61941762 Data Interprews```tation

7/27/2019 61941762 Data Interprews```tation

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Data Interpretation (DI)CATsyllabus.com | Data Interpretation

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 000s)

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.


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EXAMPLE:

NUMBER OF BOYS OF STANDARD XI PARTICIPATING IN DIFFERENT GAMES

Class

Games

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

Note:

Every student (boy or girl) of each class participates in a game.

In each class, the number of girls participating in each game is 25% of the number of boys

participating in each game.

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 = 401/4 = 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 = 32 + 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)


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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 =68

= 272. Answer: (1)0.25

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 = 96Boys 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 = 10Number of students in the school = boys + girls = 228 + 57 = 285

Percentage =10

100 = 3.51. Answer: (2)285

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)


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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, 72oRail, 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.


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EXAMPLE:

These questions are based on the diagram given below

EXPENSES OF TCY

[as a percentage of turnover]

X = Salaries + Profit

Faculty 8%

Advertising &

Promotion 31%

X 24%

Material

Preparation 10%

Printing 15%

Administration

& Miscellaneous

12%

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 = 24/1002105 = 48000

Salaries = Rs. 95000

Loss = (9500048000)/ (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 = 0.15106 = 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 = 2.5105

More amount spent on material preparation than faculty = 0.1x

0.08x = 0.022.5105= 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

x = Rs. 5.5 lacs. Answer: (3)


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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 000s)

200

150

100

50

02003 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 :

INDIAS CASHEWNUT EXPORTS

600

500

400

300 330400

500

200

100

0

150

100

150

75

150 160200

1995 1996 1997 1998 1999

Quanity in Lakh Kgs Values in crores


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1. In which year was the value per kg minimum

Sol.

(A) 1995 (B) 1996

Value per kg for the years given in options

1995 1996 1997

(C) 1997

1998

(D) 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

500 150Sol. Percentage increase in export value from1995 to 1999 =

150100 = 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 =75 100

100

= 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 = (330107)/(150105) = Rs. 220.

Value of exports in 1998 = 160105220 = 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

100

5050

60

30

100

5060

125

80

100 10090

0

2003 2004 2005 2006

Maruti Huyndai Others

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


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Example:

CONSUMPTION OF CHOCOBAR ACROSS THE COUNTRY (in 000 bars)

160

140

120

100

80

60

40

20

0

124 118128

92

134126 122

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% (B) 39.1% (C) 25% (D) 32.2%

36Sol. In 1996 the % drop =

128 100 =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

0.3[124 118 128 92 134]No. of cartons of 120 bars that has to be supplied =

120 1000 = 1490Answer: (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 = 122103

Production = 200103

No. of chocobars thrown away = 200(0.39) 0.81000 = 62,400

Answer: (D)


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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.

Hindi (80)

32 10

25

12

English (120)

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 (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) = 10Let x be the no. of students who play all the 3 games

C (20)

x

5

15

15 - x

x

F (25)

x

2

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)

10-x

10

12-x

12x 4

Now, we can find the no. of students who play cricket only, football

only & volleyball only is

V (18)


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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.

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


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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

GMAT

75

(100)

75

x50

y

CAT

25

z0

0 25 50 75 100

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 (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) Cant 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)


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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

3

Total number of students in institute x = 2 75% of T = 2 T3

Number of CAT students in institute x = 50% of 3/2T =4

T

1

Number of GMAT students in institute z = 25% of T =4

T

3 1

Required ratio =4

T :4

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 beforethis 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.


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Fixing Air-Conditioner 1

Painting Walls 1 False roofing.

Partition Systems 2 False roofing, Laying the carpet.

Laying of the carpet 1False roofing, Painting of Doors and

Windows, Painting of walls.

Painting of Doors and

Windows1 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 Name1 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

9Painting of Door and

Windows 2

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

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

TIPAlways look at theoptions. If they aresufficiently widelyspaced, you can saveprecious time.

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.


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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

500

400

300

200

100

0

1986 1987 1988 1989 1990 1991

TIPAlways set an orderof questions

sets

that you have toattempt first, second,third etc.

EXPORT IN 1990

Proportion of Zones

Middle

East

22%

W.Europe

33%

Far East

15%

U.S.

12%

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%

500

200Sol. Percentage increase in exports of Leather goods from 1986 to 1989 =

Answer: (2)

200100 = 150%

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)


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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%

33 22Required percentage = 100 = 50% Answer: (3)

22

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 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)

CATsyllabus.com | Data Interpretation

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