## Maharashtra Board 8th Class Maths Practice Set 2.3 Solutions Chapter 2 Parallel Lines and Transversals

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 2.3 8th Std Maths Answers Solutions Chapter 2 Parallel Lines and Transversals.

## Practice Set 2.3 8th Std Maths Answers Chapter 2 Parallel Lines and Transversals

Question 1.
Draw a line l. Take a point A outside the line. Through point A draw a line parallel to line l.
Solution:
Steps of construction:

1. Draw a line l and take any point A outside the line.
2. Place a set-square, such that one arm of the right angle passes through A and the other arm is on line l.
3. Place the second set-square as shown in the figure such that the vertex of the right angle is at point A.
4. Hold the two set-squares in place and draw a line parallel to line l through the edge of the second set-square. Name the line as m.

Line m is the required line parallel to line l and passing through point A.

Question 2.
Draw a line l. Take a point T outside the line. Through point T draw a line parallel to line l.
Solution:
Steps of construction:

1. Draw a line l and take any point T outside the line.
2. Place a set-square, such that one arm of the right angle passes through T and the other arm is on line l.
3. Place the second set-square as shown in the figure such that the vertex of the right angle is at point T.
4. Hold the two set-squares in place and draw a line parallel to line l through the edge of the second set-square. Name the line as m.

Line m is the required line parallel to line l and passing through point T.

Question 3.
Draw a line m. Draw a line n which is parallel to line m at a distance of 4 cm from it.
Solution:
Steps of construction:

1. Draw a line m and take any two points M and N on the line.
2. Draw perpendiculars to line m at points M and N.
3. On the perpendicular lines take points S and T at a distance 4 cm from points M and N respectively.
4. Draw a line through points S and T. Name the line as n.

Line n is parallel to line m at a distance of 4 cm from it.

## Maharashtra Board 8th Class Maths Practice Set 1.2 Solutions Chapter 1 Rational and Irrational Numbers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 1.2 8th Std Maths Answers Solutions Chapter 1 Rational and Irrational Numbers.

## Practice Set 1.2 8th Std Maths Answers Chapter 1 Rational and Irrational Numbers

Question 1.
Compare the following numbers.
i. 7, -2
ii. 0, $$\frac { -9 }{ 5 }$$
iii. $$\frac { 8 }{ 7 }$$, 0
iv. $$-\frac{5}{4}, \frac{1}{4}$$
v. $$\frac{40}{29}, \frac{141}{29}$$
vi. $$-\frac{17}{20},-\frac{13}{20}$$
vii. $$\frac{15}{12}, \frac{7}{16}$$
viii. $$-\frac{25}{8},-\frac{9}{4}$$
ix. $$\frac{12}{15}, \frac{3}{5}$$
x. $$-\frac{7}{11},-\frac{3}{4}$$
Solution:
i. 7, -2
If a and b are positive numbers such that a < b, then -a > -b.
Since, 2 < 7 ∴ -2 > -7

ii. 0, $$\frac { -9 }{ 5 }$$
On a number line, $$\frac { -9 }{ 5 }$$ is to the left of zero.
∴ 0 > $$\frac { -9 }{ 5 }$$

iii. $$\frac { 8 }{ 7 }$$, 0
On a number line, zero is to the left of $$\frac { 8 }{ 7 }$$ .
∴ $$\frac { 8 }{ 7 }$$ > 0

iv. $$-\frac{5}{4}, \frac{1}{4}$$
We know that, a negative number is always less than a positive number.
∴ $$-\frac{5}{4}<\frac{1}{4}$$

v. $$\frac{40}{29}, \frac{141}{29}$$
Here, the denominators of the given numbers are the same.
Since, 40 < 141
∴ $$\frac{40}{29}<\frac{141}{29}$$

vi. $$-\frac{17}{20},-\frac{13}{20}$$
Here, the denominators of the given numbers are the same.
Since, 17 < 13
∴ -17 < -13
∴ $$-\frac{17}{20}<-\frac{13}{20}$$

vii. $$\frac{15}{12}, \frac{7}{16}$$
Here, the denominators of the given numbers are not the same.
LCM of 12 and 16 = 48

Alternate method:
15 × 16 = 240
12 × 7 = 84
Since, 240 > 84
∴ 15 × 16 > 12 × 7

viii. $$-\frac{25}{8},-\frac{9}{4}$$
Here, the denominators of the given numbers are not the same.
LCM of 8 and 4 = 8

ix. $$\frac{12}{15}, \frac{3}{5}$$
Here, the denominators of the given numbers are not the same.
LCM of 15 and 5 = 15

x. $$-\frac{7}{11},-\frac{3}{4}$$
Here, the denominators of the given numbers are not the same.
LCM of 11 and 4 = 44

#### Maharashtra Board Class 8 Maths Solutions Chapter 1 Rational and Irrational Numbers Practice Set 1.2 Questions and Activities

Question 1.
Verify the following comparisons using a number line. (Textbook pg. no, .3)
i. 2 < 3 but – 2 > – 3
ii. $$\frac{5}{4}<\frac{7}{4}$$ but $$\frac{-5}{4}<\frac{-7}{4}$$
Solution:

We know that, on a number line the number to the left is smaller than the other.
∴ 2 < 3 and -3 < -2
i.e. 2 < 3 and -2 > -3

## Maharashtra Board 9th Class Maths Part 1 Problem Set 7 Solutions Chapter 7 Statistics

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 7 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.

## Problem Set 7 Algebra 9th Std Maths Part 1 Answers Chapter 7 Statistics

Question 1.
Write the correct alternative answer for each of the following questions.

i. Which of the following data is not primary ?
(A) By visiting a certain class, gathering information about attendance of students.
(B) By actual visit to homes, to find number of family members.
(C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.
(D) Review the cleanliness status of canals by actually visiting them.
(C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.

ii. What is the upper class limit for the class 25 – 35?
(A) 25
(B) 35
(C) 60
(D) 30
(B) 35

iii. What is the class-mark of class 25 – 35?
(A) 25
(B) 35
(C) 60
(D) 30
(D) 30

iv. If the classes are 0 – 10, 10 – 20, 20 – 30, …, then in which class should the observation 10 be included?
(A) 0 – 10
(B) 10 – 20
(C) 0 – 10 and 10-20 in these 2 classes
(D) 20 – 30
(B) 10 – 20

v. If $$\overline { x }$$ is the mean of x1, x2, ……. , xn and $$\overline { y }$$ is the mean of y1, y2, ….. yn and $$\overline { z }$$ is the mean of x1,x2, …… , xn , y1, y2, …. yn , then z = ?

x1, x2, x3, ……. , xn
∴ $$\overline{x}=\frac{\sum x}{\mathrm{n}}$$
∴ n$$\overline{x}$$ = ∑x
Similarly, n$$\overline{y}$$ = ∑y
Now,

$$\text { (A) } \frac{\overline{x}+\overline{y}}{2}$$

vi. The mean of five numbers is 50, out of which mean of 4 numbers is 46, find the 5th number.
(A) 4
(B) 20
(C) 434
(D) 66
5th number = Sum of five numbers – Sum of four numbers
= (5 x 50) – (4 x 46)
= 250 – 184
= 66
(D) 66

vii. Mean of 100 observations is 40. The 9th observation is 30. If this is replaced by 70 keeping all other observations same, find the new mean.
(A) 40.6
(B) 40.4
(C) 40.3
(D) 40.7
New mean = $$\frac { 4000-30+70 }{ 100 }$$
= 40.4
(B) 40.4

viii. What is the mode of 19, 19, 15, 20, 25, 15, 20, 15?
(A) 15
(B) 20
(C) 19
(D) 25
(A) 15

ix. What is the median of 7, 10, 7, 5, 9, 10 ?
(A) 7
(B) 9
(C) 8
(D) 10
(C) 8

x. From following table, what is the cumulative frequency of less than type for the class 30 – 40?

(A) 13
(B) 15
(C) 35
(D) 22
Cumulative frequency of less than type for the class 30 – 40 = 7 + 3 + 12 + 13 = 35
(C) 35

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

Question 3.
The mean of nine numbers is 77. If one more number is added to it, then the mean increases by 5. Find the number added in the data.
Solution:
∴ $$\text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}$$
∴ The sum of all observations = Mean x Total number of observations mean of nine numbers is 77
∴ sum of 9 numbers = 11 x 9 = 693 …(i)
If one more number is added, then mean increases by 5
mean of 10 numbers = 77 + 5 = 82
∴ sum of the 10 numbers = 82 x 10 = 820 …(ii)
∴ Number added = sum of the 10 numbers – sum of the 9 numbers = 820 – 693 … [From (i) and (ii)]
= 127
∴ The number added in the data is 127.

Question 4.
The monthly maximum temperature of a city is given in degree Celsius in the following data. By taking suitable classes, prepare the grouped frequency distribution table
29.2, 29.0, 28.1, 28.5, 32.9, 29.2, 34.2, 36.8, 32.0, 31.0, 30.5, 30.0, 33, 32.5, 35.5, 34.0, 32.9, 31.5, 30.3, 31.4, 30.3, 34.7, 35.0, 32.5, 33.5.29.0. 29.5.29.9.33.2.30.2
From the table, answer the following questions.
i. For how many days the maximum temperature was less than 34°C?
ii. For how many days the maximum temperature was 34°C or more than 34°C?
Solution:

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

Question 5.
If the mean of the following data is 20.2, then find the value of p.

Solution:

∴ 20.2 (30 + p) = 610 + 20p
∴ 606 + 20.2p = 610 + 20p
∴ 20.2p – 20p = 610 – 606
∴ 0.2p = 4
∴ p = $$\frac { 4 }{ 0.2 }$$ = $$\frac { 40 }{ 2 }$$ = 20
∴ p = 20

Question 6.
There are 68 students of 9th standard from Model Highschool, Nandpur. They have scored following marks out of 80, in written exam of mathematics.
70, 50, 60, 66, 45, 46, 38, 30, 40, 47, 56, 68,
80, 79, 39, 43, 57, 61, 51, 32, 42, 43, 75, 43,
36, 37, 61, 71, 32, 40, 45, 32, 36, 42, 43, 55,
56, 62, 66, 72, 73, 78, 36, 46, 47, 52, 68, 78,
80, 49, 59, 69, 65, 35, 46, 56, 57, 60, 36, 37,
45, 42, 70, 37,45, 66, 56, 47
By taking classes 30 – 40, 40 – 50, …. prepare the less than type cumulative frequency table. Using the table, answer the following questions:

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

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

Question 7.
By using data in example (6), and taking classes 30 – 40, 40 – 50,… prepare equal to or more than type cumulative frequency table and answer the following questions based on it.
i. How many students have scored marks 70 or more than 70?
ii. How many students have scored marks 30 or more than 30?
Solution:

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

Question 8.
There are 10 observations arranged in ascending order as given below.
45, 47, 50, 52, JC, JC + 2, 60, 62, 63, 74. The median of these observations is 53.
Find the value of JC. Also find the mean and the mode of the data.
Solution:
i. Given data in ascending order:
45,47, 50, 52, x, JC+2, 60, 62, 63, 74.
∴ Number of observations (n) = 10 (i.e., even)
∴ Median is the average of middle two observations
Here, the 5th and 6th numbers are in the middle position.

∴ 106 = 2x + 2
∴ 106 – 2 = 2x
∴ 104 = 2x
∴ x = 52
∴ The given data becomes:
45, 47, 50, 52, 52, 54, 60, 62, 63, 74.

∴ The mean of the given data is 55.9.

iii. Given data in ascending order:
45, 47, 50, 52, 52, 54, 60, 62, 63, 74.
∴ The observation repeated maximum number of times = 52
∴ The mode of the given data is 52.

Maharashtra Board Class 9 Maths Chapter 7 Statistics Problem Set 7 Intext Questions and Activities

Question 1.
To show following information diagrammatically, which type of bar- diagram is suitable?
i. Literacy percentage of four villages.
ii. The expenses of a family on various items.
iii. The numbers of girls and boys in each of five divisions.
iv. The number of people visiting a science exhibition on each of three days.
v. The maximum and minimum temperature of your town during the months from January to June.
vi. While driving a two-wheeler, number of people wearing helmets and not wearing helmet in 100 families.
(Textbook pg. no. 112)
Solution:
i. Percentage bar diagram
ii. Sub-divided bar diagram
iii. Sub-divided bar diagram
iv. Sub-divided bar diagram
v. Sub-divided bar diagram
vi. Sub-divided bar diagram

Question 2.
You gather information for several reasons. Take a few examples and discuss whether the data is primary or secondary.
(Textbook pg. no, 113)
[Students should attempt the above activity on their own.]

## Maharashtra Board 9th Class Maths Part 1 Practice Set 7.3 Solutions Chapter 7 Statistics

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.3 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.

## Practice Set 7.3 Algebra 9th Std Maths Part 1 Answers Chapter 7 Statistics

Question 1.
For class interval 20 – 25 write the lower class limit and the upper class limit.
Lower class limit = 20
Upper class limit = 25

Question 2.
Find the class-mark of the class 35-40.
Solution:
Class-mark

∴ Class-mark of the class 35 – 40 is 37.5

Question 3.
If class-mark is 10 and class width is 6, then find the class.
Solution:
Let the upper class limit be x and the lower class limit be y.
Class mark = 10 …[Given]
Class-mark

∴ x + y = 20 …(i)
Class width = 6 … [Given]
Class width = Upper class limit – Lower class limit
∴ x – y = 6 …(ii)
x + y = 20
x – y = 6
2x = 26
∴ x = 13
Substituting x = 13 in equation (i),
13 + y = 20
∴ y = 20 – 13
∴ y = 7
∴ The required class is 7 – 13.

Question 4.
Complete the following table.

Solution:
Let frequency of the class 14 – 15 be x then, from table,
5 + 14 + x + 4 = 35
∴ 23 + x = 35
∴ x = 35 – 23
∴ x = 12

Question 5.
In a ‘tree plantation’ project of a certain school there are 45 students of ‘Harit Sena.’ The record of trees planted by each student is given below:
3, 5, 7, 6, 4, 3, 5, 4, 3, 5, 4, 7, 5, 3, 6, 6, 5, 3, 4, 5, 7, 3, 5, 6, 4, 4, 3, 5, 6, 6, 4, 3, 5 ,7, 3, 4, 5, 7, 6, 4, 3, 5, 4, 4, 7.
Prepare a frequency distribution table of the data.
Solution:

Question 6.
The value of n upto 50 decimal places is given below:
3.14159265358979323846264338327950288419716939937510
From this information prepare an ungrouped frequency distribution table of digits appearing after the decimal point.
Solution:

Question 7.
In the tables given below, class-mark and frequencies is given. Construct the frequency tables taking inclusive and exclusive classes.
i.

ii.

Solution:
i. Let the Lower class limit and upper class limit of the class mark 5 be x and y respectively.

∴x + y = 10
Here, class width = 15 – 5 = 10
But, Class width = Upper class limit – Lower class limit
∴ y – x = 10
∴ -x + y = 10 …(ii)
x+ y = 10
-x + y = 10
∴ 2y = 20
∴ y = 10
Substituting y = 10 in equation (i),
∴ x + 10 = 10
∴ x = 0
∴ class with class-mark 5 is 0 – 10
Similarly, we can find the remaining classes.
∴ frequency table taking inclusive and exclusive classes.

ii. Let the lower class limit and upper class limit of the class mark 22 be x andy respectively.

∴ x + y = 44 …(i)
Here, class width = 24 – 22 = 2
But, Class width = Upper class limit – Lower class limit
∴ y – x = 2
∴ -x + y = 2 …. (ii)
x + y = 44
– x + y= 2
2y = 46
∴ y = 23
Substituting y = 23 in equation (i),
∴ x + 23 = 44
∴ x = 21
∴ class with class-mark 22 is 21 – 23
Similarly, we can find the remaining classes
∴ frequency table taking inclusive and exclusive classes.

Question 8.
In a school, 46 students of 9th standard, were told to measure the lengths of the pencils in their compass-boxes in Centimetres. The data collected was as follows:
16, 15, 7, 4.5, 8.5, 5.5, 5, 6.5, 6, 10, 12, 13,
4.5, 4.9, 16, 11, 9.2, 7.3, 11.4, 12.7, 13.9, 16,
5.5, 9.9, 8.4, 11.4, 13.1, 15, 4.8, 10, 7.5, 8.5,
6.5, 7.2, 4.5, 5.7, 16, 5.7, 6.9, 8.9, 9.2, 10.2, 12.3, 13.7, 14.5, 10
By taking exclusive classes 0-5, 5-10, 10-15,…. prepare a grouped frequency distribution table.
Solution:

Question 9.
In a village, the milk was collected from 50 milkmen at a collection center in litres as given below:
27, 75, 5, 99, 70, 12, 15, 20, 30, 35, 45, 80, 77,
90, 92, 72, 4, 33, 22, 15, 20, 28, 29, 14, 16, 20,
72, 81, 85, 10, 16, 9, 25, 23, 26, 46, 55, 56, 66,
67, 51, 57, 44, 43, 6, 65, 42, 36, 7, 35
By taking suitable classes, prepare grouped frequency distribution table.
Solution:

Question 10.
38 people donated to an organisation working for differently abled persons. The amount in rupees were as follows:
101, 500, 401, 201, 301, 160, 210, 125, 175,
190, 450, 151, 101, 351, 251, 451, 151, 260,
360, 410, 150, 125, 161, 195, 351, 170, 225,
260, 290, 310, 360, 425, 420, 100, 105, 170, 250, 100
i. By taking classes 100 – 149, 150 – 199, 200 – 249… prepare grouped frequency distribution table.
ii. From the table, find the number of people who donated ₹350 or more.
Solution:
i.

ii. Number of people who donated ₹ 350 or more = 4 + 4 + 2 + 1 = 11

Maharashtra Board Class 9 Maths Chapter 7 Statistics Practice Set 7.3 Intext Questions and Activities

Question 1.
The record of marks out of 20 in Mathematics in the first unit test is as follows:
20,6, 14, 10, 13, 15, 12, 14, 17. 17, 18, 1119,
9, 16. 18, 14, 7, 17, 20, 8, 15, 16, 10, 15, 12.
18, 17, 12, 11, 11, 10, 16, 14, 16, 18, 10, 7, 17,
14, 20, 17, 13, 15, 18, 20, 12, 12, 15, 10
Answer the following questions, from the above information.
a. How many students scored 15 marks?
b. How many students scored more than 15 marks?
c. How many students scored less than 15 marks?
d. What is the lowest score of the group?
e. What is the highest score of the group? (Textbook pg. no. 114)
Solution:
a. 5 students scored 15 marks.
b. 20 students scored more than 15 marks.
c. 25 students scored less than 15 marks.
d. 6 is the lowest score of the group.
e. 20 is the highest score of the group.

Question 2.
For the above Question prepare Frequency Distribution Table. (Textbook pg. no. 115)
Solution:

## Maharashtra Board 9th Class Maths Part 1 Practice Set 7.2 Solutions Chapter 7 Statistics

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.

## Practice Set 7.2 Algebra 9th Std Maths Part 1 Answers Chapter 7 Statistics

Question 1.
Classify following information as primary or secondary data.
i. Information of attendance of every student collected by visiting every class in a school
ii. The information of heights of students was gathered from school records and sent to the head office, as it was to be sent urgently.
iii. In the village Nandpur, the information collected from every house regarding students not attending school.
iv. For science project, information of trees gathered by visiting a forest.
i. Primary data
ii. Secondary data
iii. Primary data
iv. Primary data

## Maharashtra Board 9th Class Maths Part 1 Practice Set 7.1 Solutions Chapter 7 Statistics

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.

## Practice Set 7.1 Algebra 9th Std Maths Part 1 Answers Chapter 7 Statistics

Question 1.
The following table shows the number of Buses and Trucks in nearest lakh units. Draw percentage bar diagram. (Approximate the percentages to the nearest integer)

Solution:

Question 2.
In the table given below, the information is given about roads. Using this draw sub-divided and percentage bar diagram (Approximate the percentages to the nearest integer)

Solution:
i. Sub-divided bar diagram:

ii. Percentage bar diagram:

Maharashtra Board Class 9 Maths Chapter 7 Statistics Practice Set 7.1 Intext Questions and Activities

Question 1.
A farmer has produced Wheat and Jowar in his field. The following joint bar diagram shows the production of Wheat and Jowar. From the gken diagram answer the following questions: (Textbook pg. no. 108)
i. Which crop production has increased consistently in 3 years?
ii. By how many quintals the production ofjowar has reduced in 2012 as compared to 2011?
iii. What is the difference between the production of wheat in 2010 and 2012 ?
iv. Complete the following table using this diagram.

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

Question 2.
In the following table, the information of number of girls per 1000 boys is given for different states. Fill In the blanks and complete the table. (Textbook pg. no. 111)

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

Question 3.
For the above given activity, the information of number of girls per 1000 boys is given for five states. The literacy percentage of these five states is given below. Assam (73%), Bihar (64%), Punjab (77%), Kerala (94%), Maharashtra (83%). Think of the number of girls and the literacy percentages in the respective states. Can you draw any conclusions from it? (Textbook pg. no. 112)
Solution:
By observing the number of girls per 1000 boys and literacy percentages in the given respective states, we can conclude that the literacy rate of girls is least in Bihar and is highest in Kerala.

## Maharashtra Board 9th Class Maths Part 1 Practice Set 3.4 Solutions Chapter 3 Polynomials

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.4 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

## Practice Set 3.4 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials

Question 1.
For x = 0, find the value of the polynomial x2 – 5x + 5.
Solution:
p(x) = x2 – 5x + 5
Put x = 0 in the given polynomial.
∴ P(0) = (0)2 – 5(0) + 5
= 0 – 0 + 5
∴ p(0) = 5

Question 2.
If p(y) = y2 – 3√2 + 1, then find p( 3√2 ).
Solution:
p(y) = y2 – 3√2 y + 1
Putp= 3√2 in the given polynomial.
∴ p( 3√2 ) = (3√2 )2 – 3√2 (3√2 ) + 1
= 9 x 2 – 9 x 2 + 1
= 18 – 18 + 1
∴ p( 3√2 ) = 1

Question 3.
If p(m) = m3 + 2m2 – m + 10, then P(a) + p(-a) = ?
Solution:
p(m) = m3 + 2m2 – m + 10
Put m = a in the given polynomial.
∴ p(a) = a3 + 2a2 – a + 10 …(i)
Put m = -a in the given polynomial.
p(-a) = (-a)3 + 2(-a)2 – (-a) +10
∴ p (-a) = -a3 + 2a2 + a + 10 …(ii)
p(a) + p(-a) = (a3 + 2a2 – a + 10) + (-a3 + 2a2 + a + 10)
= a3 – a3 + 2a2 + 2a2a + a + 10 + 10
∴ p(a) + p(-a) = 4a2 + 20

Question 4.
If p(y) = 2y3 – 6y2 – 5y + 7, then find p(2).
Solution:
p(y) = 2y3 – 6y2 – 5y + 7
Put y = 2 in the given polynomial.
∴ p(2) = 2(2)3 – 6(2)2 – 5(2) + 7
= 2 x 8 – 6 x 4 – 10 + 7
= 16 – 24 – 10 + 7
∴ P(2) = -11

## Maharashtra Board 9th Class Maths Part 1 Practice Set 3.3 Solutions Chapter 3 Polynomials

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.3 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

## Practice Set 3.3 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials

Question 1.
Divide each of the following polynomials by synthetic division method and also by linear division method. Write the quotient and the remainder.
i. (2m2 – 3m + 10) ÷ (m – 5)
ii. (x4 + 2x3 + 3x2 + 4x + 5) ÷ (x + 2)
iii. (y3 – 216) ÷ (y – 6)
iv. (2x4 + 3x3 + 4x – 2x2) ÷ (x + 3)
v. (x4 – 3x2 – 8) ÷ (x + 4)
vi. (y3 – 3y2 + 5y – 1) ÷ (y – 1)
Solution:
i. Synthetic division:
(2m2 – 3m + 10) ÷ (m – 5)
Dividend = 2m² – 3m + 10
∴ Coefficient form of dividend = (2, -3, 10)
Divisor = m – 5
∴ Opposite of -5 is 5.

Coefficient form of quotient = (2, 7)
∴ Quotient = 2m + 7,
Remainder = 45
Linear division method:
2m2 – 3m + 10
To get the term 2m2, multiply (m – 5) by 2m and add 10m,
= 2m(m – 5) + 10m- 3m + 10
= 2m(m – 5) + 7m + 10
To get the term 7m, multiply (m – 5) by 7 and add 35
= 2m(m – 5) + 7(m- 5) + 35+ 10
= (m – 5) (2m + 7) + 45
∴ Quotient = 2m + 7,
Remainder = 45

ii. Synthetic division:
(x4 + 2x3 + 3x2 + 4x + 5) ÷ (x + 2)
Dividend = x4 + 2x3 + 3x2 + 4x + 5
∴ Coefficient form of dividend = (1, 2, 3, 4, 5)
Divisor = x + 2
∴ Opposite of + 2 is -2.

Coefficient form of quotient = (1, 0, 3, -2)
∴ Quotient = x3 + 3x – 2,
Remainder = 9

Linear division method:
x4 + 2x3 + 3x2 + 4x + 5
To get the term x4, multiply (x + 2) by x3 and subtract 2x3,
= x3(x + 2) – 2x3 + 2x3 + 3x2 + 4x + 5
= x3(x + 2) + 3x2 + 4x + 5
To get the term 3x2, multiply (x + 2) by 3x and subtract 6x,
= x3(x + 2) + 3x(x + 2) – 6x + 4x + 5
= x3(x + 2) + 3x(x + 2) – 2x + 5
To get the term -2x, multiply (x + 2) by -2 and add 4,
= x3(x + 2) + 3x(x + 2) – 2(x + 2) + 4 + 5
= (x + 2) (x3 + 3x – 2) + 9
∴ Quotient = x3 + 3x – 2,
Remainder – 9

iii. Synthetic division:
(y3 – 216) ÷ (y – 6)
Dividend = y3 – 216
∴ Index form = y3 + 0y3 + 0y – 216
∴ Coefficient form of dividend = (1, 0, 0, -216)
Divisor = y – 6
∴ Opposite of – 6 is 6.

Coefficient form of quotient = (1, 6, 36)
∴ Quotient = y2 + 6y + 36,
Remainder = 0

Linear division method:
y3 – 216
To get the term y3, multiply (y – 6) by y2 and add 6y2,
= y2(y – 6) + 6y2 – 216
= y2(y – 6) + 6ysup>2 – 216
To get the, term 6 y2 multiply (y – 6) by 6y and add 36y,
= y2(y – 6) + 6y(y – 6) + 36y – 216
= y2(y – 6) + 6y(y – 6) + 36y – 216
To get the term 36y, multiply (y- 6) by 36 and add 216,
= y2 (y – 6) + 6y(y – 6) + 36(y – 6) + 216 – 216
= (y – 6) (y2 + 6y + 36) + 0
Quotient = y2 + 6y + 36
Remainder = 0

iv. Synthetic division:
(2x4 + 3x3 + 4x – 2x2) ÷ (x + 3)
Dividend = 2x4 + 3x3 + 4x – 2x2
∴ Index form = 2x4 + 3x3 – 2x2 + 4x + 0
∴ Coefficient form of the dividend = (2,3, -2,4,0)
Divisor = x + 3
∴ Opposite of + 3 is -3

Coefficient form of quotient = (2, -3, 7, -17)
∴ Quotient = 2x3 – 3x2 + 7x – 17,
Remainder = 51

Linear division method:
2x4 + 3x3 + 4x – 2x2 = 2x2 + 3x3 – 2x2 + 4x
To get the term 2x4, multiply (x + 3) by 2x3 and subtract 6x3,
= 2x3(x + 31 – 6x3 + 3x3 – 2x2 + 4x
= 2x3(x + 3) – 3x3 – 2x2 + 4x

To get the term – 3x3, multiply (x + 3) by -3x2 and add 9x2,
= 2x3(x + 3) – 3x2(x + 3) + 9x2 – 2x2 + 4x
= 2x3(x + 3) – 3x2(x + 3) + 7x2 + 4x

To get the term 7x2, multiply (x + 3) by 7x and subtract 21x,
= 2x3(x + 3) – 3x2(x + 3) + 7x(x + 3) – 21x + 4x
= 2x3(x + 3) – 3x2(x + 3) + 7x(x + 3) – 17x

To get the term -17x, multiply (x + 3) by -17 and add 51,
= 2x3(x + 3) – 3x2(x + 3) + 7x(x+3) – 17(x + 3) + 51
= (x + 3) (2x3 – 3x2 + 7x- 17) + 51
∴ Quotient = 2x3 – 3x2 + 7x – 17,
Remainder = 51

v. Synthetic division:
(x4 – 3x2 – 8) + (x + 4)
Dividend = x4 – 3x2 – 8
∴ Index form = x4 + 0x3 – 3x2 + 0x – 8
∴ Coefficient form of the dividend = (1,0, -3,0, -8)
Divisor = x + 4
∴ Opposite of + 4 is -4

∴ Coefficient form of quotient = (1, -4, 13, -52)
∴ Quotient = x3 – 4x2 + 13x – 52,
Remainder = 200

Linear division method:
x4 – 3x2 – 8
To get the term x4, multiply (x + 4) by x3 and subtract 4x3,
= x3(x + 4) – 4x3 – 3x2 – 8
= x3(x + 4) – 4x3 – 3x2 – 8
To get the term – 4x3, multiply (x + 4) by -4x2 and add 16x2,
= x3(x + 4) – 4x2 (x + 4) + 16x2 – 3x2 – 8
= x3(x + 4) – 4x2 (x + 4) + 13x2 – 8
To get the term 13x2, multiply (x + 4) by 13x and subtract 52x,
= x3(x + 4) – 4x2(x + 4) + 13x(x + 4) – 52x – 8
= x3(x + 4) – 4x2(x + 4) + 13x(x + 4) – 52x – 8
To get the term -52x, multiply (x + 4) by – 52 and add 208,
= x3(x + 4) – 4x2(x + 4) + 13x(x + 4) – 52(x + 4) + 208 – 8
= (x + 4) (x3 – 4x2 + 13x – 52) + 200
∴ Quotient = x3 – 4x2 + 13x – 52,
Remainder 200

vi. Synthetic division:
(y3 – 3y2 + 5y – 1) ÷ (y – 1)
Dividend = y3 – 3y2 + 5y – 1
Coefficient form of the dividend = (1, -3, 5, -1)
Divisor = y – 1
∴Opposite of -1 is 1.

∴ Coefficient form of quotient = (1, -2, 3)
∴ Quotient = y2 – 2y + 3,
Remainder = 2

Linear division method:
y3 -3y2 + 5y – 1
To get the term y3 , multiply (y – 1) by y2 and add y2
= y2 (y – 1) + y2 – 3y2 + 5y – 1
= y2 (y – 1) – 2y2 + 5y – 1
To get the term -2y2, multiply (y – 1) by -2y and subtract 2y,
= y2 (y – 1) – 2y(y – 1) – 2y + 5y – 1
= y2 (y – 1) – 2y(y – 1) + 3y – 1
To get the term 3y, multiply (y – 1) by 3 and add 3,
= y2 (y – 1) – 2y(y – 1) + 3(y- 1) + 3 – 1
= (y – 1)(y2 – 2y + 3) + 2
∴ Quotient = y2 – 2y + 3,
Remainder = 2.

## Maharashtra Board 9th Class Maths Part 1 Practice Set 3.2 Solutions Chapter 3 Polynomials

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

## Practice Set 3.2 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials

Question 1.
Use the given letters to write the answer.
i. There are ‘a’ trees in the village Lat. If the number of trees increases every year by ’b‘. then how many trees will there be after ‘x’ years?
ii. For the parade there are y students in each row and x such row are formed. Then, how many students are there for the parade in all ?
iii. The tens and units place of a two digit number is m and n respectively. Write the polynomial which represents the two digit number.
Solution:
i. Number of trees in the village Lat = a
Number of trees increasing each year = b
∴ Number of trees after x years = a + bx
∴ There will be a + bx trees in the village Lat after x years.

ii. Total rows = x
Number of students in each row = y
∴ Total students = Total rows × Number of students in each row
= x × y
= xy
∴ There are in all xy students for the parade.

iii. Digit in units place = n
Digit in tens place = m
∴ The two digit number = 10 x digit in tens place + digit in units place
= 10m + n
∴ The polynomial representing the two digit number is 10m + n.

Question 2.
i. x3 – 2x2 – 9; 5x3 + 2x + 9
ii. -7m4+ 5m3 + √2 ; 5m4 – 3m3 + 2m2 + 3m – 6
iii. 2y2 + 7y + 5; 3y + 9; 3y2 – 4y – 3
Solution:
i. (x3 – 2x2 – 9) + (5x3 + 2x + 9)
= x3 – 2x2 – 9 + 5x3 + 2x + 9
= x3 + 5x3 – 2x2 + 2x – 9 + 9
= 6x3 – 2x2 + 2x

ii. (-7m4 + 5m3 + √2 ) + (5m4 – 3m3 + 2m2 + 3m – 6)
= -7m4 + 5m3 + √2 + 5m4 – 3m3 + 2m2 + 3m – 6
= -7m4 + 5m4 + 5m3 – 3m3 + 2m2 + 3m +√2 – 6
= -2m4 + 2m3 + 2m2 + 3m + √2 – 6

iii. (2y2 + 7y + 5) + (3y + 9) + (3y2 – 4y – 3)
= 2y2 + 7y + 5 + 3y + 9 + 3y2 – 4y – 3
= 2y2 + 3y2 + 7y + 3y – 4y + 5 + 9 – 3
= 5y2 + 6y + 11

Question 3.
Subtract the second polynomial from the first.
i. x2 – 9x + √3 ; – 19x + √3 + 7x2
ii. 2ab2 + 3a2b – 4ab; 3ab – 8ab2 + 2a2b
Solution:
i. x2 – 9x + √3 -(- 19x + √3 + 7x2)
= x2 – 9x + √3 + 19x – √ 3 – 7x2
= x2 – 7x29x + 19x + √3 – √3
= – 6x2 + 10x

ii. (2ab2 + 3a2b – 4ab) – (3ab – 8ab2 + 2a2b)
= 2ab2 + 3a2b – 4ab – 3ab + 8ab2 – 2a2b
= 2ab2 + 8ab2 + 3a2b – 2a2b 4ab – 3ab
= 10ab2 + a2b – 7ab

Question 4.
Multiply the given polynomials.
i. 2x; x2 – 2x – 1
ii. x5 – 1; x3 + 2x2 + 2
iii. 2y +1; y2 – 2y + 3y
Solution:
i. (2x) x (x2 – 2x – 1) = 2x3 – 4x2 – 2x

ii. (x5 – 1) × (x3 + 2x2 + 2)
= x5 (x3 + 2x2 + 2) -1(x3 + 2x2 + 2)
= x8 + 2x7 + 2x5 – x3 – 2x2 – 2

iii. (2y + 1) × (y2 – 2y3 + 3y)
= 2y(y2 – 2y3 + 3y) + 1(y2 – 2y3 + 3y)
= 2y3 – 4y4 + 6y2 + y2 – 2y3 + 3y
= -4y4 + 2y3 – 2y3 + 6y2 + y2 + 3y
= -4y4 + 7y2 + 3y

Question 5.
Divide first polynomial by second polynomial and write the answer in the form ‘Dividend = Divisor x Quotient + Remainder’.
i. x3 – 64; x – 4
ii. 5x5 + 4x4 – 3x3 + 2x2 + 2 ; x2 – x
Solution:
i. x3 – 64 = x3 + 0x2 + 0x – 64

∴ Quotient = x2 + 4x + 16, Remainder = 0
Now, Dividend = Divisor x Quotient + Remainder
∴ x3 – 64 = (x – 4)(x2 + 4x + 16) + 0

ii. 5x5 + 4x4 – 3x3 + 2x2 + 2 = 5x5 + 4x4 – 3x3 + 2x + 0x + 2

∴ Quotient = 5x3 + 9x2 + 6x + 8,
Remainder = 8x + 2
Now, Dividend = Divisor x Quotient + Remainder
∴ 5x5 + 4x4 – 3x3 + 2x2 + 2 = (x2 – x)(5x3 + 9x2 + 6x + 8) + (8x + 2)

Question 6.
Write down the information in the form of algebraic expression and simplify.
There is a rectangular farm with length (2a2 + 3b2) metre and breadth (a2 + b2) metre. The farmer used a square shaped plot of the farm to build a house. The side of the plot was (a2 – b2) metre. What is the area of the remaining part of the farm? [4 Marks]
Solution:
Length of the rectangular farm = (2a2 + 3b2) m
Breadth of the rectangular farm = (a2 + b2) m
Area of the farm = length x breadth = (2a2 + 3b2) x (a2 + b2)
= 2a2(a2 + b2) + 3b2(a2 + b2)
= 2a2 + 2a2b2 + 3a2b2 + 3b4
= (2a4 + 5a2b2 + 3b4) sq. m … (i)
The farmer used a square shaped plot of the farm to build a house.
Side of the square shaped plot = (a2 – b2) m
∴ Area of the plot = (side)2
= (a2 – b2)2
= (a4 – 2a2b2 + b4) sq m… .(ii)

∴ Area of the remaining farm = Area of the farm – Area of the plot
= (2a4 + 5a2b2 + 3b4) – (a4 – 2a2b2 + b4) … [From (i) and (ii)]
= 2a4 + 5a2b2 + 3b4 – a4 + 2a2b2 – b4
= 2a4 – a4 + 5a2b2 + 2a2b2 + 3b4 – b4
= a4 + 7a2b2 + 2b4
∴ The area of the remaining farm is (a4 + 7a2b2 + 2b4) sq. m.

## Maharashtra Board 9th Class Maths Part 1 Practice Set 3.1 Solutions Chapter 3 Polynomials

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

## Practice Set 3.1 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials

Question 1.
State whether the given algebraic expressions are polynomials? Justify.
i. y + $$\frac { 1 }{ y }$$
ii. 2 – 5√x
iii. x2 + 7x + 9
iv. 2m-2 + 7m – 5
v. 10
i. No, because power of v in the term 5√x is -1 (negative number).
ii. No, because the power of x in the term 5√x is
i. e. 0.5 (decimal number).
iii. Yes. All the coefficients are real numbers. Also, the power of each term is a whole number.
iv. No, because the power of m in the term 2m-2 is -2 (negative number).
v. Yes, because 10 is a constant polynomial.

Question 2.
Write the coefficient of m3 in each of the given polynomial.
i. m3
ii. $$\sqrt [ -3 ]{ 2 }$$ + m – √3m3
iii. $$\sqrt [ -2 ]{ 3 }$$m3 + 5m2 – 7m -1
i. 1
ii. -√3
iii. – $$\frac { 2 }{ 3 }$$

Question 3.
Write the polynomial in x using the given information. [1 Mark each]
i. Monomial with degree 7
ii. Binomial with degree 35
iii. Trinomial with degree 8
i. 5x7
ii. x35 – 1
iii. 3x8 + 2x6 + x5

Question 4.
Write the degree of the given polynomials.
i. √5
ii. x°
iii. x2
iv. √2m10 – 7
v. 2p – √7
vi. 7y – y3 + y5
vii. xyz +xy-z
viii. m3n7 – 3m5n + mn
i. √5 = √5 x°
∴ Degree of the polynomial = 0

ii. x°
∴Degree of the polynomial = 0

iii. x2
∴Degree of the polynomial = 2

iv. √2m10 – 7
Here, the highest power of m is 10.
∴Degree of the polynomial = 10

v. 2p – √7
Here, the highest power of p is 1.
∴ Degree of the polynomial = 1

vi. 7y – y3 + y5
Here, the highest power of y is 5.
∴Degree of the polynomial = 5

vii. xyz + xy – z
Here, the sum of the powers of x, y and z in the term xyz is 1 + 1 + 1= 3,
which is the highest sum of powers in the given polynomial.
∴Degree of the polynomial = 3

viii. m3n7 – 3m5n + mn
Here, the sum of the powers of m and n in the term m3n7 is 3 + 7 = 10,
which is the highest sum of powers in the given polynomial.
∴ Degree of the polynomial = 10

Question 5.
Classify the following polynomials as linear, quadratic and cubic polynomial. [2 Marks]
i. 2x2 + 3x +1
ii. 5p
iii. √2 – $$\frac { 1 }{ 2 }$$
iv. m3 + 7m2 + $$\sqrt [ 5 ]{ 2 }$$m – √7
v. a2
vi. 3r3
Linear polynomials: ii, iii
Cubic polynomials: iv, vi

Question 6.
Write the following polynomials in standard form.
i. m3 + 3 + 5m
ii. – 7y + y5 + 3y3 – $$\frac { 1 }{ 2 }$$+ 2y4 – y2
i. m3 + 5m + 3
ii. y5 + 2y4 + 3y3 – y2 – 7y – $$\frac { 1 }{ 2 }$$

Question 7.
Write the following polynomials in coefficient form.
i. x3 – 2
ii. 5y
iii. 2m4 – 3m2 + 7
iv. – $$\frac { 2 }{ 3 }$$
i. x3 – 2 = x3 + 0x2 + 0x – 2
∴ Coefficient form of the given polynomial = (1, 0, 0, -2)

ii. 5y = 5y + 0
∴Coefficient form of the given polynomial = (5,0)

iii. 2m4 – 3m2 + 7
= 2m4 + Om3 – 3m2 + 0m + 7
∴ Coefficient form of the given polynomial = (2, 0, -3, 0, 7)

iv. – $$\frac { 2 }{ 3 }$$
∴Coefficient form of the given polynomial = (- $$\frac { 2 }{ 3 }$$)

Question 8.
Write the polynomials in index form.
i. (1, 2, 3)
ii. (5, 0, 0, 0 ,-1)
iii. (-2, 2, -2, 2)
i. Number of coefficients = 3
∴ Degree = 3 – 1 = 2
∴ Taking x as variable, the index form is x2 + 2x + 3

ii. Number of coefficients = 5
∴ Degree = 5 – 1=4
∴ Taking x as variable, the index form is 5x4 + 0x3 + 0x2 + 0x – 1

iii. Number of coefficients = 4
∴Degree = 4 – 1 = 3
∴Taking x as variable, the index form is -2x3 + 2x2 – 2x + 2

Question 9.
Write the appropriate polynomials in the boxes.

i. Quadratic polynomial: x2; 2x2 + 5x + 10; 3x2 + 5x
ii. Cubic polynomial: x3 + x2 + x + 5; x3 + 9
iii. Linear polynomial: x + 7
iv. Binomial: x + 7; x3 + 9; 3x2 + 5x
v. Trinomial: 2x2 + 5x + 10
vi. Monomial: x2

Question 1.
Write an example of a monomial, a binomial and a trinomial having variable x and degree 5. ( Textbook pg. no. 3)