5 Mark Question - Maths STD 9 Questions - Vidyadip

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5 Mark Question

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24 questions · self-marked practice — reveal the answer and mark yourself.

Question 15 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:
$45,47, 50, 52, x, JC+2, 60, 62, 63, 74.$
$\therefore$ Number of observations $(n) = 10$ (i.e., even)
$\therefore$ Median is the average of middle two observations
Here, the 5th and 6th numbers are in the middle position.
$\therefore \text { Median }=\frac{(x)+(x+2)}{2}$
$\therefore 53=\frac{2 x+2}{2}$
$\therefore 106 = 2x + 2$
$\therefore 106 – 2 = 2x$
$\therefore 104 = 2x$
$\therefore x = 52$
\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 25 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 35 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 45 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 55 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 65 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 75 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 85 Marks

In the neighbouring figure, percentage bar-diagram is given. Percentage expenses on different items of two families are given.
Answer the following questions based on it :
(i) Write the percentage expenses of every component for each family.
(ii) Which family spends more percent of expenses on food as compared to the other and by how much ?
(iii) What are the percentage expenses on other items ?
(iv) Which family shows more percentage expenses on electricity?
(v) Which family's percentage expense is more on education ?

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Question 95 Marks

The mean of 35 observations is 20, out of which mean of first 18 observations is 15 and mean of last 18 observations is 25. Find the 18th observation.

Answer

Mean $=\frac{\text { The sum of all observations }}{\text { Total number of observations }}$
$\therefore$ The sum of all observations
$=$ Mean $\times$ Total number of observations
The mean of 35 observations is 20
$\therefore$ Sum of 35 observations $=20 \times 35=700$,..(i)
The mean of first 18 observations is 15
Sum of first 18 observations $=15 \times 18$
$=270 \ldots \text {...(ii) }$
The mean of last 18 observations is 25 Sum of last 18 observations $=25 \times 18$
$=450 \ldots \text { (iii) }$
$\therefore 18^{\text {th }}$ observation $=$ (Sum of first 18 observations + Sum of last 18 observations) - (Sum of 35 observations)
$=(270+450)-(700) \ldots[\text { From (i), (ii) and (iii) }]$
$=720-700=20$
The $18^{\text {th }}$ observation is 20 .

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Question 105 Marks

The calculated mean of $50$ observations was $80.$ It was later discovered that observation $19$ was recorded by mistake as $91.$ What Was the correct mean?

Answer

Here, mean = 80, number of observations $= 50$
Mean $=\frac{\text { The sum of all observations }}{\text { Total number of observations }}$
$\therefore$ The sum of all observations $=$ Mean $\times$ Total number of observations
$\therefore$ The sum of 50 observations $= 80 \times 50$
$= 4000$
One of the observation was $19.$ However, by mistake it was recorded as $91.$
Sum of observations after correction $=$ sum of $50$ observation $+$ correct observation – incorrect observation
$= 4000 + 19 – 91$
$= 3928$
$\therefore$ Corrected mean
$=\frac{\text { Sum of observations after correction }}{\text { Total number of observations }}$
$=\frac{3928}{50}$
$= 78.56$
$\therefore$ The corrected mean is $78.56.$

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Question 115 Marks

The following information is regarding marks in mathematics, obtained out of 40, scored by 50 students of 9th std. ¡n the first unit test.

From the table, fill in the blanks in the following statements.
i. For class interval 10 – 20 the lower class limit is _____ and upper class limit is _____
ii. How many students obtained marks less than 10? 2
iii. How many students obtained marks less than 20? 2 + ____ = 14
iv. How many students obtained marks less than 30? ______ + _____ = 34
v. How many students obtained marks less than 40? ______ + ______ =50

Answer

i. 10, 20
iii. 12
iv. 14 + 20
v. 34 + 16

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Question 125 Marks

Using the data In example (3) above, prepare less than type cumulative frequency table and answer the following questions.
i. How many students obtained less than 40 marks?
ii. How many students obtained less than 10 marks?
iii. How many students obtained less than 60 marks?
iv. Find the cumulative frequency of the class 50 – 60.

Answer

i. 24 students obtained less than 40 marks.
ii. 3 students obtained less than 10 marks.
iii. 43 students obtained less than 60 marks.
iv. Cumulative frequency of the class 50 – 60 is 43.

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Question 135 Marks

The data is given for 62 students in a certain class regarding their mathematics marks out of 100. Take the classes 0 – 10, 10 – 20,… and prepare frequency distribution
table and cumulative frequency table more than or equal to type.
55. 60, 81, 90, 45, 65, 45, 52, 30, 85, 20, 10,
75, 95, 09, 20, 25, 39, 45, 50, 78, 70, 46, 64,
42. 58. 31, 82, 27, 11, 78, 97, 07, 22, 27, 36,
35, 40, 75, 80, 47, 69, 48, 59, 32, 83, 23, 17,
77, 45, 05, 23, 37, 38, 35, 25, 46, 57, 68, 45.
47,49
From the prcparcd table, answer the following questions :
i. How many students obtained marks 40 or above 40?
ii. How many students obtained marks 90 or above 90?
iii. How many students obtained marks 60 or above 60?
iv. What is the cumulative frequency of equal to or more than type of the class 0 – 10?

Answer

i. 38 students obtained marks 40 or above 40.
ii. 3 students obtained marks 90 or above 90.
iii. 19 students obtained marks 60 or above 60.
iv. Cumulative frequency of equal to or more than type of the class 0 – 10 is 62.

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Question 145 Marks

Complete the following Cumulative Frequency Table:

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Question 155 Marks

Complete the following cumulative frequency table:

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Question 165 Marks

For the above Question prepare Frequency Distribution Table.

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Question 175 Marks

The record of marks out of 20 in Mathematics in the first unit test is as follows:
20,6, 14, 10, 13, 15, 12, 14, 17. 17, 18, 1119,
9, 16. 18, 14, 7, 17, 20, 8, 15, 16, 10, 15, 12.
18, 17, 12, 11, 11, 10, 16, 14, 16, 18, 10, 7, 17,
14, 20, 17, 13, 15, 18, 20, 12, 12, 15, 10
Answer the following questions, from the above information.
a. How many students scored 15 marks?
b. How many students scored more than 15 marks?
c. How many students scored less than 15 marks?
d. What is the lowest score of the group?
e. What is the highest score of the group?

Answer

a. 5 students scored 15 marks.
b. 20 students scored more than 15 marks.
c. 25 students scored less than 15 marks.
d. 6 is the lowest score of the group.
e. 20 is the highest score of the group.

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Question 185 Marks

In the tables given below, class-mark and frequencies is given. Construct the frequency tables taking inclusive and exclusive classes.

Answer

Let the lower class limit and upper class limit of the class mark 22 be x andy respectively.
$\text { Class mark }=\frac{\text { Lower class limit }+ \text { Upper class limit }}{2}$
$\therefore \quad 22=\frac{x+y}{2}$
$\therefore x + y = 44 …(i)$
Here, class width $= 24 – 22 = 2$
But, Class width = Upper class limit – Lower class limit
$\therefore y – x = 2$
$\therefore -x + y = 2 …. (ii)$
Adding equations (i) and (ii),
$x + y = 44$
$– x + y= 2$
$2y = 46$
$\therefore y = 23$
Substituting $y = 23$ in equation (i),
$\therefore x + 23 = 44$
$\therefore x = 21$
$\therefore$ class with class-mark $22$ is $21 – 23$
Similarly, we can find the remaining classes
$\therefore$ frequency table taking inclusive and exclusive classes.

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Question 195 Marks

In the tables given below, class-mark and frequencies is given. Construct the frequency tables taking inclusive and exclusive classes.

Answer

Let the Lower class limit and upper class limit of the class mark 5 be x and y respectively.
$\text { Class mark }=\frac{\text { Lower class limit }+ \text { Upper class limit }}{2}$
$\therefore \quad 5=\frac{x+y}{2}$
∴x + y = 10
Here, class width = 15 – 5 = 10
But, Class width = Upper class limit – Lower class limit
∴ y – x = 10
∴ -x + y = 10 …(ii)
Adding equations (i) and (ii),
x+ y = 10
-x + y = 10
∴ 2y = 20
∴ y = 10
Substituting y = 10 in equation (i),
∴ x + 10 = 10
∴ x = 0
∴ class with class-mark 5 is 0 – 10
Similarly, we can find the remaining classes.
∴ frequency table taking inclusive and exclusive classes.

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Question 205 Marks

For the above given activity, the information of number of girls per 1000 boys is given for five states. The literacy percentage of these five states is given below. Assam (73%), Bihar (64%), Punjab (77%), Kerala (94%), Maharashtra (83%). Think of the number of girls and the literacy percentages in the respective states. Can you draw any conclusions from it?

Answer

By observing the number of girls per 1000 boys and literacy percentages in the given respective states, we can conclude that the literacy rate of girls is least in Bihar and is highest in Kerala.

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Question 215 Marks

In the following table, the information of number of girls per 1000 boys is given for different states. Fill In the blanks and complete the table.

Answer

Draw percentage bar diagram from this information and discuss the findings from the diagram.

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Question 225 Marks

A farmer has produced Wheat and Jowar in his field. The following joint bar diagram shows the production of Wheat and Jowar. From the gken diagram answer the following questions:
i. Which crop production has increased consistently in 3 years?
ii. By how many quintals the production ofjowar has reduced in 2012 as compared to 2011?
iii. What is the difference between the production of wheat in 2010 and 2012 ?
iv. Complete the following table using this diagram.

Answer

i. The crop production of wheat has increased consistently in 3 years.
ii. The production of jowar has reduced by 3 quintals in 2012 as compared to 2011.
iii. The difference between the production of wheat in 2010 and 2012 = 48 – 30 = 18 quintals
iv.

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Question 235 Marks

In the table given below, the information is given about roads. Using this draw sub-divided and percentage bar diagram (Approximate the percentages to the nearest integer)

Answer

i. Sub-divided bar diagram:

ii. Percentage bar diagram:

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Question 245 Marks

The following table shows the number of Buses and Trucks in nearest lakh units. Draw percentage bar diagram. (Approximate the percentages to the nearest integer)

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