To solve data interpretation questions, we have to understand the common question types given in standard tests. We have selected 2 sets which almost cover all important models. Please study them carefully.

1. What is the total marks obtained by B in all subjects ?

Solution: As the maximum marks of all subjects are not equal to 100, we have to calculate separately for each subject.

as A% of B = B % of A, instead of calculating 69% (150) + 72 % (75) + 71% (200) + 78% ( 100) + 69% (50) + 66% (75) we calculate 150% (69) + 75% (72) + 200% (71) + 78 + 50% (69) + 75% (66)

= 69 + 69/2 + 3/4 ( 72) + 2 x 71 + 78 + 69/2 + 3/4(66)

= 138 + 54 + 142 + 78 + 49.5 = 461.5

So Option D is correct.

2. What is the average marks obtained by 6 students in Chemistry out of 75 marks ?

Solution: First we calculate average percentage = $\displaystyle\frac{{63 + 72 + 78 + 81 + 69 + 57}}{6} = \displaystyle\frac{{420}}{6} = 70$

Now 70% (75) = 52.5

3. What is the difference in the total marks obtained by C in Physics and Chemistry and that obtained by E in the same subjects ?

Solution:

Marks obtained by C in Physics and Chemistry = 82% (150) + 78% (75)

Marks obtained by E in Physics and Chemistry = 58% (150) + 69% ( 75)

Difference = 24% (150) + 9% (75) = 24 + 12 + 7.5 - 0.75 = 42.75

4. What is the percent marks obtained by A in both Maths and English ? Find up to two decimal places,.

Solution: We have to calculate the marks obtained by A in maths and English then we calculate the percentage.

His marks = 89% (200) + 72% (75) = 2 x 89 + 3/4(75) = 178 + 54 = 232

Percentage = $\displaystyle\frac{{232}}{{275}} \times 100$

Now we make 275 as 300. Now $\displaystyle\frac{{232}}{{300}} \times 100 = 77.33$

But this answer is less than actual answer as we increased the denominator. We added 25 to denominator, which is 25 / 275 = 1/11. So calculate 1/11th part of the answer and add to it. 77.33 x 1/11 = 7.03

Adding 7.03 to 77.33 we get 84.36

5. What is the average marks obtained by 6 students in Geography out of 50 marks ? Find up to two decimal places .

First we calculate average of percentages. 64+69+75+58+66+71 / 6 = 403 / 6 = 67.16

Now 67.16% (50) = 50% (67.16) = 67.16/2 = 33.58

1. The agro product which witnessed the highest growth rate in production from 1997 to 2000 is ____.

Solution: Growth rate = $\displaystyle\frac{\text{Final year value - Initial year value}}{\text{Initial year value}} \times 100$

For Wheat = $\displaystyle\frac{{131 - 100}}{{131}} \times 100 = 31\% $

For Rice = $\displaystyle\frac{{107 - 91}}{{91}} \times 100 = \displaystyle\frac{{16}}{{91}} \times 100$

We make the denominator = 100, so we get 16% as answer. But we know that this answer is less than actual answer as we increased the denominator. We added 9, which is 9/91 = 1/10th part of denominator. So we have to add 1/10th (16) = 1.6 to 16. which gives 17.6.

For sugar cane = $\displaystyle\frac{{(25 - 15)}}{{15}} \times 100 = 66.66\% $

For pulses = $\displaystyle\frac{{(88 - 71)}}{{71}} \times 100 = 23.94\% $

So Option C is correct.

2. Pulses production in 1998 is what percent of the total production of rice in the given 4-year period?

Solution: We have to calculate pulses production if how much percent of total production of Rice.

Total production of Rice for 4 years = 91 + 88 + 97 + 107 = 383

Now required percentage = $\left( {\displaystyle\frac{{75}}{{383}}} \right) \times 100 = 19.58\% $

To simplify, make the denominator 400. Then 75/400 x 100 = 18.75. But we added 17, which is 17/383 = 1/22.5 or 2/45 part. So calculate 2/45 (18.75) = 0.83. Add this to 18.75. So we get 19.58.

So Option D is correct

3. By what percent is the average wheat production more than the average sugar cane production for the given 4-year period?

Solution: Percentage is a ratio. So percentage calculation of total production and average production gives same value. *

Total Wheat production = 476

Total Sugar cane production = 79

To calculate how much percent A is more than B, we have to use $\displaystyle\frac{{(A - B)}}{A} \times 100$

Required Percentage = $\displaystyle\frac{{476 - 79}}{{79}}\,{\kern 1pt} \times \,\,100\,\, = \,\,502$

So Option D is correct.

4. The simple annual growth rate of wheat from 1997-2000 is:

Solution: To calculate simple annual growth, we have to calculate percentage increase for the given years and divide it by (n-1). Here n is number of years.

So Simple annual growth = $\displaystyle\frac{{131 - 100}}{{100}} \times \frac{1}{{(4 - 1)}} = 10.33\% $

So option C is correct.

5. What would be the actual production of wheat in 2001 if the growth in 2001 is the same as the average growth for the period?

Average growth is 10.3. So next year production is expected to increase by 10.3%.

If to increase a number by k%, we have to calculate k% (original number ) and add to the original number. Hence, the actual value of wheat in 2001 is 131 + 10.3 % (131) = 144.5 MT

**Set 1: Study the following table carefully and answer the questions given below it :**
Percent marks obtained by 6 students in different subjects.

1. What is the total marks obtained by B in all subjects ?

a. 425 | b. 542 |

c. 469.5 | d. 461.5 |

as A% of B = B % of A, instead of calculating 69% (150) + 72 % (75) + 71% (200) + 78% ( 100) + 69% (50) + 66% (75) we calculate 150% (69) + 75% (72) + 200% (71) + 78 + 50% (69) + 75% (66)

= 69 + 69/2 + 3/4 ( 72) + 2 x 71 + 78 + 69/2 + 3/4(66)

= 138 + 54 + 142 + 78 + 49.5 = 461.5

So Option D is correct.

2. What is the average marks obtained by 6 students in Chemistry out of 75 marks ?

a. 55.5 | b. 70 |

c. 52.5 | d. 64.5 |

Now 70% (75) = 52.5

3. What is the difference in the total marks obtained by C in Physics and Chemistry and that obtained by E in the same subjects ?

a. 43 | b. 38 |

c. 38.75 | d. 42.75 |

Marks obtained by C in Physics and Chemistry = 82% (150) + 78% (75)

Marks obtained by E in Physics and Chemistry = 58% (150) + 69% ( 75)

Difference = 24% (150) + 9% (75) = 24 + 12 + 7.5 - 0.75 = 42.75

4. What is the percent marks obtained by A in both Maths and English ? Find up to two decimal places,.

a. 83.46 | b. 84.36 |

c. 84.43 | d. 84.92 |

His marks = 89% (200) + 72% (75) = 2 x 89 + 3/4(75) = 178 + 54 = 232

Percentage = $\displaystyle\frac{{232}}{{275}} \times 100$

Now we make 275 as 300. Now $\displaystyle\frac{{232}}{{300}} \times 100 = 77.33$

But this answer is less than actual answer as we increased the denominator. We added 25 to denominator, which is 25 / 275 = 1/11. So calculate 1/11th part of the answer and add to it. 77.33 x 1/11 = 7.03

Adding 7.03 to 77.33 we get 84.36

5. What is the average marks obtained by 6 students in Geography out of 50 marks ? Find up to two decimal places .

a. 33.58 | b. 1.47 |

c. 38.18 | d. 36.26 |

Now 67.16% (50) = 50% (67.16) = 67.16/2 = 33.58

**Set 2: Answer the questions based on the following table.**
The table gives the production of major agricultural products in Million Tonnes (MT).

1. The agro product which witnessed the highest growth rate in production from 1997 to 2000 is ____.

a. Wheat | b. Rice |

c. Sugar Cane | d. Pulses |

For Wheat = $\displaystyle\frac{{131 - 100}}{{131}} \times 100 = 31\% $

For Rice = $\displaystyle\frac{{107 - 91}}{{91}} \times 100 = \displaystyle\frac{{16}}{{91}} \times 100$

We make the denominator = 100, so we get 16% as answer. But we know that this answer is less than actual answer as we increased the denominator. We added 9, which is 9/91 = 1/10th part of denominator. So we have to add 1/10th (16) = 1.6 to 16. which gives 17.6.

For sugar cane = $\displaystyle\frac{{(25 - 15)}}{{15}} \times 100 = 66.66\% $

For pulses = $\displaystyle\frac{{(88 - 71)}}{{71}} \times 100 = 23.94\% $

So Option C is correct.

2. Pulses production in 1998 is what percent of the total production of rice in the given 4-year period?

a. 19.13 | b. 19.58 |

c. 20.38 | d. 19.38 |

Total production of Rice for 4 years = 91 + 88 + 97 + 107 = 383

Now required percentage = $\left( {\displaystyle\frac{{75}}{{383}}} \right) \times 100 = 19.58\% $

To simplify, make the denominator 400. Then 75/400 x 100 = 18.75. But we added 17, which is 17/383 = 1/22.5 or 2/45 part. So calculate 2/45 (18.75) = 0.83. Add this to 18.75. So we get 19.58.

So Option D is correct

3. By what percent is the average wheat production more than the average sugar cane production for the given 4-year period?

a. 535 | b. 529 |

c. 629 | d. 502 |

Total Wheat production = 476

Total Sugar cane production = 79

To calculate how much percent A is more than B, we have to use $\displaystyle\frac{{(A - B)}}{A} \times 100$

Required Percentage = $\displaystyle\frac{{476 - 79}}{{79}}\,{\kern 1pt} \times \,\,100\,\, = \,\,502$

So Option D is correct.

4. The simple annual growth rate of wheat from 1997-2000 is:

a. 8.23% | b. 9.65% |

c. 10.33% | d. 11.33% |

So Simple annual growth = $\displaystyle\frac{{131 - 100}}{{100}} \times \frac{1}{{(4 - 1)}} = 10.33\% $

So option C is correct.

5. What would be the actual production of wheat in 2001 if the growth in 2001 is the same as the average growth for the period?

a. 140 MT | b. 144.5 MT |

c. 141.3 MT | d. 145 MT |

If to increase a number by k%, we have to calculate k% (original number ) and add to the original number. Hence, the actual value of wheat in 2001 is 131 + 10.3 % (131) = 144.5 MT