Free Data Handling Subjective Test 02 Practice Test - 7th grade

Question 1

What is the scale of the bar graph shown below? [1 MARK]

SOLUTION

Solution : The scale is read on the vertical axis. We can see that 1 division is 1 unit = 20 runs.

Question 2

Marks obtained by different students are represented using the bar graph. Find the difference between (a) Pratik's and Shubendu's marks (b) Rohan's and Gaurav's marks. [2 MARKS]

SOLUTION

Solution :
Solution: 1 Mark each

From the graph,

(a) Marks obtained by Pratik = 30

Marks obtained by Shubendu = 60

Difference of marks obtained by them = 60 - 30 = 30

(b) Marks obtained by Rohan = 30

Marks obtained by Gaurav = 50

Difference of marks obtained by them = 50 - 30 = 20

Question 3

Find the arithmetic mean of the following data : [2 MARKS]

(a) 2, 3, 4, 5, 6, 7, 8, 9
(b) 12,13,14,15,16,17,18,19, 20

SOLUTION

Solution :
Each part: 1 Mark

The arithmetic mean is,

$Mean = \frac{Sum of all observations}{Total number of observations}$

(a) $Mean = \frac{2 + 3 + 4 + 5 + 6 + 7 + 8 + 9}{8}$

$\Rightarrow Mean = \frac{44}{8} = 5.5$

(b) $Mean = \frac{12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20}{8}$

$\Rightarrow Mean = \frac{144}{9} = 16$

Question 4

What is the probability of drawing the black slice of the pizza? [2 MARKS]

SOLUTION

Solution :
Formula: 1 Mark
Solution: 1 Mark

The whole pizza is divided into 8 slices.

$\Rightarrow$ Total number of slices = 8

Number of black piece = 1

The probability of drawing black slice:
$= \frac{Number of black slices}{Total number of slices}$

$P(E) = \frac{1}{8}$ .

Question 5

(a) Find the median of the data: 24, 36, 46, 17, 18, 25, 35
(b) In a math quiz competition, Aditi's math quiz scores were 90, 92, 93, 88, 95, 88, 97, 87, and 98. What was the median quiz score? [2 MARKS]

SOLUTION

Solution : Each part: 1 Mark

(a) Arrange the data in ascending order
17, 18, 24, 25, 35, 36, 46.
Median is the middle observation.
Clearly, the middle observation is 25.
$\Rightarrow$ 25 is the median.

(b) Arrange the data in ascending order
87, 88, 88, 90, 92, 93, 95, 97, 98
Median is the middle observation.
$\Rightarrow$ 92 is the median.

Question 6

(a) Arrange the plants in such a way that one plant acts as the dividing point and there are equal number of plants on either side of this dividing point.

The heights of the plants are: 5, 4, 132, 173, 142, 111, 123 (mm)

(b) You have to select the players for your football team. Which of the following given datasets will help you to take the decision?

Dataset 1. Name of the players

Dataset 2. Height of the players

Dataset 3. Stamina of players

Dataset 4. Nationality of the players
[3 MARKS]

SOLUTION

Solution :
Each part: 1 Mark
Steps: 1 Mark

(a) Since we have to choose one plant which acts as the dividing point and there are equal number of plants on either side of this dividing point, we need to find the median.

On arranging them in an ascending order we get,

4, 5, 111, 123, 132, 142, 173

The middle observation is 123 mm.

$\Rightarrow$ The median of the observation is 123 mm.

(b) The performance of a football player doesn't depend on his name or nationality. Performance solely depends on stamina and height. So, the data sets which will help us take a better decision are height and stamina of the players.

Question 7

(a) A box contains marbles which are marked with first ten natural numbers. Find the probability of getting a marble marked with a multiple of 2.

(b) A bag contains 3 red balls, 6 blue balls, 1 is black and 2 balls are green. If you want to pick out a red ball what is the probability that you will get a red ball? [3 MARKS]

SOLUTION

Solution :
Part a: 1.5 Marks
Part b: 1.5 Marks

(a) First ten natural numbers = 1,2,3,4,5,6,7,8,9,10.

$\Rightarrow$ Total number of natural numbers = 10

Multiple of 2 = 2,4,6,8,10.

$\Rightarrow$ Number of multiples of 2 are = 5

Probability of getting multiple of 2:
$= \frac{Number of multiples of 2}{Total number of natural numbers}$

$\Rightarrow P(E) = \frac{5}{10} = \frac{1}{2}$

(b) Total number of balls in the bag = 3 + 6 + 1 + 2 = 12

Probability of getting a red ball:
$= \frac{Number of red balls}{Total number of balls}$

$\Rightarrow P (E) = \frac{3}{12} = \frac{1}{4}$

Question 8

Batsman 'A' scored the following number of runs in six innings
36, 35, 50, 46, 60, 55

Batsman 'B' scored the following number of runs in six innings
20, 100, 50, 35, 10, 45

Calculate the mean runs scored by them in an inning. Whose average run scored is better? [3 MARKS]

SOLUTION

Solution : Formula: 1 Mark
Solution: 2 Marks

Total runs by batsman 'A' is 36 + 35 + 50 + 46 + 60 + 55 = 282
Total runs by batsman 'B' is 20 + 100 + 50 + 40 + 10 + 50 = 270

To find the mean, we find the sum of all the observations and divide it by the number of observations.

$M e a n = \frac{S u m o f a l l o b s e r v a t i o n}{T o t a l n u m b e r o f o b s e r v a t i o n}$

$∴ M e a n f o r b a t s m a n^{'} A^{'} = \frac{282}{6} = 47$ .

Thus, the mean runs scored by the batsman 'A' in an inning is 47.

$M e a n f o r b a t s m a n^{'} B^{'} = \frac{270}{6} = 45$ .

Thus, the mean runs scored by the batsman 'B' in an inning is 45.

On an average batsman-A scored better than batsman-B.

Question 9

There are 7 observations in a set of data and their mean is 11. If each observation is multiplied by 2, find the new mean. [3 MARKS]

SOLUTION

Solution : Formula: 1 Mark
Solution: 2 Marks

$M e a n = \frac{S u m o f o b s e r v a t i o n s}{N u m b e r o f o b s e r v a t i o n s}$

Given, Mean = 11 and Number of observations = 7

$∴ 11 = \frac{S u m o f o b s e r v a t i o n}{7}$

$\Rightarrow$ Sum of observations = 77

If each observation is multiplied by 2

New sum of observation is $= 77 \times 2 = 144$

$∴$ New Mean

$= \frac{144}{7} = 22.57$

Question 10

Observe the following graph and answer the questions. [4 MARKS]

(i) In which year is the population highest?

(ii) What is the population in the year 2010?

(iii) Compare the ratio of the population of the years 2008 and 2013.

(iv) In which year is the population lowest?

SOLUTION

Solution :
Each part: 1 Mark

(i) In the year 2013, the population is highest.

(ii) The population in the year 2010 is 10 lakhs.

(iii) The population in 2008 and 2013 is 9 lakhs and 13 lakhs. Hence, the ratio will be 9:13.

(iv) In the year 2009, the population is lowest.

Question 11

A fair die is rolled once. [4 MARKS]
(i) What is the probability of getting a multiple of 2?
(ii) What is the probability of getting a prime number?

SOLUTION

Solution :
Each part: 2 Marks

When a die is rolled once, the possible outcomes are 1, 2, 3, 4, 5 and 6

$\Rightarrow$ Total numbers on the dice = 6

(i) Multiples of 2 are 2, 4, 6

$\Rightarrow$ Numbers of Multiples of 2 = 3

Probability of getting a multiple of 2 is

$P (E) = \frac{Number of multiples of 2}{Total numbers on the dice}$

$P (E) = \frac{3}{6} = \frac{1}{2}$

(ii) Prime numbers are 2, 3, 5

$\Rightarrow$ Numbers of Multiples of 2 = 3

Probability of getting a prime number is

$P (E) = \frac{Number of prime numbers}{Total numbers on the dice}$

$P (E) = \frac{3}{6} = \frac{1}{2}$

Question 12

(a) The following table gives the number of children grouped according to their favourite subjects :

Draw a bar graph using a suitable scale. [4 MARKS]

$\begin{matrix} S u b j e c t & N o . o f C h i l d r e n E n g l i s h & 32 H i n d i & 18 M a t h e m a t i c s & 28 S c i e n c e & 40 H i s t o r y & 36 G e o g r a p h y & 26 \end{matrix}$

(b) According to the graph, in which month is the number of people attending mass prayers maximum and in which month is the number of people attending lunch minimum?

SOLUTION

Solution : Each part: 2 Marks

(a) Scale: x axis - Subjects
y axis - Number of children; 1 cm = 5 units

Bar graph :

(b)

As seen from the graph, the month in which the maximum number of people attended the mass prayer is in April because the blue bar is longest for the month of April. Similarly, the month in which the minimum number of people attended lunch is in March because the orange bar is shortest for the month of March.

Question 13

Plot a double bar graph using the given data and answer the following: [4 MARKS]

(i) Which city has the largest difference in the minimum and maximum temperature?

(ii) Which is the hottest city and which is the coldest city?

(iii) Name two cities where maximum temperature of one was less than the minimum temperature of the other.

(iv) Name the city which has the least difference between its minimum and the maximum temperature.

SOLUTION

Solution : Drawing double bar graph: 2 Marks
Each part: 0.5 Marks

$\Rightarrow$ Data represented by double bar graph is as follows:

(i) Jammu has the largest difference in temperature i.e., Maximum temperature = $41^{\circ} C$ and Minimum temperature = $26^{\circ} C$

$∴$ Difference = $41^{\circ} C - 26^{\circ} C = 15^{\circ} C$

(ii) Jammu is the hottest city as the maximum temperature is the highest among all the given cities and Bangalore is the coldest city because the minimum temperature there is lowest among all the given cities.

(iii) Maximum temperature of Bangalore = $28^{\circ} C$

Minimum temperature of two cities whose minimum temperature is higher than the maximum temperature of Bangalore are Ahmedabad and Jaipur = $29^{\circ} C$

(iv) Mumbai has the least difference in temperature i.e., Maximum temperature = $32^{\circ} C$ and Minimum temperature = $27^{\circ} C$ .

$∴$ Difference = $32^{\circ} C - 27^{\circ} C = 5^{\circ} C$

Question 14

A mathematics teacher wanted to see, whether the new technique of teaching she applied after the quarterly test was effective or not. She took the scores of the 5 weak children in the quarterly test (out of 25) and in the half-yearly test (out of 25): [4 MARKS]

$\begin{matrix} S t u d e n t s & A s h i s h & A r u n & K a v i s h & M a y a & R i t a Q u a r t e r l y & 10 & 15 & 12 & 20 & 9 H a l f y e a r l y & 15 & 18 & 16 & 21 & 15 \end{matrix}$

Draw a double bar graph. Using the graph answer the following questions.
(i) Is the new technique of teaching effective?
(ii) Who has benefitted more from the new technique?

SOLUTION

Solution : Double bar graph: 2 Marks
Each part: 1 Mark

(i) Double bar graph :

Scale: y axis - 1 cm = 5 units

(i) Every student scored more in the half-yearly than the quarterly.

$\Rightarrow$ New technique of teaching is effective.

(ii) The difference of marks scored by Rita in quarterly and half yearly is maximum.

$\Rightarrow$ Rita has benefitted more from the new technique.

Question 15

(i) The birth rate per thousand of different countries over a particular period of time is shown below:

$\begin{matrix} I N D I A & U . K . & C H I N A & G E R M A N Y & S W E D E N 35 & 22 & 42 & 13 & 8 \end{matrix}$

Draw a bar-graph. [4 MARKS]

(ii) A survey of a colony was done. The data is as following:

$\begin{matrix} Favourite Sport & C r i c k e t & Basket Ball & Swimming & Hockey & Athletics W a t c h i n g & 1240 & 470 & 510 & 423 & 250 P a r t i c i p a t i n g & 620 & 320 & 320 & 250 & 105 \end{matrix}$

a) Draw a double bar graph choosing an appropriate scale. What do you get from the bar graph?

b) Which sport is most popular?

c) Which is more preferred, watching or participating in sports?

SOLUTION

Solution : Each part: 2 Marks

(i) Scale : x - axis - Countries
y - axis - Birth rate - 1 cm = 5 births

(ii) The given data is represented in a bar graph:

a) The bar graph represents the number of people who are watching and participating in their favourite sports.

b) Cricket is more popular.

c) Watching sports is more preferred.

Free Data Handling Subjective Test 02 Practice Test - 7th grade

Question 1

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Question 2

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Question 3

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Question 4

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Question 5

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Question 6

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Question 7

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Question 8

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Question 9

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Question 10

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Question 11

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Question 12

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Question 13

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Question 14

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Question 15

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