Quantitative Aptitude: Data Interpretation Set 16 (Missing DI)

Tabular Data Interpretation on Missing Values. Missing DI Questions

Direction (1-5): The table shows the Cost Price of 5 products divided in 3 costs: Production Cost, Transportation Cost and Packaging Cost, the selling price, profit/loss and profit%/loss%. Some values are missing. Find the answers based on information in table and respective questions.

1. If the percentage of profit on selling product A is 15%, then what is its cost of packaging?
   A) Rs 82.43
   B) Rs 83.50
   C) Rs 86.56
   D) Rs 71.09
   E) Rs 77.80
   
   View Answer

   Option A
   Solution:
   SP = 150, profit% = 15%
   So CP = 100/115 * 150 = Rs 130.43
   So cost of packaging = 130.43 – (40+8) = Rs 82.43
2. What is the difference between the selling price of products B and C, if the cost of transportation of C is Rs 8?
   A) Rs 26.6
   B) Rs 32.4
   C) Rs 29.8
   D) Rs 36.4
   E) Rs 12.2
   
   View Answer

   Option C
   Solution:
   CP of B = 50+10+4 = Rs 64
   30% profit
   So SP of B = 130/100 * 64 = Rs 83.2
   CP of C = 45+8+10 = Rs 63
   Profit = Rs 50
   So SP of C = 63+50= Rs 113
   So difference = 113 – 83.2 = Rs 29.8
3. Suppose all the prices are given for per kg of a product. What amount of product B will have to added to 33 kg of product E such that the resultant product costs Rs 86.
   A) 64.1 kg
   B) 51.9 kg
   C) 62.7 kg
   D) 54.3 kg
   E) 50.7 kg
   
   View Answer

   Option D
   Solution:
   CP of B = 50+10+4 = Rs 64
   For CP of E:
   SP = 100, loss% = 10%
   So CP of E = 100/90 * 110 = Rs 122.2
   Using method of allegation
   (x kg)…………………………(33 kg)
   Rs 64……………………….Rs 122.2
   ………………Rs 86
   (122.2-86)…………………….(86-64)
   36.2……………………………..22
   So
   x/33 = 36.2/22
   Solve, x = 54.3 kg
4. What is the percentage profit (approximate) on selling product D if its selling price is 80% of the selling price of B?
   A) 28%
   B) 30%
   C) 40%
   D) 35%
   E) 38%
   
   View Answer

   Option B
   Solution:
   SP of B = 130/100 * 64 = Rs 83.2
   SP of D = 80/100 * 83.2 = Rs 66.56
   CP of D = 30+6+15 = Rs 51
   So profit% = 15.56/51 * 100 = 30%
5. If 2 kg of A, 3 kg of C and 4 kg of E are sold, then what will be the final profit/loss% (approximate) on selling these given transportation cost of C as Rs 5 and profit of 5% on selling A?
   A) 22%
   B) 8%
   C) 17%
   D) 14%
   E) 12%
   
   View Answer

   Option E
   Solution:
   SP of A = 150, profit is 5%, So CP of A = 100/105 * 150 = Rs 143 (we have to find approximate)
   Given transportation cost of C is Rs 5, so total CP of C = 45+5+10 = Rs 60, profit is Rs 50, so SP of C = Rs 110
   SP of E = Rs 110, at 10% loss, CP of E = Rs 122
   So CP of (2 kg of A + 3 kg of C + 4 kg of E) = 2*143 + 3*60 + 4*122 = Rs 954
   Similarly SP of (2 kg of A + 3 kg of C + 4 kg of E) = 2*150 + 3*110 + 4*110 = Rs 1070
   So profit% = 116/954 * 100 = 12%

Direction (6-10): The table given below shows the scorecard of India during a test match. Some values are missing. Find the answers based on information in table and respective questions.

6. What is the total number of white balls face by V Kohli?
   A) 42
   B) 53
   C) 49
   D) 45
   E) 38
   
   View Answer

   Option D
   Solution:
   Runs scored by 4’s = 13*4 = 52
   Total 6’s = 10, Each player except Pujara hit at least a 6. And Kohli hit the maximum 6’s. So 6’s by Kohli = 4. So runs scored by 6’s = 4*6 = 24
   Total runs scored by 4’s and 6’s = 52+24 = 76
   For these runs balls played are 13+4 = 17 …….(1)
   So runs by 1’2 and 2’s = 148-76 = 72
   1’s and 2’s runs ratio = 1 : 5
   So 12 runs by 1’2 and 60(2*30) by 2’s
   So balls for 1’s and 2’s = 12+30 = 42 ……..(2)
   So total balls = 17+42 = 59
   So number of white balls = 104 – 59 = 45
7. What is the minimum number of balls faced by MS Dhoni?
   A) 24
   B) 32
   C) 29
   D) 36
   E) 20
   
   View Answer

   Option A
   Solution:
   Runs scored by 4’s = 11*4 = 44 ….(1)
   by 6’s = 3*6 = 18 ……(2)
   Total runs by 4’s and 6’s = 44+18 = 62
   remaining runs = 81-62 = 19
   By taking 1’s and 2’s he has scored these 19 runs. To minimize the number of balls Dhoni has to score more runs taking 2’s. So he can score 18 runs by taking 2’s and 1 by taking a 1.
   So balls for 1’s and 2’s = 9+1 = 10 …….(3)
   So total balls (minimum) = 11+3+10 = 24
8. If the number of balls faced by C Pujara to take 1’s is greater than that for 2’s, then he can score a maximum of how many runs?
   A) 24
   B) 27
   C) 32
   D) 21
   E) 34
   
   View Answer

   Option B
   Solution:
   Runs by 4’s = 2*4 = 8 ….(1)
   Number of white balls = 6
   So remaining balls = 21 – (2+6) = 13
   So, to maximize his score in 13 balls, he can take 7 one’s and 6 two’s. ……(2)
   So a maximum of 8 + 7*1 + 6*2 = 27 runs
9. Assume if only the given players score during the match for the team, then what is the minimum score of the team?
   A) 333
   B) 309
   C) 403
   D) 358
   E) 341
   
   View Answer

   Option E
   Solution:
   In order to find the minimum score of team, we have to find the minimum runs scored by Pujara.
   he takes 2 four’s so 2*4 = 8 runs
   So now 21-2 = 19 balls left. Now given that he faced 6 white balls, So now balls left = 19-6 = 13 balls. Now for minimum runs, he should score more 1’s than 2’s
   Since at least one 1’s and one 2’s is necessary so minimum runs on 13 balls is 1*12 + 2*1 
   So total min runs by Pujara =  2*4 + 1*12 + 2*1 = 22
   So minimum score of team = 148+40+37+22+13+81 = 341
10. What was the maximum possible new run rate of the team?
    A) 9.32
    B) 8.48
    C) 7.56
    D) 4.45
    E) 12.23
    
    View Answer

    Option B
    Solution:
    First find the maximum runs scored and minimum balls faced.
    Minimum number of balls faced by Dhoni = 24
    Maximum runs scored by Pujara = 2*12 + 1*1 + 4*2 = 33
    Maximum runs scored by team = 148+40+37+33+13+81 = 352
    Minimum number of balls faced = 104+55+42+21+5+24 = 251 balls or 251/6 = 41.5 overs
    So maximum run rate = 352/41.5 = 8.48