MCQ Questions for Class 11 Commerce Economics Presentation Of Data Quiz 6

3 weeks?

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

Explanation

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

$B$ .

From the graph, we can observe that after

$3$ weeks Plant

$B$ has grown

$10\space cm$

The patients temperature at 1 p.m. is:

0%
$37.0$ C
0%
$39.5$ C
0%
$36.5$ C
0%
$36.0$ C

Explanation

From the given graph, we can observe that at

$1 p.m.$ the patient temperature is

$36.5^0C$

Draw a graph for the following:

Side of square (in cm)	$2$	$3$	$3.5$	$5$	$6$
Perimeter (in cm)	$8$	$12$	$14$	$20$	$24$

State the typre of graph.

0%
Linear graph
0%
Bar graph
0%
Pie graph
0%
None of these

During which week did Plant B grow least?

0%
1st week
0%
2nd week
0%
3rd week
0%
None of these

Explanation

Growth of plant B during 1st week

$= 1 - 0 = 1.$

Growth of plant B during 2nd week

$= 7 - 1 = 6.$

Growth of plant B during 3rd week

$= 10 - 7 = 3.$

Plant B grew least in the first week.

What was the maximum forecast temperature during the week?

0%
$10$
0%
$20$
0%
$25$
0%
$35$

How much did Plant B grow from the end of the

$2^{nd}$ week to the end of the

$3^{rd}$ week?

0%
$2 cm$
0%
$3 cm$
0%
$4 cm$
0%
$5 cm$

Explanation

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

$B$ .

From the graph, we can observe that after

$2$ weeks Plant

$B$ has grown

$7\space cm$ and

after

$3$ weeks Plant

$B$ has grown

$10\space cm$

Therefore, from the end of the

$2^{nd}$ week to the end of the

$3^{rd}$ week Plant

$B$ has grown

$10-7=3\space cm$

Days when the forecast temperature the same as the actual temperature are:

0%
Tuesday, Wednesday and Thursday
0%
Tuesday, Friday and Sunday
0%
Thursday, Friday and Saturday
0%
Tuesday, Thursday and Monday

For an experiment in Botany, two different plants, plant A and plant B were grown under similar laboratory conditions. Their heights were measured at the end of each week for 3 weeks. The results are shown by the following graph.
How much did Plant A grow during the 3rd week?

0%
$2cm$
0%
$3cm$
0%
$4cm$
0%
$5cm$

Explanation

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

$A$ .

From the graph, we can observe that after

$2$ weeks Plant

$A$ has grown

$7\space cm$ and

after

$3$ weeks Plant

$A$ has grown

$9\space cm$

Therefore, during

$3^{rd}$ week Plant

$A$ has grown

$9-7=2\space cm$

The actual temperature differs the most from the forecast temperature on:

0%
Tuesday
0%
Wednesday
0%
Thursday
0%
Friday

Explanation

In the given graph, the dotted line represents the forecast temperature and the straight line represents the actual temperature.

From the graph, we can observe that the actual temperature and the forecast temperature differs mostly on Thursday i.e., the maximum difference between actual and forecast temperature (

$23.5-15=8.5$ )

During which week did Plant A grow most?

0%
First week
0%
Second week
0%
Third week
0%
Fourth week

Explanation

In the given graph dotted line represents the growth of the plant A.

In the first week, the plant A grows from

$0$ cm to

$2$ cm.

Therefore the growth of plant A in first week is

$2-0=2$ cm.

In the second week, the plant A grows from

$2$ cm to

$7$ cm.

Therefore the growth of the plant A in the second week is

$7-2=5$ cm.

In the third week, the plant A grows from

$7$ cm to

$9$ cm.

Therefore the growth of the plant A in the third week is

$9-7=2$ cm

Hence, plant A has grow mostly in second week.

Were the two plants of the same height during any week shown here? Specify.

0%
At the end of the 3rd week
0%
At the end of the 1st week
0%
At the end of the 2nd week
0%
At the end of the 4th week

How far is the place of the merchant from the town?

0%
$16$
0%
$18$
0%
$20$
0%
$22$

Explanation

It is given that the courier person travels from town to a neighboring suburban area.

From the graph, we can observe that the destination place is

$22\space k.m.$ from the town.

Find the expenditure on the labours if the total construction cost is Rs.

$5,40,000$

0%
$1,12,500$
0%
$1,22,500$
0%
$1,35,500$
0%
$1,50,000$

Explanation

Sum of all angles is

$360$ degrees
Labour

$=100$ degree
So, part of labour in total construction

$= \dfrac {100}{360}$
Total construction

$= 5,40,000$
Labour expenditure

$= 5,40,000\times \left (\dfrac {100}{360}\right)$

$= 1,50,000$

A family spends Rs.

$6,000$ p.m. to meet the monthly expenditure. The expenditure has been shown in the given pie diagram. The miscellaneous expenditure is

0%
Rs. $1,500$
0%
Rs. $2,000$
0%
Rs. $1,200$
0%
Rs. $3,000$

Explanation

Angle made by rent expenditure

$=$ angle made by education expenditure

$=60^0$ ...(vertically opposite angles)
Angle made by miscellaneous expenditure

$=360 - 90 - 30 - 60 - 60 = 120\,^{\circ}$
Miscellaneous expenditure

$= \dfrac{120 \times 6000}{360} =$ Rs.

$2000$

The given histogram shows a frequency distribution of marks obtained by

$56$ students in a subject.
Number of students securing marks between

$70$ and

$100$ is

0%
2
0%
4
0%
6
0%
8

Explanation

Frequency density

$= \dfrac{\text {Frequency in the class}}{\text {Lenght of the class interval}}$

$\therefore 0.2 = \dfrac{\text {Frequency in the class}}{100 - 70}$
Frequency in the class

$= 0.2 \times 30 = 6$

The given component bar diagram shows percentage of marks obtained by a student in different subjects in a test : The height of the line AB from x-axis is

0%
21 cm approximately
0%
38 cm
0%
86 cm approximately
0%
48 cm

Explanation

In the given diagram.

$AB$ line is above the physics and mathematics.

From the graph, physics is

$25\%$ and mathematics is

$32\%$ .

Therefore, line

$AB$ is

$25+32=57\%$ away from X-axis.

Given that scale is

$3\%=2$ cm.

$\implies 57\%=\dfrac{57\times2}{3}=\dfrac{114}{3}=38$ cm

How many students are both female and excellent ?

0%
0
0%
8
0%
16
0%
32

Explanation

In given table we can see, there are not there are not any female whose performance is excellent.

$\therefore$ The students are both female and excellent is

$0$

Allocations to various sectors of the yearly budget for Rs.

$1000$ crane are represented by the following pie diagram. The expenditure (in rupees) on agriculture is

0%
$250$ crane
0%
$150$ $ crane
0%
$300$ crane
0%
$200$ crane

Explanation

Total angle

$= 360\,^{\circ}$
Thus angle that agriculture makes

$= 360 - 18 - 90 - 72 - 54 - 54 = 72\,^{\circ}$

$\therefore$ the expenditure on agriculture

$= \dfrac{72}{360} \times 1000$ crane

$= 200$ crane.

The times, in seconds, taken by 150 atheletes to run a 110 hurdle race are tabulated below :

Class	Frequency
13.8-14	2
14-4.2	4
14.2-14.4	5
14.4-14.6	71
14.6-14.8	48
14.8-15	20

The number of atheletes who completed the race in less then 14.6 seconds is :

0%
$11$
0%
$71$
0%
$82$
0%
$130$

Explanation

The number of athletes who completed the race in

$<14.6$ seconds is

$n(13.8-14)+n(14-14.2)+n(14.2-14.4)+n(14.4-14.6)$

$=2+4+5+71=82$

When did the person stop on his way?

0%
9:00 a.m - 9.30 a.m
0%
9:30 a.m - 10.00 a.m
0%
10:00 a.m - 10.30 a.m
0%
10:30 a.m - 11.00 a.m

Explanation

The person had stopped between the time

$10$ a.m to

$10.30$ a.m.

This can be easily concluded since the curve on the graph between these times was horizontal to the x-axis which means no distance travelled.

Hence, option C is correct.

Fastest ride is from interval :

0%
8 a.m to 9 a.m
0%
9 a.m to 10 a.m
0%
10 a.m to 11 a.m
0%
11 a.m to 12 noon

Explanation

Time Period	Distance covered
8am-9am	$10 - 0 = 10 km$
9am-10am	$16 - 10 = 6 km$
10am-10.30am	$0 km$
10.30am-12am	$22 - 16= 6 km$

The person ride fastest during 8am to 9am.

Annual Development Expenditure of a state under different heads is as follows :
Agriculture

$40\%$
Education

$35\%$
Industry

$15\%$
Misc.

$10\%$
Which one of the following charts represents this information ?

Explanation

For pie diagram, find the sectorial angle in
each case as follows

		Angle
Agriculture	$40\%$	$=\cfrac{(360 \times 40)}{100}$	$=144^{\circ}$
Education	$35\%$	$=\cfrac{(360 \times 35)}{100}$	$=126^{\circ}$
Industry	$15\%$	$=\cfrac{(360 \times 15)}{100}$	$=54^{\circ}$
Miscellaneous	$10\%$	$=\cfrac{(360 \times 10)}{100}$	$=36^{\circ}$

Find the loan amount for agriculture sector.

0%
Rs. 60 crore
0%
Rs. 70 crore
0%
Rs. 80 crore
0%
Rs. 90 crore

Explanation

Total loan is

$= 180$ crore Rs.

Agriculture sector has an amount of

$120^{o}$ out of

$360^{o}$ i.e.

$\left(\dfrac{1}{3}\right)^{rd}$ part.

$\therefore$ Part of agriculture

$=\left(\dfrac{1}{3}\right)^{rd}$ of

$180$ crore

$=\dfrac{180}{3}=60$ crore Rs.

Hence, loan amount for agriculture sector is

$60$ crore Rs.

How much additional amount did industrial sector receive than agriculture sector ?

0%
$20$ crore Rs.
0%
$30$ crore Rs.
0%
$40$ crore Rs.
0%
$50$ crore Rs.

Explanation

Loan amount of agriculture

$= 60$ crore
Total loan

$= 180$ crore
Fraction of industrial sector

$=\dfrac { (360 - 120 - 40) }{360}$

$= \dfrac {200 }{ 360}$
So, loan amount of industrial sector

$= 180 \times \left (\dfrac {200 }{ 360}\right)$

$= 100$ crore
Differece between industrial sector and agriculture is

$= 100 - 60$

$= 40$ crore Rs

How many workers have their wages between Rs.

$90$ and Rs.

$120$ ?

0%
24
0%
34
0%
44
0%
54

Explanation

Number of workers

$( 90$ Rs.

$- 100$ Rs.

$) = 6$
Number of workers

$( 100$ Rs.

$- 110$ Rs.

$) = 10$
Number of workers

$( 110$ Rs.

$- 120$ Rs.

$) = 18$
Total number of workers

$( 90$ Rs.

$- 120$ Rs.

$) = 18+10+6 =34$

Express the number of workers with wages less than Rs 110 as the percent of the number of workers with wages more than Rs 100.

0%
$28$ %
0%
$38$ %
0%
$38\frac{6}{13}$ %
0%
None of these

Explanation

Answer:-

Converting the histogram into frequency distribution table-

Wages	No. of workers
80-90	4
90-100	6
100-110	10
110-120	18
120-130	16
130-140	8

No. of workers with wage less than Rs. 110 = 4 + 6 + 10 = 20

No. of workers with wage more tthan Rs. 100 = 10 + 18 + 16 + 8 = 52

Percentage =

$\cfrac{20}{52} \times 100 \% = \cfrac{500}{13} \% = 38\cfrac{6}{13}\%$

How many students have their height less than 140 cm ?

0%
$31$
0%
$32$
0%
$33$
0%
$34$

Explanation

Number of students in

$(125$ cm

$- 130$ cm

$) = 6$
Number of students in

$(130$ cm

$- 135$ cm

$) = 8$
Number of students in

$(135$ cm

$- 140$ cm)

$= 20$
So, total number of students less than

$140$ cm.

$= 20+6+8=34$

How many students have their height more than

$135$ cm and less than

$155$ cm ?

0%
$56$
0%
$66$
0%
$76$
0%
$86$

Explanation

Number of students in

$(135$ cm

$- 140$ cm

$) = 20$
Number of students in

$(140$ cm

$- 145$ cm

$) = 18$
Number of students in

$(145$ cm

$- 150$ cm

$) = 16$
Number of students in

$(150$ cm

$- 155$ cm

$) = 12$
Total number of student with height

$( 135$ cm

$- 155$ cm

$) = 20+18+16+12 =66$

Which is the item having the maximum expenditure?

0%
Bricks
0%
Labour
0%
Cement
0%
Steel

Explanation

The biggest part in circle is of labour which

$100$ out of

$360$ .
So, labour will have the maximum expenditure.

How many workers get maximum wages ?

0%
$12$
0%
$15$
0%
$18$
0%
$21$

Explanation

Given, histogram shows per day wages of workers in a factory.

From the histogram, the number of workers who get maximum wages is

$18$ (i.e., 110-120 rs are given to maximum number of workers)

CBSE Questions for Class 11 Commerce Economics Presentation Of Data Quiz 6 - MCQExams.com

0:0:1

Practice Class 11 Commerce Economics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Economics Presentation Of Data Quiz 6 - MCQExams.com

0:0:1

Practice Class 11 Commerce Economics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.