# 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.

**1. Answer : 2**

**Explanation : **Productivity = Total production/area of Agr.land

Productivity of UP = [ (35000+30000+25000) / (30/100) ] = 300000

Productivity of MP = [ (30000+37500+27500) / (25/100) ] = 380000

Productivity of Bihar = [ (22500+27500+25000) / (20/100) ] = 375000

Productivity of Odisha = [ (22500+15000+10000) / (5/100) ]= 950000

Productivity of Haryana = [ (30000+25000+35000) / (8/100) ] = 1125000

Productivity of Punjab = [ (40000+35000+30000) / (12/100) ] = 875000

The Productivity of Haryana is the maximum.

**2. Answer : 5**

**Explanation : **Production of Punjab is maximum = 105000 tonnes

**3. Answer : 3**

**Explanation : **Production of Wheat in Punjab = 40000 tones

Production of Maize in Odisha = 10000 tones

Required % = (40000 - 10000)/100 = 300%

**4. Answer : 4**

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

**5. Answer : 1**

**Explanation : **Income of MP from export of 40% of Rice at the rate of Rs.30 per kg = Rs.39crore

Income of UP from export of 30% of Rice at the rate of Rs.32 per kg = Rs.28.8 crore

Required ratio = 39 : 28.8 = 390 : 288 = 65 : 48

**1. ANSWER : C**

** 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.

**2. ANSWER : D**

**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.

**3. ANSWER : D**

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

For Section A = $\frac{14+6+64}{28+14+6+64}\times 100=\frac{84}{112}\times 100$% = 75%

For Section B =$\frac{12+17+55}{23+12+17+55}\times 100$ % = 78.5%

For Section C = $\frac{8+9+46}{17+8+9+46}\times 100$%= 78.75%

For Section D = $\frac{13+15+76}{27+13+15+76}\times 100$%= 79.39%

Clearly ,the pass percentage is maximum for Section D

**4. ANSWER : A**

**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

Success rate in annual exams in Section A= $\frac{14+64}{112}\times 100$% = 69.64%

Similarly, success rate in annual exams in:

Section B = $\frac{12+55}{107}\times 100$% = 62.62%

Section C = $\frac{8+46}{80}\times 100$% = 67.5%

Section D = $\frac{89}{131}\times 100$% = 67.94%

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

**5. ANSWER : D**

**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

Failure rate in half-yearly exams in Section A %= 37.5 %

Similarly, failure rate in half-yearly exams in:

Section B = 32.71%

Section C = 31.25%

Section D = 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.

**1. ANSWER : C**

**Explanation - **Required percentage=$\frac{9\%of5700}{8\%of8550}\times 100$ % = $\frac{9\times 5700}{8\times 8550}\times 100$ % = 75%

**2. ANSWER : B**

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

P = $\frac{18\%of5700}{22\%of8550}\times 100$ % = 54.55%

Q = $\frac{17\%of5700}{15\%of8550}\times 100$ % = 75.56%

R = $\frac{13\%of5700}{10\%of8550}\times 100$ % = 86.67%

S =$\frac{16\%of5700}{17\%of8550}\times 100$ % = 62.75%

T = $\frac{9\%of5700}{8\%of8550}\times 100$% = 75%

V= $\frac{15\%of5700}{12\%of8550}\times 100$% = 83.33%

X= $\frac{12\%of5700}{16\%of8550}\times 100$% = 50%

Highest of these is 86.67% corresponding to institute R.

**3. ANSWER : C**

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

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

= (1938 - 1539)

= 399

**4. ANSWER : B**

**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 = $\frac{30\%of5700}{25\%of8550}\times 100$ % = 80%

**5. ANSWER : C**

**Explanation - **Required ratio = $\frac{18\%of5700}{22\%of8550}$ = 6/11

**1. ANSWER : D**

**Explanation - **

Average amount of interest paid by the company during the given period

=$Rs.\frac{23.4+32.5+41.6+36.4+49.4}{5}lakhs=Rs.\frac{183.3}{5}lakhs$

= Rs. 36.66 lakhs** **

**2. ANSWER : C**

**Explanation - **

Required Percentage = $\frac{3.00+2.52+3.84+3.68+3.96}{288+342+324+336+420}\times 100$ %

= $\frac{17}{1710}\times 100$% $\approx $1%

**3. ANSWER: C**

**Explanation - **** **

Required Percentage = $\frac{288+98+3.00+23.4+83}{420+142+3.96+49.4+98}\times 100$

= $\frac{495.4}{713.36}\times 100$ $\approx $ 69.45%

**4. ANSWER: A**

**Explanation - **Total expenditure of the Company during 2000 = (324 + 101 + 3.84 + 41.6 + 74) lakhs = 544.44 lakhs

**5. ANSWER : B**

**Explanation - **

Required ratio = $\frac{83+108+74+88+98}{98+112+101+133+142}$

= $\frac{451}{586}$

= ** $\frac{1}{1.3}=\frac{10}{13}$ **

**1. ANSWER : D**

**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 1999 = Rs. *x* crores.

Then, the exports in the year 1998 = Rs. (500 - *x*) crores.

$\frac{250}{500-\mathrm{x}}=1.25=\mathrm{x}=300\mathrm{crores}$ [ Using ratio for 1998 ]

Thus, the exports in the year 1999 = Rs. 300 crores.

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

Then, Imports in the year 1999 = $\frac{\mathrm{y}}{300}=1.4=\mathrm{y}=420$= Rs. 420 crores.

**2. ANSWER : C**

**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.

**3. ANSWER : D**

**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.

Hence, the data is inadequate to answer this question.

**4. ANSWER : B**

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

Let the exports in 1996 = Rs. *x* crores.

Then , $\frac{272}{\mathrm{x}}=0.85=\mathrm{x}=320$

Exports in 1996 = Rs. 320 crores.

**5. ANSWER : D**

**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.

**1. ANSWERS : D**

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

**2. ANSWERS : C**

**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.

**3. ANSWERS : A**

**Explanation- **Average number of vehicles manufactured by Company X

= $\frac{1}{6}\times \left(119000+99000+141000+78000+120000+159000\right)$

= 119333.

**4. ANSWERS : D**

**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.

**5. ANSWERS : B**

**Explanation - **Required percentage= $\frac{128000}{78000}\times 100$ % = 164%

**1. ANSWER : D**

**Explanation - **Required difference

= (272 + 240 + 236 + 256 + 288) - (200 + 224 + 248 + 272 + 260)

= 88.

**2. ANSWER : B**

**Explanation - **Total number of Peons working in the Company in 1999

= (820 + 184 + 152 + 196 + 224) - (96 + 88 + 80 + 120)

= 1192.

**3. ANSWER : A**

**Explanation -** Number of Managers working in the Company:

In 1995 = 760.

In 2000 = (760 + 280 + 179 + 148 + 160 + 193) - (120 + 92 + 88 + 72 + 96)= 1252.

Therefore, Percentage increase in the number of Managers

=$\frac{\left(1252-760\right)}{760}\times 100$ %= 64.74%

Number of Technicians working in the Company:

In 1995 = 1200.

In 2000 = (1200 + 272 + 240 + 236 + 256 + 288) - (120 + 128 + 96 + 100 +112) = 1936.

Therefore, Percentage increase in the number of Technicians

= $\frac{\left(1936-1200\right)}{1200}\times 100$ % = 61.33%

Number of Operators working in the Company:

In 1995 = 880.

In 2000 = (880 + 256 + 240 + 208 + 192 + 248) - (104 + 120 + 100 + 112 + 144) = 1444.

Therefore, Percentage increase in the number of Operators

=$\frac{\left(1444-880\right)}{880}\times 100$ % = 64.09%

Number of Accountants working in the Company:

In 1995 = 1160.

In 2000 = (1160 + 200 + 224 + 248 + 272 + 260) - (100 + 104 + 96 + 88 + 92) = 1884.

Therefore, Percentage increase in the number of Accountants

=$\frac{\left(1884-1160\right)}{1160}\times 100$ % = 62.14%

Number of Peons working in the Company:

In 1995 = 820.

In 2000 = (820 + 184 + 152 + 196 + 224 + 200) - (96 + 88 + 80 + 120 + 104) = 1288.

Therefore, Percentage increase in the number of Peons

=$\frac{\left(1288-820\right)}{820}\times 100$ % = 57.07%

Clearly, the percentage increase is maximum in case of Managers.

**4. ANSWER : B**

**Explanation - ** Total number of employees of various categories working in the Company in 1997 are:

Managers = (760 + 280 + 179) - (120 + 92) = 1007.

Technicians = (1200 + 272 + 240) - (120 + 128) = 1464.

Operators = (880 + 256 + 240) - (104 + 120) = 1152.

Accountants = (1160 + 200 + 224) - (100 + 104) = 1380.

Peons = (820 + 184 + 152) - (96 + 88) = 972.

Therefore, Pooled average of all the five categories of employees working in the Company in 1997 = 1/5 x (1007 + 1464 + 1152 + 1380 + 972)

= 1/5 x (5975)

= 1195.

**5. ANSWER : D**

**Explanation - ** Total number of Operators who left the Company during 1995 - 2000

= (104 + 120 + 100 + 112 + 144)

= 580.

Total number of Operators who joined the Company during 1995 - 2000

= (880 + 256 + 240 + 208 + 192 + 248

= 2024.

Therefore, Required Percentage

= (580/2024)x100% = 28.66% ~= 29%.

**1. ANSWER : 2**

**Explanation - **Compare the respective pass percentage for three years : 2000, 2001 and 2002

= (140 x 100)/170 < (150 x 100)/180 and (150 x 100)/180 > (160 x 100)/200

= 82.35% < 83.33% and 83.33% > 80%

**2. ANSWER : 3**

**Explanation - **Total percentage of students who succeeded in at least one of three exams in 2000 = (140 x 100)/170= 82.35 %

**3. ANSWER : 2**

**Explanation - **Total percentage increase in the number of students in 2002 over 2000 is = (30 x 100)/170 =17.64 %

**4. ANSWER : 4**

**Explanation - ** Total percentage of students who cleared CAT in 2000 =(20 x 100)/170 = 11.76 %