3 Marks Question

Question 13 Marks

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

Years	1995	1996	1997	1998
English	350	400	450	620
Hindi	500	525	600	650

Draw a double bar graph and answer the following questions:
(i) In which year, was the difference in the sale of the two language books least?
(ii) Can you say that the demand for English books rose faster? Justify.

Answer

Double bar graph of given data is as follows

(i) The difference in the scale of the two language books in individual year is as follows

Years	Difference
1995	500 - 350 = 150
1996	525 - 400 = 125
1997	600 - 450 = 150
1998	650 - 620 = 30

Clearly, the difference in the scale of the two language books was least in the year 1998.
(ii) Yes, because the bar graph of the sale of English books became longer faster, compared to Hindi books. So, the demand for English books rose faster.

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Question 23 Marks

The bar graph in figure shows the result of a survey to test water resistant watches made by different companies. Each of these companies claimed that thelr watches were water resistant. After a test, the above results were revealed.
(I) Can you work out a fraction of the number of watches that leaked to the number tested for each company?
(ii) Could you tell on this basis which company has better watches?

Answer

(i) For Company A Number of tested watches $=40$
Number of leaked watches $=20$
Now, required fraction $=\frac{\text { Leaked watches }}{\text { Tested watches }}=\frac{20}{40}=\frac{1}{2}$
[dividing numerator and denominator by 10 ]
For Company B Number of tested watches $=40$
Number of leaked watches $=10$
$\therefore$ Required fraction $=\frac{10}{40}=\frac{1}{4}$
[dividng numerator and denominator by 10 ]
For Company C Number of tested watches $=40$
Number of leaked watches $=15$
$\therefore$ Required fraction $=\frac{15}{40}=\frac{3}{8}$
[dividing numerator and denominator by 5 ]
For Company D Number of tested watches $=40$
Number of leaked watches $=25$
$\therefore$ Required fraction $=\frac{25}{40}=\frac{5}{8}$
[dividing numerator and denominator by 5]
(ii) From above discussion, it is clear that Company $B$ has better watches because Company $B$ has least fraction of the number of watches that leaked to the number of watches that were tested.

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Question 33 Marks

Find the mean of your sleeping hours during one week.

Answer

We know that there are seven days in a week. Let the sleeping hours during a week are as follows:

Days	Hours
Monday	8 h
Tuesday	7 h
Wednesday	6 h
Thursday	7 h
Friday	6 h
Saturday	7 h
Sunday	8 h

Now, sum of the sleeping hours of one week
=8+7+6+7+6+7+8=49 h"
and number of days in a week =7
$
\therefore \quad \text { Mean }=\frac{\text { Total sleeping hours }}{\text { Total number of days }}=\frac{49}{7}=7 h
$
Hence, the mean of your sleeping hours during one week is 7 h .

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Question 43 Marks

How would you find the average of your study hours for the whole week?

Answer

We know that there are seven days in a week. Let the study hours for different days of the week are as follows:

Days	Hours
Monday	6 h
Tuesday	5 h
Wednesday	7 h
Thursday	6 h
Friday	5 h
Saturday	5 h
Sunday	7 h

Total number of study hours
$
=6+5+6+7+6+5+7=42 h
$
Number of days in a week $=7$
$
\therefore \quad \text { Average }=\frac{\text { Total study hours }}{\text { Total number of days }}=\frac{42}{7}=6 h
$
Hence, the average study hours is 6 h per day.

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Question 53 Marks

The performance ofastudent in 1st term and 2nd term is given. Draw a double bar graph choosing appropriate scale and answer the following

Subjects	English	Hindi	Maths	Science	S. Science
1st term (MM 100)	67	72	88	81	73
2nd term (MM 100)	70	65	95	85	75

(i) In which subject has the child improved his performance the most?
(ii) In which subject is the improvement the least?
(iii) Has the performance gone down in any subject?

Answer

The double bar graph is shown below:

(i) We have, marks increased in English = 70-67=3
Marks increased in Hindi = 65 – 72 = -7
Marks increased in Maths = 95 – 88 =7
Marks increased in Science = 85 - 81=4
Marks increased in S. Science =75 - 73 = 2
Hence, the child improved his performance in Maths the most, i.e. 7 marks.
(ii) From part (i), it is clear that the child has improved his performance the least in S. Science, i.e. 2 marks.
(iii) Yes, the performance has gone down in Hindi, i.e. -7 marks.

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Question 63 Marks

Number of children in six different classes are given below. Represent the data on a bar graph.

Class	Fifth	Sixth	Seventh	Eighth	Ninth	Tenth
Number of children	135	120	95	100	90	80

(i) How would you choose a scale?
(ii) Answer the following questions
(a) Which class has the maximum number of children and the minimum?
(b) Find the ratio of students of class sixth to the students of class eighth.

Answer

(i) We start the scale from 0 . The greatest value of data is 135, so end the scale at a value greater than 135 such as 140. Use equal division along axes such as increment of 10 .
Here, we take 1 unit for 10 children.

(ii) (a) From the graph, maximum number of children in class fifth i.e. 135 and minimum number of children in class tenth i.e. 80 .
(b) From the graph,
Number of students in sixth class $=120$
Number of students in eighth class $=100$
$\therefore$ The required ratio
$
\begin{array}{l}
=\frac{\text { Number of students in sixth class }}{\text { Number of students in eighth class }} \\
=\frac{120}{100}=\frac{6}{5} \text { or } 6: 5
\end{array}
$

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Question 73 Marks

The runs scored in a cricket match by 11 players are as follows :
6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15.
Find the mean, mode and median of this data. Are the three same?

Answer

We have,
$
6,15,120,50,100,80,10,15,8,10,15
$
On arranging the data in ascending order, we get
$
6,8,10,10,15,15,15,50,80,100,120
$
Now, sum of the runs $=6+8+10+10+15+15+15$
$
+50+80+100+120=429
$
$\because $ Number of players $=11$
$
\therefore \text { Mean }=\frac{\text { Sum of the runs }}{\text { Number of players }}=\frac{429}{11}=39
$
From above, ascending order of runs is as follows
$
6,8,10,10,15,15,15,50,80,100,120 .
$
Here, 15 occurs more frequently i.e. 3 times.
$
\therefore \text { Mode }=15
$
The value of the middle observation is 15 .
$
\therefore \text { Median }=15
$
Therefore, mean, mode and median are not same.

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Question 83 Marks

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

Day	Mon	Tue	Wed	Thurs	Fri	Sat	Sun
Rainfall (in mm)	0.0	12.2	2.1	0.0	20.5	5.5	1.0

(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?

Answer

(i) From the given data, it is clear that highest rainfall was on Friday i.e. 20.5 mm and lowest rainfall was on Monday and Thursday i.e, 0.0 mm.
$\therefore$Range of rainfall
= Highest rainfall - Lowest rainfall
= Rainfall on Friday - Rainfall on Monday or Thursday
= 20.5 -0.0= 20.5
(ii) Here, total rainfall
= 0.0 +12.2 + 2.1 +0.0 + 20.5 +5.5 +1.0
= 41.3 mm
Number of days = 7
$\therefore$Mean rainfall=$\frac{ Total rainfall}{Number of days}$=$\frac{41.3}{7}$=5.9 mm
Hence, the mean rainfall for the week is 5.9 mm.
(iii) The mean rainfall for the week is 5.9 mm. It is clear from the table that the rainfall on Monday,
Wednesday, Thursday, Saturday and Sunday was less than mean rainfall.
Hence, on 5 days rainfall was less than the mean rainfall.

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Question 93 Marks

The marks (out of 100 ) obtained by a group of students in a Science test are 85,76,90,85,39,48,56,95,81 and 75 . Find the
(i) highest and the lowest marks obtained by the students.
(ii) range of the marks obtained.
(iii) mean marks obtained by the group.

Answer

Given, marks obtained by a group of students in a Science test $ 85,76,90,85,39,48,56,95,81,75 $ On rearranging the marks in ascending order, we get $ 39,48,56,75,76,81,85,85,90,95 $
(i) It is clear from the ascending order of marks that highest marks $=95$ and lowest marks $=39$
(ii) Range of the marks obtained
$ \begin{array}{l} =\text { Highest marks }- \text { Lowest marks } \\ =95-39=56 \end{array} $
(iii) We have, total marks
$
\begin{array}{l}
=39+48+56+75+76+81+85+85+90+95 \\
=730
\end{array}
$
Number of students $=10$
$
\begin{aligned}
\therefore \text { Mean marks } & =\frac{\text { Total marks obtained by group }}{\text { Number of students }} \\
& =\frac{730}{10}=73
\end{aligned}
$
Hence, the mean marks obtained by the group is 73 .

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Question 103 Marks

Observe the data and answer the questions that follow $16,15,16,16,8,15,17$.
(i) Which value can be put in the data, so that the mode remains the same?
(ii) Atleast how many and which value(s) must be put into, to change the mode to 15 ?
(iii) What is the least number of values that must be put into, to change the mode to 17 ? Name them.

Answer

Arranging the given data in ascending order, we have
$
8,15,15,16,16,16,17
$
(i) As per the given data, 16 is the mode of data, since it has highest frequency i.e. 3.
Now, if 15 is added to the given data, mode will get changed to 15 and 16 , whereas if any other number i.e. 8,16 or 17 is added, mode will remain same.
(ii) Atleast two 15 's should be added to change the mode to 15 . On adding two 15 's, the frequency of 15 will be maximum i.e. 4 .
(iii) We will have to add atleast three 17 's to change the mode to 17 . On adding three 17 's, the frequency of 17 will be maximum i.e. 4 .

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Question 113 Marks

The runs scored in a cricket match by 11 players are as follows :
$
6,15,120,50,100,80,10,15,8,10,15
$
Find the mean, mode and median of this data. Are the three same?

Answer

We have,
$
6,15,120,50,100,80,10,15,8,10,15
$
On arranging the data in ascending order, we get
$
6,8,10,10,15,15,15,50,80,100,120
$
Now, sum of the runs $=6+8+10+10+15+15+15$
$
+50+80+100+120=429
$
$\because$ Number of players $=11$
$
\therefore \text { Mean }=\frac{\text { Sum of the runs }}{\text { Number of players }}=\frac{429}{11}=39
$
From above, ascending order of runs is as follows
$
6,8,10,10,15,15,15,50,80,100,120 .
$
Here, 15 occurs more frequently i.e. 3 times.
$
\therefore \text { Mode }=15
$
The value of the middle observation is 15 .
$\therefore$ Median $=15$
Therefore, mean, mode and median are not same.

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Question 123 Marks

Use the following bar graph to answer the questions.

$A=$ Orange $B=$ Mango; $C=$ Grapes; $D=$ Apple
(i) Which is the most popular fruit among the students?
(ii) Which is the least popular fruit among the students?
(iii) How many students like orange?

Answer

(i) Clearly, from the bar graph, mango is liked by 10 students. So, the most popular fruit among the students is mango.
(ii) Clearly, from the bar graph, grapes is liked by 4 students. So, the least popular fruit among the students is grapes.
(iii) Orange is liked by 8 students.

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Question 133 Marks

The median of observations $10,12,14,15,16,(x+2)$, $22,24,25,29$ and 30 arranged in ascending order is 26 . Find the value of $x$.

Answer

Given observation in ascending order
$
10,12,14,15,16,(x+2), 22,24,25,29,30
$
From above data, median is $(x+2)$.
$
\begin{aligned}
\therefore 26 & =x+2 \\
\Rightarrow 26-2 & =x \Rightarrow x=24
\end{aligned}
$

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Question 143 Marks

The mean of three numbers is 10. The mean of other four numbers is 12. Find the mean of all the numbers.

Answer

$\begin{array}{l}\text {Mean of three numbers }=\frac{\text { Sum of three numbers }}{3} \\ \Rightarrow \quad 10=\frac{\text { Sum of three numbers }}{3} \\ \text { Hence, sum of three numbers }=30 \\ \text { Also, Mean of other four numbers } \\ =\frac{\text { Sum of other four numbers }}{4} \\ \Rightarrow 12=\frac{\text { Sum of other four numbers }}{4}\end{array}$
Hence, sum of other 4 numbers $=48$
Then,
Mean of all the numbers $=\frac{\text { Sum of all the numbers }}{\text { Total numbers }}$
$=\frac{\left[\begin{array}{l}\text { Sum of first three numbers }+ \\ \text { Sum of other fourth numbers }\end{array}\right]}{7}$
$=\frac{30+48}{7}=\frac{78}{7}=11.14$
Hence, mean of all the numbers is 11.14.

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Question 153 Marks

The run scored by a batsman in 6 cricket matches are as follows 109, 92, 14, 2, 6, 89. Find the average score.

Answer

Batsman's score in 6 innings $109,92,14,2,6,89$. For mean, we have to find the sum of scores.
$
\therefore \text { Mean }=\frac{\text { Sum of observations }}{\text { Number of observations }}
$
Sum of scores $=109+92+14+2+6+89=312$
Average $/$ Mean $=\frac{312}{6}=52$
Hence, average score of the batsman is 52 in 6 cricket matches.

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Question 163 Marks

Find the value of $x$, if the mean of the following data is 36
$
8,11,19,14,13,18,26, x, 2 x, 6 .
$

Answer

Given data $8,11,19,14,13,18,26, x, 2 x, 6$
Sum of given data
$
\begin{array}{l}
=8+11+19+14+13+18+26+x+2 x+6 \\
=115+3 x
\end{array}
$$\begin{array}{l}\because \text { Mean }=\frac{\text { Sum of observations }}{\text { Number of observations }} \\ \Rightarrow 36=\frac{115+3 x}{10} \Rightarrow 36 \times 10=115+3 x \\ \Rightarrow 360-115=3 x \Rightarrow 3 x=245 \Rightarrow x=\frac{245}{3}=81.66\end{array}$

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Question 173 Marks

If the mean of 4 observations x, 2x, 3xand 4x is 10. Then, find the mean of the last three observations.

Answer

$
\begin{array}{l}
\text { Given, } \frac{x+2 x+3 x+4 x}{4}=10 \\
\Rightarrow \quad \frac{10 x}{4}=10 \Rightarrow 10 x=40 \Rightarrow x=4
\end{array}
$
$\therefore$ Mean of the last three observations
$
\begin{array}{l}
=\frac{2 x+3 x+4 x}{3}=\frac{9 x}{3} \\
=3 x=3 \times 4=12
\end{array}
$

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MATHS 3 Marks Question

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