Prasanna

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 5.2 8th Std Maths Answers Solutions Chapter 5 Expansion Formulae.

Expansion Formulae Class 8 Maths Chapter 5 Practice Set 5.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 5.2 Chapter 5 Solutions Answers

Question 1.
Expand:
i. (k + 4)³
ii. (7x + 8y)³
iii. (7x + m)³
iv. (52)³
v. (101)³
vi. \(\left(x+\frac{1}{x}\right)^{3}\)
vii. \(\left(2 m+\frac{1}{5}\right)^{3}\)
viii. \(\left(\frac{5 x}{y}+\frac{y}{5 x}\right)^{3}\)
Solution:
i. Here, a = k and b = 4
(k + 4)³ = (k)³ + 3(k)² (4) + 3(k)(4)² + (4)³
…[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= k³ + 12k² + 3(k)(16) + 64
= k³ + 12k² + 48k + 64

ii. Here, a = 7x and b = 8y
(7x + 8y)³
= (7x)³ + 3(7x)² (8y) + 3(7x) (8y)² + (8y)³
…[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= 343x³ + 3(49x²)(8y) + 3(7x)(64y²) + 512y³
= 343x³ + 1176x²y + 1344xy² + 512y³

iii. Here, a = 7 and b = m
(7 + m)³ = (7)³ + 3(7)²(m) + 3(7)(m)² + (m)³
…[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= 343 + 3(49)(m) + 3(7)(m²) + m³
= 343 + 147m + 21m² + m³

iv. (52)³ = (50 + 3)³
Here, a = 50 and b = 2
(52)³ = (50)³ + 3(50)² (2) + 3(50)(2)² + (2)³
…[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= 125000 + 3(2500)(2) + 3(50)(4) + 8
= 125000 + 15000 + 600 + 8
=140608

v. (101)³ = (100 + 1)³
Here, a = 100 and b = 1
(101)³
= (100)³ + 3(100)²(1) + 3(100)(1)² + (1)³
…[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= 1000000 + 3(10000) + 3(100) (1) + 1
= 1000000 + 30000 + 300 + 1
= 1030301

vi. Here, a = x and b = \(\frac { 1 }{ x }\)

vii. Here, a = 2m and b = \(\frac { 1 }{ 5 }\)

viii. Here, a = \(\frac { 5x }{ y }\) and b = \(\frac { y }{ 5x }\)

Std 8 Maths Digest

• Practice Set 3.1 Class 8 Answers
• Practice Set 3.2 Class 8 Answers
• Practice Set 3.3 Class 8 Answers
• Practice Set 4.1 Class 8 Answers
• Practice Set 5.1 Class 8 Answers
• Practice Set 5.2 Class 8 Answers
• Practice Set 5.3 Class 8 Answers
• Practice Set 5.4 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 4.1 8th Std Maths Answers Solutions Chapter 4 Altitudes and Medians of a Triangle.

Altitudes and Medians of a Triangle Class 8 Maths Chapter 4 Practice Set 4.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 4.1 Chapter 4 Solutions Answers

Question 1.
In ∆LMN, ___ is an altitude and __ is a median, (write the names of appropriate segments.)

Solution:
In ∆LMN, seg LX is an altitude and seg LY is a median.

Question 2.
Draw an acute angled ∆PQR. Draw all of its altitudes. Name the point of concurrence as ‘O’.
Solution:

Question 3.
Draw an obtuse angled ∆STV. Draw its medians and show the centroid.
Solution:

Question 4.
Draw an obtuse angled ∆LMN. Draw its altitudes and denote the ortho centre by ‘O’.
Solution:

Question 5.
Draw a right angled ∆XYZ. Draw its medians and show their point of concurrence by G.
Solution:

Question 6.
Draw an isosceles triangle. Draw all of its medians and altitudes. Write your observation about their points of concurrence.
Solution:

The point of concurrence of medians i.e. G and that of altitudes i.e. O lie on the same line PS which is the perpendicular bisector of seg QR.

Question 7.
Fill in the blanks.
Point G is the centroid of ∆ABC.
i. If l(RG) = 2.5, then l(GC) = ___
ii. If l(BG) = 6, then l(BQ) = ____
iii. If l(AP) = 6, then l(AG) = ___ and l(GP) = ___.

Solution:
The centroid of a triangle divides each median in the ratio 2:1.
i. Point G is the centroid and seg CR is the median.
∴ \(\frac{l(\mathrm{GC})}{l(\mathrm{RG})}=\frac{2}{1}\)
∴ \(\frac{l(\mathrm{GC})}{2.5}=\frac{2}{1}\) ……[∵ l(RG) = 2.5]
∴ l(GC) × 1 = 2 × 2.5
∴ l(GC) = 5

ii. Point G is the centroid and seg BQ is the median.
∴ \(\frac{l(\mathrm{BG})}{l(\mathrm{GQ})}=\frac{2}{1}\)
∴ \(\frac{6}{l(\mathrm{GQ})}=\frac{2}{1}\) …..[∵ l(BG) = 6]
∴ 6 × 1 = 2 × l(GQ)
∴ \(\frac { 6 }{ 2 }\) = l(GQ)
∴ 3 = l(GQ)
i.e. l(GQ) = 3
Now, l (BQ) = l(BG) + l(GQ)
∴ l(BQ) = 6 + 3
∴ l(BQ) = 9

iii. Point G is the centroid and seg AP is the median.
∴ \(\frac{l(\mathrm{AG})}{l(\mathrm{GP})}=\frac{2}{1}\)
∴ l(AG) = 2 l(GP) …..(i)
Now, l(AP) = l(AG) + l(GP) … (ii)
∴ l(AP) = 2l(GP) + l(GP) … [From (i)]
∴ l(AP) = 3l(GP)
∴ 6 = 3l(GP) ..[∵ l(AP) = 6]
∴ \(\frac { 6 }{ 3 }\) = l(GP)
∴ 2 = l(GP)
i.e. l(GP) = 2
l(AP) = l(AG) + l(GP) …[from (ii)]
∴ 6 = l(AG) + 2
∴ l(AG) = 6 – 2
∴ l(AG) = 4

Maharashtra Board Class 8 Maths Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 Intext Questions and Activities

Question 1.
Draw a line. Take a point outside the line. Draw a perpendicular from the point to the line with the help of a set-square (Textbook pg. no, 19)
Solution:
Step 1: Draw a line l and a point P lying outside it.

Step 2: By placing a set-square on line l, draw a perpendicular to the line from point P.

Question 2.
Draw an acute angled ∆ABC and all its altitudes. Observe the location of the orthocentre. (Textbook pg. no. 20)
Solution:

Point O is the orthocentre.
Orthocentre lies in the interior of ∆ABC.

Question 3.
Draw a right angled triangle and draw all its altitudes. Write the point of concurrence. (Textbook: pg, no. 20)
Solution:

Point Q is the orthocentre.
The point of concurrence of altitudes PQ, QR and QS is Q.

Question 4.
i. Draw an obtuse angled triangle and all its altitudes.
ii. Do they intersect each other?
Draw the lines containing the altitudes. Observe that these lines are concurrent. (Textbook pg. no. 20)
Solution:
i.

Point O is the orthocentre.

ii. Yes, all the altitudes intersect at point O in the exterior of ∆PQR.

Question 5.
Draw three different triangles; a right angled triangle, an obtuse angled triangle and an acute angled triangle. Draw the medians of the triangles. Note that the centroid of each of them is in the interior of the triangle. (Textbook pg. no. 21)
Solution:
i. Right angled triangle:

ii. Obtuse angled triangle:

iii. Acute angled triangle:

Question 6.
Draw a sufficiently large ∆ABC.
Draw medians; seg AR, seg BQ and seg CP of ∆ABC.
Name the point of concurrence as G.
Measure the lengths of segments from the figure and fill in the boxes in the following table.

l(AG) =	l(GR) =	l(AG): l(GR) =
l(BG) =	l(GQ) =	l(BG): l(GQ) =
l(CG) =	l(GP) =	l(CG): l(GP) =

Observe that all of these ratios are nearly 2 : 1 (Textbook pg. no. 21)
Solution:

l(AG) = 2.9	l(GR) = 1.4	l(AG): l(GR) = \(\frac{2.8}{1.4}=\frac{2}{1}\)
l(BG) = 2.4	l(GQ) = 1.2	l(BG): l(GQ) = \(\frac{2.4}{1.2}=\frac{2}{1}\)
l(CG) = 2.8	l(GP) = 1.4	l(CG): l(GP) = \(\frac{2.8}{1.4}=\frac{2}{1}\)

Question 7.
As shown in the given figure, a student drew ∆ABC using five parallel lines of a notebook. Then he found the centroid G of the triangle. How will you decide whether the location of G he found, is correct. (Textbook pg. no. 21)

Solution:
Draw seg AP ⊥ seg PE and seg EQ ⊥ seg QC.

Side AP || side EQ and AC is their transversal.
∴ ∠PAE ≅ ∠QEC …(i) [Corresponding angles]
In ∆ APE and ∆ EQC,
∠PAE ≅ ∠QEC …[From (i)]
∠APE ≅ ∠EQC
… [Each angle is of measure 90°]
side PE ≅ side QC
…. [Perpendicular distance between parallel lines]
∴ ∆ APE ≅ ∆ EQC … [By AAS test]
∴ AE = EC
… [Corresponding sides of congruent triangles]
∴ E is the midpoint of AC.
∴ seg BE is the median.
Similarly, seg CF is the median.
Since, the medians of a triangle are concurrent.
∴ G is the centroid of ∆ABC.

Question 8.
Draw an equilateral triangle. Find its circumcentre (C), incentre (I), centroid (G) and orthocentre (O). Write your observation. (Textbook pg. no. 22)
Solution:

From the figure, circumcentre (C), incentre (I), centroid (G) and orthocentre (O) of an equilateral triangle are the same.

Question 9.
Draw an isosceles triangle. Locate its centroid, orthocentre, circumcentre and incentre. Verify that they are collinear. (Textbook pg. no. 22)
Solution:

From the figure, centroid (G), orthocentre (O), circumcentre (C) and incentre (I) of an isosceles triangle lie on the same line AD.
∴ they are collinear.

Std 8 Maths Digest

• Practice Set 3.1 Class 8 Answers
• Practice Set 3.2 Class 8 Answers
• Practice Set 3.3 Class 8 Answers
• Practice Set 4.1 Class 8 Answers
• Practice Set 5.1 Class 8 Answers
• Practice Set 5.2 Class 8 Answers
• Practice Set 5.3 Class 8 Answers
• Practice Set 5.4 Class 8 Answers

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

9th Standard Maths 1 Problem Set 1 Chapter 1 Sets Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Problem Set 1 Chapter 1 Sets Questions With Answers Maharashtra Board

Question 1.
Choose the correct alternative answer for each of the following questions.
i. M= {1, 3, 5}, N= {2, 4, 6}, then M ∩ N = ?
(A) {1, 2, 3, 4, 5, 6}
(B) {1, 3, 5}
(C) φ
(D) {2, 4, 6}
Answer:
(C) φ

ii. P = {x | x is an odd natural number, 1< x ≤ 5}. How to write this set in roster form?
(A) {1, 3, 5}
(B) {1, 2, 3, 4, 5}
(C) {1, 3}
(D) {3, 5}
Answer:
(D) {3, 5}

iii. P= {1, 2, ………. , 10}. What type of set Pis?
(A) Null set
(B) Infinite set
(C) Finite set
(D) None of these
Answer:
(C) Finite set

iv. M ∪ N = {1, 2, 3, 4, 5, 6} and M = {1, 2, 4}, then which of the following represent set N ?
(A) {1, 2, 3}
(B) {3, 4, 5, 6}
(C) {2, 5, 6}
(D) {4, 5, 6}
Answer:
(B) {3, 4, 5, 6}

v. If P ⊆ M, then which of the following set represent P ∩ (P ∪ M)?
(A) P
(B) M
(C) P ∪ M
(D) P’ ∩ M
Answer:
(A) P

vi. Which of the following sets are empty sets?
(A) Set of intersecting points of parallel lines.
(B) Set of even prime numbers.
(C) Month of an english calendar having less than 30 days.
(D) P = {x | x ∈ I , – 1 < x < 1}
Answer:
(A) Set of intersecting points of parallel lines.

Hints:
v. Here, P ⊆ M
∴ P ∪ M = M
∴ P ∩ (P ∪ M) = P ∩ M
= P … [∵ P ⊆M]

Question 2.
Find the correct option for the given question.
i. Which of the following collections is a set ?
(A) Colors of the rainbow
(B) Tall trees in the school campus.
(C) Rich people in the village
(D) Easy examples in the book
Answer:
(A) Colors of the rainbow

ii. Which of the following set represent N ∩W?
(A) {1, 2, 3,….}
(B) {0, 1, 2, 3,….}
(C) {0}
(D) { }
Answer:
(A) {1, 2, 3,….}

iii. P = {x | x is an odd natural number, 1< x < 5}. How to write this set in roster form?
(A) {1, 3, 5}
(B) {1, 2, 3, 4, 5}
(C) {1, 3}
(D) {3, 5}
Answer:
(B) {1, 2, 3, 4, 5}

iv. If T = {1, 2, 3, 4, 5} and M = {3,4, 7, 8}, then T ∪ M = ?
(A) {1, 2, 3, 4, 5,7}
(B) {1, 2, 3, 7, 8}
(C) {1, 2, 3, 4, 5, 7, 8}
(D) {3, 4}
Answer:
(C) {1, 2, 3, 4, 5, 7, 8}

Hints:
i. The elements of options B, C and D cannot be definitely and clearly decided.
ii. The common elements of N and W are 1 2, 3,….

Question 3.
Out of 100 persons in a group, 72 persons speak English and 43 persons speak French. Each one out of 100 persons speak at least one language. Then how many speak only English? How many speak only French ? How many of them speak English and French both?
Solution:
i. Let U be the set of all the persons,
E be the set of persons who speak English and
F be the set of persons who speak French.
∴ n(E) = 72, n(F) = 43
Since, each one out of 100 persons speak at least one language
∴ n(U) = n(E ∪ F)= 100,

ii. n (E ∪ F) = n (E) + n (F) – n(E ∩ F)
100 = 72 + 43 – n (E ∩ F)
n (E ∩ F) = 72 + 43 – 100
∴ n(E ∩ F) = 15
Number of people who speak English and French = 15

iii. Number of people who speak only English = n(E) – n(E ∩ F)
= 72 – 15 = 57

iv. Number of
people who speak only French = n(F) – n(E ∩ F)
= 43 – 15 = 28

Alternate Method:
Let U be the set of all the persons,
E be the set of persons who speak English,
F be the set of persons who speak French and x people speak both the languages.

Since, each one out of 100 persons speak at least one language.
∴ n(U) = n(E ∪ F) = 100
∴ 72 – x + x + 43 – x = 100
∴ 115 – x= 100
∴ x = 115 – 100= 15.
Number of people who speak English and French = 15
Number of people who speak only English = 72 – x = 72 – 15 = 57
Number of people who speak only French = 43 – x = 43 – 15 = 28

Question 4.
70 trees were planted by Parth and 90 trees were planted by Pradnya on the occasion of Tree Plantation Week. Out of these 25 trees were planted by both of them together. How many trees were planted by Parth or Pradnya?
Solution:
i. Let P be the trees planted by Parth and Q be the trees planted by Pradnya
∴ n(P) = 70 and n(Q) = 90
Total number of trees planted by Parth and Pradnya = n(P ∩ Q) = 25

ii. Number of trees planted by Parth or Pradnya = n(P ∪ Q)
= n(P) + n(Q) – n(P ∩ Q)
= 70 + 90 – 25 = 135
∴ A total of 135 trees were planted by Parth or Pradnya.

Alternate Method:
Let P be the trees planted by Parth and Q be the trees planted by Pradnya

From Venn diagram
∴ Total trees planted by parth or pradnya = n(P ∪ Q)
= 45 + 25 + 65
= 135
A total of 135 trees were planted by Parth or Pradnya.

Question 5.
If n(A) = 20, n(B) = 28 and n(A ∪ B) = 36, then n(A ∩ B) = ?
Solution:
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
∴ 36 = 20 + 28 – n(A ∩ B)
∴ n(A ∩ B) = 20 + 28 – 36
∴ n(A ∩ B) = 12

Question 6.
In a class, 8 students out of 28 have a dog as their pet animal at home, 6 students have a cat as their pet animal, 10 students have dog and cat both, then how many students do not have dog or cat as their pet animal at home?
Solution:
i. Let U be the set of all the students, then n(U) = 28
Let D be the set of students who have dog as pet and C be the set of students who have cat as pet.
10 students have dog and cat as their pet animal
n(D ∩ C) = 10

From venn diagram,

ii. Number of students who have cat or dog as pet
= n(D ∪ C)
= 8 + 10 + 6
= 24

iii. Number of students who do not have dog or cat as pet = n (U) – n(D ∪ C)
= 28 – 24
= 4

Question 7.
Represent the union of two sets by Venn diagram for each of the following.
i. A = {3, 4, 5, 7},B = {1, 4, 8} l Marks
ii. P {a, b, c, e, f, Q = {l, m, n, e, b) I Markj
iii. X = {x x is a prime number between 80 and 100}
Y = { y | y is an odd number between 90 and 100}
Solution:
i. A = {3, 4, 5, 7}, B = {1, 4, 8}

ii. P = {a, b, c, e, f}, Q = {l, m, n, e, b}

iii. X = {x | x is a prime number between 80 and 100}
∴ X = {83, 89, 97}
Y = {y | y is an odd number between 90 and 100}
∴ Y = {91, 93, 95, 97, 99}

Question 8.
Write the subset relations between the following sets.
X = set of all quadrilaterals.
Y = set of all rhombuses.
S = set of all squares.
T = set of all parallelograms.
V = set of all rectangles. [3 Marks]
Solution:
i. Rhombus, square, parallelogram and rectangle all are quadrilaterals.
∴ Y ⊆ X,S ⊆ X,T ⊆ X,V ⊆ X

ii. Every square is a rhombus, parallelogram and rectangle.
∴ S ⊆ Y, S ⊆ T, S ⊆ V

iii. Every rhombus and rectangle is a parallelogram.
∴ Y ⊆ T, V ⊆ T

Question 9.
If M is any set, then write M ∪Φ and M ∩ Φ.
Solution:
Let M = {2, 3, 4, 8} and Φ = { }
∴ M ∪ Φ = {2, 3, 4, 8}
∴ M ∪ Φ = M Also, M ∩ Φ = { }
∴ M ∩ Φ = i(i

Question 10.
Observe the Venn diagram and write the given sets U, A, B, A ∪ B and A ∩ B.

U = {1,2, 3,4, 5, 7, 8, 9, 10, 11, 13}
A = {1, 2, 3, 5,7}
B = {1, 5, 8, 9, 10}
A ∪ B = {1,2, 3, 5, 7, 8, 9, 10}
A ∩ B = {1, 5}

Question 11.
If n(A) = 7, n(B) = 13, n(A ∩ B) = 4, then n(A ∪ B) = ?
Solution:
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 7 + 13 – 4
n(A ∪ B) = 16

Question 1.
Set of students in a class and set of students in the same class who can swim, are shown by the Venn diagram.

Observe the diagram and draw Venn diagrams for the following subsets.
i. a. Set of students in a class
b. Set of students who can ride bicycles in the same class

ii. A set of fruits is given as follows.
U = {guava, orange, mango, jackfruit, chickoo, jamun, custard apple, papaya, plum}
Show these subsets.
A = fruit with one seed
B = fruit with more than one seed. (Textbook pg. no. 8)
Solution:

ii. A = {mango, jamun, plum}
B = {guava, orange, jackfruit, chickoo, custard apple, papaya}

Question 2.
Every student should take 9 triangular sheets of paper and one plate. Numbers from 1 to 9 should, be written on each triangle. Everyone should keep some numbered triangles in the plate. Now the triangles in each plate form a subset of the set of numbers from 1 to 9.

Look at the plates of Sujata, Hameed, Mukta, Nandini, Joseph with the numbered triangles. Guess the thinking behind selecting these numbers. Hence write the subsets in set builder form. (Textbook pg, no. 9)
Solution:
Sujata:
S = {x | x = 2n- 1, n ∈ N, x < 9}
Hameed:
f H = {x | x = 2n, n ∈ N, x < 9}
Mukta:
M = {x | x = n², n ∈ N, x ≤ 9}
Nandini:
N = {x | x ∈ N, x ≤ 9}
Joseph:
J = {x | x is a prime number between 1 and 9}

Question 3.
Collect the following information from 20 families nearby your house.
i. Number of families subscribing for Marathi Newspaper.
ii. Number of families subscribing for English Newspaper.
iii. Number of families subscribing for both English as well as Marathi Newspaper.
Show the collected information using Venn diagram. (Textbook pg.no. 18)
[Students should attempt the above activity on their own.]

Class 9 Maths Digest

• Practice Set 1.1 Class 9 Answers
• Practice Set 1.2 Class 9 Answers
• Practice Set 1.3 Class 9 Answers
• Practice Set 1.4 Class 9 Answers
• Problem Set 1 Class 9 Answers
• Practice Set 2.1 Class 9 Answers
• Practice Set 2.2 Class 9 Answers
• Practice Set 2.3 Class 9 Answers
• Practice Set 2.4 Class 9 Answers
• Practice Set 2.5 Class 9 Answers
• Problem Set 2 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 2 Real Numbers.

9th Standard Maths 1 Practice Set 2.1 Chapter 2 Real Numbers Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 2.1 Chapter 2 Real Numbers Questions With Answers Maharashtra Board

Question 1.
Classify the decimal form of the given rational numbers into terminating and non-terminating recurring type.

Solution:
i. Denominator = 5 = 1 x 5
Since, 5 is the only prime factor denominator.
the decimal form of the rational number \(\frac { 13 }{ 5 }\) will be terminating type.

ii. Denominator = 11 = 1 x 11
Since, the denominator is other than prime factors 2 or 5.
∴ the decimal form of the rational number \(\frac { 2 }{ 11 }\) will be non-terminating recurring type.

iii. Denominator = 16
= 2 x 2 x 2 x 2
Since, 2 is the only prime factor in the denominator.
∴ the decimal form of the rational number \(\frac { 29 }{ 16 }\) will be terminating type.

iv. Denominator = 125
= 5 x 5 x 5
Since, 5 is the only prime factor in the denominator.
the decimal form of the rational number \(\frac { 17 }{ 125 }\) will be terminating type.

v. Denominator = 6
= 2 x 3
Since, the denominator is other than prime factors 2 or 5.
∴ the decimal form of the rational number \(\frac { 11 }{ 6 }\) will be non-terminating recurring type.

Question 2.
Write the following rational numbers in decimal form.





Solution:
i. \(\frac { -5 }{ 7 }\)

ii. \(\frac { 9 }{ 11 }\)

iii. √5

iv. \(\frac { 121 }{ 13 }\)

v. \(\frac { 29 }{ 8 }\)

Question 3.
Write the following rational numbers in \(\frac { p }{ q }\) form.

Solution:
i. Let x = \(0.\dot { 6 }\) …(i)
∴ x = 0.666…
Since, one number i.e. 6 is repeating after the decimal point.
Thus, multiplying both sides by 10,
10x = 6.666…
∴ 10 x 6.6 …(ii)
Subtracting (i) from (ii),
10x – x = 6.6 – 0.6
∴ 9x = 6

ii. Let x = \(0.\overline { 37 }\)
∴ x = 0.3737…
Since, two numbers i.e. 3 and 7 are repeating after the decimal point.
Thus, multiplying both sides by 100,
100x = 37.3737……
∴ 100x = \(37.\overline { 37 }\) ……(ii)
Subtracting (i) from (ii),
100x – x = \(37.\overline { 37 }\) – \(0.\overline { 37 }\)
∴ 99x = 37

iii. Letx = \(3.\overline { 17 }\) …(i)
∴ x = 3.1717…
Since, two numbers i.e. 1 and 7 are repeating after the decimal point.
Thus, multiplying both sides by 100,
100x = 317.1717…
∴ 100x= 317.17 …(ii)
Subtracting (i) from (ii),
100x – x = \(317.\overline { 17 }\) – \(3.\overline { 17 }\)
∴ 99x = 314

iv. Let x = \(15.\overline { 89 }\) …….. (i)
∴ x = 15.8989…
Since, two numbers i.e. 8 and 9 are repeating after the decimal point.
Thus, multiplying both sides by 100,
100x= 1589.8989…
∴ 100x = \(1589.\overline { 89 }\) …(ii)
Subtracting (i) from (ii),
100x – x = \(1589.\overline { 89 }\) – \(15.\overline { 89 }\)
∴ 99x = 1574

v. Let x = \(2.\overline { 514 }\)
∴ x = 2.514514…
Since, three numbers i.e. 5, 1 and 4 are repeating after the decimal point.
Thus, multiplying both sides by 1000,
1000x = 2514.514514…
1000x = \(2514.\overline { 514 }\) ….(ii)
Subtracting (i) from (ii),
1000x – x = \(2514.\overline { 514 }\) – \(2.\overline { 514 }\)
∴ 999x = 2512

Question 1.
How to convert 2.43 in \(\frac { p }{ q }\) form ? (Textbook pg. no. 20)
Solution:
Let x = 2.43
In 2.43, the number 4 on the right side of the decimal point is not recurring.
So, in order to get only recurring digits on the right side of the decimal point, we will multiply 2.43 by 10.
∴ 10x = 24.3 …(i)
∴ 10x = 24.333…
Here, digit 3 is the only recurring digit. Thus, by multiplying both sides by 10, 100x = 243.333…
∴ 100x= 243.3 …(ii)
Subtracting (i) from (ii),
100x – 10x = 243.3 – 24.3
∴ 90x = 219

Class 9 Maths Digest

• Practice Set 1.1 Class 9 Answers
• Practice Set 1.2 Class 9 Answers
• Practice Set 1.3 Class 9 Answers
• Practice Set 1.4 Class 9 Answers
• Problem Set 1 Class 9 Answers
• Practice Set 2.1 Class 9 Answers
• Practice Set 2.2 Class 9 Answers
• Practice Set 2.3 Class 9 Answers
• Practice Set 2.4 Class 9 Answers
• Practice Set 2.5 Class 9 Answers
• Problem Set 2 Class 9 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 3.3 8th Std Maths Answers Solutions Chapter 3 Indices and Cube Root.

Indices and Cube Root Class 8 Maths Chapter 3 Practice Set 3.3 Solutions Maharashtra Board

Std 8 Maths Practice Set 3.3 Chapter 3 Solutions Answers

Question 1.
Find the cube root of the following numbers.
i. 8000
ii. 729
iii. 343
iv. -512
v. -2744
vi. 32768
Solution:
i. 8000
= 2 × 2 × 2 × 10 × 10 × 10
= (2 × 10) × (2 × 10) × (2 × 10)
= (2 × 10)³
= 20³
∴ \(\sqrt[3]{8000}=20\)

ii. 729
= (3 × 3) × (3 × 3) × (3 × 3)
= (3 × 3)³
= 9³
∴ \(\sqrt[3]{729}=9\)

iii. 343
= 7 × 7 × 7
= 7³
∴ \(\sqrt[3]{343}=7\)

iv. -512
= 2 × 2 × 2 × 4 × 4 × 4
= (2 × 4) × (2 × 4) × (2 × 4)
= (2 × 4)³
= 8³
∴ – 512 = (- 8) × (- 8) × (- 8)
= (-8)³
∴ \(\sqrt[3]{-512}=-8\)

v. -2744
= 2 × 2 × 2 × 7 × 7 × 7
= (2 × 7) × (2 × 7) × (2 × 7)
= (2 × 7)³
= 14³
∴ -2744 = (-14) × (-14) × (-14)
= (-14)³
∴ \(\sqrt[3]{-2744}=-14\)

vi. 32768
= 2 × 2 × 2 × 4 × 4 × 4 × 4 × 4 × 4
= (2 × 4 × 4) × (2 × 4 × 4) × (2 × 4 × 4)
= (2 × 4 × 4)³
= 32³
∴ \(\sqrt[3]{32768}=32\)

Question 2.
Simplify:
i. \(\sqrt[3]{\frac{27}{125}}\)
ii. \(\sqrt[3]{\frac{16}{54}}\)
iii. If \(\sqrt[3]{729}=9\) then \(\sqrt[3]{0.000729}\) = ?
Solution:
i. \(\sqrt[3]{\frac{27}{125}}\)

ii. \(\sqrt[3]{\frac{16}{54}}\)

iii. \(\sqrt[3]{0.000729}\)


Note:
Here, number of decimal places in cube root = 6
∴ number of decimal places in cube of number = 2

Maharashtra Board Class 8 Maths Chapter 3 Indices and Cube Root Practice Set 3.3 Intext Questions and Activities

Question 1.
17 is a positive number. The cube of 17, which is 4913, is also a positive number. Cube of -6 is -216. Take some more positive and negative numbers and obtain their cubes. Find the relation between the sign of a number and the sign of its cube. (Textbook pg. no. 17)
Solution:
Consider, 6³ = 6 × 6 × 6 = 216 and (-4)³ = (- 4) × (- 4) × (- 4) = – 64
Thus, cube of a positive number is positive and cube of a negative number is negative.
∴ Sign of a number = sign of its cube.

Question 2.
In example 4 and 5 on textbook pg. no. 17, observe the number of decimal places in the number and number of decimal places in the cube of the number. Is there any relation between the two? (Textbook pg. no. 17)
Solution:
Yes, there is a relation between the number of decimal places in the number and its cube.
(1.2)³ = 1.728, (0.02)³ = 0.000008
No. of decimal places in 1.2 = 1
No. of decimal places in 1.728 = 3
No. of decimal places in 0.02 = 2
No. of decimal places in 0.000008 = 6
Thus, number of decimal places in cube of a number is three times the number of decimal places in that number.

Std 8 Maths Digest

• Practice Set 3.1 Class 8 Answers
• Practice Set 3.2 Class 8 Answers
• Practice Set 3.3 Class 8 Answers
• Practice Set 4.1 Class 8 Answers
• Practice Set 5.1 Class 8 Answers
• Practice Set 5.2 Class 8 Answers
• Practice Set 5.3 Class 8 Answers
• Practice Set 5.4 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 3.1 8th Std Maths Answers Solutions Chapter 3 Indices and Cube Root.

Indices and Cube Root Class 8 Maths Chapter 3 Practice Set 3.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 3.1 Chapter 3 Solutions Answers

Question 1.
Express the following numbers in index form.
i. Fifth root of 13
ii. Sixth root of 9
iii. Square root of 256
iv. Cube root of 17
v. Eighth root of 100
vi. Seventh root of 30
Solution:
i. \((13)^{\frac{1}{5}}\)
ii. \((9)^{\frac{1}{6}}\)
iii. \((256)^{\frac{1}{2}}\)
iv. \((17)^{\frac{1}{3}}\)
v. \((100)^{\frac{1}{8}}\)
vi. \((30)^{\frac{1}{7}}\)

Question 2.
Write in the form ‘nth root of a’ in each of the following numbers.
i. \((81)^{\frac{1}{4}}\)
ii. \((49)^{\frac{1}{2}}\)
iii. \((15)^{\frac{1}{5}}\)
iv. \((512)^{\frac{1}{9}}\)
v. \((100)^{\frac{1}{19}}\)
vi. \((6)^{\frac{1}{7}}\)
Solution:
i. Fourth root of 81.
ii. Square root of 49.
iii. Fifth root of 15.
iv. Ninth root of 512.
v. Nineteenth root of 100.
vi. Seventh root of 6.

Maharashtra Board Class 8 Maths Chapter 3 Indices and Cube Root Practice Set 3.1 Intext Questions and Activities

Question 1.
Using laws of indices, write proper numbers in the following boxes. (Textbook pg, no. 14)
i. \({ 3 }^{ 5 }\times { 3 }^{ 2 }={ 3 }^{ \left( \right) }\)
ii. \({ 3 }^{ 7 }\div { 3 }^{ 9 }={ 3 }^{ \left( \right) }\)
iii. \(({ 3 }^{ 4 })^{ 5 }={ 3 }^{ \left( \right) }\)
iv. \(5^{ -3 }=\frac { 1 }{ { 5 }^{ \left( \right) } }\)
v. \(5^{ 0 }=\left( \right) \)
vi. \(5^{ 1 }=\left( \right) \)
vii. \((5\times 7)^{ 2 }={ 5 }^{ \left( \right) }\times { 7 }^{ \left( \right) }\)
viii. \({ \left( \frac { 5 }{ 7 } \right) }^{ 3 }=\frac { { \left( \right) }^{ 3 } }{ { \left( \right) }^{ 3 } } \)
ix. \({ \left( \frac { 5 }{ 7 } \right) }^{ -3 }={ \left( \frac { \left( \right) }{ \left( \right) } \right) }^{ 3 }\)
Solution:
i. \({ 3 }^{ 5 }\times { 3 }^{ 2 }={ 3 }^{ 7 }\)
ii. \({ 3 }^{ 7 }\div { 3 }^{ 9 }={ 3 }^{ -2 }\)
iii. \(({ 3 }^{ 4 })^{ 5 }={ 3 }^{ 20 }\)
iv. \(5^{ -3 }=\frac { 1 }{ { 5 }^{ 3 } } \)
v. \(5^{ 0 }=1\)
vi. \(5^{ 1 }=5\)
vii. \((5\times 7)^{ 2 }={ 5 }^{ 2 }\times { 7 }^{ 2 }\)
viii. \({ \left( \frac { 5 }{ 7 } \right) }^{ 3 }=\frac { { 5 }^{ 3 } }{ { 7 }^{ 3 } } \)
ix. \({ \left( \frac { 5 }{ 7 } \right) }^{ -3 }={ \left( \frac { 7 }{ 5 } \right) }^{ 3 }\)

Std 8 Maths Digest

• Practice Set 3.1 Class 8 Answers
• Practice Set 3.2 Class 8 Answers
• Practice Set 3.3 Class 8 Answers
• Practice Set 4.1 Class 8 Answers
• Practice Set 5.1 Class 8 Answers
• Practice Set 5.2 Class 8 Answers
• Practice Set 5.3 Class 8 Answers
• Practice Set 5.4 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 3.2 8th Std Maths Answers Solutions Chapter 3 Indices and Cube Root.

Indices and Cube Root Class 8 Maths Chapter 3 Practice Set 3.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 3.2 Chapter 3 Solutions Answers

Question 1.
Complete the following table.

S.No.	Number	Power of the root	Root of the power
1.	\((225)^{\frac{3}{2}}\)	Cube of square root of 225	Square root of cube of 225
2.	\((45)^{\frac{4}{5}}\)
3.	\((81)^{\frac{6}{7}}\)
4.	\((100)^{\frac{4}{10}}\)
5.	\((21)^{\frac{3}{7}}\)

Solution:

S.No.	Number	Power of the root	Root of the power
1.	\((225)^{\frac{3}{2}}\)	Cube of square root of 225	Square root of cube of 225
2.	\((45)^{\frac{4}{5}}\)	4th power of 5th root of 45	5th root of 4th power of 45
3.	\((81)^{\frac{6}{7}}\)	6th power of 7th root of 81	7th root of 6th power of 81
4.	\((100)^{\frac{4}{10}}\)	4th power of 10th root of 100	10th root of 4th power of 100
5.	\((21)^{\frac{3}{7}}\)	Cube of 7th root of 21	7th root of cube of 21

Question 2.
Write the following numbers in the form of rational indices.
i. Square root of 5th power of 121.
ii. Cube of 4th root of 324.
iii. 5th root of square of 264.
iv. Cube of cube root of 3.
Solution:
i. \((121)^{\frac{5}{2}}\)
ii. \((324)^{\frac{3}{4}}\)
iii. \((264)^{\frac{2}{5}}\)
iv. \((3)^{\frac{3}{3}}\)

Std 8 Maths Digest

• Practice Set 3.1 Class 8 Answers
• Practice Set 3.2 Class 8 Answers
• Practice Set 3.3 Class 8 Answers
• Practice Set 4.1 Class 8 Answers
• Practice Set 5.1 Class 8 Answers
• Practice Set 5.2 Class 8 Answers
• Practice Set 5.3 Class 8 Answers
• Practice Set 5.4 Class 8 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 1.3 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 1 Sets.

9th Standard Maths 1 Practice Set 1.3 Chapter 1 Sets Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 1.3 Chapter 1 Sets Questions With Answers Maharashtra Board

Question 1.
If A = {a, b, c, d, e}, B = {c, d, e, f}, C = {b, d}, D = {a, e}, then which of the following statements are true and which are false?
i. C ⊆ 3
ii. A ⊆ D
iii. D ⊆ B
iv. D ⊆ A
V. B ⊆ A
vi. C ⊆ A
Ans:
i. C = {b, d}, B = {c, d, e ,f}
C ⊆ B
False
Since, all the elements of C are not present in B.

ii. A = {a, b, c, d, e}, D = {a, e}
A ⊆ D
False
Since, all the elements of A are not present in D.

iii. D = {a, e}, B = {c, d, e, f}
D ⊆ B
False
Since, all the elements of D are not present in B.

iv. D = {a, e}, A = {a, b, c, d, e}
D ⊆ A
True
Since, all the elements of D are present in A.

v. B = {c, d, e, f}, A = {a, b, c, d, e}
B ⊆ A
False
Since, all the elements of B are not present in A.

vi. C = {b, d}, A= {a, b, c, d, e}
C ⊆A
True
Since, all the elements of C are present in A.

Question 2.
Take the set of natural numbers from 1 to 20 as universal set and show set X and Y using Venn diagram. [2 Marks each]
i. X= {x |x ∈ N, and 7 < x < 15}
ii. Y = { y | y ∈ N, y is a prime number from 1 to 20}
Answer:
i. U = {1, 2, 3, 4, …….., 18, 19, 20}
x = {x | x ∈ N, and 7 < x < 15}
∴ x = {8, 9, 10, 11, 12, 13, 14}

ii. U = {1, 2, 3, 4, …… ,18, 19, 20}
Y = { y | y ∈ N, y is a prime number from 1 to 20}
∴ Y = {2, 3, 5, 7, 11, 13, 17, 19}

Question 3.
U = {1, 2, 3, 7, 8, 9, 10, 11, 12} P = {1, 3, 7,10}, then
i. show the sets U, P and P’ by Venn diagram.
ii. Verify (P’)’ = P
Solution:
i. Here, U = {1,2, 3, 7, 8,9, 10, 11, 12} P = {1, 3, 7, 10}
∴ P’ = {2, 8, 9, 11, 12}

II. Here, U = {1, 2, 3, 7, 8, 9, 10, 11, 12}
P = {1, 3, 7, 10} ….(i)
∴ P’= {2, 8, 9, 11, 12}
Also, (P’)’ = {1,3,7, 10} …(ii)
∴ (P’)’ = P … [From (i) and (ii)]

Question 4.
A = {1, 3, 2, 7}, then write any three subsets of A.
Solution:
Three subsets of A:
i. B = {3}
ii. C = {2, 1}
iii. D= {1, 2, 7}
[Note: The above problem has many solutions. Students may write solutions other than the ones given]

Question 5.
i. Write the subset relation between the sets.
P is the set of all residents in Pune.
M is the set of all residents in Madhya Pradesh.
I is the set of all residents in Indore.
B is the set of all residents in India.
H is the set of all residents in Maharashtra.

ii. Which set can be the universal set for above sets ?
Solution:
i.
a. The residents of Pune are residents of India.
∴ P ⊆ B
b. The residents of Pune are residents of Maharashtra.
∴ P ⊆ H
c. The residents of Madhya Pradesh are residents of India.
∴ M ⊆ B
d. The residents of Indore are residents of India.
∴ I ⊆ B
e. The residents of Indore are residents of Madhya Pradesh.
∴ I ⊆ M
f. The residents of Maharashtra are residents of India.
∴ H ⊆B

ii. The residents of Pune, Madhya Pradesh, Indore and Maharashtra are all residents of India.
∴ B can be the Universal set for the above sets.

Question 6.
Which set of numbers could be the universal set for the sets given below?
i. A = set of multiples of 5,
B = set of multiples of 7,
C = set of multiples of 12

ii. P = set of integers which are multiples of 4.
T = set of all even square numbers.
Answer:
i. A = set of multiples of 5
∴ A = {5, 10, 15, …}
B = set of multiples of 7
∴ B = {7, 14, 21,…}
C = set of multiples of 12
∴ C = {12, 24, 36, …}
Now, set of natural numbers, whole numbers, integers, rational numbers are as follows:
N = {1, 2, 3, …}, W = {0, 1, 2, 3, …}
I = {…,-3, -2, -1, 0, 1, 2, 3, …}
Q = { \(\frac { p }{ q }\) | p,q ∈ I,q ≠ 0}
Since, set A, B and C are the subsets of sets N, W , I and Q.
∴ For set A, B and C we can take any one of the set from N, W, I or Q as universal set.

ii. P = set of integers which are multiples of 4.
P = {4, 8, 12,…}
T = set of all even square numbers T = {2², 4², 6², …]
Since, set P and T are the subsets of sets N, W, I and Q.
∴ For set P and T we can take any one of the set from N, W, I or Q as universal set.

Question 7.
Let all the students of a class form a Universal set. Let set A be the students who secure 50% or more marks in Maths. Then write the complement of set A.
Answer:
Here, U = all the students of a class.
A = Students who secured 50% or more marks in Maths.
∴ A’= Students who secured less than 50% marks in Maths.

Question 1.
If A = {1, 3, 4, 7, 8}, then write all possible subsets of A.
i. e. P = {1, 3}, T = {4, 7, 8}, V = {1, 4, 8}, S = {1, 4, 7, 8}
In this way many subsets can be written. Write five more subsets of set A. (Textbook pg. no, 8)
Answer:
B = { },
E = {4},
C = {1, 4},
D = {3, 4, 7},
F = {3, 4, 7,8}

Question 2.
Some sets are given below.
A ={…,-4, -2, 0, 2, 4, 6,…}
B = {1, 2, 3,…}
C = {…,-12, -6, 0, 6, 12, 18, }
D = {…, -8, -4, 0, 4, 8,…}
I = {…,-3, -2, -1, 0, 1, 2, 3, 4, }
Discuss and decide which of the following statements are true.
a. A is a subset of sets B, C and D.
b. B is a subset of all the sets which are given above. (Textbook pg. no. 9)
Solution:
a. All elements of set A are not present in set B, C and D.
∴ A ⊆ B,
∴ A ⊆ C,
∴ A ⊆ D
∴ Statement (a) is false.

b. All elements of set B are not present in set A, C and D.
∴ B ⊆ A,
∴ B ⊆ C,
∴ B ⊆ D
∴ Statement (b) is false.

Question 3.
Suppose U = {1, 3, 9, 11, 13, 18, 19}, and B = {3, 9, 11, 13}. Find (B’)’ and draw the inference. (Textbook pg. no. 10)

Solution:
U = {1, 3, 9, 11, 13, 18, 19},
B= {3, 9, 11, 13} ….(i)
∴ B’= {1, 18, 19}
(B’)’= {3, 9, 11, 13} ….(ii)
∴ (B’)’ = B … [From (i) and (ii)]
∴ Complement of a complement is the given set itself.

Class 9 Maths Digest

• Practice Set 1.1 Class 9 Answers
• Practice Set 1.2 Class 9 Answers
• Practice Set 1.3 Class 9 Answers
• Practice Set 1.4 Class 9 Answers
• Problem Set 1 Class 9 Answers
• Practice Set 2.1 Class 9 Answers
• Practice Set 2.2 Class 9 Answers
• Practice Set 2.3 Class 9 Answers
• Practice Set 2.4 Class 9 Answers
• Practice Set 2.5 Class 9 Answers
• Problem Set 2 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 1.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 1 Sets.

9th Standard Maths 1 Practice Set 1.2 Chapter 1 Sets Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 1.2 Chapter 1 Sets Questions With Answers Maharashtra Board

Question 1.
Decide which of the following are equal sets and which are not ? Justify your answer.
A= {x | 3x – 1 = 2}
B = {x | x is a natural number but x is neither prime nor composite}
C = {x | x e N, x < 2}
Solution:
A= {x | 3x – 1 = 2}
Here, 3x – 1 = 2
∴ 3x = 3
∴ x = 1
∴ A = {1} …(i)

B = {x | x is a natural number but x is neither prime nor composite}
1 is the only number which is neither prime nor composite,
∴ x = 1
∴ B = {1} …(ii)

C = {x | x G N, x < 2}
1 is the only natural number less than 2.
∴ x = 1
∴ C = {1} …(iii)
∴ The element in sets A, B and C is identical. … [From (i), (ii) and (iii)]
∴ A, B and C are equal sets.

Question 2.
Decide whether set A and B are equal sets. Give reason for your answer.
A = Even prime numbers
B = {x | 7x – 1 = 13}
Solution:
A = Even prime numbers
Since 2 is the only even prime number,
∴ A = {2} …(i)
B= {x | 7x – 1 = 13}
Here, 7x – 1 = 13
∴ 7x = 14
∴ x = 2
∴ B = {2} …(ii)
∴ The element in set A and B is identical. … [From (i) and (ii)]
∴ A and B are equal sets.

Question 3.
Which of the following are empty sets? Why?
i. A = {a | a is a natural number smaller than zero}
ii. B = {x | x² = 0}
iii. C = {x | 5x – 2 = 0, x ∈N}
Solution:
i. A = {a| a is a natural number smaller than zero}
Natural numbers begin from 1.
∴ A = { }
∴ A is an empty set.

ii. B = {x | x² = 0}
Here, x² = 0
∴ x = 0 … [Taking square root on both sides]
∴ B = {0}
∴B is not an empty set.

iii. C = {x | 5x – 2 = 0, x ∈ N}
Here, 5x – 2 = 0
∴ 5x = 2
∴ x = \(\frac { 2 }{ 5 }\)
Given, x ∈ N
But, x = \(\frac { 2 }{ 5 }\) is not a natural number.
∴ C = { }
∴ C is an empty set.

Question 4.
Write with reasons, which of the following sets are finite or infinite.
i. A = {x | x<10, xisa natural number}
ii. B = {y | y < -1, y is an integer}
iii. C = Set of students of class 9 from your school.
iv. Set of people from your village.
v. Set of apparatus in laboratory
vi. Set of whole numbers
vii. Set of rational number
Solution:
i. A={x| x < 10, x is a natural number}
∴ A = {1,2, 3,4, 5,6, 7, 8, 9}
The number of elements in A are limited and can be counted.
∴A is a finite set.

ii. B = (y | y < -1, y is an integer}
∴ B = { …,-4, -3, -2}
The number of elements in B are unlimited and uncountable.
∴ B is an infinite set.

iii. C = Set of students of class 9 from your school.
The number of students in a class is limited and can be counted.
∴ C is a finite set.

iv. Set of people from your village.
The number of people in a village is limited and can be counted.
∴ Given set is a finite set.

v. Set of apparatus in laboratory
The number of apparatus in the laboratory are limited and can be counted.
∴ Given set is a finite set.

vi. Set of whole numbers
The number of elements in the set of whole numbers are unlimited and uncountable.
∴ Given set is an infinite set.

vii. Set of rational number
The number of elements in the set of rational numbers are unlimited and uncountable.
∴ Given set is an infinite set.

Question 1.
If A = {1, 2, 3} and B = {1, 2, 3, 4}, then A ≠ B verify it. (Textbook pg. no. 6)
Answer:
Here, 4 ∈ B but 4 ∉ A
∴ A and B are not equal sets,
i.e. A ≠ B

Question 2.
A = {x | x is prime number and 10 < x < 20} and B = {11,13,17,19}. Here A = B. Verify. (Textbook pg. no. 6)
Answer:
A = {x | x is prime number and 10 < x < 20}
∴ A = {11, 13, 17, 19}
B = {11, 13, 17, 19}
∴ All the elements in set A and B are identical.
∴ A and B are equal sets, i.e. A = B

Class 9 Maths Digest

• Practice Set 1.1 Class 9 Answers
• Practice Set 1.2 Class 9 Answers
• Practice Set 1.3 Class 9 Answers
• Practice Set 1.4 Class 9 Answers
• Problem Set 1 Class 9 Answers
• Practice Set 2.1 Class 9 Answers
• Practice Set 2.2 Class 9 Answers
• Practice Set 2.3 Class 9 Answers
• Practice Set 2.4 Class 9 Answers
• Practice Set 2.5 Class 9 Answers
• Problem Set 2 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 1.4 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 1 Sets.

9th Standard Maths 1 Practice Set 1.4 Chapter 1 Sets Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 1.4 Chapter 1 Sets Questions With Answers Maharashtra Board

Question 1.
If n(A) = 15, n(A ∪ B) = 29, n(A ∩ B) = 7, then n(B) = ?
Solution:
Here, n(A) = 15, n(A ∪ B) = 29, n(A ∩ B) = 7
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
∴ 29 = 15 + n(B) – 7
∴ 29 – 15 + 7 = n(B)
∴ n(B) = 21

Question 2.
In a hostel there are 125 students, out of which 80 drink tea, 60 drink coffee and 20 drink tea and coffee both. Find the number of students who do not drink tea or coffee.
Solution:
i. Let U be the set of students in the hostel, T be the set of students who drink tea and C be the set of students who drink coffee.
n(U) = 125, n(T) = 80, n(C) = 60,
number of students who drink Tea and Coffee = n(T ∩ C) = 20

ii. n(T ∪ C) = n(T) + n(C) – n(T ∩ C)
= 80 + 60 – 20
∴ n(T ∪ C) = 120
∴ 120 students drink tea or coffee
Also, there are 125 students in the hostel.

iii. Number of students who do not drink tea or coffee = n(U) – n(T ∪ C)
= 125 – 120
= 5
∴ 5 students do not drink tea or coffee.

Alternate Method:
Let U be the set of students in the hostel, T be the set of students who drink tea and C be the set of students who drink coffee.

From Venn diagram,
Student who drinks tea or coffee = n(T ∪ C) = 60 + 20 + 40 = 120
∴ The number of students who do not drink tea or coffee = n(U) – n(T ∪ C)
= 125 – 120 = 5
∴ 5 students do not drink tea or coffee.

Question 3.
In a competitive exam 50 students passed in English, 60 students passed in Mathematics and 40 students passed in both the subjects. None of them failed in both the subjects. Find the number of students who passed at least in one of the subjects ?
Solution:
Let U be the set of students who appeared for the exam,
E be the set of students who passed in English and
M be the set of students who passed in Maths.
∴ n(E) = 50, n(M) = 60,
40 students passed in both the subjects
∴ n(M ∩ E) = 40
Since, none of the students failed in both subjects
∴ Total students = n(E ∪M)
= n(E) + n(M) – n(E ∩ M)
= 50 + 60 – 40
= 70
∴ The number of students who passed at least in one of the subjects is 70.

Alternate Method:
Let U be the set of students who appeared for the exam,
E be the set of students who passed in English and M be the set of students who passed in Maths.

Since, none of the students failed in both subjects
∴ Total student = n(E ∪M)
= 10 + 40 + 20
= 70
∴ The number of students who passed at least in one of the subjects is 70.

Question 4.
A survey was conducted to know the hobby of 220 students of class IX. Out of which 130 students informed about their hobby as ’rock climbing and 180 students informed about their hobby as sky watching. There are 110 students who follow both the hobbies. Then how many students do not have any of the two hobbies? How many of them follow the hobby of rock climbing only? How many students follow the hobby of sky watching only?
Solution:
i. Let U be the set of students of class IX,
R be the set of students who follow the hobby of rock climbing and
S be the set of students who follow the hobby of sky watching.
∴ n (U) = 220, n (R) = 130, n (S) = 180,
110 students follow both the hobbies
∴ n (R ∩ S) = 110

ii. n(R ∪ S)=n (R) + n (S) – n (R ∩ S)
= 130 + 180 – 110
∴n (R ∪ S) = 200
∴ 200 students follow the hobby of rock climbing or sky watching.

iii. Total number of students = 220.
Number of students who do not follow the hobby of rock climbing or sky watching
= n (U) – n (R ∪ S)
= 220 – 200
= 20

iv. Number of students who follow the hobby of rock climbing only
= n (R) – n(R ∩ S)
= 130 – 110
= 20

v. Number of students who follow the hobby of sky watching only
= n (S) – n (R ∩ S)
= 180 – 110
= 70

Alternate Method:
Let U be the set of students of class IX,
R be the set of students who follow the hobby of rock climbing and
S be the set of students who follow the hobby of sky watching.

From the Venn diagram
i. Students who follow the hobby of rock climbing or sky watching
= n(R ∪ S)
= 20 + 110 + 70
= 200

ii. Number of students who do not follow the hobby of rock climbing or sky watching
= n (U) – n(R ∪S)
= 220 – 200
= 20

iii. Number of students who follow the hobby of rock climbing only
= n (R) – n(R ∩S)
= 130 – 110
= 20

iv. Number of students who follow the hobby of sky watching only
= n (S) – n (R ∩ S)
= 180 – 110
= 70

Question 5.
Observe the given Venn diagram and write the following sets.

i. A
ii. B
iii. A ∪ B
iv. U
v. A’
vi. B’
vii. (A ∪B )’
Ans:
i. A = {x, y, z, m, n}
ii. B = {p, q, r, m, n}
iii. A ∪ B = {x, y, z, m, n, p, q, r }
iv. U = {x, y, z, m, n, p, q, r, s, t}
v. A’ = {p, q, r, s, t}
vi. B’ = {x, y, z, s, t}
vii. (A ∪ B )’ = {s, t}

Question 1.
Take different examples of sets and verify the above mentioned properties. (Textbook pg.no. 12)
Solution:
i. Let A = {3, 5}, B= {3, 5, 8, 9, 10}
A ∩ B = B ∩ A = {3, 5}

ii. Let A = {3, 5}, B = {3, 5, 8, 9, 10}
Since, all elements of set A are present in set B.
∴ A ⊆ B
Also, A ∩ B = {3, 5} = A
∴ If A ⊆ B, then A ∩B = A.

iii. Let A = {2, 3, 8, 10}, B = {3,8}
A ∩ B = {3, 8} = B
Also, all the elements of set B are present in set A
∴ B ⊆ A
∴ If A ∩ B = B, then B ⊆ A.

iv. Let A = {2, 3, 8, 10}, B = {3, 8}, A ∩B = {3, 8}
Since, all the elements of set A n B are present in set A and B
A ∩ B ⊆ A and A ∩B ⊆B

v. Let U= {3, 4, 6, 8}, A = {6, 4}
∴ A’ = {3, 8}
∴ A ∩ A’= { } = φ

vi. A ∩ φ = { } = φ

vii. Let A = {6, 4}
∴ A ∩ A = {6, 4}
∴ A ∩ A = A

Question 2.
Observe the set A, B, C given by Venn diagrams and write which of these are disjoint sets. (Textbook pg. no. 12)

Solution:
Here, A = {1, 2, 3, 4, 5, 6, 7}
B = {3, 6, 8, 9, 10, 11, 12}
C = {10, 11, 12}
Now, A ∩ C = φ
∴ A and C are disjoint sets.

Question 3.
Let the set of English alphabets be the Universal set. The letters of the word ‘LAUGH’ is one set and the letter of the word ‘CRY’ is another set. Can we say that these are two disjoint sets? Observe that intersection of these two sets is empty. (Textbook pg. no. 13)

Solution:
Let A = {L, A, U, G, H}
B = {C, R, Y}
Now, A ∩ B = φ
∴ A and B are disjoint sets.

Question 4.
Fill in the blanks with elements of that set.
U = {1, 3, 5, 8, 9, 10, 11, 12, 13, 15}
A = {1,11, 13}
B = {8,5, 10, 11, 15}
A’ = { }
B’ = { }
A ∩ B = { }
A’ ∩ B’ = { }
A ∪ B = { }
A’ ∪ B’= { }
(A ∩ B)’ = { }
(A ∪ B)’ = { }
Verify: (A ∩ B)’ = A’ u B’, (A u B)’ = A’ ∩ B’ (Textbook pg. no, 18)
Solution:
U = {1, 3, 5, 8, 9, 10, 11, 12, 13, 15}
A = {1, 11, 13}
B = {8, 5, 10, 11, 15}
A’ = {3, 5, 8, 9, 10, 12, 15}
B’ = {1, 3, 9, 12, 13}
A∩ B= {11}
A’ ∩ B’= {3, 9, 12} …(i)
A ∪ B = {1, 5, 8, 10, 11, 13, 15}
A’ ∪ B’ = { 1, 3, 5, 8, 9, 10, 12, 13, 15} …(ii)
(A ∩ B)’= { 1, 3, 5, 8, 9, 10, 12, 13,15} …(iii)
(A ∪ B)’ = {3, 9, 12} ,..(iv)
(A ∩ B)’ = A’ ∪ B’ … [From (ii) and (iii)]
(A ∪ B)’ = A’ ∩ B’ … [From (i) and (iv)]

Question 5.
A = {1,2,3, 5, 7,9,11,13}
B = {1,2,4, 6, 8,12,13}
Verify the above rule for the given set A and set B. (Textbook pg. no. 14)
Solution:
A = {1, 2, 3, 5, 7, 9, 11, 13}
B = {1, 2, 4, 6, 8, 12, 13}
A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13}
A ∩ B= {1, 2, 13}
n(A) = 8, n(B) = 7,
n(A ∪ B) = 12, n(A ∩ B) = 3
n(A ∩ B) = 12 …(i)
n(A) + n(B) – n(A ∩ B) = 8 + 7 – 3 = 12 …(ii)
∴ n(A ∪ B) = n(A) + n(B) – n(A ∩ B) … [From (i) and (ii)]

Question 6.
Verify the above rule for the given Venn diagram. (Textbook pg. no. 14)

Solution:
n(A) = 5 , n(B) = 6
n(A ∪ B) = 9 , n(A ∩ B) = 2
Now, n(A ∪ B) = 9 …(i)
n(A) + n(B) – n(A ∩ B) = 5 + 6 – 2 = 9 …(ii)
∴ n(A ∪ B) = n(A) + n(B) – n(A ∩ B). …[From (i) and (ii)]

Class 9 Maths Digest

• Practice Set 1.1 Class 9 Answers
• Practice Set 1.2 Class 9 Answers
• Practice Set 1.3 Class 9 Answers
• Practice Set 1.4 Class 9 Answers
• Problem Set 1 Class 9 Answers
• Practice Set 2.1 Class 9 Answers
• Practice Set 2.2 Class 9 Answers
• Practice Set 2.3 Class 9 Answers
• Practice Set 2.4 Class 9 Answers
• Practice Set 2.5 Class 9 Answers
• Problem Set 2 Class 9 Answers