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CBSE Questions for Class 11 Commerce Applied Mathematics Descriptive Statistics Quiz 6 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Descriptive Statistics
Quiz 6
Find the standard deviation of $$210, 240, 250, 260, 220, 230$$ and $$270$$.
Report Question
0%
$$18$$
0%
$$20$$
0%
$$22$$
0%
$$25$$
Explanation
Mean of the given numbers is $$\bar x =\dfrac{210+240+250+260+220+230+270}{7}=\dfrac{1680}{7}=240$$
total number of values $$n=7$$
Standard deviation is given by $$SD=\sqrt{\dfrac{\sum_{i=1}^n (x_i-\bar x)^2}{n}}$$
$$\implies SD=\sqrt{\dfrac{(210-240)^2+(240-240)^2+.....+(270-240)^2}{7}}$$
$$\implies SD=\sqrt{\dfrac{900+0+100+400+400+100+900}{7}}$$
$$\implies SD=\sqrt{\dfrac{2800}{7}}$$
$$\implies SD=\sqrt{400}=20$$
Therefore the standard deviation for the given values is $$20$$
The ages (in years) of a family of 6 members are 1, 5, 12, 15, 38 andThe standard deviation is found to be 15.9. After 10 years the standard deviation is
Report Question
0%
increased
0%
decreased
0%
remains same
0%
none of these
Explanation
Then mean of six members=$$\frac{1+5+12+15+38+40}{6}=\frac{111}{6}=18.5$$
Then Variance $$\alpha ^{2}=\frac{(18.5-1)^{2}+(18.5-5)^{2}+(18.5-12)^{2}+(18.5-15)^{2}+(18.5-38)^{2}+(18.5-40)^{2}}{5}$$
$$=\frac{(17.5)^{2}+(13.5)^{2}+(6.5)^{2}+(3.5)^{2}+(-19.5)^{2}+(21.5)^{2}}{6}$$
$$=\frac{306.25+182.25+42.25+12.25+380.25++462.25}{6}=\frac{1385.50}{6}=230.91$$
So standard deviation $$\alpha =15.19$$
Given after 10 year The
standard deviation is 15.9
Then
standard deviation remain same
Find the quartile deviation of $$4, 6, 9, 12, 18, 20, 23, 27, 34, 48$$ and $$53$$
Report Question
0%
$$12.5$$
0%
$$13$$
0%
$$12$$
0%
$$11.5$$
Explanation
Given series of numbers is already arranged in ascending order.
Thus median is the centre value $$ = 20 $$
First Quartile $$ {Q}_{1} $$ will be the mid value of the observations to the left of $$ 20 $$ which is $$ 9 $$.
Similarly, Third Quartile $$ {Q}_{3} $$ will be the mid value of the observations to the right of $$ 20 $$ which is $$ 34 $$.
And Quartile deviation will be $$ \dfrac {{Q}_{3} - {Q}_{1}}{2} = \dfrac {34-9}{2} = \dfrac {25}{2} = 12.5 $$.
The line plot below shows how students scored on last week's maths test.
How many students scored 95 or higher on the test?
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0%
5 students
0%
7 students
0%
12 students
0%
16 students
Explanation
It is given that $$1$$ star $$= 1$$ student.
Number of students who scored $$95$$ marks or more than $$95$$ = number of students who scored $$95$$ marks + number students who scored $$100$$ marks
Students who scored $$95$$ marks $$=7$$
Students who scored $$100$$ marks $$=5$$
Total students $$=7+5=12$$
The standard deviation of $$3x + 5, 3x + 7, 3x + 9, 3x + 11, 3x + 13, 3x + 15$$ and $$3x + 17$$ is ...........
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0%
$$2$$
0%
$$3$$
0%
$$4$$
0%
$$1$$
Explanation
Mean of the given numbers is $$\bar x =\dfrac{(3x+5)+(3x+7)+(3x+9)+(3x+11)+(3x+13)+(3x+15)+(3x+17)}{7}=\dfrac{21x+77}{7}=3x+11$$
total number of values $$n=7$$
Standard deviation is given by $$SD=\sqrt{\dfrac{\sum_{i=1}^n (x_i-\bar x)^2}{n}}$$
$$\implies SD=\sqrt{\dfrac{(5-11)^2+(7-11)^2+(9-11)^2+(11-11)^2+(13-11)^2+(15-11)^2+(17-11)^2}{7}}$$
$$\implies SD=\sqrt{\dfrac{36+16+4+0+4+16+36}{7}}$$
$$\implies SD=\sqrt{\dfrac{112}{7}}$$
$$\implies SD=\sqrt{16}=4$$
Therefore the standard deviation for the given values is $$4$$
What is the total number of persons who made the journey during the week?
(Assume that one represents 10 persons)
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0%
200
0%
400
0%
600
0%
900
Explanation
Given that one symbol represents 10 persons and the total number of symbols in picture =$$9+4+6+4+8+6+2= 40$$
Total number of persons who made the journey during the week = $$ \displaystyle 10 \times 40 = 400 $$
If the lower and upper quartiles of a data are 4 and
12 then the quartile deviation is_____
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0%
2
0%
4
0%
6
0%
8
Explanation
Lower Quartile $$ {Q}_{1} = 4 $$
Upper Quartile $$ {Q}_{3} = 12 $$
And Quartile deviation will be $$ \frac {{Q}_{3} - {Q}_{1}}{2} = \frac {12-4}{2} = \frac {8}{2} = 4 $$
If the lower and upper quartiles of a data are $$4$$ and $$12$$, then the quartile deviation is ____
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0%
$$2$$
0%
$$4$$
0%
$$6$$
0%
$$8$$
Explanation
Quartile deviation is the half of the inter quartile deviation.
Inter quartile deviation is the difference between upper quartile and lower quartile.
Given that, lower quartile is $$4$$ and upper quartile is $$12$$
Therefore, Quartile deivation is $$\dfrac{\text{upper quartile - lower quartile}}{2}$$
$$\implies \text{quartile deviation} =\dfrac{12-4}2=\dfrac 82 =4$$
Read the graph and answer the following:
The horizontal line in the graph is called:
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0%
X-axis
0%
Y-axis
0%
line
0%
none
Explanation
Horizontal line represents X-axis in the graph.
So, the horizontal line in the graph is called X-axis.
Read the graph and answer the following:
The vertical line in the graph is called:
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0%
Y-axis
0%
X-axis
0%
ray
0%
none of the above
Explanation
In the graph, vertical line always represents Y-axis.
So, the vertical line in the graph is called Y-axis.
How many school teachers does simile represent?
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0%
$$500$$
0%
$$600$$
0%
$$700$$
0%
$$800$$
Explanation
In the table Hyderabad town has $$4200$$ teachers which represented by $$7$$ similes in the pictograph.
Therefore, one simile represent $$\dfrac{4200}{7} = 600$$ school teachers.
How many school teachers does each smile represent?
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0%
500
0%
400
0%
700
0%
none of these
Explanation
In the table Hyderabad town has $$4200$$ teachers which represented by $$7$$ similes in the pictograph.
Therefore, one simile represent $$\dfrac{4200}{7} = 600$$ school teachers.
If the median of the data 6,7,x-2,x,18,21 written in ascending order is 16, then the variance of that data is
Report Question
0%
$$30\dfrac{1}{5}$$
0%
$$31\dfrac{1}{3}$$
0%
$$32\dfrac{1}{2}$$
0%
$$33\dfrac{1}{3}$$
Explanation
$$M=\dfrac{x-2+x}{2}=16$$
Variance = $$\dfrac{1}{n}\sum (x_i-\overline{x})^{2}$$
Required variance=$$31\dfrac{1}{3}$$
Find the mean for the following data using step deviation method.
Report Question
0%
$$4$$
0%
$$5$$
0%
$$6$$
0%
$$7$$
Explanation
Answer:- By shortcut Method
Class interval width (w) = 2
$$X_i$$
$$F_i$$
$$d=\cfrac{X-A}{i}$$
$$F_i d_i$$
2
10
-1
-10
4=A
20
0
0
6
30
1
30
8
40
2
80
$$\Sigma f = 100$$
$$\Sigma fd = 100$$
Mean = $$A+\cfrac{\Sigma fd}{\Sigma f} \times w=4+\cfrac{100}{100} \times 2 ={4+2}=6$$
C)6
From the bar graph
What is total no. of students in class VIII?
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0%
$$45$$
0%
$$185$$
0%
$$215$$
0%
$$245$$
Explanation
As seen in the graph, we can say the following things.
The total number of students in class VIII are $$40+48+52+45+30=215$$.
Hence, option C is correct.
Choose the formula used for arithmetic mean of grouped data by step deviation method is
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0%
$$\bar { x } =A-\dfrac { \sum { fd\prime } }{ \sum { f } } \times i$$
0%
$$\bar { x } =A\times \dfrac { \sum { fd\prime } }{ \sum { f } } \times i$$
0%
$$\bar { x } =A+\dfrac { \sum { fd\prime } }{ \sum { f } } \times i$$
0%
$$\bar { x } =A+\dfrac { \sum { fd\prime } }{ \sum { f } } +i$$
Explanation
The formula used for arithmetic mean of groped data by step deviation method is $$\bar { x } =A+\dfrac { \sum { fd\prime } }{ \sum { f } } \times i$$
$$A =$$ Assumed mean of the given data
$$\sum { f } =$$ Summation of the frequencies given in the grouped data
$$\sum { fd\prime } =$$ Summation of the frequencies and deviation of a given mean data
$$d\prime =\dfrac { x-A }{ i } $$
$$i =$$ Class interval width
$$\bar { x } =$$ Arithmetic mean
From the given bar graph
In which year the difference of the values of export and import is maximum?
Report Question
0%
$$1986 - 87$$
0%
$$1982 - 83$$
0%
$$1984 - 85$$
0%
$$1983 - 84$$
Explanation
We can see the graph of years versus Export and Import.
The difference of the values of export and import is maximum in the year $$1986-87$$.
From the given bar graph.
In which year the export is minimum
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0%
$$1984 - 85$$
0%
$$1986 - 87$$
0%
$$1983 - 84$$
0%
$$1982 - 83$$
Explanation
The graph of years versus Export and Import in shown on $$x$$ and $$y$$ axis respectively.
As seen in the graph, we can say in year $$1982-83$$, the export is minimum.
From the bar graph, answer the following question
What is the total no. of students in section $$A, B$$ and $$C$$`.
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0%
$$40$$
0%
$$140$$
0%
$$52$$
0%
$$215$$
Explanation
As seen from the graph, the total number of students in sections $$A,B$$ and $$C$$ are $$A+B+C$$
$$=40+48+52=140$$
Hence, option B is correct.
Study the bar graph and answer the question given below:
What was the percentage decline in the production of oil from $$2007$$ to $$2008$$?
Report Question
0%
$$30\%$$
0%
$$33\%$$
0%
$$33.33\%$$
0%
$$36.3\%$$
Explanation
Percentage we can get easily by getting the difference between two and then by finding that difference is what percent of the initial value.
So Percentage $$= \dfrac{20 - 30}{30}\times 100\%$$
$$= -33.33%$$
So there was decline of $$33.33\%$$ from the year $$2007$$ and $$2008$$
Study the bar graph and answer the question given below:
In how many of the given years was the production of oil more than average production of the given years?
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0%
$$1$$
0%
$$2$$
0%
$$3$$
0%
$$4$$
Explanation
Average production (in $$1000$$ tonnes) over the given years will be
$$\cfrac{1}{7}[15 + 30 + 20 + 40 + 50 + 30 + 60]$$
= $$\cfrac{245}{7} = 35$$
So, the production in the years $$2009, 2010, 2012$$ are more than the average production.
Study the above graph and calculate the difference between the number of cars manufactured by company $$Y$$ in $$2001$$ and $$2000$$?
Report Question
0%
$$20000$$
0%
$$30000$$
0%
$$40000$$
0%
$$50000$$
Explanation
Number of cars manufactured by company $$Y$$ in $$2001$$ $$=80000$$
Number of cars manufactured by company $$Y$$ in $$2000$$ $$= 100000$$
The difference is $$ 100000-80000 = 20000$$
Find the mean for the following data using step deviation method.
Report Question
0%
$$9$$
0%
$$10$$
0%
$$11$$
0%
$$12$$
Explanation
Answer:- By shortcut Method
Class interval width (w) = 3
X
F
$$d=\cfrac{X-A}{i}$$
Fd
3
5
-1
-5
6=A
10
-0
-0
9
15
1
15
12
20
2
40
$$\Sigma f = 50$$
$$\Sigma fd = 50$$
Mean = $$A+\cfrac{\Sigma fd}{\Sigma f} \times i=6+\left(\cfrac{50}{50} \times 3\right) = 6+3 = 9$$
A) 9
Find the mean for the following data using step deviation method.
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0%
$$43$$
0%
$$44$$
0%
$$45$$
0%
$$46$$
Explanation
Answer:- By shortcut Method
Class interval width (i)= 24-12 = 12
X
F
$$d=\cfrac{x-A}{i} $$
Fd
12=A
1
0
0
24
2
1
2
36
3
2
6
48
4
3
12
60
5
4
20
$$\Sigma f = 15$$
$$\Sigma fd = 40$$
Mean = $$A+\cfrac{\Sigma fd}{\Sigma f} \times i=12+\left(\cfrac{40}{15} \times 12\right)={12+32}=44$$
B)44
Study the above graph and calculate
the difference between the number of cars manufactured by Company $$X$$ in $$2001$$ and $$2002$$?
Report Question
0%
$$20000$$
0%
$$30000$$
0%
$$40000$$
0%
$$50000$$
Explanation
Number of cars manufacturd by $$X$$ in $$2001 = 120000$$
Number of cars manufactured by $$ X$$ in $$2002=80000$$
The difference is $$120000-80000=40000$$
If the mean of the data : $$7,8,9,7,8,7 , \lambda , 8$$
is $$8 ,$$ then the variance of this data is :
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0%
$$\dfrac { 7 } { 8 }$$
0%
$$1$$
0%
$$\dfrac { 9 } { 8 }$$
0%
$$2$$
Study the bar graph and answer the question given below:
What was the percent increase in production of oil in $$2006$$ compared to that in $$2012$$?
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0%
$$100\%$$
0%
$$200\%$$
0%
$$300\%$$
0%
$$400\%$$
Explanation
Percentage we can get easily by getting the difference between two and then by finding that difference is what percent of the initial value.
So Percentage $$= \dfrac{60 - 15}{15}\times 100\%$$
$$= 300%$$
So there was an increase of $$300\%$$ from the year $$2006$$ and $$2012$$
Study the bar graph and answer the question given below:
The average production of $$2007$$ and $$2009$$ was exactly to the average production of which of the following pair of the years?
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0%
$$2007$$ and $$2006$$
0%
$$2007$$ and $$2008$$
0%
$$2009$$ and $$2011$$
0%
$$2011$$ and $$2012$$
Explanation
Average production of $$2007$$ and $$2009$$ (in $$1000$$ tonnes) was
$$\dfrac{30 + 40}{2}= 35$$
Now for other given years it was
Option $$A$$ for $$2007$$ and $$2006 = \cfrac{30 + 15}{2}= 22.5$$
Option $$B$$ for $$2007$$ and $$2008 = \cfrac{30 + 20}{2}= 25$$
Option $$C$$ for $$2009$$ and $$2011 = \cfrac{40 + 30}{2}= 35$$
Option $$A$$ for $$2011$$ and $$2012 = \cfrac{30 + 60}{2}= 45$$
So it is equal to the year $$2009$$ and $$2011$$
Study the bar graph and answer the question given below:
What was the percentage increase in the production of oil from $$2009$$ to $$2010$$?
Report Question
0%
$$10\%$$
0%
$$15\%$$
0%
$$20\%$$
0%
$$25\%$$
Explanation
Percentage we can get easily by getting the difference between two and then by finding that difference is what percent of the initial value.
So Percentage $$= \dfrac{50- 40}{40}\times 100\%$$
$$= 25\%$$
So there was increase of $$25\%$$ from the year $$2009$$ and $$2010$$
Find the mean for the following data using step deviation method.
Report Question
0%
$$15.833$$
0%
$$16.833$$
0%
$$17.833$$
0%
$$18.833$$
Explanation
Answer:- By shortcut Method
Class interval width (i) = $$5-0 = 5$$
X
C. I.
mid point
F
$$d=\cfrac{X-A}{i}$$
Fd
0-5
2.5
4
-1
-4
5-10
7.5=A
8
0
0
10-15
12.5
12
1
12
15-20
17.5
16
2
32
20-25
22.5
20
3
60
$$\Sigma F = 60$$
$$\Sigma Fd=100$$
Mean = $$A+\cfrac{\Sigma fd}{\Sigma f} \times i=7.5+\left(\cfrac{100}{60} \times 5\right)={7.5+8.33}=15.83$$
A) $$15.83$$
If the interest on loans amounted to Rs. $$2.5$$ lakhs then the total amount of expenditure on advertisement, transport and infrastructure is $$5.625$$ lakhs.
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0%
True
0%
False
Explanation
Let the total amount of expenditure on advertisement, transport and infrastructure is $$x$$.
Given, the expenditure on loans= 2.5 lakhs
So the ratio of
total expenditure on advertisement, transport and infrastructure to the
expenditure on loans
$$=$$ $$\dfrac{x}{2.5}$$
Also the ratio of
total expenditure on advertisement, transport and infrastructure to the
expenditure on loans according to the bar graph
$$=$$ $$\dfrac{25+10+10}{20}$$
$$=$$ $$\dfrac{45}{20}$$
$$\therefore$$
$$\dfrac{x}{2.5}$$=
$$\dfrac{45}{20}$$
$$\Rightarrow x=\dfrac{2.5\times 45}{20}=5.625$$ lakhs
So, t
he
total amount of expenditure on advertisement, transport and infrastructure is
$$= 5.625$$ lakhs
The ratio of the total expenditure on R & D and transport to the total expenditure on infrastructure and salary is $$17:30$$
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0%
True
0%
False
Explanation
Let the total amount of expenditures be $$x$$.
Then, the total expenditure on R & D and transport:
$$=$$ $$\dfrac{24+10}{10+14+24+10+25+20+40}\times x$$
$$= \dfrac{34 x}{143}$$
And total expenditure on infrastructure and salary:
$$=$$ $$\dfrac{10+40}{10+14+24+10+25+20+40}\times x$$
$$= \dfrac{50 x}{143}$$
Required ratio $$=$$ $$\dfrac{\dfrac{34x}{1143}}{\dfrac{50x}{143}}$$
$$=$$ $$\dfrac{17}{25}$$
The total amount of expenditures of the company is $$10.21$$ times of expenditure on Taxes
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0%
True
0%
False
Explanation
Let the total expenditures be $$x$$.
Then, the expenditure on Taxes
$$=$$ $$\dfrac{14}{10+14+24+10+25+20+40}\times x$$
$$=$$ $$\dfrac{14}{143}\times x$$
Ratio of the total expenditure to the expenditure on Taxes:
$$=$$ $$\dfrac{x}{x\times 0.0979}$$
$$= 10.214$$
Then, the total expenditure is $$10.21$$ times the expenditure of Taxes.
Study the above graph and calculate
the average numbers of cars manufactured by Company $$X$$ over the given period.
Report Question
0%
$$200{,}000$$
0%
$$100{,}000$$
0%
$$300{,}000$$
0%
$$400{,}000$$
Explanation
Average number of cars manufactured by Company $$X$$
$$=\cfrac{1}{6}[100000 + 80000 + 60000 + 120000 + 80000 + 160000]$$
$$=\cfrac{1}{6}[600000]$$
$$= 100,000$$
Study the above graph and calculate the average numbers of cars manufactured by Company $$Y$$ over the given period.
Report Question
0%
$$96{,}666.66$$
0%
$$86{,}666.66$$
0%
$$76{,}666.66$$
0%
$$66{,}666.66$$
Explanation
Average number of cars manufactured by Company $$Y$$
$$=\cfrac{1}{6}[80000 + 60000 + 100000 + 80000 + 120000 + 140000]$$
$$= \cfrac{1}{6}[580000]$$
$$= 96{,}666.66$$
for two data sets,each of size 5, the variances are given to be 4 and 5 andthe correspondings means are given to be 2 and 4 respectively. the variances of the combined data set is
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0%
11/2
0%
6
0%
13/2
0%
5/2
How many bottles of pickles were exported on Friday and Tuesday altogether?
Report Question
0%
$$4500$$
0%
$$1500$$
0%
$$5000$$
0%
$$5500$$
Explanation
$$1$$ pickle jar $$= 500$$ bottles
Friday $$= 6 \times 500 = 3000$$
Tuesday $$= 5 \times 500 = 2500$$
Altogether $$= 3000 + 2500 = 5500$$.
Subtracting two of the officers earnings, the difference is $$1800$$. The name of officers are
Report Question
0%
Amit and Bala
0%
Frank and Thomas
0%
Thomas and Amit
0%
Amit and Madesh
Explanation
Given: Subtracting two of the officers earnings, the difference is $$1800$$.
Amit $$= 6$$
Thomas $$= 2$$
So, the difference $$= 6 - 2 = 4$$
1 money bag $$= 450$$
Then, $$4 \times 450 = 1800$$
Therefore, the two officers are Thomas and Amit.
How much money did Frank earn in a month?
Report Question
0%
$$2450$$
0%
$$900$$
0%
$$4050$$
0%
$$3400$$
Explanation
From the figure, we cay say
Frank earnings in a month is $$= 9 \times 450 = 4050$$
If the expenditure on advertisement is $$5.5$$ lakhs then the difference between the transport and R & D is $$3.8$$ lakhs
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0%
True
0%
False
Explanation
Let the total expenditures be $$x$$.
Then, the expenditure on advertisement
$$=$$ $$\dfrac{25}{10+14+24+10+25+20+40}\times x$$
$$=$$ $$\dfrac{25}{143}\times x$$
$$ = 5.5$$ lakhs
$$\Rightarrow x=\dfrac{5.5\times 143}{25}=31.46$$ lakhs
So, the difference between the expenditures on transport and R&D is,
$$= \dfrac{24 - 10}{143}\times 31.46$$ lakhs
$$= 3.08$$ lakhs
His earning is one-half of what Bala earns. Who is he?
Report Question
0%
Amit
0%
Madesh
0%
Thomas
0%
Frank
Explanation
Bala earnings $$= 4 \times 450 = 1800$$
Likewise Thomas earnings $$= 2 \times 450 = 900$$
So, Thomas is the officers who earns one-half of what Bala earns.
Therefore, Thomas is the correct officer.
Using the line graph, find the average income of sportsman during the period $$2000-2003$$.
Report Question
0%
$$187.5 $$ crores
0%
$$287.5$$ crores
0%
$$387.5$$ crores
0%
$$487.5$$ crores
Explanation
Average income os sportsman from the year $$2000-2003 =$$ $$\dfrac{210 + 440 + 300 + 200}{4}$$
$$=$$ $$\dfrac{1150}{4}$$
$$= 287.5$$ crores
The number of washing machines purchased in the past $$35$$ weeks is given in a line graph. Find the total number of washing machines purchased in $$20$$ weeks.
Report Question
0%
$$140$$
0%
$$150$$
0%
$$160$$
0%
$$170$$
Explanation
The total number of washing machines purchased in $$20$$ weeks $$= 10 + 20 + 40 + 90 = 160$$
How many books were lent in first $$3$$ days?
Report Question
0%
$$15$$
0%
$$5$$
0%
$$200$$
0%
$$225$$
Explanation
First $$3$$ days books were lent $$= 6 + 5 + 4 = 15$$
$$1$$ book $$= 15$$ books
So, $$15 \times 15 = 225$$
How much more money did Frank earn than Amit?
Report Question
0%
$$1350$$
0%
$$1200$$
0%
$$1690$$
0%
$$1540$$
Explanation
Frank earnings $$= 9 \times 450 = 4050$$
Amit earnings $$= 6 \times 450 = 2700$$
So, the difference $$= 4050 - 2700$$
$$= 1350$$
How many books were checked out on Friday?
Report Question
0%
$$75$$`
0%
$$90$$
0%
$$60$$
0%
$$6$$
Explanation
Books were checked out on Friday $$= 6 \times 15 = 90$$ books ($$1$$ book $$= 15$$ books)
Which day fewest books were checked out?
Report Question
0%
Friday
0%
Tuesday
0%
Saturday
0%
Wednesday
Explanation
From the pictograph, on Saturday, fewest books were checked out.
How many fewer books were checked out on Saturday than Tuesday?
Report Question
0%
$$15$$
0%
$$30$$
0%
$$45$$
0%
$$10$$
Explanation
Tuesday $$= 5$$
Saturday $$= 3$$
So, the difference is $$5 - 3 = 2 \times 15 = 30$$ books were checked out on Tuesday than Saturday.
How much less money did Thomas earn than Bala?
Report Question
0%
$$1800$$
0%
$$900$$
0%
$$1000$$
0%
$$1200$$
Explanation
Thomas earnings $$= 2 \times 450 = 900$$
Bala earnings $$= 4 \times 450 = 1800$$
So, the difference $$= 1800 - 900$$
$$= 900$$
In which day, where the number of checkouts are $$75$$ books?
Report Question
0%
Thursday
0%
Friday
0%
Tuesday
0%
Wednesday
Explanation
In figure, there are total $$5$$ book on Tuesday.
$$\therefore$$ on Tuesday the number of checkouts $$= 5 \times 15 = 75$$ books.
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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