Free Data Handling Subjective Test 01 Practice Test - 7th grade

Question 1

Find the mode of 2, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14. [1 MARK]

SOLUTION

Solution : Given Data is: 2, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14.

Arranging the observations in the ascending order.

2, 10, 12, 14, 14, 14, 14, 14, 14, 16, 16, 18.

Clearly, the most repeated observation is 14.

$\Rightarrow$ 14 is mode.

Question 2

Find the mean of 9, 10, 8, 9, 12, 10. [2 MARKS]

SOLUTION

Solution :
Formula: 1 Mark
Result: 1 Mark

$M e a n = \frac{Sum of all observation}{Total number of observation}$

$M e a n = \frac{9 + 10 + 8 + 9 + 12 + 10}{6} = 9.66$ .

The mean is 9.66.

Question 3

The runs scored by 11 players in a cricket match are 12, 32, 43, 32, 12, 32, 12, 32, 36, 54,11. Find the mode of the runs scored by the players. [2 MARKS]

SOLUTION

Solution :
Steps: 1 Mark
Result: 1 Mark

Given runs are, 12, 32, 43, 32, 12, 32, 12, 32, 36, 54,11.

Arranging the numbers in ascending order:

11, 12, 12, 12, 32, 32, 32, 32, 36, 43, 54.

Mode of a set of observations is the observation which occurs most often.

Clearly, here, 32 runs are scored by most of the batsman.

So, the mode of the observation is 32.

Question 4

Find the mean of first five multiples of 5. [2 MARKS]

SOLUTION

Solution :
Formula: 1 Mark
Solution: 1 Mark

First five multiples of 5 are 5,10,15, 20 and 25.

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

$Mean = \frac{5 + 10 + 15 + 20 + 25}{5} = 15$

Question 5

Marks obtained by students out of 20 in a class test are written on the board. What is the range of the marks?

12 ,14 , (-1), (-6), 15, 0, 8, 2. [2 MARKS]

SOLUTION

Solution :
Formula: 1 Mark
Result: 1 Mark

Range is defined as the difference between the highest and the lowest observation.

Here, after arranging marks in ascending order : ( -6 ),( -1 ) , 0 , 2, 8 , 12 , 14 , 15.

Lowest observation: (-6)

Highest observation: 15

Range = Highest observation - Lowest observation

Range = 15 - (-6) = 21

Question 6

Calculate the Mean, Median, Mode of the following data: 5,10,10,12, 13

Are these three equal? [3 MARKS]

SOLUTION

Solution :
Mean, Median, Mode: Each 1 Mark

$M e a n = \frac{Sum of all observation}{Total number of observation}$

$\Rightarrow Mean = \frac{5 + 10 + 10 + 12 + 13}{5} = \frac{50}{5} = 10$

Median is the middle most observation of the given observations.

$\Rightarrow Median = 10$

The mode is the most repeated observation of the given observations.

$\Rightarrow Mode = 10$

Hence, for this observation, Mean, Median, Mode are equal.

Question 7

Given below are the weights of 9 different students of class 10^th: 54 kg , 56 kg , 54 kg , 57 kg , 54 kg , 52 kg , 54 kg , 55 kg , 59 kg. Find the mode and median of the given observations. [3 MARKS]

SOLUTION

Solution :
Steps: 2 Marks
Result: 1 Mark

Given,

Weights of 9 different students : 54kg, 56kg , 54kg , 57kg , 54kg , 52kg , 54kg , 55kg , 59kg

Arranging the data in ascending order,

52kg, 54kg, 54kg, 54kg, 54kg, 55kg, 56kg, 57kg, 59kg

Median is the middle most observation of the given observations

$\Rightarrow$ Median = 54kg

Mode is the most repeated observation of the given observations

Clearly, 54kg is the most repeated observation

$\Rightarrow$ Mode = 54kg

Question 8

Following cards are facing down: [3 MARKS]

Find the probability of drawing:

1. Vowel

2. Consonant

SOLUTION

Solution :
Formula: 1 Mark
Application: 1 Mark
Calculation: 1 Mark

Here, we have 8 cards A, E, I, O, U, B, G, H

Vowels = A, E, I, O, U

$\Rightarrow$ Number of vowel cards = 5

So, probability of drawing vowel

$P (E) = \frac{No.of vowels}{Total number of cards} = \frac{5}{8}$

Consonants = B, G, H

$\Rightarrow$ Number of Consonant cards = 3

Probability of drawing consonant:

$P (E) = \frac{No.of consonants}{Total number of cards} = \frac{3}{8}$

Question 9

The ages in years of 10 teachers of a school are:
34, 43, 31, 56, 37, 28, 25, 35, 40, 42

(i) What is the range of the ages of the teachers?

(ii) What is the mean age of these teachers? [3 MARKS]

SOLUTION

Solution : Formulas: 1 Mark
Each part: 1 Mark

(i) Arranging the ages in ascending order, we get:

25, 28, 30, 34, 35, 37, 40, 42, 43, 56

We find that the age of the oldest teacher is 56 years and the age of the youngest teacher is 25 years.

Range of the ages of the teachers = (56 - 25) years = 31 years

(ii) Mean age of the teachers:

$= \frac{25 + 28 + 30 + 34 + 35 + 37 + 40 + 42 + 43 + 56}{10} y e a r s$

$= \frac{370}{10} y e a r s = 37 y e a r s$

Question 10

Following table shows the points of each player scored in four games:

$\begin{matrix} P l a y e r & G a m e & G a m e & G a m e & G a m e 1 & 2 & 3 & 4 S u r a j & 10 & 16 & 14 & 10 A b h i j i t & 8 & 0 & 4 & 6 H a r s h i t h & 13 & D i d n o t & 11 & 8 P l a y \end{matrix}$

Based on the above table answer the following questions:

(i) Find the mean to determine Suraj's average number of points scored per game.

(ii) To find the mean number of points per game of Harshith, would you divide the total points by 3 or by 4? Why?

(iii) Abhijit played in all the four games. How would you find the mean?

(iv) Who is the best performer? [4 MARKS]

SOLUTION

Solution : Solution: 1 Mark each

(i) Suraj's average number of points.

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

$M e a n = \frac{10 + 16 + 14 + 10}{4} = \frac{50}{4}$

$\Rightarrow M e a n = 12.5$

(ii) To find the mean number of points per game of Harshith, mean should be divided by 3 because he played in only 3 games.

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

$M e a n = \frac{13 + 11 + 8}{3} = \frac{32}{3}$

$\Rightarrow M e a n = 10.66$

(iii) For Abhijit's game mean is:

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

$M e a n = \frac{8 + 0 + 4 + 6}{4} = \frac{18}{4}$

$\Rightarrow M e a n = 4.5$

(iv) Comparing everybody's mean score, Suraj has highest mean.

$\Rightarrow$ Suraj is best performer.

Question 11

There are 10 marbles in a box which are marked with first 10 multiples of 2.
(a) What is the probability of drawing a multiple of 4?
(b) What is the probability of drawing a multiple of 6? [4 MARKS]

SOLUTION

Solution :
Formula: 1 Mark
Steps: 1 Mark
Answer: Each part 1 Mark

First 10 multiples of 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20.

$\Rightarrow$ Numbers of multiples of 2 = 10

In these number which are multiples of 4 are 4, 8,12,16, 20

$\Rightarrow$ Numbers of multiples of 4 = 5

In these number which are multiples of 6 are 6,12,18

$\Rightarrow$ Numbers of multiples of 6 = 3

i) Probability of drawing a multiple of 4 is:

$P(E) = \frac{Numbers of multiples of 4}{Total numbers of marbles}$

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

ii) Probability of drawing a multiple of 6 is:

$P(E) = \frac{Numbers of multiples of 6}{Total numbers of marbles}$

$\Rightarrow P(E) = \frac{3}{10}$

Question 12

Sale of English and Hindi books in the years 1995, 1996, 1997 and 1998 are given below:

$\begin{matrix} Y e a r s & 1995 & 1996 & 1997 & 1998 E n g l i s h & 350 & 400 & 450 & 620 H i n d i & 500 & 525 & 600 & 650 \end{matrix}$

Draw a double bar graph and answer the following questions:

(a) In which year was the difference in the sale of the two language books least?

(b) Can you say that the demand for english books increased? Justify. [4 MARKS]

SOLUTION

Solution : Double Bar graph: 2 Marks
Answers: 1 Mark each

Double bar graph:

(a) In the year 1998, the difference in the sale of the two language books was least.

(b) Yes, the demand for English books increased. From the graph, it is evident that the difference in the demand for English books from 1995 to 1998 is 620 - 350 = 270.

Whereas for Hindi books the difference is 650 - 500 = 150.

Comparatively, the demand for English books increased.

Question 13

The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:

$\begin{matrix} D a y & M o n & T u e & W e d & T h u r s & F r i & S a t & S u n R a i n f a l l & 0.0 & 12.2 & 2.1 & 0.0 & 20.5 & 5.5 & 1.0 (i n m m) \end{matrix}$

(i) Find the range of the rainfall in the above data.

(ii) Find the mean rainfall for the week.

(iii) On how many days was the rainfall less than the mean rainfall? [4 MARKS]

SOLUTION

Solution : Range: 1 Mark
Mean: 2 Marks
Answer: 1 Mark

(i) Range = Highest observation - Lowest observation

Range = 20.5 - 0.0 = 20.5

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

$M e a n = \frac{0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.5 + 1.0}{7}$

$M e a n = \frac{41.3}{7} = 5.9$

(iii) On Mon, Wed, Thurs, Sat, and Sun the rainfall was less than mean rainfall i.e. 5.9 mm.

$\Rightarrow$ On 5 days the rainfall was less than the mean rainfall.

Question 14

(a) The profits of a soft drink firm during certain months of a year are given alongside. Draw a Bar graph. [4 MARKS]

$\begin{matrix} M o n t h s & P r o f i t s (i n t h o u s a n d s) J a n u a r y & 2.4 F e b r u a r y & 1.8 M a r c h & 3.0 A p r i l & 4.0 M a y & 5.2 J u n e & 4.0 J u l y & 2.0 \end{matrix}$

(b) The following bar graph shows the number of men and women visiting a sports complex. Can you spot the day when the number of men was equal to the number of women and on which day is the number of women 20 more than the number of men?

SOLUTION

Solution : (a) 2 Marks
(b) 2 Marks

Scale : x - axis - Months
y - axis - Profits in thousands - 1 cm = 1 Thousand

(b)

Here, the height of the bar graph gives us the number of men and women. For Thursday, the height of both the graphs is same. So, on Thursday, the number of men was equal to the number of women.
On Monday the number of women is 20 more than the number of men. (No. of men = 40 , No. of women = 60)

Question 15

(a) The number of boys taking part in different extracurricular activities is shown in the table given below. Represent the data on a bar graph. [4 MARKS]

$\begin{matrix} Activities & No. of boys Scouts & 25 Debate & 13 Science Club & 7 Inst. Music & 30 S.U.P.W & 15 \end{matrix}$

(b) What is the difference between the number of mangoes sold on Sunday and the number of custard apples sold on Wednesday?

SOLUTION

Solution : (a) Scale: 1 Mark
Bar graph: 1 Mark
(b) Solution: 2 Marks

(a) Scale: x axis - extracurricular activities
y axis - Number of boys; 1 cm = 5 units

(b) Number of mangoes sold on Sunday = 85
Number of custard apples sold on Wednesday = 20

Difference = 65

Free Data Handling Subjective Test 01 Practice Test - 7th grade

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

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Question 8

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Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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