Data Interpretation

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Data Interpretation - DI 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) 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

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.

EXAMPLE: NUMBER OF BOYS OF STANDARD XI PARTICIPATING IN DIFFERENT GAMES Class

XI A

XI B

XI C

XI D

XI E

Total

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

Games

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

¾

Each student (boy or girl) participated in one and only one game.

participating in each 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 or

Boys –> 5;

Girls –> 1

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?

Sol.

(1) Table Tennis & Hockey

(2) Badminton & Table Tennis

(3) Chess & Hockey

(4) Hockey & Foot ball

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

Sol.

(2) 560

(3) 656

(4) 340

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 =

4.

68 = 272. Answer: (1) 0.25

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

Sol.

(2) XIA & XIB

(3) XI A & XI D

(4) None of these

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

Sol.

(2) 3.51

(3) 10.52

(4) 13.5

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 =

10 × 100 = 3.51. 285

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.

Road 20%

Rail 20%

Ship 10%

Pipeline 50%

The above pie chart can also be represented as below

Road, 72o

Rail, 72o

Ship, 36o Pipeline, 180o

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.

3.

The total oil transported, if the oil transported through any particular mode is known.

EXAMPLE: These questions are based on the diagram given below EXPENSES OF TCY [as a percentage of turnover] X = Salaries + Profit

Faculty 8%

X 24%

Advertising & Promotion 31%

Administration & Miscellaneous 12% Material Preparation 10% Printing 15%

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%

Sol.

(2) 19.03%

(3) 47.5%

(4) 26.7%

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

Sol.

(2) Rs. l.51acs

0.24x = 120000 + 0.12x, where x is total turnover Printing charges = 0.15x = 0.15 ×

3.

106

(3) Rs. 1 lac ⇒x=

(4) Rs. 75000

106

= 1.5 lac. Answer: (2)

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?

Sol.

(1) Rs. 2500

(2) Rs. 6000

0.31x – 0.15x = 40000

⇒ x = 2.5 ×

(3) Rs. 7500

(4) Rs. 5000

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

Sol.

(2) Rs. 5 lacs

0.24x = 132000, where x is total turnover ⇒ x = Rs. 5.5 lacs. Answer: (3)

(3) Rs. 5.5 lacs

(4) None of these

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)

200 150 100 50 0 2003

2004 Maruti

2005 Hyundai

2006 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 600 500

500

400

400 330

300 200 100

150 100

150 75

150

160

200

0 1995

1996

1997

Quanity in Lakh Kgs

1998 Values in crores

1999

1.

In which year was the value per kg minimum

Sol.

Value per kg for the years given in options

(A) 1995

(B) 1996

(C) 1997

1995

1996

1997

1998

150/100

150/15

360/150

400/160

(D) 1998

From the above values it is clear that value per kg is minimum for the year 1995. 2.

What was the difference in volume exported in 1997 and 1998? (A) 10000 kg

(B) 1000 kg 105

(C) 100000 kg

(D) 1000000 kg

= 1000000 kg

Answer: (D)

Sol.

Difference = (160 – 150)

3.

What was the approximate percentage increase in export value from 1995 to 1999? (A) 350

Sol.

Answer: (A)

(B) 330

(C) 430

Percentage increase in export value from1995 to 1999 =

(D) 230

500 − 150 × 100 = 230% approx. 150

Answer: (D) 4.

What was the percentage drop in export quantity from 1995 to 1996? (A) 75%

(B) 31/3%

(C) 25%

(D) 0%

75 − 100 = 25% 100

Sol.

Percentage decrease in export quantity from 1995 to 1996 =

5.

If in 1998 cashew nuts were exported at the same rate per kg. as that in 1997what would be the value of

Answer: (C)

exports in 1998 (A) Rs. 400 Crores Sol.

(B) Rs. 352 Crores

(C) Rs. 375 Crores

(D) Rs. 330 Crores

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. 200

175

150

125 100

100 50

100

100

80 60

50

50

90

60

30

0 2003

2004 Maruti

2005 Huyndai

2006

Others

Here also we can deduce all the parameters as we could do in the case of two-variable graph.

Example: CONSUMPTION OF CHOCOBAR ACROSS THE COUNTRY (in ‘000 bars)

160 140 120 100 80 60 40 20 0

124

118

134

128

126

122

92

1993 1994 1995 1996 1997 1998 1999 YEARS

1.

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%

Sol.

In 1996 the % drop =

(B) 39.1%

(C) 25%

(D) 32.2%

36 × 100 = 28.1% 128 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

Sol.

(B) 4998

(C) 4967

(D) 1490

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 =

0.3[124 + 118 + 128 + 92 + 134 ] × 1000 = 1490 120 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

Sol.

(B) 59,536

61% of production in 1999 = 122 ×

(C) 38,000

(D) 62,400

103

⇒ Production = 200 × 103 ∴ No. of chocobars thrown away = 200(0.39) 0.8 × 1000 = 62,400 Answer: (D)

5.

The least percentage decrease recoded was (A) 3.14

Sol.

(B) 3.19

(C) 3.22

(D) 3.17

By observation, least percentage decrease is from 1998 – 99, =

126 − 122 × 100 = 3.17% 126 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. English (120)

Hindi (80) 32

10 25

12

Punjabi (125)

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

Sol.

(B) 7

(C) 2

(D) 3

Let C, F & V denote the sets of no of students who play cricket, football & volleyball respectively.

15

C (20)

F (25)

∴ n(C) = 20, n(F) = 25, n(V) = 18 n(C ∩ F) = 15. n(F ∩ V) = 12, n(C ∩ V) = 10

x – 5 15 - x x – 2

Let ‘x’ be the no. of students who play all the 3 games

x

∴ 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-x

12-x

12

10 x–4 V (18)

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

Sol.

∴ No. of students who play all 3 games = 7.

3.

Who play any two games? (A) 16

Sol.

(B) 18

(C) 2

(D) 3

(C) 7

(D) 14

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

Sol.

(B) 16

(C) 10

(D) 5

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. i.e. no. of students who play all 3 games = 7. Now we can find the others as in the previous solutions.

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. 0 25

(100)

50

75

GMAT 75

x

(100)

y

25 z

0

25

50

CAT

50

75

0

100

GRE

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

Sol.

(2) 1 : 2

(3) 2 : 1

(4) None of these

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

2.

If there are 132 GRE students in institute ‘y’, how many GMAT students are there in the same institute? (1) 132

Sol.

Answer: (3)

(2) 264

(3) 396

(4) Can’t say

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)

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

Sol.

(2) 2 : 1

(3) 1 : 3

(4) 3 : 1

Let the total number of students in institute z be T Total number of students in institute x = 2

Number of CAT students in institute x = 50% of 3/2T = Number of GMAT students in institute z = 25% of T = Required ratio =

3 T 2

75% of T =

3 T 4

1 T 4

3 1 T: T=3:1 4 4

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

Sol.

(2) 1 : 3 : 2

(3) 2 : 3 : 1

(4) 3 : 2 : 1

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

Other activities to be completed before

(in week)

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

Painting of Doors and

False roofing, Painting of Doors and Windows, Painting of walls.

1

Windows

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 1

1

False roofing

1

2

Making Furniture

1

3

Fixing Furniture

4

Fixing Venetian Blends

5

Fixing Air-Conditioner

6

Painting Walls

7

Partition System

8

Laying Carpet

9

Painting Windows

of

3

4

5

6

7

5 3 1 2 4

4

3 Door

and

2

TIP 

As can be seen from the chart the entire work can be completed by the 7th

Always look at the

week. In this chart we could also have shown in another column, the

options. If they are

"prerequisite" activities to be completed for any activity to be taken up.

sufficiently

widely

spaced, you can save

From the chart, we can also easily take up rescheduling of activities

precious time.

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.

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) 600

TIP

500

Always set an order

400

of questions – sets

300 200

that

100

attempt first, second,

0 1986

1987

1988

1989

1990

you

have

third etc.

1991

EXPORT IN 1990 Proportion of Zones Middle East 22%

Far East 15%

U.S. 12%

W.Europe 33%

Africa 18%

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 =

2. Sol.

500 − 200 × 100 = 150% 200

Answer: (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 Total value of the Leather goods exported from India to Africa in 1990 = 18% of 400 = 72 crores. Answer: (2)

to

3. Sol.

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% Value of exports from India to W.Europe = 33% Value of exports from India to Middle East = 22% Required percentage =

33 − 22 × 100 = 50% 22

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.

Road 22%

Airfreight 11% Ship 9%

Rail 12%

Road 6%

Airfreight 7%

Ship 10%

Rail 9% Pipeline 49%

Pipeline 65%

Chart 1: Volume Transported 1.

The cost in rupees per tonne of oil moved by rail and road happens to be roughly (1) 3

2.

3.

Chart 2: Cost of Transportation

(2) 1.5

(3) 4.5

From the charts given, it appears that the cheapest mode of transport is (1) Road (2) Rail (3) Pipeline

(4) 8

(4) Ship

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 Answer: (2) 1. Answer: (1) 2. Answer: (3) 3.