Data Interpretation Capsules - Edwin Jose, CAT DILR 100 Percentiler - Set 8
DIRECTIONS: Answer the questions on the basis of the information given below. A team of 5 players Arpit, Bimal, Chatur, Dinu and Elan participated in a ‘Freaket’ tournament and played four matches (1 to 4). The following table gives partial information about their individual scores and the total runs scored by the team in each match.
- Each column has two values missing. These are the runs scored by the two lowest scorers in that match.
- None of the two missing values is more than 10% of the total runs scored in that match.
(1) What is the maximum possible percentage contribution of Arpit in the total runs scored in the four matches?
(2) If the absolute difference between the total runs scored by Arpit and Chatur in the four matches is minimum possible then what is the absolute difference between total runs scored by Bimal and Elan in the four matches?
(d) Cannot be determined
(3) The players are ranked 1 to 5 on the basis of the total runs scored by them in the four matches, with the highest scorer getting Rank 1. If it is known that no two players scored the same number of total runs, how many players are there whose rank can be exactly determined?
Let us analyse the scores of Match-1 first: Runs scored by Bimal, Dinu and Elan = 88 + 72 + 60 = 220 Hence, runs scored by Arpit and Chatur = 270 – 220 = 50 Also 10% of 270 = 27 So, both Arpit and Chatur can score a maximum of 27 runs but the sum of their scores should be 50. Arpit’s score’s range in Match-1 is 23–27 and subsequently Chatur’s score’s range in Match-1 is 27–23. Similarly Chatur and Dinu scored 30 runs each in Match-2. In Match-3 even though 10% of 240 = 24, as Dinu scored 20 runs, both Arpit and Bimal can score a maximum of 19 runs, but the sum of their scores should be 32. In Match-4, Chatur and Elan combined scored 200 – 53 – 52 –56 = 39 runs. As 10% of 200 = 20, one of Chatur or Elan scores 20 runs and the other scores 19 runs. The table can be re-written as:
(1) Option A
Maximum possible runs scored by Arpit in Match-1
Maximum possible runs scored by Arpit in Match-3
Maximum possible percentage contribution: = (27+ 100+19+53)/(270+ 300+240+ 200) × 100%
=199/1010 ×100 %
= 19.17 %
(2) Option B
Maximum possible total runs scored by Chatur in the four matches = 27 + 30 + 110 + 20 = 187.
In such a case minimum possible total runs scored by Arpit in the four matches = 23 + 100 + 13 + 53 = 189.
Difference = 189 – 187 = 2 (minimum possible)
Subsequently total runs scored by Bimal in the four matches = 88 + 65 + 19 + 52 = 224.
Also, total runs scored by Elan in the four matches = 60 + 30 + 78 + 19 = 187
Absolute difference = 224 – 187 = 37
(3) Option C
Individual ranges for total score:
Arpit -> 189 199
Bimal ->218 224
Chatur ->182 187
Elan ->187 188
Least total will be of Chatur (Rank 5)
2nd least will be Elan (Rank 4)
Rank 3 must be of Arpit.
It is not possible to determine the exact ranks of Bimal and Dinu.
DIRECTIONS: Answer the questions on the basis of the information given below: The annual sugarcane production (in million tonnes) in Meethagaon for the period 2000-2006 is shown in the bar graph given below.
(1) What is the approximate average annual sugarcane production (in million tonnes) in Meethagaon for the period 2000-2005?
(2) The sugarcane production in Meethagaon in the year 2007 increases by 15% over the year 2006. What is the approximate compounded annual growth rate of sugarcane production in Meethagaon over the period 2004-2007?
(3) Out of the following, which year has shown the highest percentage increase in sugarcane production in Meethagaon compared to the previous year?
(1) Option C
Average = (295.96 297.21 287.38 233.86+237.09 281.17)/6 = 272.11
(2) Option D
Sugarcane production in 2007 = 337.41×115/100 = 388.02 million
Let the required value be X%
(Production in 2007)/(Production in 2004) ((100+X)/100)^3
(3) Option D
For 2001 % change
For 2004 % change
= ((237.09- 233.86)/233.86)×100=1.38%
For 2005 % change
= ((281.17- 237.09)/237.09)×100=18.59%
For 2006 % change
= ((337.41- 281.17)/295.96)×100=20%
Directions: Answer the questions on the basis of the information given below.
The following tables show the batting performance of the Australian Cricket Team in a match. Table 1 indicates the score of the team at the fall of each wicket (from 1 to 10). Table 2 gives the runs scored by the 11 batsmen and the order in which they appeared in the batting line up.
• At any point there are two batsmen on the field, till the fall of the 10th wicket. Whenever the team loses a wicket, the new batsman comes as per the batting order. E.g. If one of the openers gets out, the no. 3 batsman takes the field.
• A partnership between any two batsmen is the number of runs scored while both of them are batting.
(1) How many batsmen lost their wicket between Hayden’s and Hussey’s dismissal?
(d) More than 2
(2) How many runs were scored by the batsman who was the 9th to be dismissed?
(d) Cannot be determined
(3) What was the percentage contribution to the second highest partnership of the batsman to be dismissed first in that partnership?
(d) None of these
(4) The Australian total comprised only ‘Singles’ and ‘Fours’. The number of Fours scored cannot exceed
The first batsman to get out can be either Hayden or Gilchrist. Table 1 states that the first wicket falls at score 25. As Hayden himself scored 28 which is greater than 25, we can conclude that Gilchrist was the first to get out. At that time Hayden must have scored 25 – 7 = 18 runs. Similar analysis leads to the following table:
(1) Option B
Hodge lost his wicket between Hayden and Hussey.
(2) Option D
It can either be Clark or Lee. (See table).
(3) Option A
The second highest partnership was for the fourth wicket between Hayden and Hussey for 15 runs. Percentage contribution of Hayden (who got dismissed first):
(4) Option C
Refer to the table. The maximum number of fours that could have been scored between the fall of any two wickets can be summarised as:
0 to 1st : 5 fours
1st – 2nd : 2 fours
2nd – 3rd : 2 fours
3rd – 4th : 3 fours
4th – 5th : 0 fours
5th – 6th : 3 fours
6th – 7th : 1 four
7th – 8th : 1 four
8th – 9th : 2 fours
9th – 10th : 2 fours
So in all 21 fours could have been scored.
The rest of the score is by virtue of singles.