NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability

NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability

Komal MiglaniUpdated on 28 Mar 2025, 09:06 PM IST

Every day we make decisions based on data: whether checking cricket scores, tracking business patterns, or knowing if we need an umbrella for the day. That is the role of Statistics—it helps us arrange, interpret, and conclude based on data, especially in understanding patterns and trends. With the help of central tendencies such as mean, mode, and median, as well as data representation through graphs and distribution of frequencies, we can analyse and process the data to obtain the desired result. Probability determines the chances of the occurrence of an event. In this chapter, you will learn about the types of events, like certain events, impossible events, and sample spaces. Through the solution of the NCERT exemplar class 10 chapter 13 on statistics and probability, you will be able to do such an analysis.

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  1. NCERT Exemplar Class 10 Maths Solutions Chapter 13:
  2. NCERT Class 10 Maths Exemplar Solutions for Other Chapters:
  3. Importance of Solving NCERT Exemplar Class 10 Maths Solutions Chapter 13
  4. NCERT solutions of class 10 - Subject Wise
  5. NCERT Notes of class 10 - Subject Wise
  6. NCERT Books and NCERT Syllabus
  7. NCERT Class 10 Exemplar Solutions - Subject Wise

To understand this topic, students must try the exemplar questions from NCERT, past board questions, and sample papers from previous years. The better students understand the concepts, the better they will problem-solve in an exam and the real world. Remember to stay within the CBSE Class 10 Maths syllabus, and if you practice every day, you will build up the confidence to solve both statistics and probability.

NCERT Exemplar Class 10 Maths Solutions Chapter 13:

Class 10 Maths Chapter 13 exemplar solutions Exercise: 13.1
Page number: 157-161
Total questions: 26

Question:1

In the formula $\bar{x}= a+\frac{\sum f_{i}d_{i}}{\sum f_{i}}$
For finding the mean of grouped data, $d_i$’s are deviations from

(A) lower limits of the classes
(B) upper limits of the classes
(C) midpoints of the classes
(D) frequencies of the class marks

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

Question:2

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

Question:3

If xi’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: [A]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate the 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, ui =
(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 the mean. First of all, add up all the observations and then divide by the total number of observations.
Also we know that di = xi – a and $u_{i}= \frac{d_{i}}{h}$
$u_{i}= \frac{d_{i}}{h}$
put di = xi – 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 three above

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 a graph of less than type and of more than type grouped data and find the intersection point, then the value at the 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 the lower limits of the median class and the modal class is
(A) 15
(B) 25
(C) 30
(D) 35

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.

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 the maximum frequency is the modal class, which is 15 – 20
The lower limit of the median class = 10
The lower limit of the 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 ) 17.5 (C) 18 (D) 18.5

Answer: [B]

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 is 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
The upper limit of the 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 the 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: [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 the maximum frequency is the 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: [C]
Solution: Frequency:- The number of times an event occurs in 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: [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. [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 = $\frac{Number \, of\, favourable\, case}{Total\, number\, of\, cases}$
$\Rightarrow$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 the 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 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
$\Rightarrow$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 a percentage of an event A is between 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 does not represent the probability of event A because the probability of an event can never be less than 0.
(B) P(A) > 1
It does not represent the probability of event A because the probability of an event can never be greater than 1.
(C) 0 ≤ P(A) ≤ 1
It represents the probability of event A because the probability of an event always lies between 0 to 1.
(D) –1 ≤ P(A) ≤ 1
It does not represent the probability of event A because the 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 selection of a card from 52 cards.
The probability that it is a red card is p(A)
$P\left ( A \right )= \frac{Number\, of\, favourable\, cases}{Total\, number\, of\, cases}$
$\Rightarrow$$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 contains 53 Sundays, then the 1 day of the year must be Sunday.
But there are a total of 7 days.
Hence, the total number of favourable 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:

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 no. of cases = 6
odd number less than 3 = 1
Number of favorable cases = 1
Probability = $= \frac{1}{6}$

Question:21

A card is drawn from a deck of 52 cards. The event E is that the 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 favourable 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 be to get a bad egg.
So, p (A) = 0.035 (given)
P(A) = $\frac{Number\, of\, favorable\ cases }{Total\, number\, of\, cases}$
$\Rightarrow$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) $= \frac{Number\, of\, tickets\, the\, bought}{Total\, tickets}$
$\Rightarrow$0.08 × 6000 = Number of tickets the bought
Number of tickets they 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 a number multiple of 5.
p(A) = $\frac{Number\, of\, tickets\, the\, bought}{Total\, tickets}$
$\Rightarrow$$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 the rest from house E. A single student is selected at random to be the class monitor. The probability that the selected 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
$p\left ( A \right )= \frac{Student\, from\, C,D}{Total\, students}$
$\Rightarrow$$p\left ( A \right )= \frac{6}{23}$

Class 10 Maths Chapter 13 exemplar solutions Exercise: 13.2
Page number: 161-163
Total questions: 14

Question:1

The median of 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 a reason.

Answer:

Answer. [False]
Solution. Grouped data are data formed by aggregating individual observations of a variable into groups.
Ungrouped data:- The data that is not grouped is called ungrouped data.
The median is the middle number in the grouped data, but when the data is ungrouped, the median is also changed.
Hence, the median is not the same for 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. Is 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 the mean. First of all, add up all the numbers, then divide by how many numbers there are.
This last statement is not correct because a can be any point in the grouped data; it is not necessary that a must be the midpoint.
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 the 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 the highest frequency it can happen that the median class is of the highest frequency.
So the given statement is false, the median class and mode class can be the same.

Question:5

In a family having three children, there may be no girls, 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 contains 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
The 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 $\frac{1}{2}$. Is this 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 the probability of no heads is $\frac{1}{4}$. What is wrong with this conclusion?

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 cases with no head = TTT
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of no head =$\frac{1}{8}$
The conclusion that the probability of no head is $\frac{1}{4}$ is wrong because, as we calculated above, it comes out to $\frac{1}{8}$. 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, which means that we never get a tail. But this is not true because we have both heads and tails on a coin. Hence probability of getting a head is 1 is false.

Question:12

Sushma tosses a coin 3 times and gets tails each time. Do you think that the outcome of the 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 tails on the fourth toss. She can also get a head.

Question:13

If I toss a coin 3 times and get heads each time, should I expect a tail to have a higher chance in the 4th toss? Give a 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, the probability of both outcomes is $\frac{1}{2}$ .
Hence tail does not have a higher chance than the head.
Both have an 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 an 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 $\frac{1}{2}$.

Class 10 Maths Chapter 13 exemplar solutions Exercise: 13.3
Page number: 166-174
Total questions: 42

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 the mean by following the given steps:

  1. Find the midpoint of each interval.

  2. Multiply the frequency of each interval by its midpoint.

  3. Get the sum of all the frequencies (f) and sum of all the (fx)

  4. Now divide the sum of (fx) by the sum of (f)

    Class

    Marks (xi)

    Frequency(fi)

    fixi

    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 the mean by following the given steps:

  1. Find the midpoint of each interval.

  2. Multiply the frequency of each interval by its midpoint.

  3. Get the sum of all the frequencies (f) and the sum of all the (fx)
    4 Now divide the sum of (fx) by the sum of (f)

Marks

xi

No.of students fi

fixi

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 the mean by following the given steps:

  1. Find the midpoint of each interval.

  2. Multiply the frequency of each interval by its midpoint.

  3. Get the sum of all the frequencies (f) and the sum of all the (fx)

  4. Now divide the sum of (fx) by the sum of (f)

Class

xi

fi

fi xi

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 the mean by following the given steps:

  1. Find the midpoint of each interval.

  2. Multiply the frequency of each interval by its midpoint.

  3. Get the sum of all the frequencies (f) and sthe um of all the (fx)

  4. Now divide the sum of (fx) by the sum of (f)

    No.of pages written per day

    no.of days(fi)

    (xi)

    fixi

    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 the mean by following the given steps:

  1. Find the midpoint of each interval.

  2. Multiply the frequency of each interval by its midpoint.

  3. Get the sum of all the frequencies (f) and the sum of all the (fx)

  4. Now divide the sum of (fx) by the sum of (f)

    Income (in Rs )

    xi

    No.of employees

    fixi

    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 the mean by following the given steps:

  1. Find the midpoint of each interval.

  2. Multiply the frequency of each interval by its midpoint.

  3. Get the sum of all the frequencies (f) and the sum of all the (fx)

  4. Now divide the sum of (fx) by the sum of (f)

Number of seats

Frequency fi

xi

fixi

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 the mean by following the given steps:

  1. Find the midpoint of each interval.

  2. Multiply the frequency of each interval by its midpoint.

  3. Get the sum of all the frequencies (f) and the sum of all the (fx)

  4. Now divide the sum of (fx) by the sum of (f)

    Weight

    fi

    xi

    fixi

    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 the mean by following
1. Find the midpoint of each interval.
2. Multiply the frequency of each interval by its midpoint.
3. Get the sum of all the frequencies (f) and the sum of all the (fx)
4. Now divide the sum of (fx) by the sum of (f)

MIleage (km/I)

No.of cars (fi)

xi

fixi

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:

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.

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:

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

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:

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.

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 )
$\Rightarrow$12 + b = 25
$\Rightarrow$b = 25 – 12
$\Rightarrow$b = 13
$\Rightarrow$25 + 10 = c
$\Rightarrow$35= c
$\Rightarrow$c + d = 43
$\Rightarrow$35 + d = 43
$\Rightarrow$d = 43 – 35
$\Rightarrow$d=8
$\Rightarrow$43 + e = 48
$\Rightarrow$e = 48 – 43
$\Rightarrow$e =5
$\Rightarrow$48+2 = f
$\Rightarrow$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:

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

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:

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

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
$\frac{n}{2}= \frac{600}{2}= 300$
$\iota$= 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 = $\frac{8}{999}\approx 0\cdot 008$

Importance of Solving NCERT Exemplar Class 10 Maths Solutions Chapter 13

NCERT Exemplar Class 10 Maths Solutions Chapter 13 pdf downloads are available through online tools for the students to access this content in an offline version, so that no breaks in continuity are faced while practicing NCERT Exemplar Class 10 Maths Chapter 13.

  • These Class 10 Maths NCERT exemplar chapter 13 solutions provide a basic knowledge of statistics and probability, which has great importance in higher classes.

  • The questions based on statistics and probability can be practised in a better way, along with these solutions.

  • The NCERT exemplar Class 10 Maths chapter 13 solution statistics and probability has a good amount of problems for practice and is sufficient for a student to solve the questions of other books.

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NCERT solutions of class 10 - Subject Wise

Here are the subject-wise links for the NCERT solutions of class 10:

NCERT Notes of class 10 - Subject Wise

Given below are the subject-wise NCERT Notes of class 10 :

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and NCERT syllabus for class 10:

NCERT Class 10 Exemplar Solutions - Subject Wise

Given below are the subject-wise exemplar solutions of class 10 NCERT:

Frequently Asked Questions (FAQs)

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

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.

Q: Are these solutions accessible offline?
A:

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.

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

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

  1. Direct Method, Assumed Mean Method, and Step-Deviation Method for determining the mean of grouped data.
  2. Finding the mode of the given data.
  3. Calculating the median of grouped data.
  4. Graphical representation of cumulative frequency distribution.
Q: In Chapter 13 of NCERT Exemplar Solutions for Class 10 Maths, can you elucidate the concept of mean?
A:

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.

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Hello,

It appears you are asking if you can fill out a form after passing your 10th grade examination in the 2024-2025 academic session.

The answer depends on what form you are referring to. Some forms might be for courses or examinations where passing 10th grade is a prerequisite or an eligibility criteria, such as applying for further education or specific entrance exams. Other forms might be related to other purposes, like applying for a job, which may also have age and educational requirements.

For example, if you are looking to apply for JEE Main 2025 (a competitive exam in India), having passed class 12 or appearing for it in 2025 are mentioned as eligibility criteria.

Let me know if you need imformation about any exam eligibility criteria.

good wishes for your future!!

Hello Aspirant,

"Real papers" for CBSE board exams are the previous year's question papers . You can find these, along with sample papers and their marking schemes , on the official CBSE Academic website (cbseacademic.nic.in).

For notes , refer to NCERT textbooks as they are the primary source for CBSE exams. Many educational websites also provide chapter-wise revision notes and study material that align with the NCERT syllabus. Focus on practicing previous papers and understanding concepts thoroughly.