The class marks in a frequency table are given to be $$5,10,15,20,25,30,35,40,45,50$$. The class limits of the first classes are

$$1.5-8.5,8.5-11.5,11.5-18.5,18.5-21.5,21.5-28.5$$

MCQ Questions for Class 11 Commerce Economics Presentation Of Data Quiz 5

Choose the correct answer from the alternatives given.
What is the difference in the amount estimated by family on electricity bill and cell bill?

0%
Rs. $1920$
0%
Rs. $4160$
0%
Rs. $6080$
0%
Rs. $8000$

Explanation

Difference between electricity bill and cell bill

= (19 - 6) % = 13 %

$= (19 - 6)\% = 13\%$
So, required difference

= 13 % o f 32, 000 = R s . 4, 160

$=13\%\ of\ 32,000 = Rs.\ 4,160$

Which of the following is advantage of classification of data?

0%
It arranges and presents data in condensed and meaningful form.
0%
It facilitates comparison of data.
0%
It highlights significant features of the data at a glance.
0%
All of above

Year-wise production of rice, wheat and maize for the last ten years can be displayed by ________.

0%
simple column chart
0%
multiple column chart
0%
broken column chart
0%
subdivided column chart

Choose the correct answer from the alternatives given.
The family actually paid Rs.

4672

$4672$ on grocery. What is the difference in the amount budgeted and spent on grocery?

0%
Rs. $1738$
0%
Rs. $1672$
0%
Rs. $1728$
0%
Rs. $1038$

Explanation

Amount budgeted on grocery

= 20 % o f 32, 000 =

$= 20\%\ of\ 32,000 =$ Rs.

6, 400

$6,400$
So, required difference

= 6, 400 - 4, 672 =

$= 6,400 - 4,672 =$ Rs.

1, 728

$1,728$

_______ is the tendency of data to increase or decrease or stagnate over a long period of tiem.

0%
Random variations
0%
Irregular variations
0%
Secular trend
0%
Data trend

A survey was made to find the type of music that a certain group of young people linked in a city. Above pie chart shows the findings of this survey.
From this pie chart answer which type of music is liked by the maximum number of people?

0%
$\text{Semi Classical}$
0%
$\text{Classical}$
0%
$\text{Light}$
0%
$\text{Folk}$

Explanation

Light music is liked by

40 %

$40\%$ people which is maximum among all the other three music types.

Hence, option

C

$C$ is correct.

The data may be said to be classified 'qualitatively', when data relates to ____________.

0%
time or period
0%
character or attributes
0%
place, area or region
0%
numerical values or magnitudes

If data are classified for registration of students for Company secretary chip course region wise in Eastern, Northern, Southern, Western. It is knows as _____________________.

0%
chronological classification
0%
qualitative classification
0%
geographical classification
0%
quantitative classification

The data may be said to be classified 'chronologically', when data relates to __________________.

0%
time or period
0%
character or attributes
0%
place, area or region
0%
numerical values or magnitudes

The central angle of a sector in a pie chart cannot be more than

180^{0}

$180^0$

0%
True
0%
False

Explanation

The central angle in a pie chart is more than

180^{0}

$180^0$ , but not more than

360^{0}

$360^0$

The class marks in a frequency table are given to be

5, 10, 15, 20, 25, 30, 35, 40, 45, 50

$5,10,15,20,25,30,35,40,45,50$ . The class limits of the first classes are

0%
$3-7,7-17,13-17,17-23,23-27$
0%
$2.5-7.5,7.5-12.5,12.5-17.5,17.5-22.5,22.5-27.5$
0%
$1.5-8.5,8.5-11.5,11.5-18.5,18.5-21.5,21.5-28.5$
0%
$2-8,8-12,12-18,18-22,22-28$

Explanation

2.5 - 7.5, 7.5 - 12.5, 12.5 - 17.5, 17.5 - 22.5, 22.5 - 27.5

$2.5 - 7.5, 7.5 - 12.5, 12.5 - 17.5, 17.5 - 22.5, 22.5 - 27.5$

Given,

5, 10, 15, 20, 25, 30, 35, 40, 45, 50

$5,10,15,20,25,30,35,40,45,50$

the class limit is given by,

\frac{5}{2} = 2.5

$\dfrac{5}{2}=2.5$

\frac{5 + 10}{2} = 7.5

$\dfrac{5+10}{2}=7.5$

\frac{10 + 15}{2} = 12.5

$\dfrac{10+15}{2}=12.5$

\frac{15 + 20}{2} = 17.5

$\dfrac{15+20}{2}=17.5$

\frac{20 + 25}{2} = 22.5

$\dfrac{20+25}{2}=22.5$

\frac{25 + 30}{2} = 27.5

$\dfrac{25+30}{2}=27.5$

Data represented using circle is known as

0%
Bar Graph
0%
Histogram
0%
Pictograph
0%
Pie chart

Explanation

A pie chart represents the data in circular form.

Hence, option

D

$D$ is correct.

A graph that displays data that changes continuously over periods of time is

0%
Bar graph
0%
pie chart
0%
Histogram
0%
Line graph

Look at the histogram and answer the following .

How many workers earn more than Rs. 850?

0%
7
0%
8
0%
9
0%
10

Explanation

From the graph, we get the following table.

$\text{Wages (Rs.)}$	$\text{Number of workers}$
800-810	3
810-820	2
820-830	1
830-840	9
840-850	5
850-860	1
860-870	3
870-880	1
880-890	1
890-900	4

The workers who earn more than

850

$850$ are

n (850 - 860) + n (860 - 870) + n (870 - 880) + n (880 - 890) + n (890 - 900) = 1 + 3 + 1 + 1 + 4 = 10

$n(850-860)+n(860-870)+n(870-880)+n(880-890)+n(890-900)=1+3+1+1+4=10$

For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are the:

0%
upper limits of the classes
0%
lower limits of the classes
0%
class marks of the classes
0%
upper limits of perceeding classes

Explanation

For drawing a frequency polygon of a continuous frequency distribution,

we draw the class marks of the classes on the abscissae and the frequency of the classes on the ordinates.

We know, class marks are the mean of lower and upper limit of the class intervals.

Hence, option

C

$C$ is correct.

Class mark of a particular class is 10.5 and class size is 7, then class interval is :

0%
10.5 -17.5
0%
3.5 -10.5
0%
7 -17.5
0%
7 -14

Explanation

Class mark is the mid point of the class interval.

Class size is the upper limit minus lower limit.

Given that class mark of a class is

10.5

$10.5$ and

class size is

7

$7$

By option verification,

Opition A class mark is

\frac{10.5 + 17.5}{2} = \frac{28}{2} = 14

$\dfrac{10.5+17.5}{2}=\dfrac{28}{2}=14$ False

Option B class mark is

\frac{10.5 + 3.5}{2} = \frac{14}{2} = 7

$\dfrac{10.5+3.5}{2}=\dfrac{14}{2}=7$ False

Option C class mark is

\frac{17.5 + 7}{2} = \frac{24.5}{2} = 12.25

$\dfrac{17.5+7}{2}=\dfrac{24.5}{2}=12.25$ False

Option D class size is

14 - 7 = 7

$14-7=7$ and

Class mark is

\frac{14 + 7}{2} = \frac{21}{2} = 10.5

$\dfrac{14+7}{2}=\dfrac{21}{2}=10.5$ True

Therefore the class interval is

7 - 14

$7-14$

Class mark of class interval 60 - 70 is

0%
60
0%
70
0%
65
0%
75

Explanation

Class mark is the mid point of the class interval.

Therefore, the class mark for the interval

60 - 70

$60-70$ is

\frac{60 + 70}{2} = \frac{130}{2} = 65

$\dfrac{60+70}{2}=\dfrac{130}{2}=65$

Find the central angle of pie chart for winter.

0%
$90$
0%
$120$
0%
$150$
0%
None of these

Explanation

Central Angle for Summer

= \frac{90}{360} \times 360 = 90

$= \dfrac { 90 }{ 360 } \times 360 = 90$ degree

Central Angle for Rainy

= \frac{120}{360} \times 360 = 120

$= \dfrac { 120 }{ 360 } \times 360 = 120$ degree

Central Angle for Winter

= \frac{150}{360} \times 360 = 150

$= \dfrac { 150 }{ 360 } \times 360 = 150$ degree

If the class marks in a frequency distribution are 19.5, 26.5, 33.5, 40.5, then the class corresponding to the class mark 33.5 is :

0%
16 - 23
0%
23-30
0%
30-37
0%
37-41

Explanation

Class mark is the mid point of the class interval.

Class size is the upper limit minus lower limit.

Given that class mark of a class is

6.5

$6.5$ and

the class size is

26.5 - 19.5 = 33.5 - 26.5 = 40.5 - 33.5 = 7

$26.5-19.5=33.5-26.5=40.5-33.5=7$

By option verification,

Opition A class mark is

\frac{16 + 23}{2} = \frac{39}{2} = 19.5

$\dfrac{16+23}{2}=\dfrac{39}{2}=19.5$ False

Option B class mark is

\frac{30 + 23}{2} = \frac{53}{2} = 26.5

$\dfrac{30+23}{2}=\dfrac{53}{2}=26.5$ False

Option C class mark is

\frac{30 + 37}{2} = \frac{67}{2} = 33.5

$\dfrac{30+37}{2}=\dfrac{67}{2}=33.5$ and

class size is

37 - 30 = 7

$37-30=7$ True

Option D class mark is

\frac{37 + 41}{2} = \frac{78}{2} = 39

$\dfrac{37+41}{2}=\dfrac{78}{2}=39$ False

Therefore the class interval is

30 - 37

$30-37$

The class mark of the class 29.5

-

$-$ 30.5 is :

0%
30
0%
30.5
0%
31.5
0%
31

Explanation

Class mark is the mid point of the class interval.

Therefore, the class mark for the interval

29.5 - 30.5

$29.5-30.5$ is

\frac{29.5 + 30.5}{2} = \frac{60}{2} = 30

$\dfrac{29.5+30.5}{2}=\dfrac{60}{2}=30$

The class mark of the class 100

-

$-$ 150:

0%
$100$
0%
$125$
0%
$130$
0%
$150$

Explanation

Class mark is the mid point of the class interval.

Therefore the class mark of the class

100 - 150

$100-150$ is

\frac{100 + 150}{2} = \frac{250}{2} = 125

$\dfrac{100+150}{2} = \dfrac{250}{2}=125$

How much distance did the car cover during the period 7.30 a.m. to 8 a.m?

0%
$20$ km
0%
$40$ km
0%
$80$ km
0%
$100$ km

Explanation

Using the information provided in the table, we can plot the graph shown above, taking the time on the

x - a x i s

$x-axis$ and the distance (in km) on the

y - a x i s

$y-axis$ .

From the above graph, we can observe that at

7 : 30 a m

$7:30\ am$ , the distance traveled is

100 k m

$100\space km$ and at

8 : 00 a m

$8:00\ am$ , the distance traveled is

120 k m

$120 \space km$ .

Therefore, the distance covered during the period

7 : 30

$7:30$ and

8 : 00

$8:00$ is

120 - 100 = 20 k m

$120-100=20\space km$

Hence, option A is correct.

Draw a graph for the given information with suitable scales on the axes and answer the following question.

The time when the car had covered a distance of

100

$100$ km since its start is:

0%
$7:00$ am
0%
$7:30$ am
0%
$8:00$ am
0%
$8:30$ am

To get an interest of Rs

280

$280$ per year, how much money should be deposited?

0%
$2500$
0%
$3000$
0%
$3500$
0%
$400$

Explanation

From the above graph, we can notice that, to get an interest of

280

$280$ rupees per year, the money to be deposited is

3500

$3500$ rupees (approximately).

Use the graph to find the interest on Rs

2500

$2500$ for a year.

0%
$Rs. 100$
0%
$Rs.200$
0%
$Rs.300$
0%
$Rs.400$

Explanation

From the above graph, we can observe that simple interest on

R s . 2500

$Rs. 2500$ rupees is

R s . 200

$Rs. 200$ rupees which is marked as black point in the graph.

3 weeks

0%
$7cm$
0%
$8cm$
0%
$9cm$
0%
$10cm$

Explanation

In the given graph, The dotted line represents the growth of the Plant

A

$A$ .

From the graph, we can observe that after

3

$3$ weeks Plant

A

$A$ has grown

9 c m

$9\space cm$

2 weeks

0%
$7cm$
0%
$8cm$
0%
$9cm$
0%
$10cm$

Explanation

In the given graph, The dotted line represents the growth of the Plant

A

$A$ .

From the graph, we can observe that after

2

$2$ weeks Plant

A

$A$ has grown

7 c m

$7\space cm$

During which periods did the patient's temperature showed an upward trend?

0%
9 a.m. to 10 a.m., 10 a.m. to 11 a.m., 2 p.m. to 3 p.m.
0%
9 a.m. to 10 a.m., 10.30 a.m. to 11.30 a.m., 2 p.m. to 3 p.m.
0%
9 a.m. to 10 a.m., 10 a.m. to 11.30 a.m., 1 p.m. to 2 p.m.
0%
9 a.m. to 10 a.m., 10 a.m. to 11 a.m., 1 p.m. to 2 p.m.

Explanation

Whenever there is an increase in the value of

y - a x i s

$y - axis$ as the time(on

x - a x i s

$x - axis$ ) passes by, there is an upward trend.
Thus, the patient shows an upward trend between

9 a m - 10 a m

$9 am - 10 am$ ,

10 a m - 11 a m

$10 am - 11 am$ and

2 p m - 3 p m

$2 pm - 3 pm$

Hence, option A is correct.

2 weeks

0%
$7cm$
0%
$8cm$
0%
$9cm$
0%
$10cm$

Explanation

In the given graph, The straight line represents the growth of the Plant

B

$B$ .

From the graph, we can observe that after

2

$2$ weeks Plant

B

$B$ has grown

7 c m

$7\space cm$

When was the patients temperature

38.5

$38.5$ C?

0%
$9$
0%
$10$
0%
$11$
0%
$12$

Explanation

From the given graph, we can observe that the patient temperature

{38.5}^{0} C

$38.5^0C$ was at

12

$12$ noon

CBSE Questions for Class 11 Commerce Economics Presentation Of Data Quiz 5 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Economics Presentation Of Data Quiz 5 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

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