4 Mark Question

Question 14 Marks

There are $10$ observations arranged in ascending order as given below.
$45, 47, 50, 52, JC, JC + 2, 60, 62, 63, 74.$ The median of these observations is $53.$
Find the value of JC. Also find the mean and the mode of the data.

Answer

i. Given data in ascending order:
$45,47, 50, 52, x, JC+2, 60, 62, 63, 74.$
∴ Number of observations $(n) = 10$ (i.e., even)
∴ Median is the average of middle two observations
Here, the 5th and 6th numbers are in the middle position.
$\therefore \quad \text { Median }=\frac{(x)+(x+2)}{2}$
$\therefore \quad 53=\frac{2 x+2}{2}$
$\therefore 106 = 2x + 2$
$\therefore 106 – 2 = 2x$
$\therefore 104 = 2x$
$\therefore x = 52$
$\therefore$ The given data becomes:
$45, 47, 50, 52, 52, 54, 60, 62, 63, 74.$
$\text { ii. Mean }=\frac{\text { The sum of all observations in the data }}{\text { Total number of observations }}$
$=\frac{45+47+50+52+52+54+60+62+63+74}{10}$
$=\frac{559}{10}=55.9$
$\therefore$ The mean of the given data is 55.9.
iii. Given data in ascending order:
$45, 47, 50, 52, 52, 54, 60, 62, 63, 74.$
$\therefore$ The observation repeated maximum number of times $= 52$
$\therefore$ The mode of the given data is $52.$

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Question 24 Marks

By using data in example (6), and taking classes 30 – 40, 40 – 50,… prepare equal to or more than type cumulative frequency table and answer the following questions based on it.
i. How many students have scored marks 70 or more than 70?
ii. How many students have scored marks 30 or more than 30?

Answer

i. 11 students have scored marks 70 or more than 70.
ii. 68 students have scored marks 30 or more than 30.

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Question 34 Marks

There are 68 students of 9th standard from Model Highschool, Nandpur. They have scored following marks out of 80, in written exam of mathematics.
70, 50, 60, 66, 45, 46, 38, 30, 40, 47, 56, 68,
80, 79, 39, 43, 57, 61, 51, 32, 42, 43, 75, 43,
36, 37, 61, 71, 32, 40, 45, 32, 36, 42, 43, 55,
56, 62, 66, 72, 73, 78, 36, 46, 47, 52, 68, 78,
80, 49, 59, 69, 65, 35, 46, 56, 57, 60, 36, 37,
45, 42, 70, 37,45, 66, 56, 47
By taking classes 30 – 40, 40 – 50, …. prepare the less than type cumulative frequency table. Using the table, answer the following questions:

i. How many students have scored marks less than 80?
ii. How many students have scored marks less than 40?
iii. How many students have scored marks less than 60?

Answer

Class

i. 66 students have scored marks less than 80.
ii. 14 students have scored marks less than 40.
iii. 45 students have scored marks less than 60.

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Question 44 Marks

If the mean of the following data is 20.2, then find the value of p.

Answer

$\begin{array}{r}\operatorname{Mean}(\bar{x})=\frac{\sum f_i x_i}{\sum f_i} \\ \therefore \quad 20.2=\frac{610+20 p }{30+ p }\end{array}$
∴ 20.2 (30 + p) = 610 + 20p
∴ 606 + 20.2p = 610 + 20p
∴ 20.2p – 20p = 610 – 606
∴ 0.2p = 4
$\therefore p=\frac{4}{0.2}=\frac{40}{2}=20$
∴ p = 20

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Question 54 Marks

The monthly maximum temperature of a city is given in degree Celsius in the following data. By taking suitable classes, prepare the grouped frequency distribution table
29.2, 29.0, 28.1, 28.5, 32.9, 29.2, 34.2, 36.8, 32.0, 31.0, 30.5, 30.0, 33, 32.5, 35.5, 34.0, 32.9, 31.5, 30.3, 31.4, 30.3, 34.7, 35.0, 32.5, 33.5.29.0. 29.5.29.9.33.2.30.2
From the table, answer the following questions.
i. For how many days the maximum temperature was less than 34°C?
ii. For how many days the maximum temperature was 34°C or more than 34°C?

Answer

i. Number of days for which the maximum temperature was less than 34°C
= 8 + 8 + 8 = 24
ii. Number of days for which the maximum temperature was 34°C or more than 34°C
= 5 + 1 = 6

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Question 64 Marks

The mean of nine numbers is 77. If one more number is added to it, then the mean increases by 5. Find the number added in the data.

Answer

$\therefore$ Mean $=\frac{\text { The sum of all observations }}{\text { Total number of observations }}$
∴ The sum of all observations = Mean x Total number of observations mean of nine numbers is 77
∴ sum of 9 numbers = 11 x 9 = 693 …(i)
If one more number is added, then mean increases by 5
mean of 10 numbers = 77 + 5 = 82
∴ sum of the 10 numbers = 82 x 10 = 820 …(ii)
∴ Number added = sum of the 10 numbers – sum of the 9 numbers = 820 – 693 … [From (i) and (ii)]
= 127
∴ The number added in the data is 127.

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Question 74 Marks

The mean salary of 20 workers is ₹10,250. If the salary of office superintendent is added, the mean will increase by ₹ 750. Find the salary of the office superintendent.

Answer

Mean $=\frac{\text { The sum of all observations }}{\text { Total number of observations }}$
∴ The sum of all observations = Mean x Total number of observations
The mean salary of 20 workers is ₹ 10,250.
∴ Sum of the salaries of 20 workers
= 20 x 10,250
= ₹ 2,05,000 …(i)
If the superintendent’s salary is added, then mean increases by 750
new mean = 10, 250 + 750 = 11,000
Total number of people after adding superintendent = 20 + 1 = 21
∴ Sum of the salaries including the superintendent’s salary = 21 x 11,000 = ₹ 2,31,000 …(ii)
∴ Superintendent salary = sum of the salaries including superintendent’s salary – sum of salaries of 20 workers
= 2, 31,00 – 2,05,000 …[From (i) and (ii)]
= 26,000
∴ The salary of the office superintendent is ₹ 26,000.

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Question 84 Marks

There are 40 students in a class, out of them 15 are boys. The mean of marks obtained by boys is 33 and that for girls is 35. Find out the mean of all students in the class.

Answer

Total number of students = 40
Number of boys =15
∴ Number of girls = 40 – 15 = 25
The mean of marks obtained by 15 boys is 33
Here, sum of the marks obtained by boys
= 33 x 15
= 495 …(i)
The mean of marks obtained by 25 girls is 35 Sum of the marks obtained by girls = 35 x 25
= 875 …(ii)
Sum of the marks obtained by boys and girls = 495 + 875 … [From (i) and (ii)]
= 1370
∴ Mean of all the students

= 34.25
∴ The mean of all the students in the class is 34.25.

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Question 94 Marks

The mean of 5 observations is 50. One of the observations was removed from the data, hence the mean became 45. Find the observation which was removed.

Answer

Mean $=\frac{\text { The sum of all observations }}{\text { Total number of observations }}$
∴ The sum of all observations = Mean x Total number of observations
The mean of 5 observations is 50
Sum of 5 observations = 50 x 5 = 250 …(i)
One observation was removed and mean of remaining data is 45.
Total number of observations after removing one observation = 5 – 1 = 4
Now, mean of 4 observations is 45.
∴ Sum of 4 observations = 45 x 4 = 180 …(ii)
∴ Observation which was removed
= Sum of 5 observations – Sum of 4 observations = 250 – 180 … [From (i) and (ii)]
= 70
∴ The observation which was removed is 70.

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Question 104 Marks

A hockey player has scored following number of goals in $9$ matches: $5, 4, 0, 2, 2, 4, 4, 3,3.$
Find the mean, median and mode of the data.

Answer

i. Given data: $5, 4, 0, 2, 2, 4, 4, 3, 3.$
Total number of observations = $9$
$\begin{aligned} \text { Mean }= & \frac{\text { The sum of all observations in the data }}{\text { Total number of observations }} \\ & =\frac{5+4+0+2+2+4+4+3+3}{9} \\ & =\frac{27}{9}\end{aligned}$∴ The mean of the given data is $3$.
ii. Given data in ascending order:
$0,2, 2, 3, 3, 4, 4, 4,5$
∴ Number of observations(n) = $9$ (i.e., odd)
∴ Median is the middle most observation
Here, the $5^{th}$ number is at the middle position, which is $3$.
∴ The median of the given data is $3$.
iii. Given data in ascending order:
$0,2, 2, 3, 3, 4, 4, 4,5$
Here, the observation repeated maximum number of times = 4
∴ The mode of the given data is $4$.

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Question 114 Marks

If class-mark is 10 and class width is 6, then find the class.

Answer

Let the upper class limit be x and the lower class limit be y.
Class mark = 10 …[Given]
Class-mark
Class-mark
$
\begin{aligned}
= & \frac{\text { Lower class limit }+ \text { Upper class limit }}{2} \\
\therefore \quad 10 & =\frac{x+y}{2}
\end{aligned}
$
∴ x + y = 20 …(i)
Class width = 6 … [Given]
Class width = Upper class limit – Lower class limit
∴ x – y = 6 …(ii)
Adding equations (i) and (ii),
x + y = 20
x – y = 6
2x = 26
∴ x = 13
Substituting x = 13 in equation (i),
13 + y = 20
∴ y = 20 – 13
∴ y = 7
∴ The required class is 7 – 13.

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Maths 4 Mark Question

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