Data plays an important role in day to day life. If data is too large, it can be represented in precise form in a number of ways. Once data is represented in precise form, the user of that data has to understand it properly. The process of interpreting the data from its precise form is called Data Interpretation.
Data Interpretation is a part of every MBA entrance exam. So, we will discuss different ways of representing data and we will see how we can extract the data from the given representations.
Different ways of representing data:
1. Data Tables 2. Pie Charts
3. Two-Variable Graphs 4. Bar Charts
5. Venn Diagrams 6. Three-Variable Graphs
7. PERT Chart 8. Combination of 2 or more charts
Now, we shall study these methods in detail.
1. Data Table:
Here the entire data is represented in the form of a table. The data can be represented in a single table or in combination of tables. To understand it better, look at the following example.
Population of different cities (in 000’s)
From the above table, we can find the following:
1. Population of a particular city with respect to that in any other city for a given year. 2. Percentage change in the population of any city from one year to another.
3. The rate of growth of population of any city in any given year over the previous year. 4. The city, which has maximum percentage population growth in the given period.
5. For a given city, finding out the year in which the percentage increase in the population over the previous year was the highest.
6. Rate of growth of the population of all the cities together in any given year over the previous year.
Year Hyderabad Mumbai Chennai Bangalore Delhi
2002 2000 4000 3700 1650 3850 2003 2400 4800 4300 1760 4160 2004 3000 5500 5150 2325 4750 2005 3500 6450 6070 2810 4800 2006 3750 7210 6910 3020 5110 2007 4500 7800 7430 4010 6500 2008 8000 9560 8150 6000 8050
CATsyllabus.com
Data Interpretation - DI
EXAMPLE:
NUMBER OF BOYS OF STANDARD XI PARTICIPATING IN DIFFERENT GAMES
Class XI A XI B XI C XI D XI E Total Games Chess 8 8 8 4 4 32 Badminton 8 12 8 12 12 52 Table Tennis 12 16 12 8 12 60 Hockey 8 4 8 4 8 32 Football 8 8 12 12 12 52 Note:
¾ Every student (boy or girl) of each class participates in a game.
¾ In each class, the number of girls participating in each game is 25% of the number of boys participating in each game.
¾ Each student (boy or girl) participated in one and only one game.
1. All the boys of class XI D passed at the annual examination but a few girls failed. If all the boys and girls who passed XI D and entered XII D are in the ratio of 5 : 1, how many girls failed in XI D ?
(1) 8 (2) 5 (3) 2 (4) 1
Sol. Note: Before solving these questions note that the table is given for the number of boys and not for the total number of students.
The number of boys in XI D are 40 ∴ Girls in XI D = 40 × 1/4 = 10
Number of boys who passed XI D and entered XII D = 40 Ratio in XII D = 5 : 1
∴ In XII D Boys –> 5; Girls –> 1 or Boys 40; Girls 8
Girls in XI D = 10 Girls in XII D = 8 ∴ 10 – 8 = 2 girls failed. ∴ Answer: (3)
2. Girls playing which of the following games need to be combined to yield a ratio of boys to girls of 4 : 1 if all boys playing chess and badminton are combined?
(1) Table Tennis & Hockey (2) Badminton & Table Tennis (3) Chess & Hockey (4) Hockey & Foot ball Sol. Number of boys playing chess and badminton = 52 + 32 = 84 boys
Since girls are 25% of boys,
To yield a ratio of 4:1, number of girls should be 21
∴Girls playing Hockey and Football = ¼ × 32 + ¼ × 52 = 8 + 13 = 21 girls
∴ Girls of hockey + football have to be combined to give a ratio of 4 : 1 if boys playing chess & badminton are combined. Answer: (4)
3. What should be the total number of students in the school if all the boys of class XI A together with all the girls of class XI B and class XI C were to be equal to 25% of the total number of students?
(1) 272 (2) 560 (3) 656 (4) 340
Sol. Boys of XI A = 44
Boys of XI B = 48 Girls of XI B = 12 Boys of XI C = 48 Girls of XI C = 12
Total 68
We are given that (44 + 12 + 12) = 68 is 25% of total students in the school. ∴ Total students =
25 . 0
68 = 272. Answer: (1)
4. Boys of which of the following classes need to be combined to equal four times the number of girls in class XI B and class XI C
(1) XID & XIE (2) XIA & XIB (3) XI A & XI D (4) None of these Sol. Number of girls in XI B + XI C = 24
4 times = 96
∴ Boys of XI B and XI E have to be combined. Hence Answer: (4)
5. If boys of class XI E participating in chess together with girls of class XI B and class XI C participating in Table Tennis & hockey respectively are selected for a course at the college of sports, what percentage of the students will get this advantage approximately?
(1) 4.38 (2) 3.51 (3) 10.52 (4) 13.5
Sol. Boys of XI E playing chess = 4 Girls of XI B playing Table Tennis = 4 Girls of XI C playing Hockey = 2
∴ Number of student selected = 4 + 4 + 2 = 10
Number of students in the school = boys + girls = 228 + 57 = 285 ∴ Percentage =
285
10 × 100 = 3.51. Answer: (2)
Note: Number of students in the school should not be taken as 272 – that figure is valid only for Q.3 6. If for social work every boy of class XI D and XI C is paired with a girl of the same class, what
percentage of boys of these two classes cannot participate in social work?
(1) 88 (2) 66 (3) 60 (4) 75
Sol. Since girls are only25% of the boys only 25% of the boys can participate and 75% of the boys cannot participate in social work. Hence Answer: (4)
2. Pie Chart
In this, the total quantity is distributed over one complete circle. This circle is made into various parts for various elements. Each part represents share of the corresponding element as portion of the total quantity. These parts can be represented in terms of percentage or in terms of angle.
Look at the following Pie-chart representing crude oil transported through different modes over a specific period of time.
Ship 10% Rail 20% Road 20% Pipeline 50%
The above pie chart can also be represented as below
Rail, 72o Road, 72o
Pipeline, 180o
Ship, 36o
We can find the following from the above pie chart.
1. The oil that has been transported through any mode if the total transported amount is known. 2 The proportion of oil transported through any mode with respect to any other mode.
Administration & Miscellaneous 12% X 24% Faculty 8% Advertising & Promotion 31% Material Preparation 10% Printing 15% EXAMPLE:
These questions are based on the diagram given below
EXPENSES OF TCY [as a percentage of turnover]
X = Salaries + Profit
1. If the turnover of TCY was Rs. 2 lakhs this year and the salaries to be paid were Rs. 95000, what is the loss this year as a percentage of turnover?
(1) 23.5% (2) 19.03% (3) 47.5% (4) 26.7% Sol. From the pie chart we can say that X = 24%
X = 24/100 × 2 × 105 = 48000 Salaries = Rs. 95000
Loss = (95000 – 48000)/ (200000) × 100 = 23.5%. Answer: (1)
2. If total salaries are Rs. 1,20,000 per year and 12% profit on turnover is made, what will be the printing charges that year ?
(1) Rs. 10 lacs (2) Rs. l.51acs (3) Rs. 1 lac (4) Rs. 75000 Sol. 0.24x = 120000 + 0.12x, where x is total turnover ⇒ x = 106
Printing charges = 0.15x = 0.15 × 106 = 1.5 lac. Answer: (2)
3. If TCY had spent Rs. 40000 more for Advertising and Promotion than for printing, how much more would they have spent for material preparation than for faculty?
(1) Rs. 2500 (2) Rs. 6000 (3) Rs. 7500 (4) Rs. 5000 Sol. 0.31x – 0.15x = 40000 ⇒ x = 2.5 × 105
More amount spent on material preparation than faculty = 0.1x – 0.08x = 0.02 × 2.5 × 105 = Rs. 5000. Answer: (4)
4. If TCY has to pay total salaries of Rs. 1.32 lacs, what should be the turnover of TCY so that there is no profit no loss?
(1) Rs. 6 lacs (2) Rs. 5 lacs (3) Rs. 5.5 lacs (4) None of these Sol. 0.24x = 132000, where x is total turnover
3. Two-Variable Graphs
Here the data will be represented in the form of a graph. Generally it represents the change of one variable with respect to the other variable.
Look at the following graph.
Car sales in India in different years (in 000’s)
0 50 100 150 200 2003 2004 2005 2006
Maruti Hyundai Others
From the above graph, we can calculate.
1. Percentage change in the sales of any brand in any year over the previous year. 2. Rate of growth of total sales of the cars (all the brands) in a given period.
3. Proportion of the sales of any brand with respect to those of any other brand in the given year.
Example :
INDIA’S CASHEWNUT EXPORTS
75 150 160 150 200 100 500 400 330 150 0 100 200 300 400 500 600 1995 1996 1997 1998 1999
1. In which year was the value per kg minimum
(A) 1995 (B) 1996 (C) 1997 (D) 1998
Sol. Value per kg for the years given in options
1995 1996 1997 1998
150/100 150/15 360/150 400/160
From the above values it is clear that value per kg is minimum for the year 1995. Answer: (A)
2. What was the difference in volume exported in 1997 and 1998?
(A) 10000 kg (B) 1000 kg (C) 100000 kg (D) 1000000 kg
Sol. Difference = (160 – 150) 105 = 1000000 kg Answer: (D)
3. What was the approximate percentage increase in export value from 1995 to 1999?
(A) 350 (B) 330 (C) 430 (D) 230
Sol. Percentage increase in export value from1995 to 1999 = 150
150 500−
× 100 = 230% approx.
Answer: (D) 4. What was the percentage drop in export quantity from 1995 to 1996?
(A) 75% (B) 31/3% (C) 25% (D) 0%
Sol. Percentage decrease in export quantity from 1995 to 1996 = 100
100 75−
= 25% Answer: (C)
5. If in 1998 cashew nuts were exported at the same rate per kg. as that in 1997what would be the value of exports in 1998
(A) Rs. 400 Crores (B) Rs. 352 Crores (C) Rs. 375 Crores (D) Rs. 330 Crores Sol. Rate per kg of cashew nut in 1998 = (330 × 107)/(150 × 105) = Rs. 220.
Value of exports in 1998 = 160 × 105 × 220 = Rs. 352 crores. Answer: (B)
Bar Chart
Bar Chart is also one of the ways to represent data.
The data given in the above graph can also be represented in the form of bar chart as shown below.
Here also we can deduce all the parameters as we could do in the case of two-variable graph. 50 100 125 175 30 50 80 100 60 60 100 90 0 50 100 150 200 2003 2004 2005 2006
Example:
CONSUMPTION OF CHOCOBAR ACROSS THE COUNTRY (in ‘000 bars)
124
118
128
92
134
126
122
0
20
40
60
80
100
120
140
160
1993 1994 1995 1996 1997 1998 1999
YEARS1. Which of the following statements is true regarding the consumption of chocobar?
(A) the percentage change in consumption of chocobar over the previous year is the same every year. (B) The rate of fall of consumption chocobar is increasing steadily.
(C) The steepest increase in the consumption of chcocbar follows the steepest fall in consumption (D) The consumption is falling and increasing in alternate years.
Sol. In 1997 the rise was 42 = It is the steepest rise and in 1996 the fall is 36, it is the steepest fall.
Answer: (C)
2. The highest percent fall in the consumption of chocobar s equal to
(A) 28.1% (B) 39.1% (C) 25% (D) 32.2%
Sol. In 1996 the % drop =
128 36
× 100 = 28.1%
Answer: (A)
3. If 30% of the consumption of chocobars for the first five years was in marriage parties, then find the number of cartons of chocobar supplied to marriage parties given that each carton has 120 bars.
(A) 1590 (B) 4998 (C) 4967 (D) 1490
Sol. Consumption of the chocobars for the first five years = (124 + 118 + 128 + 92 + 134 + 126 + 122) × 1000 No. of cartons of 120 bars that has to be supplied =
120 ] 134 92 128 118 124 [ 3 . 0 + + + + × 1000 = 1490 Answer: (D)
4. If only 61% of the production for the year 1999 was consumed and of the rest 20% was stored and the rest had to be thrown away, then the number of chocobars that had to be thrown away is
(A) 40,260 (B) 59,536 (C) 38,000 (D) 62,400 Sol. 61% of production in 1999 = 122 × 103
⇒ Production = 200 × 103
∴ No. of chocobars thrown away = 200(0.39) 0.8 × 1000 = 62,400
5. The least percentage decrease recoded was
(A) 3.14 (B) 3.19 (C) 3.22 (D) 3.17
Sol. By observation, least percentage decrease is from 1998 – 99, = 126 122 126 − × 100 = 3.17% Answer: (D) 5. Venn Diagrams
If the information comes under more than one category, we represent such data in the form of a Venn diagram.
The following Venn diagram represents the number of people who speak different languages.
From the above Venn diagram, we can find
1. the number of people who can speak only English. 2. the number of people who can speak only Punjabi.
3. the number of people who can speak both Punjabi and Hindi. 4. the number of people who can speak all the three languages. 5. the number of people who can speak exactly one or two languages.
Example:
In a class of 33 students, 20 play cricket, 25 football, & 18 volleyball, 15 play both cricket & football, 12 football & volleyball, 10 cricket & volleyball. If each student plays at least one game, find the number of students:
1. Who play only cricket?
(A) 5 (B) 7 (C) 2 (D) 3
Sol. Let C, F & V denote the sets of no of students who play cricket, football & volleyball respectively. ∴ n(C) = 20, n(F) = 25, n(V) = 18
n(C ∩ F) = 15. n(F ∩ V) = 12, n(C ∩ V) = 10
Let ‘x’ be the no. of students who play all the 3 games
∴ No. of students who like cricket & football but not volleyball = (15 – x) Similarly, no. of students playing F & V but not cricket = (12 – x)
No. of students playing C & V but not football = (10 – x)
Now, we can find the no. of students who play cricket only, football only & volleyball only is
10 25 12 English (120) Hindi (80) Punjabi (125) 32 C (20) F (25) V (18) 12 10 15 15 - x 10-x x x – 4 12-x x – 5 x – 2
n(C) only = 20 – (15 – x + x + 10 – x) = x – 5 n(V) only = 18 – (10 – x + x + 12 – x) = x – 4 & n(F) only = 25 – (15 – x + x + 12 – x) = x – 2
∴ 33 = (x – 5) + 15 – x + x + 10 – x + 12 – x + x – 4 + x – 2 33 = x + 26 ∴ x = 7.
∴ No. of students who play only cricket = 7 – 5 = 2.
Answer. (C) 2. Who play all the three games?
(A) 5 (B) 7 (C) 2 (D) 3
Sol. ∴ No. of students who play all 3 games = 7.
3. Who play any two games?
(A) 16 (B) 18 (C) 7 (D) 14
Sol. No. of who play any 2 games
= Total – [students who play all 3 games + Students who play only 1 game]. = 33 – [7 + 10] = 16.
Answer. (A) 4. Who play only one game?
(A) 18 (B) 16 (C) 10 (D) 5
Sol. No. of students who play only one game = No. who play (C only + V only + F only)
= 2 + 3 + 5 = 10. Answer. (C)
OR
We can also use the formula
n(C ∪ F ∪ V) = n(C) + n(F) + n(V) – n(C ∩ F) – n(F ∩ V) – n(C ∩ V) + n (C ∩ F ∩ V) ∴ 33 = 20 + 25 + 18 – 15 – 12 – 10 + x.
∴ x = 33 – 26 = 7.
6. Three-Variable Graphs
Look at the following example to understand the concept. The graph represents percentage of GRE, GMAT and CAT students in three institutes x, y, z.
The above diagram gives the percentage of students of each category (GRE, GMAT, CAT) in each of the institutes x, y, z.
EXAMPLE :
1. In institute ‘x’, what is the ratio of the number of CAT students to that of GMAT students? (1) 1 : 1 (2) 1 : 2 (3) 2 : 1 (4) None of these
Sol. Number of CAT students in institute x = 50% of total Number of GMAT students in institute x = 25% of total Therefore, required ratio = 2 : 1 Answer: (3)
2. If there are 132 GRE students in institute ‘y’, how many GMAT students are there in the same institute?
(1) 132 (2) 264 (3) 396 (4) Can’t say
Sol. Let the total number of students in institute y be T Percentage of GRE students = 25%
25% of T = 132 T = 132 4 = 528
Number of GMAT students in institute y = 50% of 528 = 264 Answer: (2) 25 50 0 (100) 75 50 25 0 (100) 75 100 x y z GRE 75 50 25 GMAT CAT 0
3. The total number of students in institute ‘x’ is twice the number of GRE students in institute ‘z’, what is the ratio of the number of CAT students of institute ‘x’ to the number of GMAT students of institute z? (1) 1 : 2 (2) 2 : 1 (3) 1 : 3 (4) 3 : 1
Sol. Let the total number of students in institute z be T Total number of students in institute x = 2 75% of T =
2 3
T
Number of CAT students in institute x = 50% of 3/2T = 4 3
T
Number of GMAT students in institute z = 25% of T = 4 1 T Required ratio = 4 3 T : 4 1 T = 3 : 1 Answer: (4)
4. If the ratio of the number of students of institutes x, y, z is 1 : 2 : 3 respectively, what is the ratio of the CAT, GRE, GMAT students (in all the institutes together)?
(1) 1 : 2 : 3 (2) 1 : 3 : 2 (3) 2 : 3 : 1 (4) 3 : 2 : 1
Sol. Let the total number of students in institutes x, y, z y be T, 2T and 3T respectively. Number of CAT students in all the institutes = 50% T + 25% 2T + 0% 3T = T Number of GRE students in all the institutes = 25% T + 25% 2T + 75% 3T = 3T Number of GMAT students in all the institutes = 25% T + 50% 2T + 25% 3T = 2T Required ratio = T : 3T : 2T = 1 : 3 : 2
Answer: (2) 7. PERT Charts
The word PERT stands for "Project Evaluation and Review Technique". The progress of any project is monitored and the execution of various activities is scheduled keeping in mind resource constraints (like labour) and time constraints. For the purpose of Data Interpretation questions, the data may be given in the form of a table or a chart.
We will take a table and draw a PERT chart from the table.
INTERIOR DECORATION OF AN OFFICE ROOM
The interior decoration work of an office is taken up. The activities involved, along with the time taken by each activity is given below:
Activity Duration
(in week)
Other activities to be completed before this activity can be taken up.
False roofing 2 –––––
Making Furniture 1 –––––
Fixing Furniture 1 False roofing, Partition systems. Fixing Venetian Blinds 1 Painting of Doors and Windows.
Fixing Air-Conditioner 1 ––––– Painting Walls 1 False roofing.
Partition Systems 2 False roofing, Laying the carpet.
Laying of the carpet 1 False roofing, Painting of Doors and Windows, Painting of walls.
Painting of Doors and
Windows 1 False roofing.
We will now represent the above data pictorially making sure each activity will start only after other" prerequisite" activities are completed.
No. Weeks
Activity Name 1 2 3 4 5 6 7
1 False roofing 1 1
2 Making Furniture 1
3 Fixing Furniture 5
4 Fixing Venetian Blends 3 5 Fixing Air-Conditioner 1
6 Painting Walls 2
7 Partition System 4 4
8 Laying Carpet 3
9 Painting of Door and
Windows 2
As can be seen from the chart the entire work can be completed by the 7th week. In this chart we could also have shown in another column, the "prerequisite" activities to be completed for any activity to be taken up.
From the chart, we can also easily take up rescheduling of activities depending on the "slack" available. For example, the activity "making
furniture" can be taken up in the second week without delaying the project. These types of decisions may be important form the point of view of resources and manpower availability.
T I P
Always look at the options. If they are sufficiently widely spaced, you can save precious time.
8. Combination of 2 or more charts
Other forms of representation of data include cases/caselets as well as combination of two or more of above forms of data-representation.
EXAMPLE:
BAR CHART AND PIE CHART:
The chart given below gives export figures for various years from 1986 to 1991 while the Pie chart gives us share of different geographical zones in the world for the year 1990.
EXPORT OF LEATHER GOODS in (Rs. Crores)
EXPORT IN 1990 Proportion of Zones Middle East 22% W.Europe 33% Africa 18% Far East 15% U.S. 12%
1. What is the percentage increase in exports of Leather goods from 1986 to 1989?
(1) 50% (2) 150% (3) 250% (4) 300%
Sol. Percentage increase in exports of Leather goods from 1986 to 1989 = 100 200 200 500 × − = 150% Answer: (2)
2. What is the total value of the Leather goods exported from India to Africa in 1990?
(1) 18 crores (2) 72 crores (3) 90 crores (4) 180 crores Sol. Total value of the Leather goods exported from India to Africa in 1990 = 18% of 400 = 72 crores.
Answer: (2) 0 100 200 300 400 500 600 1986 1987 1988 1989 1990 1991
T I P
Always set an order of questions – sets that you have to attempt first, second, third etc.
3. By what percentage, the exports from India to W.Europe is more than that to Middle East in 1990?
(1) 11% (2) 25% (3) 50% (4) 100%
Sol. Value of exports from India to W.Europe = 33% Value of exports from India to Middle East = 22% Required percentage = 100 22 22 33 × − = 50% Answer: (3) EXAMPLE:
Chart 1 shows the distribution of twelve million tones of crude oil transported through different modes over a specific period of time. Chart 2 shows the distribution of the cost of transporting this crude oil. The total cost was Rs. 30 million.
Chart 1: Volume Transported Chart 2: Cost of Transportation
1. The cost in rupees per tonne of oil moved by rail and road happens to be roughly
(1) 3 (2) 1.5 (3) 4.5 (4) 8
2. From the charts given, it appears that the cheapest mode of transport is
(1) Road (2) Rail (3) Pipeline (4) Ship
3. If the costs per tonne of transport by ship, air and road are represented by P, Q and R respectively, which of the following is true?
(1) R > Q > P (2) P > R > Q (3) P > Q > R (4) Q > P > R ANSWERS 1. Answer: (2) 2. Answer: (1) 3. Answer: (3) Pipeline 49% Rail 9% Road 22% Airfreight 11% Ship 9% Pipeline 65% Rail 12% Road 6% Airfreight 7% Ship 10%
