# Quantitative Aptitude - Data Interpretation Questions

## What is Quantitative Aptitude- Data Interpretation?

Data Interpretation is the process of analysing data, inspecting the elements in data and Interpreting to extract maximum information from the given set of data or information. The data is given in the form of charts, tables and graphs. Data Interpretation has no particular syllabus, this section tests one's ability in analysing data, decision making capability and speed. Data Interpretation looks simple and easy but the calculations are time consuming. For solving the data interpretation problems efficiently one should analyse the given data and focus on aspects of the data that are necessary to answer the questions, before attending data interpretation section one should be very comfortable with numbers, calculations, percentages, fractions, averages and ratios to increase the calculation speed.

We come across Data Interpretation Questions in many competitive exams and Entrance Tests like Bank Exams (SBI PO), MBA entrance exams (CAT, MAT), HPAS, APPSC group1, HR executives, UPSC CPF (AC),IBPS ,UP Police constable exams, TNPSC VAO, WBSC, PPSC, HAL Results, NDA, Lokhsabha secretariat, Rajyasabha secretariat exams and more

Thorough practice of different papers on data interpretation allows you to solve different kinds of data interpretation and can help improve your logic in solving problems.

We have a large database of questions on Quantitative Aptitude (Data Interpretation) for you to practice and score high.

• #### Table Charts

Q:

Study the given bar graph and pie chart to answer the following questions.

The bar graph shows the production (in thousand tonnes) of Wheat, Rice and Maize in different states.

The pie-chart shows the percentage of agricultural land in the given six states.

Productivity = $\inline \dpi{100} \frac{Total\: Production}{Area\: of\: agricultural\: land}$

1.The productivity of which state is the maximum ?

1. Bihar       2. Haryana     3. Punjab     4. UP       5. MP

2.The production of which state is the maximum ?

1. Bihar       2. MP         3. Haryana       4. UP       5. Punjab

3.The prodution of wheat in punjab is what per cent more than the production of Maize in odisha ?

1. 350%       2. 250%       3. 300%       4. 200%       5. 400%

4.What is the ratio of the production of Rice in Bihar to the production of Wheat in Haryana ?

1. 2:3           2. 3:2           3. 2:1           4. 1:1             5. 1:2

5.If MP exports 40% of Rice at the rate of Rs.30 per kg and UP exports 30% of Rice at the rate of Rs.32 per kg, then what is the ratio of the income from the exports ?

1. 65:48       2. 31:42       3. 43:54       4. 57:62       5. 1:2

Explanation : Productivity = $\inline \dpi{100} \fn_cm \frac{Toatal\: Production}{Area\: of \: agricultural\: land}$

Productivity of UP = $\inline \dpi{100} \fn_cm \frac{(35+30+25)\times 1000}{2lakh\times \frac{30}{100}}=\frac{90000}{60000}=1.5\: tonnes\: per\: sq.km$

Productivity of MP = $\inline \dpi{100} \fn_cm \frac{(30+32.5+27.5)\times 1000}{2lakh\times \frac{25}{100}}=\frac{90000}{50000}=1.8\: tonnes\: per\: sq.km$

Productivity of Bihar = $\inline \dpi{100} \fn_cm \frac{(22.5+25.5+27.5)\times 1000}{2lakh\times \frac{20}{100}}=\frac{75000}{40000}=1.875\: tonnes\: per\: sq.km$

Productivity of Odisha = $\inline \dpi{100} \fn_cm \frac{(22.5+15+10)\times 1000}{2lakh\times \frac{5}{100}}=\frac{47500}{10000}=4.75\: tonnes\: per\: sq.km$

Productivity of Haryana = $\inline \dpi{100} \fn_cm \frac{(25+35+30)\times 1000}{2lakh\times \frac{8}{100}}=\frac{90000}{16000}=5.625\: tonnes\: per\: sq.km$

Productivity of Punjab = $\inline \dpi{100} \fn_cm \frac{(40+30+35)\times 1000}{2lakh\times \frac{12}{100}}=\frac{105000}{24000}=4.375\: tonnes\: per\: sq.km$

$\inline \dpi{100} \fn_cm \therefore$ Productivity of Haryana is the maximum.

Explanation : Production of Punjab is maximum = 105000 tonnes

Explanation : Production of Wheat in Punjab = 40000 tones

Production of Maize in Odisha = 10000 tones

$\inline \dpi{100} \fn_cm \therefore$ Required % = $\inline \dpi{100} \fn_cm \frac{40000-10000}{10000}\times 100 = 300$%

Explanation : The ratio of prodution of Rice in Bihar to the production of Wheat in Haryana = 25000 tonnes : 25000 tonnes = 1 : 1

Explanation : Income of MP from export of 40% of Rice at the rate of Rs.30 per kg = $\inline \dpi{100} \fn_cm 32500\times \frac{40}{100}\times 1000\times 30$ = Rs.39 crore

Income of UP from export of 30% of Rice at the rate of Rs.32 per kg = $\inline \dpi{100} \fn_cm 30000\times \frac{30}{100}\times 1000\times 32$ = Rs.28.8 crore

$\inline \dpi{100} \fn_cm \therefore$ Required ratio = 39 : 28.8 = 390 : 288 = 65 : 48

4835
Q:

A school has four sections A, B, C, D of Class IX students.

1. If the number of students passing an examination be considered a criteria for comparision of difficulty level of two examinations, which of the following statements is true in this context?

A. Half yearly examinations were more difficult.
B. Annual examinations were more difficult.
C. Both the examinations had almost the same difficulty level.
D. The two examinations cannot be compared for difficulty level.

2. How many students are there in Class IX in the school?

A. 336                      B.189                       C. 335                         D. 430

3. Which section has the maximum pass percentage in at least one of the two examinations?

A. A Section            B. B Section            C. C Section               D. D Section

4. Which section has the maximum success rate in annual examination?

A. A Section             B. B Section            C. C Section               D. D Section

5. Which section has the minimum failure rate in half yearly examination?

A. A section             B. B section             C. C section               D. D section

Half yearly examinations were more difficult.
If the number of students passing an examination be considered a criteria for comparision of difficulty level of two examinations, which of the following statements is true in this context? - See more at: http://www.theonlinetestcentre.com/table-charts7.html#sthash.slrcbjro.dpuf

Explanation -

Number of students who passed half-yearly exams in the school

= (Number of students passed in half-yearly but failed in annual exams) + (Number of students passed in both exams)

= (6 + 17 + 9 + 15) + (64 + 55 + 46 + 76)

= 288.

Also, Number of students who passed annual exams in the school

= (Number of students failed in half-yearly but passed in annual exams) + (Number of students passed in both exams)

= (14 + 12 + 8 + 13) + (64 + 55 + 46 + 76)

= 288.

Since, the number of students passed in half-yearly = the number of students passed in annual exams. Therefore, it can be inferred that both the examinations had almost the same difficulty level.

Thus Statements (a), (b) and (d) are false and Statement (c) is true.

Explanation -

Since the classification of the students on the basis of their results and sections form independent groups, so the total number of students in the class:

= (28 + 23 + 17 + 27 + 14 + 12 + 8 + 13 + 6 + 17 + 9 + 15 + 64 + 55 + 46 + 76)

= 430.

Explanation-

Pass percentages in at least one of the two examinations for different sections are:

For Section A = $\inline&space;\left&space;[&space;\frac{(14+6+64)}{(28+14+6+64)}\times&space;100&space;\right&space;]$% = $\inline&space;\left&space;[&space;\frac{84}{112}&space;\times&space;100\right&space;]$%= 75%

For Section B = $\inline&space;\left&space;[&space;\frac{(12+17+5)}{(23+12+17+55)}\times&space;100&space;\right&space;]$% = $\inline&space;\left&space;[&space;\frac{84}{107}&space;\times&space;100\right&space;]$%= 78.5%

For Section C = $\inline&space;\left&space;[&space;\frac{(8+9+46)}{(17+8+9+46)}\times&space;100&space;\right&space;]$% = $\inline&space;\left&space;[&space;\frac{63}{80}&space;\times&space;100\right&space;]$%= 78.75%

For Section D =$\inline&space;\left&space;[&space;\frac{(13+15+76)}{(27+13+15+76)}&space;\times&space;100\right&space;]$% = $\inline&space;\left&space;[&space;\frac{104}{131}\times&space;100&space;\right&space;]$%= 79.39%

Clearly ,the pass percentage is maximum for Section D

Explanation -

Total number of students passed in annual exams in a section

= [ (No. of students failed in half-yearly but passed in annual exams) + (No. of students passed in both exams) ] in that section

$\inline&space;\therefore$ Success rate in annual exams in Section A

$\inline&space;\left&space;[&space;\frac{No\:&space;of\:&space;students\:&space;of\:&space;sectionA\:&space;passed\:&space;in\:&space;annual\:&space;exams}{Total\:&space;number\:&space;of\:&space;students\:&space;in\:&space;sectionA}\times&space;100&space;\right&space;]$ %

$\inline&space;\left&space;[&space;\frac{(14+64)}{(28+14+6+64)}\times&space;100&space;\right&space;]$ %

$\inline&space;\left&space;[&space;\frac{78}{112}\times&space;100&space;\right&space;]$ %

= 69.64%

Similarly, success rate in annual exams in:

Section B = $\inline&space;\left&space;[&space;\frac{(12+55)}{(23+12+17+55)}\times&space;100&space;\right&space;]$ %= $\inline&space;\left&space;[&space;\frac{67}{107}&space;\times&space;100\right&space;]$ % =  62.62%

Section C = $\inline&space;\left&space;[&space;\frac{(84+46)}{(17+8+9+46)}\times&space;100&space;\right&space;]$ % =  $\inline&space;\left&space;[&space;\frac{54}{80}\times&space;100\right&space;]$ % = 67.5%

Section D = $\inline&space;\left&space;[&space;\frac{(13+76)}{(27+13+15+76)}&space;\times&space;100\right&space;]$ % = $\inline&space;\left&space;[&space;\frac{89}{131}\times&space;100&space;\right&space;]$ % = 67.94%

Clearly, the success rate in annual examination is maximum for Section A.

Explanation -

Total number of failures in half-yearly exams in a section

= [ (Number of students failed in both exams) + (Number of students failed in half-yearly but passed in Annual exams) ] in that section

$\therefore$ Failure rate in half-yearly exams in Section A

$\inline&space;\left&space;[&space;\frac{Number\:&space;of\:&space;students\:&space;of\:&space;sectionA\:&space;failed\:&space;in\:&space;half\:&space;yearly}{Total\:&space;number\:&space;of\:&space;students\:&space;in\:&space;sectionA}&space;\times&space;100\right&space;]$%

= $\inline&space;\left&space;[&space;\frac{(28+14)}{(28+14+6+64)}\times&space;100&space;\right&space;]$%

$\inline&space;\left&space;[&space;\frac{42}{112}\times&space;100&space;\right&space;]$ %

= 37.5 %

Similarly, failure rate in half-yearly exams in:

Section B = $\inline&space;\left&space;[&space;\frac{(23+12)}{(23+12+17+55)}\times&space;100&space;\right&space;]$% = 32.71%

Section C = $\inline&space;\left&space;[&space;\frac{(17+8)}{(17+8+9+46)}\times&space;100&space;\right&space;]$% = 31.25%

Section D = $\inline&space;\left&space;[&space;\frac{(27+13)}{(27+13+15+76)}\times&space;100&space;\right&space;]$% = 30.53%

Clearly, the failure rate is minimum for Section D.

= (Number of students passed in half-yearly but failed in annual exams) + (Number of students passed in both exams)

= (6 + 17 + 9 + 15) + (64 + 55 + 46 + 76)

= 288.

Also, Number of students who passed annual exams in the school

= (Number of students failed in half-yearly but passed in annual exams) + (Number of students passed in both exams)

= (14 + 12 + 8 + 13) + (64 + 55 + 46 + 76)

= 288.

Since, the number of students passed in half-yearly = the number of students passed in annual exams. Therefore, it can be inferred that both the examinations had almost the same difficulty level.

Thus Statements (a), (b) and (d) are false and Statement (c) is true.

2756
Q:

The following line graph gives the ratio of the amounts of imports by a company to the amount of exports from that company over the period from 1995 to 2001.

Ratio of Value of Imports to Exports by a Company Over the Years.

1. If the imports in 1998 was Rs. 250 crores and the total exports in the years 1998 and 1999 together was Rs. 500 crores, then the imports in 1999 was ?

A. Rs. 250 cr         B. Rs. 300 cr         C. Rs. 357 cr         D. Rs. 420 cr

2. The imports were minimum proportionate to the exports of the company in the year ?

A. 1995                 B. 1996                 C. 1997                 D. 2000

3. What was the percentage increase in imports from 1997 to 1998 ?

A. 72                     B. 56                     C. 28                     D. Data Inadequate

4. If the imports of the company in 1996 was Rs. 272 crores, the exports from the company in 1996 was ?

A. Rs. 370 cr         B. Rs. 320 cr         C. Rs. 280 cr         D. Rs. 275 cr

5. In how many of the given years were the exports more than the imports ?

A. 1                       B. 2                       C. 3                       D. 4

Explanation - The ratio of imports to exports for the years 1998 and 1999 are 1.25 and 1.40 respectively.

Let the exports in the year 1998 = Rs. x crores.

Then, the exports in the year 1999 = Rs. (500 - x) crores.

$\inline&space;\fn_cm&space;\therefore&space;1.25=\frac{250}{x}\Rightarrow&space;x=\frac{250}{1.25}=200$        [ Using ratio for 1998 ]

Thus, the exports in the year 1999 = Rs. (500 - 200) crores = Rs. 300 crores.

Let the imports in the year 1999 = Rs. y crores.

Then, $\inline&space;\fn_cm&space;1.40=\frac{y}{300}\Rightarrow&space;y=(300\times&space;1.40)=420$

Imports in the year 1999 = Rs. 420 crores.

Explanation - The imports are minimum proportionate to the exports implies that the ratio of the value of imports to exports has the minimum value.

Now, this ratio has a minimum value 0.35 in 1997, i.e., the imports are minimum proportionate to the exports in 1997.

Explanation - The graph gives only the ratio of imports to exports for different years. To find the percentage increase in imports from 1997 to 1998, we require more details such as the value of imports or exports during these years.

Explanation -  Ratio of imports to exports in the year 1996 = 0.85.

Let the exports in 1996 = Rs. x crores.

Then , $\inline&space;\fn_cm&space;\frac{272}{x}=0.85\Rightarrow&space;x=\frac{272}{0.85}=320$

Exports in 1996 = Rs. 320 crores.

Explanation - The exports are more than the imports imply that the ratio of value of imports to exports is less than 1.Now, this ratio is less than 1 in years 1995, 1996, 1997 and 2000.

Thus, there are four such years.

1971
Q:

Study the following graph carefully and answer the questions given below:

Distribution of candidates who were enrolled for MBA entrance exam and the candidates (out of those enrolled) who passed the exam in different institutes:

1. What percentage of candidates passed the Exam from institute T out of the total number of candidates enrolled from the same institute?

A. 50%             B. 62.5%           C. 75%             D. 80%

2. Which institute has the highest percentage of candidates passed to the candidates enrolled?

A. Q                  B. R                  C. V                  D. T

3. The number of candidates passed from institutes S and P together exceeds the number of candidates enrolled from institutes T and R together by:

A. 228              B. 279              C. 399              D. 407

4. What is the percentage of candidates passed to the candidates enrolled for institutes Q and R together?

A. 68%             B. 80%             C. 74%             D. 65%

5. What is the ratio of candidates passed to the candidates enrolled from institute P?

A. 9:11             B. 14:17           C. 6:11             D. 9:17

Explanation - Required percentage=$\inline \fn_cm \frac{9\; percent\; of\; 5700}{8\; percent\; of\; 8550} \times 100$  % =  $\inline \fn_cm \frac{9\; \times 5700}{8\;\times 8550} \times 100$% = 75%

Explanation - The percentage of candidates passed to candidates enrolled can be determined for each institute as under:

P =  $\inline \fn_cm \frac{18\; percent \; of\; 5700}{22\;percent\; of\; 8550} \times 100$%  = 54.55%

Q = $\inline \fn_cm \frac{17\; percent \; of\; 5700}{15\;percent\; of\; 8550} \times 100$% = 75.56%

R = $\inline \fn_cm \frac{13\; percent \; of\; 5700}{10\;percent\; of\; 8550} \times 100$% = 86.67%

S = $\inline \fn_cm \frac{16\; percent \; of\; 5700}{17\;percent\; of\; 8550} \times 100$% = 62.75%

T = $\inline \fn_cm \frac{9\; percent \; of\; 5700}{8\;percent\; of\; 8550} \times 100$% = 75%

V=$\inline \fn_cm \frac{15\; percent \; of\; 5700}{12\;percent\; of\; 8550} \times 100$  % = 83.33%

X= $\inline \fn_cm \frac{12\; percent \; of\; 5700}{16\;percent\; of\; 8550} \times 100$% = 50%

Highest of these is 86.67% corresponding to institute R.

Explanation -

Required difference = [(16% + 18%) of 5700] - [(8% + 10%) of 8550]

= [(34% of 5700) - (18% of 8550)]

= (1938 - 1539)

= 399

Explanation

Candidates passed from institutes Q and R together    =  [(13% + 17%) of 5700] = 30% of 57000.

Candidates enrolled from institutes Q and R together   = [(15% + 10%) of 8550] = 25% of 8550.

Required Percentage = $\inline \fn_cm \frac{30\; percent \; of\; 5700}{25\;percent\; of\; 8550} \times 100$% = 80%

Explanation -  Required ratio =  $\inline \fn_cm \frac{18\; percent \; of\; 5700}{22\;percent\; of\; 8550}$ = $\inline \fn_cm \frac{6}{11}$

1906
Q:

Study the following line graph and answer the questions based on it.

Number of Vehicles Manufactured by Two companies ove the Years (Number in Thousands)

1. What is the difference between the number of vehicles manufactured by Company Y in 2000 and 2001 ?

A. 50000                   B. 42000                   C. 33000                   D. 21000

2. What is the difference between the total productions of the two Companies in the given years ?

A. 19000                   B. 22000                   C. 26000                   D. 28000

3. What is the average numbers of vehicles manufactured by Company X over the given period ? (rounded off to nearest integer)

A. 119333                 B. 113666                 C. 112778                 D. 111223

4. In which of the following years, the difference between the productions of Companies X and Y was the maximum among the given years ?

A. 1997                     B. 1998                     C. 1999                     D. 2000

5. The production of Company Y in 2000 was approximately what percent of the production of Company X in the same year ?

A. 173                       B. 164                       C. 132                       D. 97

Explanation-   Required difference = (128000 - 107000) = 21000.

Explanation-  From the line-graph it is clear that the productions of Company X in the   years 1997, 1998, 1999, 2000, 2001 and 2002 are 119000, 99000, 141000, 78000, 120000 and 159000 and those of Company Y are 139000, 120000,100000, 128000, 107000 and 148000 respectively.

Total production of Company X from 1997 to 2002

= 119000 + 99000 + 141000 + 78000 + 120000 + 159000

= 716000.

and total production of Company Y from 1997 to 2002

= 139000 + 120000 + 100000 + 128000 + 107000 + 148000

= 742000.

Difference = (742000 - 716000) = 26000.

Explanation-  Average number of vehicles manufactured by Company X

$\inline&space;\fn_cm&space;\frac{1}{6}\times&space;(119000&space;+&space;99000&space;+&space;141000&space;+&space;78000&space;+&space;120000&space;+&space;159000)$

= 119333.

Explanation-   The difference between the productions of Companies X and Y in various years are:

For 1997 (139000 - 119000)  = 20000.

For 1998 (120000 - 99000)  = 21000.

For 1999 (141000 - 100000)  = 41000.

For 2000 (128000 - 78000) = 50000.

For 2001 (120000 - 107000) = 13000.

For 2002 (159000 - 148000) = 11000.

Clearly, maximum difference was in 2000.

Explanation  - Required percentage= $\inline&space;\fn_cm&space;{&space;\left&space;(&space;\frac{128000}{78000}\times&space;100&space;\right&space;)}$% = 164%