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

7th Grade > Mathematics

DATA HANDLING MCQs

Total Questions : 114 | Page 8 of 12 pages

Question 71.

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

Are these three equal? [3 MARKS]

Discuss Question

Answer: Option A. ->
:

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 72.

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]

Discuss Question

Answer: Option A. ->
:
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 73.

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]

Discuss Question

Answer: Option A. ->
:
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 74.

Following cards are facing down: [3 MARKS]

Find the probability of drawing:
1. Vowel
2. Consonant

Discuss Question

Answer: Option A. ->
:

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 75.

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]

Discuss Question

Answer: Option A. ->
:
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 76.

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]

Discuss Question

Answer: Option A. ->
:

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 77.

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]

Discuss Question

Answer: Option A. ->
:
Double Bar graph: 2 Marks
Answers: 1 Mark each
Double bar graph:

Sale Of English And Hindi Books In The Years 1995, 1996, 199...

(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 78.

(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?

Discuss Question

Answer: Option A. ->
:
(a) 2 Marks
(b) 2 Marks
Scale : x - axis - Months
y - axis - Profits in thousands - 1 cm = 1 Thousand

(a) The Profits Of A Soft Drink Firm During Certain Months O...

(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 79.

(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?

Discuss Question

Answer: Option A. ->
:
(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