed of policeman with respect to thief
    = (9 – 5) kmph
    = 4 kmph
    4*\( \frac{5}{18}\)
    \( \frac{10}{9}\)m/sec
    Time taken in overtaking the thief
    \( \frac{400}{\frac{10}{9}}\)
    = 360 second
    Distance covered by thief
    = Speed × Time
    =5*\( \frac{5}{18}\)*360
    = 500 metre





Ans .

(4) 50 m


    Explanation :

    Let the length of each train be x metre.
    Relative speed = (46 – 36) kmph
    = 10 kmph
    10*\( \frac{5}{18}\)
    =\( \frac{25}{9}\)m/sec
    \( \frac{2x}{\frac{25}{9}}\)=36
    x=50 metre



TYPE-9



Ans .

(4) 50 minutes


    Explanation :

    Time taken to cover 20 km at the speed of 5km/hr = 4 hours.
    \ Fixed time = 4 hours – 40 min- utes
    = 3 hour 20 minutes
    Time taken to cover 20 km at the speed of 8 km/hr =\( \frac{20}{8}\)=2 hours 30 minutes
    Required time = 3 hours 20 minutes – 2 hours 30 minutes = 50 minutes





Ans .

(1) 2.


    Explanation :

    Since man walks at \( \frac{2}{3}\)of usual speed, time taken wil be \( \frac{3}{2}\) usual time.
    =usual time + 1 hour.
    \( \frac{3}{2}\)-1 of usual time = 1
    usual time = 2 hours.





Ans .

(3) 5 km


    Explanation :

    Let x km. be the required dis- tance.
    Difference in time
    = 2.5 + 5 = 7.5 minutes
    =\( \frac{7.5}{60}\)=\( \frac{1}{8}\)hr
    \( \frac{x}{8}\)-\( \frac{x}{10}\)=\( \frac{1}{8}\)
    x=\( \frac{40}{8}\)= 5 km.





Ans .

(4) 40


    Explanation :

    Let the distance be x km and initial speed be y kmph. According to question,
    \( \frac{x}{y}\)-\( \frac{x}{y+3}\)=\( \frac{40}{60}\).....(1)and
    \( \frac{x}{y-2}\)-\( \frac{x}{y}\)=\( \frac{40}{60}\)......(2)
    From equations (i) and (ii),
    \( \frac{x}{y}\)-\( \frac{x}{y+3}\)=\( \frac{x}{y-2}\)-\( \frac{x}{y}\)
    3 (y – 2) = 2 (y + 3)
    Þ 3y – 6 = 2y + 6
    Þ y = 12
    From equation (i),\( \frac{x}{12}\)-\( \frac{x}{15}\)=\( \frac{40}{60}\)
    x=40
    Distance = 40 km.





Ans .

(3) 19 .


    Explanation :

    If the distance be x km, then
    \( \frac{x}{40}\)-\( \frac{x}{50}\)=\( \frac{6}{60}\)
    x = 20 km.
    Required time
    \( \frac{20}{40}\)hr-11 mimnutes
    = 19 minutes





Ans .

(1) 1.75 km


    Explanation :

    Let the required distance be x km.
    Difference of time
    = 6 + 6 = 12 minutes = \( \frac{1}{5}\) hr
    According to the question,
    \( \frac{x}{\frac{5}{2}}\)-\( \frac{x}{\frac{7}{2}}\)=\( \frac{1}{5}\)
    \( \frac{14x-10x}{35}\)=\( \frac{1}{5}\)
    x=\( \frac{7}{4}\)=1.75 km





Ans .

(4) 6 km


    Explanation :

    Let the required distance be x km.
    According to the question,
    \( \frac{x}{4}\)-\( \frac{x}{5}\)=\( \frac{18}{60}\)
    x=\( \frac{3}{10}\)* 20= 6 km





Ans .

(2) 40 km/hour.


    Explanation :

    Let the initial speed of the car be x kmph and the distance be y km.
    Then,y=\( \frac{9}{2}\)x
    and, y = 4 (x + 5)
    9x = 8x + 40
    x = 40 kmph





Ans .

(3) 22 km


    Explanation :

    Let the distance of office be x km
    \( \frac{x}{24}\)-\( \frac{x}{30}\)=\( \frac{11}{60}\)
    \( \frac{x}{120}\)=\( \frac{11}{60}\)
    x=22 km





Ans .

(3) 3 km.


    Explanation :

    Let the required distance be x km.
    \( \frac{x}{3}\)-\( \frac{x}{5}\)=\( \frac{24}{60}\)
    \( \frac{2x}{3}\)=2
    2x = 2 × 3
    x = 3 km





Ans .

(2) 4


    Explanation :

    Let the required distance be x
    km \( \frac{x}{\frac{5}{2}}\)-\( \frac{x}{3}\)=\( \frac{16}{60}\)
    \( \frac{6x-5x}{15}\)=\( \frac{4}{15}\)
    x = 4 km.





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.