Data Handling(7th Grade > Mathematics ) Questions and answers for exam Preparation

7th Grade > Mathematics

DATA HANDLING MCQs

Total Questions : 114 | Page 9 of 12 pages

Question 81.

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

Discuss Question

Answer: Option A. ->
:
The scale is read on the vertical axis. We can see that 1 division is 1 unit = 20 runs.

Question 82.

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]

Discuss Question

Answer: Option A. ->
:

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 83.

(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]

Discuss Question

Answer: Option A. ->
:

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 84.

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

Discuss Question

Answer: Option A. ->
:

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 85.

(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]

Discuss Question

Answer: Option A. ->
:

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 86.

(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]

Discuss Question

Answer: Option A. ->
:
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 87.

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]

Discuss Question

Answer: Option A. ->
:
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 88.

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]

Discuss Question

Answer: Option A. ->
:
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 89.

(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?

Discuss Question

Answer: Option A. ->
:
Each part: 2 Marks
(a) Scale: x axis - Subjects
y axis - Number of children; 1 cm = 5 units
Bar graph :

(a) The Following Table Gives The Number Of Children Grouped...

(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 90.

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?

Discuss Question

Answer: Option A. ->
:

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.