Quantitative Aptitude: Time , Speed and Distance Set 9

 
Time Speed Distance Questions for SBI PO, IBPS PO, IBPS RRB, NIACL, LIC, BOI, Bank of Baroda and other competitive exams.

  1. The speed of a boat in still water is 8kmph and the speed of current is 5kmph . The boat starts from point P and rows to point Q and comes back to point P. It takes 16 hours during this journey .Find the distance between the P and Q .
    A) 42km
    B) 28km
    C) 31km
    D) 39km
    E) 47km
    View Answer
      Option D
    Solution:
    Between point P and Q = x/(8-5)  + x/(8+5) = 16
    => x = 39km
  2. To reach from point A to point B at 4pm , Anuja will have to travel at an average speed of 18kmph . She will reach the point B at 3pm if she travels at an average speed of 24kmph . What will be the  average speed of Anuja to reach point B at 2pm ?
    A) 55kmph
    B) 36kmph
    C) 45kmph
    D) 30kmph
    E) 28kmph
    View Answer
      Option B
    Solution:
    From the ques. we get to know,
    x/18 – x/24 = 1
    => x = 72km
    Time taken at 18kmph = 72/18 = 4 hours
    Therefore ,
    speed to cover 72km in 2 hours  = 72/2 =  36 kmph
  3. Minal and Dhiraj begin together writing out a novel containing 8190 line. Minal starts with the first line , writing at the rate of 200lines per hour , and Dhiraj starts with the last line , then writes 8189th line and so on , proceeding backward at the rate of 150 lines per hour. At what line will they meet?
    A) 5000
    B) 4150
    C) 4680
    D) 5780
    E) 5600
    View Answer
      Option C
    Solution:
    Duration of time of their meet  = 8190/(200+150)  = 23.4 hr.
    Their line of meet = 200 * 23.4 = 4680 line
  4. A person starts by a car from Kollam to Trivandram and at the same time another person starts from Trivandram to Kollam by a car . After passing each other  they complete their journey in 2 hours and (1/2)hour resp. At what rate does the second person drives the car if the first car runs at a speed of 40 kmph ?
    A) 80kmph
    B) 75kmph
    C) 90kmph
    D) 110kmph
    E) 60kmph
    View Answer
      Option A
    Solution:
    Ratio of speeds = √(1/2) : √2
    => S1 : S2 = 1 : 2
    If 1 === 40
    then, 2 === 80
    Therefore , S2 = 80 kmph
  5. Suppose the telegraph poles on a railway track are 30 m apart , how many poles will be passed by a train in 2 hours if the speed of the train is 90 km an hour ?
    A) 7540
    B) 8750
    C) 6000
    D) 5240
    E) 6250
    View Answer
      Option C
    Solution:
    Train travels the dist. = 90 * 2 = 180 km = 180000m
    Therefore,
    the no. of poles = 180000/30 = 6000 poles
  6. The speeds of Vijaya and Keshav are 30kmph and 40kmph . Initially, Keshav is at a point A and Vijaya is at a place B . The distance between A and B is 650 km . Vijaya started her journey 3 hours earlier than Keshav to meet each other . If they meet each other at a place C somewhere in between A and B , then find the distance  between C and B.
    A) 450km
    B) 785km
    C) 527km
    D) 470km
    E) 330km
    View Answer
      Option E
    Solution:
    In the first 3 hours Vijaya covers 90km, so rest dist. = 560km
    Now, Vijaya and Keshav travels together , towards each other.
    Time = Dist./Speed = 560/70 = 8hours
    Thus ,Vijaya travels total = 3 + 8 = 11 hours
    Thus , the dist. traveled by Vijaya  = 11 * 30 = 330km
  7. A small aeroplane can travel at 400kmph in still air . The wind is blowing at a constant speed of 40kmph . The total time for a journey against the wind is 120min. What will be the time in minutes for the return journey with the wind?
    A) 98.18min.
    B) 220min.
    C) 114min.
    D) 80min.
    E) 194min.
    View Answer
      Option A
    Solution:
    400 – 40 = 360 kmph
    Let distance be x km
    Take time in hours
    => 120/60 = x/360
    => x = 720 km
    Speed of aeroplane with the wind = 440 kmph
    Therefore ,
    Time taken by aeroplane with the wind = (720/440 ) * 60 = 98.18 min.
  8. There are 50 poles with a constant distance between each pole . A car takes 20 sec. to reach the 10th pole . How much more time will it take to reach the last pole ?
    A) 120.11 sec.
    B) 108.88 sec.
    C) 88.8 sec.
    D) 125.4 sec.
    E) 157.17 sec.
    View Answer
      Option C
    Solution:
    To reach the 10th pole, the car need to travel 9 poles
    9 poles 20 seconds
    1 pole (20/9) seconds
    To reach the last (20th) pole, the car needs to travel 19 poles.
    49 pole 49 x (20/9) seconds = 108.88 sec.
    Therefore,88.8 sec more time required  to reach the last pole.
  9. A bike travels without stoppages at the rate of 60kmph and it travels with stoppages at the rate of 52kmph. How many minutes does the bike stop?
    A) 11 mins.
    B) 10 mins.
    C) 5 mins.
    D) 8 mins.
    E) 15 mins.
    View Answer
      Option D
    Solution:
    Due to stoppages, the bike can cover 8 km less per hour 60 -52 = 8
    Time taken to cover 8 km =(8/60) x 60 = 8 minutes
  10. A cat is 50 of its own leaps behind a rat. The cat takes 5 leaps per minute to the rat’s 4 leaps. If the cat and the rat cover 8m and 5m per leap resp., what distance will the cat have to run before it catches the rat?
    A) 800m
    B) 1100m
    C) 900m
    D) 600m
    E) 500m
    View Answer
      Option A
    Solution:
    Speed of cat = 40m/min.
    Speed of rat =20m/min.
    Relative speed = 40 – 20 = 20m/min.
    Diff. in dist. = 50 * 8 = 400m
    Time in catching the rat  = 400/20 = 20min.
    Dist. traveled in 20min. = 20 * 40 = 800m

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19 Thoughts to “Quantitative Aptitude: Time , Speed and Distance Set 9”

  1. Pawan

    please explain in 7th question in second case it should b 440kph with the wind ??

    1. .-•~¹°” Ambi the akdduu ˆ”°¹~•

      yaa..that part has some error..720/440 hoga

    2. Banya

      yes .. corrected .

  2. .-•~¹°” Ambi the akdduu ˆ”°¹~•

    8th ques plz rectify banya mam 🙂

    ques says how much more time but answer is total time….and explanation too

  3. anurag agarwal

    Mam please explain 4 th one

    1. Banya

      Ratio of the speed of first car and second car resp.
      √(1/2) *( √2/√2) : √2 = √2/2 : √2 = 1/2 : 1 = 1 : 2
      If first car’s speed is 40kmph
      then from the ratio we can get , the speed of second car = 80kmph

      1. Oliver Queen

        ma’am koi bhi distance lo and ushe 40km se 2hr lgenge and same distance se 2nd car se 80km/hr ki speed se 1.2hr kaise lgega?
        let distance 80km first car is runing 40km.hr then it will complete tha journey in 2hr, to 2nd car ko ish distance ko 30min mein complete karne ke lie 160km/hr se chalna padega na?

        1. Banya

          let distance 80km first car is runing 40km.hr then it will complete tha journey in 2hr, to 2nd car ko ish distance ko 30min mein complete karne ke lie 160km/hr se chalna padega na?———–That’s ok
          But according to the question.
          we know , S * T = D
          S1 * T1 = S2 * T2
          S1 : S2 = 1 : 2 comes only
          so , from our ques. we can conclude 1 =====40kmph
          2=====80kmph

  4. .-•~¹°” Ambi the akdduu ˆ”°¹~•

    4th ques me 1:4 nhi hoga ??

    1. Banya

      Nahi ..
      Bcoz denominator mein kabhi bhi root mein nahi hota ..pehle usse simplify krte h
      like √(1/2) *( √2/√2) : √2 = √2/2 : √2 = 1/2 : 1 = 1 : 2

      1. .-•~¹°” Ambi the akdduu ˆ”°¹~•

        ohhh :))
        achha jii ..okkk madam ty ty 🙂

        1. Banya

          wc dear :)))))

  5. Jellyfish

    7/10 ty mam

  6. kumkum ahuja

    ty mam:)

  7. Ravi Upadhyay

    10/10… sb easy the bt last wala thda theek tha

  8. ครђ

    Thankyou Banya Ma’am 🙂

  9. Ruchiie

    Mam plzz explain q-10

    1. Banya

      Speed of cat = (8 * 5) = 40m/min.
      Speed of rat = (5 * 4) =20m/min.
      Relative speed = 40 – 20 = 20m/min.
      Diff. in dist. = 50 * 8 = 400m (since cat is 50 leaps behind the rat and from the second line we get the cat covers 8 m per leap so we multiple here 50 * 8 )
      Time in catching the rat = 400/20 = 20min.
      Total Dist. traveled in 20min. by the cat = 20 * 40 = 800m

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