15Mar, 2017
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.

[su_table]

Products Product-
ion Cost		 Transport-
ation Cost		 Packag-
ing Cost		 Selling
Price		 Profit/
-Loss		 Profit%/
Loss%
A Rs 40 Rs 8 – Rs 150 – –
B Rs 50 Rs 10 Rs 4 – – 30%profit
C Rs 45 – Rs 10 – Rs 50 –
D Rs 30 Rs 6 Rs 15 – – –
E Rs 60 Rs 10 – Rs 110 – 10% loss

[/su_table]

  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
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] 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 [/su_spoiler]
  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
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] 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[/su_spoiler]
  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
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] 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 [/su_spoiler]
  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%
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] 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% [/su_spoiler]
  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%
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] 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% [/su_spoiler]

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.

[su_table]

Player Runs Balls Faced 4’s 6’s
V Kohli 148 104 13 –
S Raina 40 55 3 –
R Ashwin 37 42 1 1
C Pujara – 21 2 0
R Jadeja 13 5 0 1
MS Dhoni 81 – 11 3
Total – – – 10

[/su_table]
Further information is:
(i)     Total runs scored by V Kohli from those scored by 1’s and 2’s are in the ratio of 1 : 5
(ii)    Number of white balls faced by C Pujara is 6.
(iii)   A white ball is defined as the ball on which no run is scored.
(iv)   During the entire match only 1’s, 2’s, 4’s and 6’s were taken by the batsmen.
(v)    V Kohli hits the maximum number of 6’s among all the batsmen.
(vi)   Beside C Pujara every player has hitted at least a six.
(vii)  Every player has taken at least a 1 and a 2.

  1. What is the total number of white balls face by V Kohli?
    A) 42
    B) 53
    C) 49
    D) 45
    E) 38
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] 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[/su_spoiler]
  2. What is the minimum number of balls faced by MS Dhoni?
    A) 24
    B) 32
    C) 29
    D) 36
    E) 20
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] 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[/su_spoiler]
  3. 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
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] 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 [/su_spoiler]
  4. 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
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] 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 [/su_spoiler]
  5. 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
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] 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 [/su_spoiler]