Quantitative Aptitude: Data Interpretation Set 34 (SBI PO Mains Memory Based)

Below are 2 sets of Data Interpretation Questions –  SBI PO Mains 2017 Memory Based DI 

Directions (1 – 5): Study the following table carefully and answer the questions that follow:

The table shows the discount % given by stores on different items. The Market Price of any article on all stores is same. Some values are missing. Answer the questions on the basis of given table and information in question.

  1. If the average SP of article II by in all the stores is Rs 2568, Find the MP of article II.
    A) Rs 3200
    B) Rs 4500
    C) Rs 3600
    D) Rs 4300
    E) Rs 4000
    View Answer
    Option C
    Solution:

    SP by store A = (100-28)/100 * MP = 72% of MP, by B = 82% of MP, by C = 60% of MP
    So (72+82+60)/3 * MP/100 = 2568
    Solve, MP = Rs 3600
  2. Difference between SP of article I by stores A and B is Rs 486, Find the SP of same article by store C.
    A) Rs 3506
    B) Rs 4005
    C) Rs 4808
    D) Rs 4104
    E) Rs 3205
    View Answer
    Option D
    Solution:

    Difference in SP = 93% of MP – 84% of MP
    So 9% of MP = 486
    Solve, MP = 5400
    So SP by store C = 76% of 5400 = Rs 4104
  3. Average SP of article III by stores A and B is Rs 3608, by stores B and C is Rs 3300. Find the SP of article III by store C.
    A) Rs 2984
    B) Rs 3122
    C) Rs 3080
    D) Rs 2764
    E) Rs 3452
    View Answer
    Option C
    Solution:

    Let x% discount by store B
    So [84 + (100-x)]/2*100 * MP = 3608
    And [70 + (100-x)]/2*100 * MP = 3300
    Put value of (100-x) from 1 equation to another and solve for MP
    MP = Rs 4400
    So SP by store C = 70/100 * 4400 = Rs 3080
  4. Store A earned 10% profit by selling product V. If CP of articles at all articles is same, find the ratio of profits by B and C in selling V.
    A) 6 : 15
    B) 9 : 16
    C) 8 : 15
    D) 6 : 13
    E) 9 : 14
    View Answer
    Option E
    Solution:

    MP = x
    SP by A = 88% of x
    So CP by A = 100/110 * 88/100 *x = 4x/5
    Now CP is same
    SP by B = 89x/100, so profit of B = 89x/100 – 4x/5 = 9x/100
    SP by C = 94x/100, so profit of B = 94x/100 – 4x/5 = 14x/100
    So required ratio = 9 : 14
  5. Ratio of discounts on article IV by stores A and B is 2/3. Difference in SP of article IV by stores A and C is Rs 432. If SP of article IV by store A is Rs 216 more than that by store B, find the SP of article IV by store C?
    A) Rs 2138
    B) Rs 2687
    C) Rs 2736
    D) Rs 2522
    E) Rs 2544
    View Answer
    Option C
    Solution:

    x/18 = 2/3
    So discount by A = 12%
    Now
    88% of MP = 82% of MP + 216
    Solve, MP = Rs 3600
    Now:
    let y% discount by store C
    [88 – (100-y)]/100 * MP = 432
    MP = Rs 3600
    Solve, y = 24%
    So SP by C = 76% of 3600 = Rs 2736

Directions (6 – 10): Study the following pie-chart and table carefully and answer the questions that follow:
Some values are missing. Answer the questions on the basis of given table and information in question. Speed of stream is same for both upstream and downstream distance on respective days

  1. Time taken to cover the upstream distance on Friday is same as time taken to cover the downstream distance on Thursday. Total speed of still water on Thursday and Friday is 10 km/hr. Find the ratio of speed of still water on Thursday and Friday.
    A) 6 : 13
    B) 4 : 11
    C) 7 : 13
    D) 4 : 15
    E) 7 : 12
    View Answer
    Option C
    Solution:

    Distances upstream:
    Monday = 18/100 * 150 = 27 km, Tuesday = 12/100 * 150 = 18 km. Wednesday = 39 km, Thursday = 45 km, Friday = 21 km
    Similarly Distances downstream:
    Monday = 27 km, Tuesday = 45 km. Wednesday = 18 km, Thursday = 36 km, Friday = 54 km
    Let speed of still water upstream on Friday = x, then downstream on Thursday = (10-x)
    Now
    21/(x-3) = 36/[(10-x)+2.5)]
    Solve, x = 6.5
    So required ratio is 3.5 : 6.5 = 7 : 13
  2. On Monday, the boat takes a total of 4 hrs 30 minutes to cover both upstream and downstream distance. Ratio of speed of boat in still water in going upstream to downstream is 4 : 5. Find the speed of boat in still water while going downstream.
    A) 13 km/hr
    B) 15 km/hr
    C) 9 km/hr
    D) 10 km/hr
    E) 12 km/hr
    View Answer
    Option B
    Solution:

    speeds – 4x and 5x
    So on Monday
    27/(4x-3) + 27/(5x+3) = 9/2
    Solve, x = 3
    So downstream speed = 5x = 15 km/hr
  3. On Tuesday, Ratio of speed of boat in still water in going upstream to downstream is 3 : 8. Also difference in speed of boat in still water in going upstream and downstream is 5 km/hr. If the total time taken by boat to cover upstream and downstream distance is 14 hours on Tuesday, find the speed of stream.
    A) 2 km/hr
    B) 1 km/hr
    C) 3 km/hr
    D) 2.5 km/hr
    E) 1.5 km/hr
    View Answer
    Option B
    Solution:

    8x and 3x, Also 8x – 3x = 5
    So x = 1, speeds are 8 and 3 km/hr
    So for Tuesday
    18/(3-b) + 45/(8+b) = 14
    Solve, b = 1 km/hr
  4. On Wednesday, Ratio of speed of boat in still water in going upstream to downstream is 4 : 5. The difference between time to cover upstream distance and downstream distance is 5 hours, find the total time taken to cover upstream distance and downstream distance.
    A) 5 hours
    B) 4 hours
    C) 7 hours
    D) 8 hours
    E) 6 hours
    View Answer
    Option D
    Solution:

    On Wednesday
    39/(4x-2) – 18/(5x+2) = 10
    Solve
    Solve, x = 2 km/hr
    So required distance :
    39/(8-2) + 18/(10+2) = 8 hours
  5. Time taken to cover the upstream distance on Thursday is 12 hours more than time taken to cover the downstream distance on Friday. Total speed of still water on Thursday and Friday is 11 km/hr. Find the ratio of speed of still water on Thursday and Friday.
    A) 6 : 7
    B) 4 : 5
    C) 7 : 9
    D) 5 : 6
    E) 7 : 8
    View Answer
    Option D
    Solution:

    Let speed of still water upstream on Thursday = x, then downstream on Friday = (11-x)
    Now
    45/(x-2.5) – 54/[(11-x)+3)] = 12
    15/(x-2.5) – 18/(14-x) = 4
    The time is integral value, so by putting values higher than 2.5, x can be found
    x = 5
    So required ratio is 5 : 6 

 

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25 Thoughts to “Quantitative Aptitude: Data Interpretation Set 34 (SBI PO Mains Memory Based)”

  1. vikash mishra

    thanx asz

  2. Marry..

    thanx a lot

  3. Marry..

    why in ques 1…. 3 is used

    1. we have to take average, There are 3 stores so divided by 3

      1. Marry..

        yaa i got it…..ty

      2. victor....

        mam pls aisa roz dijie….plsssssssss///for ibps po////// …./.

  4. SRIVASTAVA

    Aspirantzone is awesome site first memory based

  5. Jellyfish

    1st wala DI itna easy tha

    1. Yes it was

      Questions 1, 2, 3 were same
      4 n 5 were different

  6. Sachin Shukla

    same to same qsn……sir kaha paa jate h exam k pehle qq nai pate yee type qsn???

    1. hehe

      ye exam k bad hi milte hai

      1. Sachin Shukla

        hehehehe…mam pehle dhudha kariye aise qsn…

  7. Sanathoi Maibam

    Most of the content is same but not fully

    1. Yes the data is same. Some questions are also same

      1. Sanathoi Maibam

        But not all questions were that easy though few questions were.

        1. Yes yes. Agree. 3 were same. 2 were lengthy in which conditions were given that Had it been so and so

  8. Bijli Ka Taar...

    Good one! Most questions wese similar!
    Ty aspirantszone!!

  9. ???Vedu ???

    Thank u plz also provide quizzes like SBI PO fr upcoming exams

  10. Ap

    Plz explain ques 3rd

  11. sourabh jain

    these questions were not asked in sbi po mains exam 2017. they were totally different.

    1. You can read the below comments which says that most questions were same. They also wrote the exam.

    2. in search of happiness..

      han rrb clerk pre me puche gye the same…

  12. Sangita

    In question no. 9 GIVEN,The difference between time to cover upstream distance and downstream distance is 5 hours but in solution the difference is taken as 10. which one is correct?

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