NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability

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NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability

Edited By Ravindra Pindel | Updated on Apr 26, 2023 03:00 PM IST | #CBSE Class 10th

NCERT exemplar Class 10 Maths solutions chapter 13 is provided to students to prepare for CBSE exam. It is an extension to the learnings of statistics and probability done in Class 9. The chapter on Statistics and Probability is of great importance for the examinations as well as future career prospects for a student. All the data analytics and artificial intelligence-related verticals require a sound knowledge of Statistics and Probability. The NCERT exemplar Class 10 Maths chapter 13 solutions are curated by our highly experienced content development team which enables the students to study and practice NCERT Class 10 Maths effectively.

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NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability: Exercise-13.1
Question:1
NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability: Exercise-13.2
Question:1
NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability: Exercise-13.3
Question:1
NCERT Exemplar Solutions Class 10 Maths Chapter 13 Important Topics:
NCERT Class 10 Solutions for Other Subjects:
NCERT Class 10 Maths Exemplar Solutions for Other Chapters:
Features of NCERT Exemplar Class 10 Maths Solutions Chapter 13:

These NCERT exemplar Class 10 Maths chapter 13 solutions follow the CBSE 10 Maths Syllabus Also, in this Chapter 13, there are exercises that revolve around various statistical measures, including mean, mode, and median. Through these exercises, students will gain an understanding of how to solve these types of problems. Additionally, the chapter delves into the topic of cumulative frequency distribution and cumulative frequency curves, providing further explanation and insight.

NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability: Exercise-13.1

Question:1

In the formula $\bar{x}= a+\frac{\sum f_{i}d_{i}}{\sum f_{i}}$
for finding the mean of grouped data di’s are deviations from
(A) lower limits of the classes
(B) upper limits of the classes
(C) mid points of the classes
(D) frequencies of the class marks

Answer:

Answer. [C]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate mean. First of all, add up all the observations and then divide
by the total number of observations.
We know that d_i = x_i – a
where x_i is data and a is mean
So, di are the derivative from mid-point of the classes.

Question:2

While computing mean of grouped data, we assume that the frequencies are
(A) evenly distributed over all the classes
(B) centred at the class marks of the classes
(C) centred at the upper limits of the classes
(D) centred at the lower limits of the classes

Answer:

Answer. [B]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate mean. First of all, add up all the observations and then divide by the total number of observations.
Hence while computing mean of grouped data, we assume that the frequencies are centered at the class marks of the classes

Question:3

If x_i’s are the mid points of the class intervals of grouped data, fi’s are the corresponding frequencies and $\bar{x}$ is the mean, then $\Sigma\left(f_{i} x_{i}-\bar{x}\right)$ is equal to
(A) 0 (B) –1 (C) 1 (D) 2

Answer:

Answer. [A]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate mean. First of all, add up all the observations and then divide by the total number of observations.
That is mean $\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{n}$
By cross multiplication we get
$\sum f_{i}x_{i}= \bar{x}n\cdots (1)$
$\sum \left ( f_{i}x_{i} -\bar{x}\right )= \sum f_{i}x_{i} -\sum \bar{x}$
$= n\bar{x}-\sum \bar{x}$ (from equation (1))
$= n\bar{x}-n\bar{x}$
= 0

Question:4

In the formula $\bar{x}= a+h\left ( \frac{\sum f_{i}u_{i}}{\sum f_{i}} \right )$ for finding the mean of grouped frequency distribution, u_i =
(A) $\frac{x_{i}+a}{h}$ (B) h(xi – a) (C) $\frac{x_{i}-a}{h}$ (D) $\frac{a-x_{i}}{h}$

Answer:

Answer. [C]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate mean. First of all, add up all the observations and then divide by the total number of observations.
Also we know that d_i = x_i – a and $u_{i}= \frac{d_{i}}{h}$
$u_{i}= \frac{d_{i}}{h}$
put d_i = x_i – a
$u_{i}= \frac{x_{i}-a}{h}$
Hence option C is correct

Question:5

The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its
(A) mean (B) median (C) mode (D) all the three above

Answer:

Answer. [B]
Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.
If we make graph of less than type and of more than type grouped data and find the intersection point then the value at abscissa is the median of the grouped data.
Hence option (B) is correct.

Question:6

For the following distribution :

Class	0-5	5-10	10-15	15-20	20-25
Frequency	10	15	12	20	9

the sum of lower limits of the median class and modal class is
(A) 15 (B) 25 (C) 30 (D) 35

Answer:

Class	Frequency	Cumulative Frequency (C.F)
0-5	10	10
5-10	15	(10+15=25)
10-15	12	(25+12=37)
15-20	20	(37+20=57)
20-25	9	(57+9=66)
	N=66

$\frac{N}{2}=\frac{66}{2}=33$ which ies in the class 10-15.
Hence the median class is 10 – 15
The class with maximum frequency is modal class which is 15 – 20
The lower limit of median class = 10
The lower limit of modal class = 15
Sum = 10 + 15 = 25

Question:7

Consider the following frequency distribution:

Class	0-5	6-11	12-17	18-23	24-29
Frequency	13	10	15	8	11

The upper limit of the median class is
(A) 17 (B) 17.5 (C) 18 (D) 18.5

Answer:

Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.
Class in not continuous. So we have to make 1t continuous first.

Class	Frequency	Cumulative Frequency
0-5.5	13	13
5.5-11.5	10	(13+10=23)
11.5-17.5	15	(23+15=38)
17.5-23.5	8	(38+8=46)
23.5-29.5	11	(46+11=57)
	N = 57

$\frac{N}{2}= \frac{57}{2}= 28\cdot 5$
Here the median class is 15.5 – 17.5
upper limit of median class is 17.5

Question:8

For the following distribution :

Marks	Number of students
Below 10 Below 20 Below 30 Below 40 Below 50 Below 60	3 12 27 57 75 80

the modal class is
(A) 10-20 (B) 20-30 (C) 30-40 (D) 50-60

Answer:

Answer. [C]
Solution.
Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to
summarize categorical variables.

Marks	Frequency	Cumulative Frequency
0-10	3	3
10-20	12-3=9	12
20-30	27-12=15	27
30-40	57-27=30	57
40-50	75-57=18	75
50-60	80-75=5	80
	=80

The class with highest frequency is 30-40
Hence 30 – 40 is the modal class.

Question:9

Consider the data :

Class	65-85	85-105	105-125	125-145	145-165	165-185	185-205
Frequency	4	5	13	20	14	7	4

The difference of the upper limit of the median class and the lower limit of the modal class is
(A) 0 (B) 19 (C) 20 (D) 38

Answer:

Answer. [C]

Solution. Frequency distribution: It tells how frequencies are distributed overvalues in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

Class	Frequency	Cumulative Frequency
65-85	4	4
85-105	5	(4+5=9)
105-125	13	(9+13=22)
125-145	20	(22+20=42)
145-165	14	(42+14=56)
165-185	7	(56+7=63)
185-205	4	(63+4=67)
	N = 67

$\frac{N}{2}= \frac{67}{2}= 33\cdot 5$
Median class = 125 – 145
upper limit of median = 145
The class with maximum frequency is modal class which is 125 – 145
lower limit of modal class = 125
Difference of the upper limit of median and lower limit of modal = 145 – 125 = 20

Question:10

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

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

The number of athletes who completed the race in less than 14.6 seconds is :
(A) 11 (B) 71 (C) 82 (D) 130

Answer:

Answer. [C]
Solution. Frequency:- The number of times a event occurs is a specific period is called frequency.
The number of athletes who are below 14.6 = frequency of class (13.8-14) + frequency of class (14- 14.2) +
frequency of class (14.2-14.4) + frequency of class (14.4-14.6)
= 2 + 4 + 5 + 71 = 82
Hence the frequency of race completed in less than 14.6 = 82

Question:11

Consider the following distribution :

Marks obtained

Number of students

More than or equal to 0
More than or equal to 10
More than or equal to 20
More than or equal to 30
More than or equal to 40
More than or equal to 50

63
58
55
51
48
42

the frequency of the class 30-40 is
(A) 3 (B) 4 (C) 48 (D) 51

Answer:

Answer. [A]
Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use
frequency distribution to summarize categorical variables.

Marks obtained	Cumulative Frequency	Frequency
0-10	63	5
10-20	58	3
20-30	55	4
30-40	51	3
40-50	48	6
50-60	42	42

So the frequency of class 30 – 40 is 3.

Question:12

If an event cannot occur, then its probability is
(A) 1 (B) $\frac{3}{4}$ (C) $\frac{1}{2}$ (D) 0

Answer:

Answer. [D]
Solution. Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Here number of favorable cases is 0.
Probability = $\frac{Number \, of\, favourable\, case}{Total\, number\, of\, cases}$
Probability = $\frac{0}{\left ( Total\, cases \right )}= 0$

Question:13

Which of the following cannot be the probability of an event?
(A) $\frac{1}{3}$ (B) 0.1 (C) 3% (D) $\frac{17}{16}$

Answer:

Answer. [D]
Solution. Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
(A) 1/3
Here 0 < 1/3 < 1
Hence it can be the probability of an event.
(B) 0.1
Here 0 < 0.1 < 1
Hence it can be the probability of an event.
(C) 3% = 3/100 = 0.03
Here 0 < 0.03 < 1
Hence it can be the probability of an event.
(D)17/16
Here $\frac{17}{16}> 1$
Hence $\frac{17}{16}$ is not a probability of event
Hence option (D) is correct answer.

Question:14

An event is very unlikely to happen. Its probability is closest to
(A) 0.0001 (B) 0.001 (C) 0.01 (D) 0.1

Answer:

Answer. [A]
Solution. Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
The descending order of option (A), (B), (C), (D) is
0.1 > 0.01 > 0.001 > 0.0001 that is (D) > (C) > (B) > (A)
We can also say that it is the order of happening of an event.
Here 0.0001 it is the smallest one.
Hence 0.0001 is very unlikely to happen

Question:15

If the probability of an event is p, the probability of its complementary event will be
(A) p – 1 (B) p (C) 1 – p (D) $1-\frac{1}{p}$

Answer:

Answer. [C]
Solution. Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
The probability of an event = p
Let the probability of its complementary event = q
We know that total probability is equal to 1.
Hence, p + q = 1
q = 1 – p

Question:16

The probability expressed as a percentage of a particular occurrence can never be
(A) less than 100
(B) less than 0
(C) greater than 1
(D) anything but a whole number

Answer:

Answer. [B]
Solution. Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
The probability expressed as presentage of an event A is btewwen 0 to 100.
Hence we can say that probability can never be less than 0.
Hence option (B) is correct.

Question:17

If P(A) denotes the probability of an event A, then
(A) P(A) < 0 (B) P(A) > 1 (C) 0 ≤ P(A) ≤ 1 (D) –1 ≤ P(A) ≤ 1

Answer:

Answer. [C]
Solution. Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
(A) P(A) < 0
It is not represent the probability of event A because probability of an event can never be less than 0.
(B) P(A) > 1
It is not represent the probability of event A because probability of an event can never be greater than 1.
(C) 0 ≤ P(A) ≤ 1
It represent probability of event A because probability of an event is always lies from 0 to 1.
(D) –1 ≤ P(A) ≤ 1
It is not represent the probability of event A because probability of an event can never be equal to -1.
Hence option (C) is correct .

Question:18

A card is selected from a deck of 52 cards. The probability of its being a red face card is
(A) $\frac{3}{26}$ (B) $\frac{3}{13}$ (C) $\frac{2}{13}$ (D) $\frac{1}{2}$

Answer:

Answer. [A]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total number of cases = 52
Red face cards = 6
Favorable cases = 6
Let event A is to select a card from 52 card.
Probability that it is a red card is p(A)
$P\left ( A \right )= \frac{Number\, of\, favourable\, cases}{Total\, number\, of\, cases}$
$P\left ( A \right )=\frac{6}{52}= \frac{3}{26}$

Question:19

The probability that a non leap year selected at random will contain 53 Sundays is
(A) $\frac{1}{7}$ (B) $\frac{2}{7}$ (C) $\frac{3}{7}$ (D) $\frac{5}{7}$

Answer:

Answer. [A]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
In 365 days there are 52 weeks and 1 day.
If it contain 53 sunday then the 1 day of the year must be sunday.
But there are total 7 days.
Hence total number of favorable cases = 1
Hence probability of 53 sunday = $\frac{Number\, of\, favourable\, cases}{Total\, cases}$
$= \frac{1}{7}$

Question:20

When a die is thrown, the probability of getting an odd number less than 3 is
(A) $\frac{1}{6}$ (B) $\frac{1}{3}$ (C) $\frac{1}{2}$ (D) 0

Answer:

Question:21

A card is drawn from a deck of 52 cards. The event E is that card is not an ace of hearts. The number of outcomes favourable to E is
(A) 4 (B) 13 (C) 48 (D) 51

Answer:

Answer. [D]
Solution. Total number of cards = 52
Ace of hearts = 1
The card is not an ace of hearts = 52 – 1 = 51
The number of outcomes favourabe to E = 51

Question:22

The probability of getting a bad egg in a lot of 400 is 0.035. The number of bad eggs in the lot is
(A) 7 (B) 14 (C) 21 (D) 28

Answer:

Answer. [B]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Let event A is to get a bad egg.
So, p (A) = 0.035 (given)
P(A) = $\frac{Number\, of\, favorable\ cases }{Total\, number\, of\, cases}$
0.035 = $\frac{Number\, of\, favorable\ cases }{400}$
Number of favourable cases = $\frac{35}{1000}\times 400= \frac{140}{10}= 14$

Question:23

A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, how many tickets has she bought?
(A) 40 (B) 240 (C) 480 (D) 750

Answer:

Answer. [C]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total cases = 6000
Probability of getting first prize (p(A)) = 0.08
p(A) $= \frac{Number\, of\, tickets\, the\, bought}{Total\, tickets}$
0.08 × 6000 = Number of tickets the bought
$\frac{8}{100}\times 6000$ = Number of tickets the bought
Number of tickets the bought = 480.

Question:24

One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number which is a multiple of 5 is
(A) $\frac{1}{5}$ (B) $\frac{3}{5}$ (C) $\frac{4}{5}$ (D) $\frac{1}{3}$

Answer:

Answer. [A]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total tickets = 40
Number of tickets multiple of 5 = 5, 10, 15, 20, 25, 30, 35, 40
Total favourable cases = 8
Let A be the event of getting a ticket with number multiple of 5.
p(A) = $\frac{Number\, of\, tickets\, the\, bought}{Total\, tickets}$
$p\left ( A \right )= \frac{8}{40}= \frac{1}{5}$

Question:25

Someone is asked to take a number from 1 to 100. The probability that it is a prime is
(A) $\frac{1}{5}$ (B) $\frac{6}{25}$ (C) $\frac{1}{4}$ (D) $\frac{13}{50}$

Answer:

Answer. [C]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total number of cases = 100
prime number from 1 to 100 = 2, 3, 5, 7, 9, 11, 13, 17, 19,
23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 83, 89, 97
Total prime numbers from 1 to 100 = 25
Probability of getting prime number = $\frac{prime\, no.\, from\, 1\, to\, 100}{Total\, number}$
$\\=\frac{25}{100}\\\\=\frac{1}{4}$

Question:26

A school has five houses A, B, C, D and E. A class has 23 students, 4 from house A, 8 from house B, 5 from house C, 2 from house D and rest from house E. Asingle student is selected at random to be the class monitor. The probability that theselected student is not from A, B and C is
(A) $\frac{4}{23}$ (B) $\frac{6}{23}$ (C) $\frac{8}{23}$ (D) $\frac{17}{23}$

Answer:

Answer. [B]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a
random event. The value is expressed from zero to one.
Total students = 23
Students in A, B, C = 4 + 8 + 5 = 17
Students in C, D = 23 – 17 = 6
Number of favourable cases = 6
Let A be the event that the student is not from A, B, C
$p\left ( A \right )= \frac{Student\, from\, C,D}{Total\, students}$
$p\left ( A \right )= \frac{6}{23}$

NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability: Exercise-13.2

Question:1

The median of an ungrouped data and the median calculated when the same data is grouped are always the same. Do you think that this is a correct statement? Give reason.

Answer:

Answer. [False]
Solution. Grouped data are data formed by aggregating individual observations of a variable into groups.
Ungrouped data:- The data which is not grouped is called ungrouped data.
The median is the middle number in the grouped data but when data is ungrouped the median is also changed.
Hence the median is not same of grouped and ungrouped data

Question:2

In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula
$\bar{x}= a+\frac{\sum f_{i}d_{i}}{\sum f_{i}}$ where a is the assumed mean. a must be one of the mid-points of the classes. the last statement correct? Justify your answer.

Answer:

Answer. [False]
Solution. Grouped data are data formed by aggregating individual observations of a variable into groups.
Mean : It is the average of the given numbers. It is easy
to calculate mean. First of all add up all the numbers then divide by how many numbers are there.
This last statement is not correct because a can be any point in the grouped data it is not necessary that a must be mid-point.
Hence the statement is false.

Question:3

Is it true to say that the mean, mode and median of grouped data will always be different? Justify your answer.

Answer:

Answer. [False]
Solution. Mean : It is the average of the given numbers. It is easy to calculate mean. First of all add up all the numbers then divide by how many numbers are there.
Grouped data are data formed by aggregating individual observations of a variable into groups.
The mean, mode and median of grouped data can be the same it will depend on what type of data is given.
Hence the statement is false.

Question:4

Will the median class and modal class of grouped data always be different? Justify your answer.

Answer:

Answer. [False]
Solution. Grouped data are data formed by aggregating individual observations of a variable into groups.
The median is always the middle number and the modal class is the class with highest frequency it can be happen that the median class is of highest frequency.
So the given statement is false median class and mode class can be same.

Question:5

In a family having three children, there may be no girl, one girl, two girls or three girls. So, the probability of each is $\frac{1}{4}$ . Is this correct? Justify your answer.

Answer:

Answer. [False]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total children = 3
Cases – GGG, GGB, GBG, BGG, BBB, BBG, BGB, GBB were G is girl and B is boy.
Probability = $\frac{Number\, of\, favorable\ cases, }{Total\, number\, of\, cases}$
Probability of 0 girl = $\frac{1}{8}$
Probability of 1 girl = $\frac{3}{8}$
Probability of 2 girl = $\frac{3}{8}$
Probability of 3 girl = $\frac{1}{8}$
Here they are not equal to $\frac{1}{4}$

Question:6

A game consists of spinning an arrow which comes to rest pointing at one of the regions (1, 2 or 3) (Fig.). Are the outcomes 1, 2 and 3 equally likely to occur? Give reasons

Answer:

Answer. [False]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Here 3 contain 50% of the region and 1, 2, contain 25%, 15% of the region.
$Probability= \frac{Number\, of\, favorable\ cases }{Total\, number\, of\, cases}$
$Probability\, of\, 3= \frac{50}{100}= \frac{1}{2}$
$Probability\, of\, 1= \frac{25}{100}= \frac{1}{4}$
$probability\, of\, 2= \frac{25}{100}= \frac{1}{4}$
All probabilities are not equal. So the given statement is false.

Question:7

Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has the better chance of getting the number 36? Why

Answer:

Answer. [Peehu]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
As apoorv throws two dice total cases = 36
Product is 36 when he get = (6, 6)
Number of favourable cases = 1
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that Apoorv get 36 = $\frac{1}{36}$
Peehu throws are die total cases = 6
Square of 6 is 36
Hence case = 1
Probability that Peehu get 36 = $\frac{1}{6}$
Hence Peehu has better cases to get 36.

Question:8

When we toss a coin, there are two possible outcomes - Head or Tail. Therefore, the probability of each outcome is $\frac{1}{2}$ . Justify your answer.

Answer:

Answer. [True]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total cases when we toss a coin = 2(H, T)
Probability = $\frac{Number\, of\, favourable\ cases, }{Total\, number\, of\, cases}$
Probability of head = $\frac{1}{2}$
Probability of tail = $\frac{1}{2}$
Hence the probability of each outcome is $\frac{1}{2}$ .

Question:9

A student says that if you throw a die, it will show up 1 or not 1. Therefore, the probability of getting 1 and
the probability of getting ‘not 1’ each is equal to $\frac{1}{2}$ . Isthis correct? Give reasons.

Answer:

Answer. [False]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Here total cases = 6
Number of favourable cases in getting 1 = 1
Probability = $\frac{Number\, of\, favourable\ cases, }{Total\, number\, of\, cases}$
Probability of getting $1= \frac{1}{6}$
Number of favourable cases 'not 1' = 5 (2, 3, 4, 5, 6)
Probability of not 1 = $\frac{5}{6}$
Hence they are not equal to $\frac{1}{2}$

Question:10

I toss three coins together. The possible outcomes are no heads, 1 head, 2 heads and 3 heads. So, I say that probability of no heads is $\frac{1}{4}$ . What is wrong with thisconclusion?

Answer:

Answer. [ $\frac{1}{8}$ ]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total cases in tossing three coins = 8(HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
Number of case with no head = TTT
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of no head = $\frac{1}{8}$
The conclusion that probability of no head is $\frac{1}{4}$ is wrong because as we calculate it above, it comes out $\frac{1}{8}$ . Hence the probability of no head is $\frac{1}{8}$

Question:11

If you toss a coin 6 times and it comes down heads on each occasion. Can you say that the probability of getting a head is 1? Give reasons.

Answer:

Answer. [False]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
The probability of getting a head is 1, means that we never get tail. But this is not true because we have both head and tail in a coin. Hence probability of getting head is 1 is false.

Question:12

Sushma tosses a coin 3 times and gets tail each time. Do you think that the outcome of next toss will be a tail? Give reasons.

Answer:

Answer. [False]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
No, because when we toss a coin we can get either tail or head and the probability of each is $\frac{1}{2}$ .
So, it is not necessary that she gets tail at fourth toss. She can get head also.

Question:13

If I toss a coin 3 times and get head each time, should I expect a tail to have a higher chance in the 4th toss? Give reason in support of your answer.

Answer:

Answer. [False]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
No, because we get head or tail after tossing a coin that is the probability of both outcomes is $\frac{1}{2}$ .
Hence tail is not have higher chance than head.
Both are have equal chance.

Question:14

A bag contains slips numbered from 1 to 100. If Fatima chooses a slip at random from the bag, it will either be an odd number or an even number. Since this situation has only two possible outcomes, so, the probability of each is $\frac{1}{2}$ . Justify.

Answer:

Answer. [True]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total slips = 100
Slips with even number = 50
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of slip with even number = $\frac{50}{100}= \frac{1}{2}$
Slips with odd number = 50
Probability of slip with odd number = $\frac{50}{100}= \frac{1}{2}$
Hence the probability of each is $\frac{1}{2}$ .

NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability: Exercise-13.3

Question:1

Find the mean of the distribution :

Class	1-3	3-5	5-7	7-10
Frequency	9	22	27	17

Answer:

Answer. [5.5]
Solution. Here we calculate mean by following the given steps:

Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
Now divide sum of (fx) by sum of (f)
Class
Marks (x_i)
Frequency(f_i)
f_ixi
1-3
2
9
18
3-5
4
22
88
5-7
6
27
162
7-10
8.5
17
144.5

$\sum f_{i}= 75$
$\sum f_{i}x_{i}= 412\cdot 5$

$mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{412\cdot 5}{75}= 5\cdot 5$

Question:2

Calculate the mean of the scores of 20 students in a mathematics test :

Marks	10-20	20-30	30-40	40-50	50-60
Number of students	2	4	7	6	1

Answer:

Answer. [35]
Solution. Here we calculate mean by following the given steps:

Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
4 Now divide sum of (fx) by sum of (f)

Marks	x_i	No.of students fi	f_ixi
10-20	15	2	30
20-30	25	4	100
30-40	35	7	245
40-50	45	6	270
50-60	55	1	55

$\sum f_{i}= 20$ $\sum f_{i}x_{i}= 700$
$mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{700}{20}= 35$

Question:3

Calculate the mean of the following data

Class	4-7	8-11	12-15	16-19
Frequency	5	4	9	10

Answer:

Answer. [12.93]
Solution. Here we calculate mean by following the given steps:

Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
Now divide sum of (fx) by sum of (f)

Class	x_i	fi	f_{i x}i
4-7	5.5	5	275
8-11	9.5	4	38
12-15	13.5	9	121.5
16-19	17.5	10	175
		$\sum f_{i}= 28$	$\sum f_{i}x_{i}= 362$

$mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{362}{28}= 12\cdot 93$

Question:4

The following table gives the number of pages written by Sarika for completing her own book for 30 days :

no.of pages written per day	16-18	19-21	22-24	25-27	28-30
no.of days	1	3	4	9	13

Find the mean number of pages written per day.

Answer:

Answer. [26]
Solution. Here we calculate mean by following the given steps:

Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
Now divide sum of (fx) by sum of (f)
No.of pages written per day
_{no.of days(fi)}
(xi)
f_ixi
16-18
1
17
17
19-21
3
20
60
22-24
4
23
92
25-27
9
26
234
28-30
13
29
377

$\sum f_{i}= 30$

$\sum f_{i}x_{i}= 780$

$mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{780}{30}= 26$

Question:5

The daily income of a sample of 50 employees are tabulated as follows :

Income (in Rs)	1-200	201-400	401-600	601-800
Number of employees	14	14	14	7

Find the mean daily income of employees.

Answer:

Answer. [356.5]
Solution. Here we calculate mean by following the given steps:

Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
Now divide sum of (fx) by sum of (f)
Income (in Rs )
xi
No.of employees
f_ixi
1-200
100.5
14
1407
201-400
300.5
15
4507.5
401-600
500.5
14
7007
601-800
700.5
7
4903.5

$\sum f_{i}= 50$
$\sum f_{i}x_{i}= 17825$
$mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{17825}{50}= 356.5$

Question:6

An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given in the following table :

no.of seats	100-104	104-108	108-112	112-116	116-120
Frequency	15	20	32	18	15

Determine the mean number of seats occupied over the flights.

Answer:

Answer. [109]
Solution. Here we calculate mean by following the given steps:

Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
Now divide sum of (fx) by sum of (f)

Number of seats	Frequency fi	xi	f_ixi
100-104	15	102	1530
104-108	20	106	2120
108-112	32	110	3520
112-116	18	114	2052
116-120	15	118	177065268
	$\sum f_{i}= 100$		$\sum f_{i}x_{i}= 10992$

$mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{10992}{100}= 109\cdot 92$
number of seats = 109

Question:7

The weights (in kg) of 50 wrestlers are recorded in the following table :

Weight (in Kg)	100-110	110-120	120-130	130-140	140-150
Number of wrestlers	4	14	21	8	3

Find the mean weight of the wrestlers.

Answer:

Answer. [123.4]
Solution. Here we calculate mean by following the given steps:

Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
Now divide sum of (fx) by sum of (f)
Weight
fi
xi
f_ixi
100-110
4
105
420
110-120
14
115
1610
120-130
21
125
2625
130-140
8
135
1080
140-150
3
145
435

$\sum f_{i}= 50$

$\sum f_{i}x_{i}=6170$

$mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{6170}{50}= 123\cdot 4 \ \ kg$

Question:8

The mileage (km per litre) of 50 cars of the same model was tested by a manufacturer and details are tabulated as given below :

MIleage (km/I)	10-20	12-14	14-16	16-18
Number of cars	7	12	18	13

Find the mean mileage.
The manufacturer claimed that the mileage of the model was 16 km/litre. Do you agree with this claim?

Answer:

Answer. [14.48]
Solution. Here we calculate mean by following
1.Find the mid point of each interval.
2.Multiply the frequency of each interval by its mid point.
3.Get the sum of all the frequencies (f) and sum of all the (fx)
4.Now divide sum of (fx) by sum of (f)

MIleage (km/I)	No.of cars (fi)	xi	f_ixi
10-12	7	11	77
12-14	12	13	156
14-16	18	15	270
16-18	13	17	221
	$\sum f_{i}= 50$		$\sum f_{i}x_{i}= 724$

$mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{724}{50}= 14\cdot 48 \ \ km/L$

Question:9

The following is the distribution of weights (in kg) of 40 persons :

Weight (in kg)	40-45	45-50	50-55	55-60	60-65	65-70	70-75	70-75
Number of person	4	4	13	5	6	5	2	1

Construct a cumulative frequency distribution (of the less than type) table for the data above.

Answer:

CI.	f	cf
40-45	4	4
45-50	4	8
50-55	13	21
55-60	5	26
60-65	6	32
65-70	5	37
70-75	2	39
75-80	1	40

Question:10

The following table shows the cumulative frequency distribution of marks of 800 students in an examination:

Marks	Number of students
Below 10 Below 20 Below 30 Below 40 Below 50 Below 60 Below 70 Below 80 Below 90 Below 100	10 50 130 270 440 570 670 740 780 800

Construct a frequency distribution table for the data above.

Answer:

Marks	cf	f
0-10	10	10
10-20	50	50-10=40
20-30	130	130-50=80
30-40	270	270-130=140
40-50	440	440-270=170
50-60	570	570-440=130
60-70	670	670-570=100
70-80	740	740-670=70
80-90	780	780-740=40
90-100	800	800-780=20

Question:11

Form the frequency distribution table from the following data :

Marks (out of 90)	Number of candidates
More than or equal to 80 More than or equal to 70 More than or equal to 60 More than or equal to 50 More than or equal to 40 More than or equal to 30 More than or equal to 20 More than or equal to 10 More than or equal to 0	4 6 11 17 23 27 30 32 34

Answer:

class	f
0-10	34-32=2
10-20	32-30=2
20-30	30-27=3
30-40	27-23=4
40-50	23-17=6
50-60	17-11=6
60-70	11-6=5
70-80	6-4-=2
80-90	4

Question:12

Find the unknown entries a, b, c, d, e, f in the following distribution of heights of students in a class :

Height (in cm)	Frequency	Cumulative frequency
150-155 155-160 160-165 165-170 170-175 175-180	12 b 10 d e 2	a 25 c 43 48 f
Total	50

Answer:

Solution. Frequency distribution: It tells how frequencies are distributed overvalues in a frequency distribution. However mostly we use frequency distribution to
summarize categorical variables.
$a= 12$ ( because first term of frequency and cumulative frequency is same )
12 + b = 25
b = 25 – 12
b = 13

25 + 10 = c
35= c
c + d = 43
35 + d = 43
d = 43 – 35
d=8

43 + e = 48
e = 48 – 43
e =5
48+2 = f
50 = f
Ans. a = 12, b = 13, c = 35, d = 8, e = 5, f = 50

Question:13

The following are the ages of 300 patients getting medical treatment in a hospital on a particular day :

Age (in yeras)	10-20	20-30	30-40	40-50	50-60	60-70
Number of patients	60	42	55	70	53	20

(i) Less than type cumulative frequency distribution.
(ii) More than type cumulative frequency distribution

Answer:

Age (in year)	No.of patients
less than 10	0
less than 20	60+0 = 60
less than 30	42+60 = 102
less than 40	102+55 =157
less than 50	157+70 = 227
less than 60	227+53 =280
less than 70	280 +20 =300

Age (in year)	No.of patients
More than or equal to 10 More than or equal to 20 More than or equal to 30 More than or equal to 40 More than or equal to 50 More than or equal to 60	60+42+55+70+53+20 = 300 42+55+70+53+20 = 240 55+70+53+20= 198 70+53+20 = 143 53+20 = 73 20

Question:14

Given below is a cumulative frequency distribution showing the marks secured by 50 students of a class :

Marks	Below 20	Below 40	Below 60	Below 80	Below 100
Number of students	17	22	29	37	50

Form the frequency distribution table for the data

Answer:

Marks	Number of students CF	f
0- 20	17	17
20- 40	22	22-17 = 5
40- 60	29	29 -22 = 7
60- 80	37	37-29 = 8
80-100	50	50 -37 = 13

Question:15

Weekly income of 600 families is tabulated below :

Weekly income (in Rs)	Number of families
0-1000 1000-2000 2000-3000 3000-4000 4000-5000 5000-6000	250 190 100 40 15 5
Total	600

Compute the median income.

Answer:

Answer. [1263.15]
Solution. n = 600
$\frac{n}{2}= \frac{600}{2}= 300$
l $\iota$ = 1000, l = 1000, cf = 250, f = 190
median = $\iota +\left ( \frac{\frac{n}{2}-cf}{f} \right )\times h$
$= 1000+\frac{\left ( 300-250 \right )}{190}\times 1000$
$= 1000+\frac{50}{190}\times 1000$
$= 1000+\frac{5000}{19}$
$= \frac{19000+5000}{19}= \frac{24000}{19}= 1263\cdot 15$
Median = 1263.15

Question:16

The maximum bowling speeds, in km per hour, of 33 players at a cricket coaching centre are given as follows :

Speed (km/h)	85-100	100-115	115-130	130-145
Number of players	11	9	8	5

Calculate the median bowling speed.

Answer:

Answer. [109.16]
Solution. Here n = 33
$\frac{n}{2}= \frac{33}{2}= 16\cdot 5$
$\iota = 100$
h = 15, f = 9 , cf = 11
Median $= \iota +\left ( \frac{\frac{n}{2}-cf}{f} \right )\times h$
$=100+\frac{\left ( 16\cdot 5-11 \right )\times 15}{9}$
$=100+\left ( \frac{55}{10\times 9} \right )\times 15$
$=100+\frac{55\times 5}{30}$
$=100+\frac{275}{30}$
= 100 + 9.16 $\Rightarrow$ 109.16

Question:17

The monthly income of 100 families are given as below :

Income (in Rs)	Number of families
0-5000 5000-10000 10000-15000 15000-20000 20000-25000 25000-30000 30000-35000 35000-40000	8 26 41 16 3 3 2 1

Calculate the modal income.

Answer:

Answer. [11875]
Solution. Here l = 10000, f₁ = 41, f₀ = 26, f₂ = 16, h = 5000
Mode = $\iota +\left ( \frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}} \right )\times h$
= $1000+\left ( \frac{41-26}{2\times 41-26-16} \right )\times 5000$
= $1000+\frac{15}{40}\times 5000$
= 10000 + 1875 = 11875
Modal income is 11875 Rs.

Question:18

The weight of coffee in 70 packets are shown in the following table :

Weight (in g)	Number of packets
200-201 201-202 202-203 203-204 204-205 205-206	12 26 20 9 2 1

Determine the modal weight.

Answer:

Answer. [201.7 g]
Solution.
Here l = 201, f1 = 26, f0 = 12, f2 = 20, h = 1
$\Rightarrow 201+\left ( \frac{26-12}{2\times 26-12-20} \right )\times 1$
$\Rightarrow 201+\left ( \frac{14}{52-32} \right )$
$\Rightarrow 201+\frac{14}{20}$
$\Rightarrow 201+0\cdot 7= 201\cdot 7 g$

Question:19

Two dice are thrown at the same time.
(i) Find the probability of getting same number on both dice.
(ii) Find the probability of getting different numbers on both dice.

Answer:

Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number of cases after thrown of two dice = 36
(i) Same number = (1, 1), (2, 2), (3, 3), (4. 4), (5, 5), (6, 6)
Same number cases = 6
Let A be the event of getting same number.
Probability [p(A)] = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
$p\left ( A \right )= \frac{6}{36}= \frac{1}{6}$
(ii) Different number cases = 36 – same number case
= 36 –6 = 30
Let A be the event of getting different number
Probability [p(A)]= $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
$p\left ( A \right )= \frac{30}{36}= \frac{5}{6}$

Question:20

Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is
(i) 7?
(ii) a prime number?
(iii) 1?

Answer:

(i) Answer. [1/6]
Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases after throwing of two dice = 36
Cases when total is 7 = (1, 6), (6, 1), (3, 4), (4, 3), (2, 5), (5, 2)
Total cases = 6
Let A be the event of getting total 7
Probability [p(A)]= $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 7 = $\frac{6}{36}= \frac{1}{6}$

(ii) Answer. [5/12]
Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
Prime number as a sum = (1, 1), (1, 2), (2, 1), (1, 4),
(4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (3, 4), (4, 3), (2, 5), (5, 2), (6, 5), (5, 6)
Cases = 15
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that sum is a prime number = $\frac{15}{36}= \frac{5}{12}$
(iii) Answer. [0]
Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
pairs from which we get sum 1 = 0
Cases = 0
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 1 = $\frac{0}{36}= 0$

Question:21

Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is
(i) 6
(ii) 12
(iii) 7

Answer:

(i) Answer. [1/9]
Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
For getting product 6 = (1, 6,), (6, 1), (2, 3), (3, 2)
Cases = 4
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting product 6 = $\frac{4}{36}= \frac{1}{9}$

(ii) Answer. [1/9]
Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
product 12 = (2, 6), (6, 2), (3, 4), (4, 3)
Cases = 4
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting product 12 = $\frac{4}{36}= \frac{1}{9}$

(iii) Answer. [0]
Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
Product 7 = 0 (case)
Cases = 0
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting product 7 = $\frac{0}{36}= 0$

Question:22

Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is less than 9.

Answer:

Answer. [4/9]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases in throwing two dice = 36
Product less than 9 cases = (1, 1). (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (4, 1), (4, 2), (5, 1), (6, 1)
Number of favourable cases = 16
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting product less than 9 = $\frac{16}{36}= \frac{4}{9}$

Question:23

Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3, respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.

Answer:

Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number of cases = 36
case of getting sum 2 = ( 1 , 1 ) ( 1 , 1 )
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 2 = $\frac{2}{36}= \frac{1}{18}$
case of getting sum 3 = (1, 2), (1, 2), (2, 1), (2, 1)
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 3= $\frac{4}{36}= \frac{1}{9}$
case of getting sum 4 = (1, 3), (1, 3), (2, 2), (2, 2), (3, 1), (3, 1)
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 4= $\frac{6}{36}= \frac{1}{6}$
case of getting sum 5 = (2, 3), (2, 3), (4, 1),(4,1) (3, 2), (3, 2)
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 5 = $\frac{6}{36}= \frac{1}{6}$
case of getting sum 6 = (3, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1)
probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 6= $\frac{6}{36}= \frac{1}{6}$
case of getting sum 7 = (4, 3), (4, 3), (5, 2), (5, 2), (6, 1), (6, 1)
probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 7= $\frac{6}{36}= \frac{1}{6}$
case of getting sum 8 = (5, 3), (5, 3), (6, 2), (6, 2)
probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 8= $\frac{4}{36}= \frac{1}{9}$
case of getting sum 9 = (6, 3), (6, 3)
probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 9= $\frac{2}{36}= \frac{1}{18}$

Question:24

A coin is tossed two times. Find the probability of getting at most one head.

Answer:

Answer. [3/4]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 4 (HH, TT, HT, TH)
Cases of at most 1 head = HT, TH, TT
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting at most 1 head = $\frac{3}{4}$

Question:25

A coin is tossed 3 times. List the possible outcomes. Find the probability of getting
(i) all heads
(ii) at least 2 heads

Answer:

(i) Answer. [1/8]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Possible outcomes = (HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
Total cases = 8
Cases of getting all heads = (HHH)
Number of favourable cases = 1
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting all heads = $\frac{1}{8}$

(ii) Answer.[1/2]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Possible outcomes = 8 (HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
Cases of getting at least 2 heads = (HHH, HHT, HTH, THH)
Favorable cases = 4
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting at least 2 heads = $\frac{4}{8}= \frac{1}{2}$

Question:26

Two dice are thrown at the same time. Determine the probability that the difference of the numbers on the two dice is 2.

Answer:

Answer. [2/9]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Cases of getting difference 2 = (1, 3), (3, 1), (2, 4), (4, 2), (3, 5), (5, 3), (4, 6), (6, 4)
Favourable cases = 8
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting difference 2 = $\frac{8}{36}= \frac{2}{9}$

Question:27

A bag contains 10 red, 5 blue and 7 green balls. A ball is drawn at random. Find the probability of this ball being a
(i) red ball
(ii) green ball
(iii) not a blue ball

Answer:

(i) Answer. [5/11]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total balls = 10 + 5 + 7 = 22
Red balls = 10
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting red ball = $\frac{10}{22}= \frac{5}{11}$

(ii) Answer. [7/22]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total balls = 10 + 5 + 7 = 22
Green balls = 7
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting green ball = $\frac{7}{22}$

(iii) Answer. [17/22]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total balls = 10 + 5 + 7 = 22
Not a blue ball = 22 – (blue ball)
= 22 – 5 = 17
robability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting not a blue ball = $\frac{17}{22}$

Question:28

The king, queen and jack of clubs are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is
(i) a heart
(ii) a king

Answer:

(i) Answer. [13/49]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 3 ( three cards are removed)
= 49
Total hearts = 13
Favourable cases = 13
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting a heart = $\frac{13}{49}$

(ii) Answer. [3/49]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 3 ( three cards are removed)
= 49
Total king = 4 – 1 = 3 ( 1 king is removed)
favourable cases = 3
bability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting a King= $\frac{3}{49}$

Question:29

The king, queen and jack of clubs are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. What is the probability that the card is
(i) a club
(ii) 10 of hearts

Answer:

(i) Answer. [10/49]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 3 = 49 ( three cards are removed)
Total club = 13 – 3 = 10 ( 3 club cards are removed)
favourable cases = 10
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting a club = $\frac{10}{49}$

(ii) Answer. [1/49]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 3 = 49 ( three cards are removed)
10 of heart = 1
favourable cases = 1
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting a heart = $\frac{1}{49}$

Question:30

All the jacks, queens and kings are removed from a deck of 52 playing cards. The remaining cards are well shuffled and then one card is drawn at random. Giving ace a value 1 similar value for other cards, find the probability that the card has a value
(i) 7
(ii) greater than 7
(iii) less than 7

Answer:

(i) Answer. [1/10]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 12 = 40 ( 12 cards are removed)
card with number 7 = 4
favourable cases = 4
probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting card 7= $\frac{4}{10}= \frac{1}{10}$

(ii) Answer. [3/10]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 12 = 40 ( $\mathbb{Q}$ 12 cards are removed)
Cards greater than 7 =8,9,10 (3 × 4 = 12)
favourable cases = 12
probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting card 7= $\frac{12}{40}= \frac{3}{10}$
(iii) Answer. [3/5]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 12 = 40 ( $\because$ 12 cards are removed)
Cards less than 7 = 1, 2, 3, 4, 5, 6 (6 × 4 = 24)
favourable cases = 24
probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting card 7= $\frac{24}{40}= \frac{6}{10}= \frac{3}{5}$

Question:31

An integer is chosen between 0 and 100. What is the probability that it is
(i) divisible by 7 ?
(ii) not divisible by 7?

Answer:

(i) Answer. [14/99]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number = 99 (between 0 to 100)
Number divisible by 7 = (7, 14, 21, 28,35, 42, 49, 56, 63, 70, 77, 84, 91, 98)
Favourable cases = 14
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting number divisible by 7 = $\frac{14}{99}$

(ii) Answer. [85/99]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number between 0 to 100 = 99
Number divisible by 7 = (7, 14, 21, 28,35, 42, 49, 56, 63, 70, 77, 84, 91, 98)
Favourable cases = 14
robability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting number divisible by 7 = $\frac{14}{99}$
$= 1-\frac{14}{99}$
$= \frac{99-14}{99}= \frac{85}{99}$

Question:32

Cards with numbers 2 to 101 are placed in a box. A card is selected at random. Find the probability that the card has
(i) an even number
(ii) a square number

Answer:

(i) Answer. [1/2]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total numbers from 2 to 101 = 100
Total even numbers from 2 to 101 = 50
Favourable cases = 50
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that card is with even number = $\frac{50}{100}= \frac{1}{2}$

(ii) Answer. [9/100]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number from 2 to 101 = 100
Square numbers from 2 to 101 = (4, 9, 16, 25, 36, 49, 64, 81, 100)
Favourable cases = 9
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that the card is with a square number = $\frac{9}{100}$

Question:33

A letter of English alphabets is chosen at random. Determine the probability that the letter is a consonant.

Answer:

Answer. [21/26]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total alphabets = 26
Total consonant = 21
Favourable cases = 21
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that alphabet is consonant = $\frac{21}{26}$

Question:34

There are 1000 sealed envelopes in a box, 10 of them contain a cash prize of Rs 100 each, 100 of them contain a cash prize of Rs 50 each and 200 of them contain a cash prize of Rs 10 each and rest do not contain any cash prize. If they are well shuffled and an envelope is picked up out, what is the probability that it contains no cash prize ?

Answer:

Answer. [0.69]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total envelopes = 1000
Envelopes with no cash prize = Total envelopes – envelopes with cash prize
= 1000 – 10 – 100 – 200 = 690
Favourable cases = 690
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that the envelope is no cash prize = $\frac{690}{1000}= \frac{69}{100}= 0\cdot 69$

Question:35

Box A contains 25 slips of which 19 are marked Rs 1 and other are marked Rs 5 each. Box B contains 50 slips of which 45 are marked Rs 1 each and others are marked Rs 13 each. Slips of both boxes are poured into a third box and reshuffled. A slip is drawn at random. What is the probability that it is marked other than Rs 1 ?

Answer:

Answer. [11/75]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total slips = 25 + 50 = 75
Slips marked other than 1 = Rs. 5 slips + Rs. 13 slips
= 6 + 5 = 11
Favourable cases = 11
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that slips is not marked 1 $= \frac{11}{75}$

Question:36

A carton of 24 bulbs contain 6 defective bulbs. One bulbs is drawn at random. What is the probability that the bulb is not defective? If the bulb selected is defective and it is not replaced and a second bulb is selected at random from the rest, what is the probability that the second bulb is defective?

Answer:

Answer. [5/23]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total bulbs = 24
Defective = 6
not defective = 18
Probability that the bulb is not defective = $\frac{18}{24}= \frac{3}{4}$
Let the bulbs is defective and it is removed from 24 bulb.
Now bulbs remain = 23
In 23 bulbs, non-defective bulbs = 18
defective = 5
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Now probability that the bulb is defective = $\frac{5}{23}$ .

Question:37

A child’s game has 8 triangles of which 3 are blue and rest are red, and 10 squares of which 6 are blue and rest are red. One piece is lost at random. Find the probability that it is a
(i) triangle
(ii) square
(iii) square of blue colour
(iv) triangle of red colour

Answer:

(i) Answer. [4/9]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total piece = 8 + 10 = 18
Total triangles = 8
Favourable cases = 8
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that piece is a triangle $= \frac{8}{18}= \frac{4}{9}$

(ii) Answer. [5/9]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total piece = 8 + 10 = 18
Total square = 10
Favourable cases = 10
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that the piece is a square = $\frac{10}{18}= \frac{5}{9}$

(iii) Answer. [1/3]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total piece = 10 + 8 = 18
Square of blue color = 6
favourable cases = 6
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that piece is a square of blue color = $\frac{6}{18}= \frac{1}{3}$

(iv) Answer. [5/18]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total piece = 10 + 8 = 18
triangle of red color = 8 – 3 = 5
favourable cases = 5
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that piece is a triangle of red colour $= \frac{5}{18}$

Question:38

In a game, the entry fee is Rs 5. The game consists of a tossing a coin 3 times. If one or two heads show, Shweta gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise she will lose. For tossing a coin three times, find the probability that she
(i) loses the entry fee.
(ii) gets double entry fee.
(iii) just gets her entry fee.

Answer:

(i) Answer. [1/8]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 8(HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
case in which the lose entry = 8 – (in which she gets entry book + in which she gets double)
= 8 – 6 (HHT, HTH, THH, TTH, THT, HTT) – 1(HHH)
= 8 – 7 = 1
Favourable cases = 1
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that she will lose money = $\frac{1}{8}$

(ii) Answer. [1/8]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 8(HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
case in which she gets double entry = HHH
favourable cases = 1
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that she gets double entry fee = $\frac{1}{8}$

(iii) Answer. [3/4]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 8(HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
case in which she gets entry book = 6(HHT, HTH, THH, TTH, THT, HTT)
favourable cases = 6
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that she gets entry fees = $\frac{6}{8}= \frac{3}{4}$

Question:39

A die has its six faces marked 0, 1, 1, 1, 6, 6. Two such dice are thrown together and the total score is recorded.
(i) How many different scores are possible?
(ii) What is the probability of getting a total of 7?

Answer:

(i) Answer. [6]
Solution. Count the number of sums we can notice by using two dice of (0, 1, 1, 1, 6, 6) type.
We can get a sum of 0 = (0,0)
We can get a sum of 1 = (0,1) , (1,0)
We can get a sum of 2 = (1,1)
We can get a sum of 6 = (0,6) , (6,0)
We can get a sum of 7 = (6,1) , (1,6)
We can get a sum of 12 = (6,6)
We can get a score of 0, 1, 2, 6, 7, 12
Hence we can get 6 different scores.

(ii) Answer. [4/9]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
Case of getting sum 7 = (1, 6), (1, 6), (1, 6), (1, 6), (1, 6), (1, 6), (6,1), ( 6,1), (6,1), ( 6,1), (6,1), (6,1),
Number of favourable cases = 12
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting a total 7 = $\frac{12}{36}= \frac{1}{3}$

Question:40

A lot consists of 48 mobile phones of which 42 are good, 3 have only minor defects and 3 have major defects. Varnika will buy a phone if it is good but the trader will only buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is
(i) acceptable to Varnika?
(ii) acceptable to the trader?
Answer:

(i) Answer. [7/8]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total mobiles = 48
Minor defective = 3
major defective = 3
good = 42
Varnika buy only good so favourable cases = 42
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that acceptable to Varnika = $\frac{42}{48}= \frac{7}{8}$
(ii) Answer. [15/16]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
total mobiles = 48
good = 42
minor defect = 3
major defect = 3
trader accept only good and minor defect.
So favourable cases = 42 + 3 = 45
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that trader accept $\frac{45}{48}= \frac{15}{16}$

Question:41

A bag contains 24 balls of which x are red, 2x are white and 3x are blue. A ball is selected at random. What is the probability that it is
(i) not red?
(ii) white?

Answer:

(i) Answer. [5/6]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total balls = red + white + blue
24 = x + 2x + 3x
6x = 24
x = 4
Red balls = x = 4
White balls = 2x = 2 × 4 = 8
Blue balls = 3x = 3 × 4 = 12
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that ball is not red = $\frac{blue+white}{24}$
$= \frac{8+12}{24}= \frac{20}{24}= \frac{5}{6}$

(ii) Answer. [1/3]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total balls = red + white + blue
24 = 6x
x = 4
white balls = 2x = 2 × 4 = 8
Favourable cases = 8
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting on white ball = $\frac{8}{24}= \frac{1}{3}$

Question:42

At a fete, cards bearing numbers 1 to 1000, one number on one card, are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square greater than 500, the player wins a prize. What is the probability that
(i) the first player wins a prize?
(ii) the second player wins a prize, if the first has won?

Answer:

(i) Answer. [0.009]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 1000
Player wins prize with cards = (529, 576, 625, 676, 729, 784, 841, 900, 961)
Favourable cases = 9
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that player wins = $\frac{9}{1000}= 0\cdot 009$

(ii) Answer. [0.008]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Now the total cards are = 1000 – 1 = 999
Now the total winning cards = 9 – 1 = 8
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that second player wins after first = $\frac{8}{999}\approx 0\cdot 008$

Question:1

Find the mean marks of students for the following distribution

Marks	Number of students
0 and above 10 and above 20 and above 30 and above 40 and above 50 and above 60 and above 70 and above 80 and above 90 and above 100 and above	80 77 72 65 55 43 28 16 10 8 0

Answer:

Answer. [51.75]
Solution. Here we calculate mean by following the given steps:

Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)

Now divide sum of (fx) by sum of (f)

Marks	xi	cf	fi	fixi
0-10	5	80	80-77 = 3	15
10-20	15	77	787-72 = 5	75
20-30	25	72	72-65 =7	175
30-40	35	65	65-55 = 10	350
40-50	45	55	55-43 = 12	540
50-60	55	43	43-28 = 15	825
60-70	65	28	28-16 = 12	780
70-80	75	16	16-10 =6	450
80-90	85	10	10-8 = 2	170
90-100	95	8	8-0 = 8	760
			$\sum f_{i}= 80$	$\sum f_{i}x_{i}= 4140$

$mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{4140}{80}= 51\cdot 75$

Question:2

Determine the mean of the following distribution :

Marks	Number of students
Below 10 Below 20 Below 30 Below 40 Below 50 Below 60 Below 70 Below 80 Below 90 Below 100	5 9 17 29 45 60 70 78 83 85

Answer:

Answer. [48.4]
Solution. Here we calculate mean by following the given steps:

Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)

Now divide sum of (fx) by sum of (f)

Marks	xi	cf	fi	fixi
0-10	5	5	5	15
10-20	15	9	9-5 =4	75
20-30	25	17	17-9 = 8	175
30-40	35	29	29-17 = 12	350
40-50	45	45	45-29 = 16	540
50-60	55	60	60-45 = 15	825
60-70	65	70	70-60 = 10	780
70-80	75	78	78-70 = 8	450
80-90	85	83	83-78 = 5	170
90-100	95	85	85-83 = 2	760
			$\sum f_{i}= 85$	$\sum f_{i}x_{i}= 4115$

$mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{4115}{85}= 48\cdot 4$

Question:3

Find the mean age of 100 residents of a town from the following data :

Age equal and above (in years)	0	10	20	30	40	50	60	70
Number of Persons	100	90	75	50	25	15	5	0

Answer:

Answer. [31]
Solution. Here we calculate mean by following the given steps:

Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
Now divide sum of (fx) by sum of (f)
Marks
xi
fi
0-10
5
100-90 =10
50
10-20
15
90-75 = 15
225
20-30
25
75-50 = 25
625
30-40
35
50-25 =25
875
40-50
45
25-15 =10
450
50-60
55
15-5 = 10
550
60-70
65
5-0 = 5
325

$\sum f_{i}= 100$
$\sum f_{i}x_{i}= 3100$
$mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{3100}{100}= 31$

Question:4

The weights of tea in 70 packets are shown in the following table :

Weight (in gram)	Number of packets
200-201 201-202 202-203 203-204 204-205 205-206	13 27 18 10 1 1

Find the mean weight of packets.

Answer:

Answer. [201.95]
Solution. Here we calculate mean by following the given steps:

Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
Now divide sum of (fx) by sum of (f)
Weight
xi
fi
fixi
200-201
200.5
13
2606.5
201-202
201.5
27
5440.5
202-203
202.5
18
3645.0
203-204
203.5
10
2035.0
204-205
204.5
1
204.5
205-206
205.5
1
205.5

$\sum f_{i}= 70$
$\sum f_{i}x_{i}= 14137$

$mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{14137}{70}= 201\cdot 95$

Question:5

The weights of tea in 70 packets are shown in the following table :

Weight (in gram)	Number of packets
200-201 201-202 202-203 203-204 204-205 205-206	13 27 18 10 1 1

Draw the less than type ogive for this data and use it to find the median weight.

Answer:

Answer. [201.8]
Solution.

Weight	cf
Less than 201	13
Less than 202	27+13=40
Less than 203	40+18=58
Less than 204	58+10=68
Less than 205	68+1 =69
Less than 206	69+1 = 70

$n = 70,\frac{n}{2}= \frac{70}{2}= 35$
Hence the median is 201.8

Question:6

The weights of tea in 70 packets are shown in the following table :

Weight (in gram)	Number of packets
200-201 201-202 202-203 203-204 204-205 205-206	13 27 18 10 1 1

Draw the less than type and more than type ogives for the data and use them to find the median weight.

Answer:

Answer. [201.8]
Solution.

Less than type		More than type
Weight	Number of packets	Number of packets	Number of students
Less than 200	0	More than or equal to 200	70
Less than 201	13	More than or equal to 201	70-13 = 57
Less than 202	40	More than or equal to 202	57-27 =30
Less than 203	58	More than or equal to 203	30-18 =12
Less than 204	68	More than or equal to 204	12-10 = 2
Less than 205	69	More than or equal to 205	2-1 = 1
Less than 206	70	More than or equal to 206	1-1 = 0

$Here\, \, n = 70,\frac{n}{2}= \frac{70}{2}= 35$
Hence median = 201.8

Question:7

The table below shows the salaries of 280 persons.

Salary(in thousand (Rs))	Number of persons
5-10 10-15 15-20 20-25 25-30 30-35 35-40 40-45 45-50	49 133 63 15 6 7 4 2 1

Calculate the median and mode of the data.

Answer:

Solution.

Salary	fi	cf
5-10	49	49
10-15	133	49+133=182
15-20	63	182+63=245
20-25	15	245+15 = 260
25-30	6	260+6 = 266
30-35	7	266+7 = 273
35-40	4	273+4 = 277
40-45	2	277+2 = 279
45-50	1	279+1 = 280

$n= 280,\frac{n}{2}= \frac{280}{2}= 140$
f₁ = 49, f_m= 133, f₂= 63, cf = 49, f = 133
l = 10, h = 5
median = $\iota +\frac{\left ( \frac{n}{2}-cf \right )}{f}\times h$
= $10+\frac{\left ( 140-49 \right )}{133}\times 5$
= $10+\frac{91\times 5}{133}$
= $10+\frac{455}{133}= 10+3\cdot 421$
$= 13\cdot 421$
In thousands = 13.421 × 1000 = 13421 Rs.
Mode = $\iota +\left [ \frac{f_{m}-f_{i}}{2f_{m}-f_{i}-f_{2}} \right ]\times h$

= $10+\left [ \frac{133-49}{2\times 133-49-63} \right ]\times 5$

= $10+\frac{84\times 5}{266-112}=10+\frac{84\times 5}{154}$
=10 + 2.727
=12.727
In thousands = 12.727 × 1000 = 12727 Rs.

Question:8

The mean of the following frequency distribution is 50, but the frequencies f1 and f2 in classes 20-40 and 60-80, respectively are not known. Find these frequencies, if the sum of all the frequencies is 120.

Class	0-20	20-40	40-60	60-80	80-100
Frequency	17	f_i	32	f₂	19

Answer:

Solution.

Class	(f_i)	x_i	$\mu _{c}= \frac{\left ( x_{i}-a \right )}{h}$	f_i
0-20	17	10	-2	-34
20-40	f₁	30	-1	-f₁
40-60	32	50=a	0	0
60-80	f₂	70	1	f₂
80-100	19	90	2	38
	$\sum f_{i}= 68+f_{i}+f_{2}$

Sum of all frequencies = 120
$\Rightarrow$ 68 + f₁ + f₂ = 120
f₁ + f₂ = 52 …(1)
a = 50, h = 20
mean = $a+\frac{\sum f_{i}{\mu _{i}}}{\sum f_{i}}\times h$
50= 50 + $\frac{\left ( 4+f_{2}-f_{1} \right )\times 20}{20}$
0= (4 + f₂ – f₁)
–f₂ + f₁ = 4 …(2)
add (1) and (2) we get
2f₁ = 56 $\Rightarrow f_{1}= 28$
Put f₁ = 28 in equation (1)
f₂ = 52 – 28 $\Rightarrow f_{2}= 24$

Question:9

The median of the following data is 50. Find the values of p and q, if the sum of all the frequencies is 90.

Marks	Frequency
20-30 30-40 40-50 50-60 60-70 70-80 80-90	p 15 25 20 q 8 10

Answer:

Solution.

marks	Frequency	Cummulative frequency
20-30	1	p
30-40	15	15+p
40-50	25	40+p = cf
50-60	20=f	60+p
60-70	q	68+p+q
70-80	8	68+p+q
80-90	10	78+p+q

n = 90, $\frac{n}{2}= 45$
l = 50, f = 20, cf = 40 + p, h = 10
median = $l+\frac{\left ( \frac{n}{2}-cf \right )}{f}\times h$
$50= 50+\frac{\left ( 45-40-p \right )}{20}\times 10$
$0= \frac{5-p}{2}$
5 – p = 0
p = 5
78 + 5 + q = 90
q = 90 – 83
q = 7

Question:10

The distribution of heights (in cm) of 96 children is given below :

Height (in cm)	Number of children
124-128 128-132 132-136 136-140 140-144 144-148 148-152 152-156 156-160 160-164	5 8 17 24 16 12 6 4 3 1

Draw a less than type cumulative frequency curve for this data and use it to compute median height of the children.

Answer:

Answer. [139]
Solution.

Height	Number of children
less than 124 less than 128 less than 132 less than 136 less than 140 less than 144 less than 148 less than 152 less than 156 less than 160 less than 164	0 5 13 30 54 70 82 88 92 95 96

$\frac{n}{2}= \frac{96}{2}= 48$
Hence the median is = 139

Question:11

Size of agricultural holdings in a survey of 200 families is given in the following table:

Size of agricultural holdings (in ha)	Number of families
0-5 5-10 10-15 15-20 20-25 25-30 30-35	10 15 30 80 40 20 5

Compute median and mode size of the holdings

Answer:

Answer. [17.77]
Solution.
$mode= \iota +\left [ \frac{f_{m}-f_{1}}{2f_{m}-f_{1}-f_{2}} \right ]\times h$

Size of agricultural holdings	fi	cf
0-5	10	10
5-10	15	25
10-15	30	55
15-20	80	135
20-25	40	175
25-30	20	195
30-35	5	200

(i) Here n = 200
$\frac{n}{2}= \frac{200}{2}= 100$ which lies in interval (15 – 20)
l = 15, h = 5, f = 80 and cf = 55
$median = l+\frac{\left ( \frac{n}{2}-cf \right )}{4}\times h= 15+\frac{\left ( 100-55 \right )\times 5}{80}$
= $15+\frac{45}{16}= 15+2\cdot 81= 17\cdot 81$
l = 15, f_m = 80, f₁ = 30, f₂ = 40 and h = 5
$mode= \iota +\left [ \frac{f_{m}-f_{1}}{2f_{m}-f_{1}-f_{2}} \right ]\times h$
$= 15+\left [ \frac{80-30}{2\times 80-30-40} \right ]\times 5$
$= 15+\left [ \frac{50}{160-70} \right ]\times 5$
$= 15+\left [ \frac{50}{90} \right ]\times 5$
= $15+\frac{25}{9}$
=15 + 2.77 = 17.77

Question:12

The annual rainfall record of a city for 66 days is given in the following table.

Rainfall (in cm)	0-10	10-20	20-30	30-40	40-50	50-60
Number of days	22	10	8	15	5	6

Calculate the median rainfall using ogives (of more than type and of less than type)

Answer:

Answer. [20]
Solution.

(i) less than type		(ii) more than type
Rain fall	No.of days	Rain fall	Number of days
less than 0	0	more than or equal to 0	66
less than 10	0+22 = 22	more than or equal to 10	66-22 = 44
less than 20	22+10 = 32	more than or equal to 20	44-10 = 34
less than 30	32+8 = 40	more than or equal to 30	34-8 = 26
less than 40	40+15 = 55	more than or equal to 40	26-15 = 11
less than 50	55+5 =60	more than or equal to 50	11 - 5 =6
less than 60	60+6 =66	more than or equal to 60	6-6 =0

Now let us draw ogives of more than type and of less than type then find the median

$N= 66,\frac{N}{2}= \frac{66}{2}= 33$
Here median is 20

Question:13

The following is the frequency distribution of duration for100 calls made on a mobile phone :

Duration (in seconds)	Number of calls
95-125 125-155 155-185 185-215 215-245	14 22 28 21 15

Calculate the average duration (in sec) of a call and also find the median from cumulative frequency curve.

Answer:

Answer. [170]
Solution.

Duration	fi	xi	$u_{i}\frac{\left ( x_{i}-a \right )}{h}$	f_iu_i
95-125	14	110	-2	-28
125-155	22	140	-1	-22
155=185	28	170 = a	0	0
185-215	21	200	1	21
215-245	21	230	2	30
	$\sum f_{i}= 100$			$\sum f_{i}u_{i}= 1$

a = 170, h = 30
Average = $a+\frac{\sum f_{i}u_{i}}{\sum f_{i}}\times h= 170+\frac{1}{100}\times 30= 170+0\cdot 3= 170\cdot 3$

less than type
Duration	Number of calls
less than 95 less than 125 less than 155 less than 185 less than 215 less than 245	0 0+14 = 14 14+22 =36 36 + 28 = 64 64 + 21 = 85 85 + 15 =100

n = 100
$\frac{n}{2}= \frac{100}{2}= 50$
median is 170

Question:14

50 students enter for a school javelin throw competition. The distance (in metres) thrown are recorded below :

Distance (in m)	0-20	20-40	40-60	60-80	80-100
Number of students	6	11	17	12	4

(i) Construct a cumulative frequency table.
(ii) Draw a cumulative frequency curve (less than type) and calculate the median distance thrown by using this curve.
(iii) Calculate the median distance by using the formula for median.
(iv) Are the median distance calculated in (ii) and (iii) same ?

Answer:

(i) Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

Distance	f_i	CF
0-20	6	6
20-40	11	6+11 = 17
40-60	17	17+17 = 34
60-80	12	34 + 12 = 46
80-100	4	46 + 4 =50

(ii) Answer. [49.41]
Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables

Distance	Cumulative frequency (C.F)
0 less than 20 less than 40 less than 60 less than 80 less than 100	0 6 17 34 46 50

n = 50
$\frac{n}{2}= \frac{50}{2}= 25$

median = 49.41

(iii)Answer. [49.41]
Solution. n = 50
$\frac{n}{2}= \frac{50}{2}= 25$

which lies in interval 40 – 60
l = 40, h = 20, CF = 17 and f = 17
$median = l+\frac{\left ( \frac{n}{2}-cf \right )}{f}\times h$
= $40+\frac{\left ( 25-17 \right )}{17}\times 20$
= $40+\frac{8\times 20}{17}$
=40 + 9.41
= 49.41

(iv) Yes, the median distance calculated in (ii) and (iii) are same.

NCERT Exemplar Solutions Class 10 Maths Chapter 13 Important Topics:

NCERT exemplar Class 10 Maths solutions chapter 13 deals with a wide range of concepts mentioned below:

How to find out the central tendency is of any given data.
Methods to find out the mean of any group data.
The direct method, step deviation method and assumed mean method to find out the mean of any group data.
NCERT exemplar Class 10 Maths solutions chapter 13 discusses cumulative frequency distribution and its graphical representation.
How to find out the mode and median of the given data.
Different techniques to find out medians.
Conditional probabilities, independent events, and Bayes theorem.

NCERT Class 10 Solutions for Other Subjects:

NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

Chapter 1 Real Numbers

Chapter 2 Polynomials

Chapter 3 Pair of Linear Equations in Two Variables

Chapter 4 Quadratic Equations

Chapter 5 Arithmetic Progressions

Chapter 6 Triangles

Chapter 7 Coordinate Geometry

Chapter 8 Introduction to Trigonometry & Its Equations

Chapter 9 Circles

Chapter 10 Constructions

Chapter 11 Areas related to Circles

Chapter 12 Surface Areas and Volumes

Features of NCERT Exemplar Class 10 Maths Solutions Chapter 13:

These Class 10 Maths NCERT exemplar chapter 13 solutions emphasise the methods to find out mean, median, and mode. In this chapter, students will understand the experimental and statistical approach of probability. Students will learn the condition for multiplying probability to find out the probability of any composite event. Statistics and Probability based practice problems can be easily studied and practiced using these Class 10 Maths NCERT exemplar solutions chapter 13 Statistics and Probability.

The students can comfortably sail through the NCERT Class 10 Maths, RD Sharma Class 10 Maths, A textbook for Mathematics by Monica Kapoor, and RS Aggarwal Class 10 Maths et cetera.

Check Chapter-Wise Solutions of Book Questions

Chapter No.	Chapter Name
Chapter 1	Real Numbers
Chapter 2	Polynomials
Chapter 3	Pair of Linear Equations in Two Variables
Chapter 4	Quadratic Equations
Chapter 5	Arithmetic Progressions
Chapter 6	Triangles
Chapter 7	Coordinate Geometry
Chapter 8	Introduction to Trigonometry
Chapter 9	Some Applications of Trigonometry
Chapter 10	Circles
Chapter 11	Constructions
Chapter 12	Areas Related to Circles
Chapter 13	Surface Areas and Volumes
Chapter 14	Statistics
Chapter 15	Probability

Must, Read NCERT Solution Subject Wise

Check NCERT Notes Subject Wise

Also, Check NCERT Books and NCERT Syllabus here

Frequently Asked Question (FAQs)

1. Make a List of topics and sub-topics covered in Chapter 13 of NCERT Exemplar Solutions for Class 10 Maths.

Below is a list of topics and subtopics covered in Chapter 13 of NCERT Exemplar Solutions for Class 10 Maths:

Direct Method, Assumed Mean Method, and Step-Deviation Method for determining the mean of grouped data.
Finding the mode of the given data.
Calculating the median of grouped data.
Graphical representation of cumulative frequency distribution.

2. In Chapter 13 of NCERT Exemplar Solutions for Class 10 Maths, can you elucidate the concept of mean?

The mean is the arithmetic average of a given set of values, signifying an equal distribution of values in the dataset. Central tendency refers to the statistical measure that identifies a single value to represent the entire distribution, providing an accurate description of the whole data. This value is unique and represents the collected data. Mean, median, and mode are the three frequently used measures of central tendency.

3. What makes NCERT Exemplar Solutions for Class 10 Maths Chapter 13 advantageous for students to secure good grades in the board exam?

The subject experts at CAreers360 have developed the NCERT Exemplar Solutions for Chapter 13 of Class 10 Maths, keeping in mind the learning abilities of students. The solution PDF module is downloadable from BYJU'S website according to the students' needs. All crucial concepts are explained in plain language to aid students in achieving exam success with confidence. The solutions cover all problems in the NCERT textbook, allowing students to cross-check their answers and identify their areas of weakness.

4. Are these solutions accessible offline?

Yes, the NCERT exemplar Class 10 Maths solutions chapter 13 pdf download feature provided this solution for students practicing NCERT exemplar Class 10 Maths chapter 13.

5. List out the topics and sub-topics covered in Chapter 13 NCERT Exemplar Solutions for Class 10 Maths.

Below is a list of the topics and sub-topics included in Chapter 13 of NCERT Exemplar Solutions for Class 10 Maths:

Calculation of mean for grouped data using direct method, assumed mean method and step-deviation method
Determination of mode for the given dataset
Computation of median for grouped data
Graphical representation of cumulative frequency distribution.

Also Read

NCERT Solutions for Exercise 12.3 Class 10 Maths Chapter 12 - Areas related to circles

NCERT Solutions for Exercise 12.2 Class 10 Maths Chapter 12 - Areas related to circles

NCERT Solutions for Exercise 12.1 Class 10 Maths Chapter 12 - Areas related to circles

NCERT Solutions for Class 10 Science Chapter 6 Life Processes

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

my son's dob is 13/02/2009 is hi eligible for 10th board exam in 2024,,from cbse. olz reply me

The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.

Read Complete Answer

Shubhangi 29 July,2022

my dob is 16/06/2010 I can give board exam of cbse class 10th

According to CBSE norms, a student must be 14 years old by the end of the year in which the exam will be held in order to sit for the 10th board exam. If your age is greater than 14, however, there are no limits. Therefore, based on your DOB, you will be 12 years and 6 months old by the end of December 2022. After the actual 16 June 2024, you'll be qualified to take your 10th board.

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Shubhangi 29 July,2022

Cbse class 10th term 2 result date 2022

Hello aspirant,

Central Board Of Secondary Education(CBSE) is likely to declare class 10 and 12 terms 2 board result 2022 by July 15. The evaluation process is underway. Students are demanding good results and don't want to lack behind. They are requesting the board to use their best scores in Term 1 and Term 2 exams to prepare for the results.

CBSE concluded board exams 2022 for 10, and 12 on June 15 and May 24. Exams for both classes began on April 26. A total of 35 lakh students including 21 lakh class 10 students and 14 lakh class 12 students appeared in exams and are awaiting their results.

You can look for your results on websites- cbse.gov (//cbse.gov) .in, cbseresults.nic.in

Thank you

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Nisha yadav 01 July,2022

I have one doubt I forgot to write question paper code and set no on answer sheet is their will be any problem in checking or my copy will be rejected .plz reply(cbse board)

Sir, did you get any problem in result
I have also done this same mistake in board 2023 exam
Please reply because I am panic regarding this problem

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harsh200725 05 March,2023

sir my daughter date of birth 27.02. 2010 hai march2022 admission CBSE bord school 9th me ho jaygga ki nhi. 10th ke liye paresani to nhi hogi

Hello SIR CBSE board 9th class admission 2022 Kara Raha hu entrance ki problem hai please show the Entrance for Baal vidya mandir school sambhal

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Popular CBSE Class 10th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms^-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×10⁷J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms⁻² :

Option 1)

2.45×10⁻³ kg

Option 2)

6.45×10⁻³ kg

Option 3)

9.89×10⁻³ kg

Option 4)

12.89×10⁻³ kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

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Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

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Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol^-1) is

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t²) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m² , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1) less than 3	Option 2) more than 3 but less than 6
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Class	Marks (x_i)	Frequency(f_i)	f_ixi
1-3	2	9	18
3-5	4	22	88
5-7	6	27	162
7-10	8.5	17	144.5
		$\sum f_{i}= 75$	$\sum f_{i}x_{i}= 412\cdot 5$

No.of pages written per day	_{no.of days(fi)}	(xi)	f_ixi
16-18	1	17	17
19-21	3	20	60
22-24	4	23	92
25-27	9	26	234
28-30	13	29	377
	$\sum f_{i}= 30$		$\sum f_{i}x_{i}= 780$

Marks	xi	fi
0-10	5	100-90 =10	50
10-20	15	90-75 = 15	225
20-30	25	75-50 = 25	625
30-40	35	50-25 =25	875
40-50	45	25-15 =10	450
50-60	55	15-5 = 10	550
60-70	65	5-0 = 5	325
		$\sum f_{i}= 100$	$\sum f_{i}x_{i}= 3100$

Weight	xi	fi	fixi
200-201	200.5	13	2606.5
201-202	201.5	27	5440.5
202-203	202.5	18	3645.0
203-204	203.5	10	2035.0
204-205	204.5	1	204.5
205-206	205.5	1	205.5
		$\sum f_{i}= 70$	$\sum f_{i}x_{i}= 14137$

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NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability

NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability: Exercise-13.1

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NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability: Exercise-13.3

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