Quantitative Aptitude: Data Interpretation Questions Set 59

Data Interpretation on Boats and Streams, Partnership. Chapter based DIs based on SBI PO 2017 Mains Pattern – for IBPS PO/RRB Mains 2017 Exams

Direction (1-5): Study the following bar graph and answer the questions that follow:

  1. On a particular day, the time taken by boat A to cover 255 km downstream is 2 hours more than the time taken by boat D to cover 195 km upstream. Find the ratio of upstream speed to downstream speed in case of boat A on that particular day.
    A) 4 : 11
    B) 3 : 5
    C) 15 : 22
    D) 8 : 19
    E) 7 : 17
    View Answer
    Option E
    Solution
    :
    Let x km/hr is the speed of stream on that particular day.
    So 255/(12+x) = 195/(20-x) + 2
    Solve, x = 5 km/hr
    So in case of boat A, ratio of speed upstream : downstream = (12-5) : (12+5) = 7 : 17
  2. Boat B covered 756 km downstream on Monday for which it took 6 hours less than that in which it covered half distance upstream on Tuesday. On Tuesday, speed of stream was 2 km/hr more than that on Monday. Find the downstream speed of boat B on Tuesday.
    A) 21 km/hr
    B) 19 km/hr
    C) 23 km/hr
    D) 11 km/hr
    E) 15 km/hr
    View Answer
      Option C
    Solution
    :
    Let on Monday, speed of stream is x km/hr, then on Tuesday it is (x+2) km/hr
    On Monday it covered 756 km, so on Tuesday it covered 756/2 =  378 km
    So 756/(16+x) = 378/(16-(x+2)) – 6 ………………(1)
    126/(16+x) = 63/(14-x) – 1
    OR SOlve with options in
    378/[16 -(y+2 )] – 756/(16+y )=6
    63 [1/(14-y) – 2/(16+y) ] = 1
    Solve, x = 5 km/hr
    So speed of stream on Tuesday = (5+2) = 7 km/hr
    So downstream speed of boat B = (16+7) = 23 km/hr  
  3. For boat C, its upstream speed is 6 km/hr on a particular day. Find the difference in time in covering 360 km by boats A and C on that particular day.
    A) 8 hours
    B) 6 hours
    C) 4 hours
    D) 7 hours
    E) 5 hours
    View Answer
      Option B
    Solution
    :
    Upstream speed of C = 6 km/hr, so speed of stream: 9 – x = 6, x = 3 km/hr
    Downstream speed of boat C = 9+3 = 12 km/hr
    Downstream speed of boat A = 12+3 = 15 km/hr
    So difference in timings = 360/12 – 360/15 = 30 – 24 = 6 hours  
  4. On a particular day, ratio of upstream speed to downstream speed of boat D is 3 : 7. It took 20 hours more to cover a distance upstream than same distance downstream by boat D. On that particular day, boat B covered same distance in how much time?
    A) 17.5 hours
    B) 15 hours
    C) 19 hours
    D) 19.5 hours
    E) 18.5 hours
    View Answer
      Option A
    Solution
    :
    Let speed of stream on that day = x km/hr
    So (20-x)/(20+x) = 3/7
    Solve, x = 8 km/hr
    So y/(20-8) – y/(20+8) = 20
    Solve, y = 420 km
    So required time = 420/(16+8) = 420/24 = 17.5 hours  
  5. Upstream sped of boat E is 9 km/hr. How many more hours will it take to cover a distance of 315 km upstream than same distance downstream?
    A) 18 hours
    B) 20 hours
    C) 22 hours
    D) 25 hours
    E) 15 hours
    View Answer
    Option B
    Solution
    :
    Upstream speed = 9, speed of boat = 15 km/hr, so speed of stream = 15-9 = 6 km/hr
    Downstream speed = 15+6 = 21 km/hr
    So required time = 315/9 – 315/21 = 20 hours

Direction (6-10): Study the following table and answer the questions that follow:

  1. In 2012, if out of the total profit, R got Rs 9750 as his share, what is the number of months for which R invested his money?
    A) 7 months
    B) 8 months
    C) 6 months
    D) 4 months
    E) 5 months
    View Answer
    Option C
    Solution
    :
    Ratio of share of profit of P : Q : R is
    4*4 : 3*6 : 5*x
    16 : 18 : 5x
    So 5x/(34+5x) * 20800 = 9750
    Solve, x = 6 months
  2. In 2013, if out of the total profit, P and R got a share of Rs 10,400 and Rs 19,500 respectively, what is the ratio of investment of P, Q and R respectively?
    A) 3 : 3 : 5
    B) 4 : 3 : 2
    C) 2 : 5 : 5
    D) 2 : 3 : 5
    E) 2 : 5 : 3
    View Answer
    Option D
    Solution
    :
    x*8 : 3*5 : y*6
    8x : 15 : 6y
    So 8x/(8x+15+6y) * 39650 = 10400
    Solve, 225x – 60y = 150
    15x – 4y = 10……………………………………(1)
    Also
    6y/(8x+15+6y) * 39650 = 19500
    Solve, 61y = 75 + 40x + 30y
    40x – 31y = -75………………………………….(2)
    Solve equations (1) and (2)
    x = 2, y = 5
    So ratio of investment is x : 3 : y = 2 : 3 : 5
  3. In 2014, out of the total profit, Q got Rs 12,000 as his share. R invested Rs 3600. If ratio of share of R in total profit to total profit is 3 : 10, find the total profit.
    A) Rs 40, 000
    B) Rs 35, 000
    C) Rs 45, 000
    D) Rs 50, 000
    E) Rs 30, 000
    View Answer
    Option E
    Solution
    :
    In 2014, P invested Rs 5x, Q – 4x and R 3600
    So ratio of share of profits of P, Q and R:
    5x*3: 4x*5 : 3600*5
    3x : 4x : 3600
    Let total profit = Rs y
    So 4x/(7x+3600) * y = 12000………………………(1)
    xy/(7x+3600) = 3000
    Also,
    [3600/(7x+3600) * y]/y = 3/10
    y gets cancelled
    Solve, x = 1200
    Put in (1)
    Solve, y = Rs 30, 000 = total profit
  4. In 2015, P and Q invested Rs 3900 and Rs 2600 respectively. Also P and Q invested for same number of months. If difference in the shares of P and Q out of total profit is Rs 2750, find the ratio of number of months of investment of P to investment of R.
    A) Other than given in options
    B) 2 : 265
    C) 1 : 260
    D) 1 : 1040
    E) 2 : 1045
    View Answer
    Option D
    Solution
    :
    Let investment of R is 4x, and months of investment of P = Q = y months
    So ratio of profit share of P : Q : R is
    3900*y : 2600*y = 4x*4
    975y : 650y : 4x
    So (975y-650y)/(975y+650y+4x) * 22550 = 2750
    325y/(1625y+4x) = 2750/22550
    1625y + 4x = 2665y
    4x = 1040y
    So y/4x = 1/1040
  5. The total investment of P, Q and R in 2016 is Rs 16,800. If difference in shares of profit of Q and P is Rs 5250, find the investment of Q.
    A) Rs 3600
    B) Rs 7200
    C) Rs 6000
    D) Rs 4800
    E) Rs 2400
    View Answer
    Option B
    Solution
    :
    Ratio of share of profit of P : Q : R is
    3*4 : x*7 : 5*6
    12 : 7x : 30
    So (7x-12)/(7x+42) * 14700 = 5250
    (7x-12)*14 = (7x+42)*5
    98x – 168 = 35x + 210
    Solve, x = 6
    So ratio of investments of P : Q : R is 3 : x : 5 = 3 : 6 : 5
    So investment of Q = 6/(3+6+5) * 16800 = Rs 7200

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54 Thoughts to “Quantitative Aptitude: Data Interpretation Questions Set 59”

  1. Jellyfish

    6/10
    Thx mam pls post similar pattern DI

  2. %%%

    Thanku mam …high level di

  3. %%%

    jo 9th ques acche se nhi padhega ..wo jaroor fasega jase mere sath hua…..dimag ka sara jung hta diya isne…TQ mam :))

  4. think tank

    Kisi ne first 2 questions Ki equation solve Ki Ho to plz ek bar detail me solve Kr de

    1. %%%

      378/16 -(y+2 ) – 756/(16+y )=6
      63 [1/(14-y) – 2/(16+y) ] = 1
      ab optn me jo value di h usse karo
      jaise

      A) 21 km/hr
      21 = 16 +y
      y =5…..now check satisfy or not

      1. ~CASIMIRO~ NITROUS~ OXIDE~

        TQ

        1. %%%

          🙁 ye tq vq mt bola karo …….aise hi method aate h bs…..isse accha h to btao

          1. ~CASIMIRO~ NITROUS~ OXIDE~

            sums cahiye ja solution 😛

          2. %%%

            nhi issi ka pucha h ….koi or way ho to

          3. ~CASIMIRO~ NITROUS~ OXIDE~

            kk bta duga wese mene yeh sums kiye nhi 😛
            ab kruga nikla koi new to bta duga 🙂

          4. ~CASIMIRO~ NITROUS~ OXIDE~

            🙂

          5. ~CASIMIRO~ NITROUS~ OXIDE~

            option se hi easy hoga :p

          6. ~CASIMIRO~ NITROUS~ OXIDE~

            fine u say and ur prep kessi gng

          7. ~CASIMIRO~ NITROUS~ OXIDE~

            exam date urs 10 and 24 ?

          8. ~CASIMIRO~ NITROUS~ OXIDE~

            po bhar lena ty pe :p

          9. ~CASIMIRO~ NITROUS~ OXIDE~

            :))

          10. ~CASIMIRO~ NITROUS~ OXIDE~

            ta ta

          11. %%%

            suno issi dp me tum Godzilla lg rhe ho…..

          12. ~CASIMIRO~ NITROUS~ OXIDE~

            badia hai phir do sabko darauga 😛

          13. %%%

            hmm vesse bhi punjab wale ajkl sbko shocked kr rhe h :p

          14. ~CASIMIRO~ NITROUS~ OXIDE~

            o-o essa nhi hum bhut sareef hai wese 😛

          15. %%%

            hehe tumse compare ni kiya wese :)………going now bye

          16. ~CASIMIRO~ NITROUS~ OXIDE~

            kk atb ta ta :))

          17. %%%

            hehe not compare with u

          18. ~CASIMIRO~ NITROUS~ OXIDE~

            yes yes LP

          19. ~CASIMIRO~ NITROUS~ OXIDE~

            :p KRNA THA LP CHLA GYA

          20. %%%

            ok ….wese kisi ne mjk kiya hoga apke sath ….ki ap shareef ho :/

          21. ~CASIMIRO~ NITROUS~ OXIDE~

            nhi ji sab kehte hai yeh :

          22. ~CASIMIRO~ NITROUS~ OXIDE~

            :)) /

      1. think tank

        Ok mam I will try next time through option

  5. hi

    Mam/sir……
    what i do for such type di boat and stream pattern my basic concept is not clear….
    Suggest me please_/_….

    1. Solve basic questions
      It has only basic concepts::::::::::

      time = distance/speed
      speed of boat = b, speed of stream = s
      upstream speed, u = b – s
      downstream speed, u = b + s
      just these formulas, nothing else – Questions are based only on these.

  6. Divaker

    Baht calculation Hai yr koi tarika calculation kam karne ke 1 St and 2nd one mein

    1. use options

      378/[16 -(y+2 )] – 756/(16+y )=6

      63 [1/(14-y) – 2/(16+y) ] = 1.

    2. Parag Chavda

      use the options dude, avoid quadratic method, quadratic is time consuming and lengthy..

  7. Ashish Chouhan

    Thank you

  8. hope@IBPS po..

    how to solve question 1,2 equation facing difficulty in solving..@@Shubhra_AspirantsZone:disqus

    1. use options

      378/[16 -(y+2 )] – 756/(16+y )=6

      63 [1/(14-y) – 2/(16+y) ] = 1

      1. heena

        mam 1st vala kaise kare..bahut calculative h..So 255/(12+x) = 195/(20-x) + 2

  9. Adamant

    ty ma’am.
    Post more questions like this..

  10. jaga

    thank u mam….in q9 P=Q= y not 5 🙂

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