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9th Standard Maths 1 Practice Set 2.3 Chapter 2 Real Numbers Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.3 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 2 Real Numbers.

Class 9 Maths Part 1 Practice Set 2.3 Chapter 2 Real Numbers Questions With Answers Maharashtra Board

Question 1.
State the order of the surds given below.

Answer:
i. 3, ii. 2, iii. 4, iv. 2, v. 3

Question 2.
State which of the following are surds Justify. [2 Marks each]

Answer:
i. \(\sqrt [ 3 ]{ 51 }\) is a surd because 51 is a positive rational number, 3 is a positive integer greater than 1 and \(\sqrt [ 3 ]{ 51 }\) is irrational.

ii. \(\sqrt [ 4 ]{ 16 }\) is not a surd because

= 2, which is not an irrational number.

iii. \(\sqrt [ 5 ]{ 81 }\) is a surd because 81 is a positive rational number, 5 is a positive integer greater than 1 and \(\sqrt [ 5 ]{ 81 }\) is irrational.

iv. \(\sqrt { 256 }\) is not a surd because

= 16, which is not an irrational number.

v. \(\sqrt [ 3 ]{ 64 }\) is not a surd because

= 4, which is not an irrational number.

vi. \(\sqrt { \frac { 22 }{ 7 } }\) is a surd because \(\frac { 22 }{ 7 }\) is a positive rational number, 2 is a positive integer greater than 1 and \(\sqrt { \frac { 22 }{ 7 } }\) is irrational.

Question 3.
Classify the given pair of surds into like surds and unlike surds. [2 Marks each]

Solution:
If the order of the surds and the radicands are same, then the surds are like surds.

Here, the order of 2\(\sqrt { 13 }\) and 5\(\sqrt { 13 }\) is same and their radicands are also same.
∴ \(\sqrt { 52 }\) and 5\(\sqrt { 13 }\) are like surds.

Here, the order of 2\(\sqrt { 17 }\) and 5\(\sqrt { 3 }\) is same but their radicands are not.
∴ \(\sqrt { 68 }\) and 5\(\sqrt { 3 }\) are unlike surds.

Here, the order of 12\(\sqrt { 2 }\) and 7\(\sqrt { 2 }\) is same and their radicands are also same.
∴ 4\(\sqrt { 18 }\) and 7\(\sqrt { 2 }\) are like surds.

Here, the order of 38\(\sqrt { 3 }\) and 6\(\sqrt { 3 }\) is same and their radicands are also same.
∴ 19\(\sqrt { 12 }\) and 6\(\sqrt { 3 }\) are like surds.

v. 5\(\sqrt { 22 }\), 7\(\sqrt { 33 }\)
Here, the order of 5\(\sqrt { 22 }\) and 7\(\sqrt { 33 }\) is same but their radicands are not.
∴ 5\(\sqrt { 22 }\) and 7\(\sqrt { 33 }\) are unlike surds.

Here, the order of 5√5 and 5√3 is same but their radicands are not.
∴ 5√5 and √75 are unlike surds.

Question 4.
Simplify the following surds.

Solution:

Question 5.
Compare the following pair of surds.


Solution:

Question 6.
Simplify.

Solution:

Question 7.
Multiply and write the answer in the simplest form.

Solution:

Question 8.
Divide and write form.

Solution:

Question 9.
Rationalize the denominator.

Solution:

Question 1.
\(\sqrt { 9+16 }\) ? + \(\sqrt { 9 }\) + \(\sqrt { 16 }\) (Texbookpg. no. 28)
Solution:

Question 2.
\(\sqrt { 100+36 }\) ? \(\sqrt { 100 }\) + \(\sqrt { 36 }\) (Textbook pg. no. 28)
Solution:

Question 3.
Follow the arrows and complete the chart by doing the operations given. (Textbook pg. no. 34)

Solution:

Question 4.
There are some real numbers written on a card sheet. Use these numbers and construct two examples each of addition, subtraction, multiplication and division. Solve these examples. (Textbook pg. no. 34)

Solution:

Maths Solution Class 9 Maharashtra Board 

• Real Numbers Practice Set 2.1 Class 9 Maths Solutions
• Real Numbers Practice Set 2.2 Class 9 Maths Solutions
• Real Numbers Practice Set 2.3 Class 9 Maths Solutions
• Real Numbers Practice Set 2.4 Class 9 Maths Solutions
• Real Numbers Practice Set 2.5 Class 9 Maths Solutions
• Real Numbers Problem Set 2 Class 9 Maths Solutions

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

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

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.]

Maharashtra Board Class 9 Maths Solutions

• Sets Practice Set 1.1 Class 9 Maths Solutions
• Sets Practice Set 1.2 Class 9 Maths Solutions
• Sets Practice Set 1.3 Class 9 Maths Solutions
• Sets Practice Set 1.4Class 9 Maths Solutions
• Sets Problem Set 1 Class 9 Maths Solutions

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

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.

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

Maths Solution Class 9 Maharashtra Board 

• Real Numbers Practice Set 2.1 Class 9 Maths Solutions
• Real Numbers Practice Set 2.2 Class 9 Maths Solutions
• Real Numbers Practice Set 2.3 Class 9 Maths Solutions
• Real Numbers Practice Set 2.4 Class 9 Maths Solutions
• Real Numbers Practice Set 2.5 Class 9 Maths Solutions
• Real Numbers Problem Set 2 Class 9 Maths Solutions

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

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.

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.

Maharashtra Board Class 9 Maths Solutions

• Sets Practice Set 1.1 Class 9 Maths Solutions
• Sets Practice Set 1.2 Class 9 Maths Solutions
• Sets Practice Set 1.3 Class 9 Maths Solutions
• Sets Practice Set 1.4Class 9 Maths Solutions
• Sets Problem Set 1 Class 9 Maths Solutions

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

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.

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

Maharashtra Board Class 9 Maths Solutions

• Sets Practice Set 1.1 Class 9 Maths Solutions
• Sets Practice Set 1.2 Class 9 Maths Solutions
• Sets Practice Set 1.3 Class 9 Maths Solutions
• Sets Practice Set 1.4Class 9 Maths Solutions
• Sets Problem Set 1 Class 9 Maths Solutions

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

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.

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)]

Maharashtra Board Class 9 Maths Solutions

• Sets Practice Set 1.1 Class 9 Maths Solutions
• Sets Practice Set 1.2 Class 9 Maths Solutions
• Sets Practice Set 1.3 Class 9 Maths Solutions
• Sets Practice Set 1.4Class 9 Maths Solutions
• Sets Problem Set 1 Class 9 Maths Solutions

9th Standard Maths 1 Practice Set 1.1 Chapter 1 Sets Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 1.1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 1 Sets.

Class 9 Maths Part 1 Practice Set 1.1 Chapter 1 Sets Questions With Answers Maharashtra Board

Question 1.
Write the following sets in roster form.
i. Set of even natural numbers
ii. Set of even prime numbers from 1 to 50
iii. Set of negative integers
iv. Seven basic sounds of a sargam (sur)
Answer:
i. A = { 2, 4, 6, 8,….}
ii. 2 is the only even prime number
∴ B = { 2 }
iii. C = {-1, -2, -3,….}
iv. D = {sa, re, ga, ma, pa, dha, ni}

Question 2.
Write the following symbolic statements in words.
i. \(\frac { 4 }{ 3 }\) ∈ Q
ii. -2 ∉ N
iii. P = {p | p is an odd number}
Answer:
i. \(\frac { 4 }{ 3 }\) is an element of set Q.
ii. -2 is not an element of set N.
iii. Set P is a set of all p’s such that p is an odd number.

Question 3.
Write any two sets by listing method and by rule method.
Answer:
i. A is a set of even natural numbers less than 10.
Listing method: A = {2, 4, 6, 8}
Rule method: A = {x | x = 2n, n e N, n < 5}

ii. B is a set of letters of the word ‘SCIENCE’. Listing method : B = {S, C, I, E, N}
Rule method: B = {x \ x is a letter of the word ‘SCIENCE’}

Question 4.
Write the following sets using listing method.
i. All months in the Indian solar year.
ii. Letters in the word ‘COMPLEMENT’.
iii. Set of human sensory organs.
iv. Set of prime numbers from 1 to 20.
v. Names of continents of the world.
Answer:
i. A = {Chaitra, Vaishakh, Jyestha, Aashadha, Shravana, Bhadrapada, Ashwina, Kartika, Margashirsha, Paush, Magha, Falguna}
ii. X = {C, O, M, P, L, E, N, T}
iii. Y = {Nose, Ears, Eyes, Tongue, Skin}
iv. Z = {2, 3, 5, 7, 11, 13, 17, 19}
v. E = {Asia, Africa, Europe, Australia, Antarctica, South America, North America}

Question 5.
Write the following sets using rule method.
i. A = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
ii. B= {6, 12, 18,24, 30,36,42,48}
iii. C = {S, M, I, L, E}
iv. D = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
v. X = {a, e, t}
Answer:
i. A = {x | v = n², n e N, n < 10}
ii. B = {x j x = 6n, n e N, n < 9}
iii. C = {y j y is a letter of the word ‘SMILE’} [Other possible words: ‘SLIME’, ‘MILES’, ‘MISSILE’ etc.]
iv. D = {z | z is a day of the week}
v. X = {y | y is a letter of the word ‘eat’}
[Other possible words: ‘tea’ or ‘ate’]

Question 1.
Fill in the blanks given in the following table. (Textbook pg. no. 3)
Answer:

Maharashtra Board Class 9 Maths Solutions

• Sets Practice Set 1.1 Class 9 Maths Solutions
• Sets Practice Set 1.2 Class 9 Maths Solutions
• Sets Practice Set 1.3 Class 9 Maths Solutions
• Sets Practice Set 1.4Class 9 Maths Solutions
• Sets Problem Set 1 Class 9 Maths Solutions

9th Standard Maths 1 Problem Set 6 Chapter 6 Financial Planning Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 6 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 6 Financial Planning.

Class 9 Maths Part 1 Problem Set 6 Chapter 6 Financial Planning Questions With Answers Maharashtra Board

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

i. For different types of investments what is the maximum permissible amount under section 80C of income tax ?
(A) ₹ 1,50,000
(B) ₹ 2,50,000
(C) ₹ 1,00,000
(D) ₹ 2,00,000
Answer:
(A) ₹ 1,50,000

ii. A person has earned his income during the financial year 2017-18. Then his assessment year is….
(A) 2016 – 17
(B) 2018 – 19
(C) 2017 – 18
(D) 2015 – 16
Answer:
(B) 2018 – 19

Question 2.
Mr. Shekhar spends 60% of his income. From the balance he donates ₹ 300 to an orphanage. He is then left with ₹ 3,200. What is his income ?
Solution:
Let the income of Shekhar be ₹ x.
Shekhar spends 60% of his income.
∴ Shekhar’s expenditure = 60% of x
∴ Amount remaining with Shekhar = (100 – 60)% of x
= 40% of x
= \(\frac { 1 }{ 2 }\) × x
= 0.4x
From the balance left, he donates ₹ 300 to an orphanage.
∴ Amount left with Shekhar = 0.4x – 300
Now, the amount left with him is ₹ 3200.
∴ 3200 = 0.4x- 300
∴ 0.4x = 3500

∴ The income of Mr. Shekhar is ₹ 8750.

Question 3.
Mr. Hiralal invested ₹ 2,15,000 in a Mutual Fund. He got ₹ 3,05,000 after 2 years. Mr. Ramniklal invested ₹ 1,40,000 at 8% compound interest for 2 years in a bank. Find out the percent gain of each of them. Whose investment was more profitable ?
Solution:
Mr. Hiralal:
Amount invested by Mr. Hiralal in mutual fund = ₹ 2,15,000
Amount received by Mr. Hiralal = ₹ 3,05,000
∴ Mr. Hiralal’s profit = Amount received – Amount invested
= 305000 – 215000 = ₹ 90000
Mr. Hirala’s percentage of profit
= \(\frac { 90000 }{ 215000 }\) × 100
= 41.86%

Mr. Ramniklal:
P = ₹ 140000, R = 8%, n = 2 years
∴ Compound interest (I)
= A – P

= 140000 [(1 + 0.08)² – 1]
= 140000 [ (1.08)² – 1]
= 140000(1.1664 – 1)
= 140000 x 0.1664
= ₹ 23296
∴ Mr. Ramniklal’s percentage of profit
= \(\frac { 23296 }{ 140000 }\) × 100
= 16.64%
∴ The percentage gains of Mr. Hiralal and Mr. Ramniklal are 41.86% and 16.64% respectively, and hence, Mr. Hiralal’s investment was more profitable.

Question 4.
At the start of a year there were ₹ 24,000 in a savings account. After adding ₹ 56,000 to this the entire amount was invested in the bank at 7.5% compound interest. What will be the total amount after 3 years ?
Solution:
Here, P = 24000 + 56000
= ₹ 80000
R = 7.5%, n = 3 years
Total amount after 3 years

= 80000 (1 + 0.075)³
= 80000 (1.075)³
= 80000 x 1.242297
= 99383.76
∴ The total amount after 3 years is ₹ 99383.76.

Question 5.
Mr. Manohar gave 20% of his income to his elder son and 30% to his younger son. He gave 10% of the balance income as donation to a school. He still had ₹ 1,80,000 for himself. What was Mr. Manohar’s income ?
Solution:
Let the income of Mr. Manohar be ₹ x.
Amount given to elder son = 20% of x
Amount given to younger son = 30% of x
Total amount given to both sons = (20 + 30)% of x = 50% of x
∴ Amount remaining with Mr. Manohar = (100 – 50)% of x
= 50% of x 50
= \(\frac { 50 }{ 100 }\) × 100
= 0.5 x
He gave 10% of the balance income as donation to a school.
Amount donated to school = 10% of 0.5x
= \(\frac { 10 }{ 100 }\) × 0.5x
= 0.05x
∴ Amount remaining with Mr. Manohar after donating to school = 0.5x – 0.05x
= 0.45x
Mr. Manohar still had 1,80,000 for himself after donating to school.
∴ 180000 = 0.45x
∴ \(x=\frac{180000}{0.45}=\frac{180000 \times 100}{0.45 \times 100}=\frac{18000000}{45}=400000\)
∴ The income of Mr. Manoliar is ₹4,00,000.

Question 6.
Kailash used to spend 85% of his income. When his income increased by 36% his expenses also increased by 40% of his earlier expenses. How much percentage of his earning he saves now ?
Solution:
Let the income of Kailash be ₹ x.
Kailash spends 85% of his income.
∴ Kailash’s expenditure = 85% of x
= \(\frac { 85 }{ 100 }\) × x = 0.85 x
Kailash’s income increased by 36%.
∴ Kailash’s new income = x + 36% of x
= x + \(\frac { 36 }{ 100 }\) × x
= x + 0.36x
= 1.36x
Kailash’s expenses increased by 40%.
∴ Kailash’s new expenditure = 0.85x + 40% of 0.85x
= 0.85x + \(\frac { 40 }{ 100 }\) × 0.85 × 100
= 0.85x + 0.4 × 0.85x
= 0.85x (1 + 0.4)
= 0.85x × 1.4
= 1.19x
∴ Kailash’s new saving = Kailash’s new income – Kailash’s new expenditure
= 1.36x – 1.19x
= 0.17x
Percentage of Kailash’s new saving
= \(\frac { 0.17x }{ 1.36x }\) × 100
= 12.5%
∴ Kailash saves 12.5% of his new earning.

Question 7.
Total income of Ramesh, Suresh and Preeti is ₹ 8,07,000. The percentages of their expenses are 75%, 80% and 90% respectively. If the ratio of their savings is 16 : 17 : 12, then find the annual saving of each of them.
Solution:
Let the annual income of Ramesh, Suresh and Preeti be ₹ x, t y and ₹ z respectively.
Total income of Ramesh, Suresh and Preeti = ₹ 8,07,000
∴ x + y + z = 807000 …(i)

∴ Savings of Ramesh = 25% of x
= ₹ \(\frac { 25x }{ 100 }\) ..(ii)
Savings of Suresh = 20% of y
= ₹\(\frac { 20y }{ 100 }\) …(iii)
Savings of Preeti = 10% of z
= ₹\(\frac { 10z }{ 100 }\) …..(iv)

Ratio of their savings = 16 : 17 : 12
Let the common multiple be k.
Savings of Ramesh = ₹ 16 k … (v)
Savings of Suresh = ₹ 17 k … (vi)
Savings of Preeti = ₹ 12 k .. .(vii)

∴ z = 120k …(x)
From (i), (viii), (ix) and (x), we get
64k + 85k + 120k = 807000
269k = 807000
k = \(\frac { 807000 }{ 269 }\)
k = 3000
∴ Annual saving of Ramesh = 16k
= 16 x 3000
= ₹ 48,000
Annual saving of Suresh = 17k
= 17 x 3000
= ₹ 51,000
Annual saving of Preeti = 12k
= 12 x 3000
= ₹ 36,000
The annual savings of Ramesh, Suresh and Preeti are ₹ 48,000, ₹ 51,000 and ₹ 36,000 respectively.

Question 8.
Compute the income tax payable by following individuals.
i. Mr. Kadam who is 35 years old and has a taxable income of ₹13,35,000.
ii. Mr. Khan is 65 years of age and his taxable income is ₹4,50,000.
iii. Miss Varsha (Age 26 years) has a taxable income of ₹2,30,000.
Solution:
i. Mr. Kadam is 35 years old and his taxable income is ₹13,35,000.
Mr. Kadam’s income is more than ₹ 10,00,000.
∴ Income tax = ₹1,12,500 + 30% of (taxable income -10,00,000)
= ₹ 1,12,500 + 30% of (13,35,000 – 10,00,000)
= 112500+ \(\frac { 30 }{ 100 }\) x 335000 100
= 112500+ 100500
= ₹ 213000
Education cess = 2% of income tax
= \(\frac { 2 }{ 100 }\) x 213000
= ₹ 4260.
Secondary and Higher Education cess
= 1% of income tax
= \(\frac { 1 }{ 100 }\) x 213000 100
= 2130
Total income tax = Income tax + Education cess + Secondary and higher education cess
= 213000 + 4260 + 2130 = ₹ 2,19,390
∴ Mr. Kadam will have to pay income tax of ₹ 2,19,390.

ii. Mr. Khan is 65 years old and his taxable income is ₹ 4,50,000.
Mr. Khan’s income falls in the slab ₹ 3,00,001 to ₹ 5,00,000.
∴ Income tax
= 5% of (taxable income – 300000)
= 5% of (450000 – 300000)
= \(\frac { 5 }{ 100 }\) x 150000 100
= ₹ 7500
Education cess = 2% of income tax
= \(\frac { 2 }{ 100 }\) x 7500
= ₹ 150
Secondary and Higher Education cess = 1 % of income tax
= \(\frac { 1 }{ 100 }\) x 7500
= 75
Total income tax = Income tax + Education cess + Secondary and higher education cess
= 7500+ 150 + 75
= ₹ 7725
Mr. Khan will have to pay income tax of ₹7725.

iii. Taxable income = ₹2,30,000
age = 26 years
The yearly income of Miss Varsha is less than ₹ 2,50,000.
Hence, Miss Varsha will not have to pay income tax.

Maharashtra Board Class 9 Maths Chapter 6 Financial Planning Problem Set 6 Intext Questions and Activities

Question 1.
With your parent’s help write down the income and expenditure of your family for one week. Make 7 columns for the seven days of the week. Write all expenditure under such heads as provisions, education, medical expenses, travel, clothes and miscellaneous. On the credit side write the amount received for daily expenses, previous balance and any other new income. (Textbook pg. no. 98)

Question 2.
In the holidays, write the accounts for the whole month. (Textbook pg. no. 98)

Question 3.
What is a tax? Which are different types of taxes? Find out more information on following websites
www.incometaxindia.gov.in,
www.mahavat.gov.in
www.gst.gov.in (Textbook pg. no. 99)

Question 4.
Obtain more information about different types of taxes from employees and professionals who pay taxes. (Textbook pg. no. 99)

Question 5.
Obtain information about sections 80C, 80G, 80D of the Income Tax Act. (Textbook pg. no. 103)

Question 6.
Study a PAN card and make a note of all the information it contains. (Textbook pg.no. 103)

Question 7.
Obtain information about all the devices and means used for carrying out cash minus transactions. (Textbook pg, no, 103)

Question 8.
Visit www.incometaxindia.gov.in which is a website of the Government of India. Click on the ‘incometax calculator’ menu. Fill in the form that gets downloaded using an imaginary income and imaginary deductible amounts and try to compute the income tax payable for this income. (Textbook pg.no. 107)
[Students should attempt the above activities on their own.]

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Maharashtra Board 9th Maths Solution 

• Financial Planning Practice Set 6.1 Class 9 Maths Solutions
• Financial Planning Practice Set 6.2 Class 9 Maths Solutions
• Financial Planning Problem Set 6 Class 9 Maths Solutions

10th Standard Maths 2 Problem Set 6 Chapter 6 Trigonometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 6 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 6 Trigonometry.

Class 10 Maths Part 2 Problem Set 6 Chapter 6 Trigonometry Questions With Answers Maharashtra Board

Question 1.
Choose the correct alternative answer for the following questions.

i. sin θ.cosec θ = ?
(A) 1
(B) 0
(C) \(\frac { 1 }{ 2 } \)
(D) \(\sqrt { 2 }\)
Answer:
(A)

ii. cosec 45° = ?
(A) \(\frac{1}{\sqrt{2}}\)
(B) \(\sqrt { 2 }\)
(C) \(\frac{\sqrt{3}}{2}\)
(D) \(\frac{1}{\sqrt{3}}\)
Answer:
(B)

iii. 1 + tan² θ = ?
(A) cot² θ
(B) cosec² θ
(C) sec² θ
(D) tan² θ
Answer:
(C)

iv. When we see at a higher level, from the horizontal line, angle formed is ______
(A) angle of elevation.
(B) angle of depression.
(C) 0
(D) straight angle.
Answer:
(A)

Question 2.
If sin θ = \(\frac { 11 }{ 61 } \), find the value of cos θ using trigonometric identity.
Solution:
sin θ = \(\frac { 11 }{ 61 } \) … [Given]
We know that,
sin² θ + cos² θ = 1


…[Taking square root of both sides]

Question 3.
If tan θ = 2, find the values of other trigonometric ratios.
Solution:
tan θ = 2 …[Given]
We know that,
1 + tan² θ = sec⁷ θ
∴ 1 + (2)⁷ = sec⁷ θ
∴ 1 + 4 = sec⁷ θ
∴ sec⁷ θ = 5
∴ sec θ = \(\sqrt { 5 }\) …[Taking square root of both sides]

Question 4.
If sec θ = \(\frac { 13 }{ 12 } \), find the values of other trigonometric ratios.
Solution:
sec θ = \(\frac { 13 }{ 12 } \) … [Given]
We know that,
1 + tan² θ = sec² θ


∴ sin θ = \(\frac { 5 }{ 13 } \), cos θ = \(\frac { 12 }{ 13 } \), tan θ = \(\frac { 5 }{ 12 } \), cot θ = \(\frac { 12 }{ 5 } \), cosec θ = \(\frac { 13 }{ 5 } \)

Question 5.
Prove the following:
i. sec θ (1 – sin θ) (sec θ + tan θ) = 1
ii. (sec θ + tan θ) (1 – sin θ) = cos θ
iii. sec² θ + cosec² θ = sec² θ × cosec² θ
iv. cot² θ – tan² θ = cosec² θ – sec² θ
v. tan⁴ θ + tan² θ = sec⁴ θ – sec² θ
vi. \(\frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}\) = 2 sec2 θ
vii. sec⁶ x – tan⁶ x = 1 + 3 sec² x × tan² x

Proof:
i. L.H.S. = sec θ (1 – sin θ) (sec θ + tan θ)

∴ sec θ (1 – sin θ) (sec θ + tan θ) = 1

ii. L.H.S. = (sec θ + tan θ) (1 – sin θ)

∴ (sec θ + tan θ) (1 – sin θ) = cos θ

iii. L.H.S. = sec² θ + cosec² θ

∴ sec² θ + cosec² θ = sec² θ × cosec² θ

iv. L.H.S. = cot² θ – tan² θ
= (cosec² θ – 1) – (sec² θ – 1)
[∵ tan² θ = sec² θ – 1,
cot² θ = cosec² θ – 1]
= cosec² θ – 1 – sec² θ + 1
cosec² θ – sec² θ
= R.H.S.
∴ cot² θ – tan² θ = cosec² θ – sec² θ

v. L.H.S. = tan⁴ θ + tan² θ
= tan² θ (tan² θ + 1)
= tan² θ. sec² θ
…[∵ 1 + tan² θ = sec² θ]
= (sec² θ – 1) sec² θ
…[∵ tan² θ = sec² θ – 1]
= sec⁴ θ – sec² θ
= R.H.S.
∴ tan⁴ θ + tan² θ = sec⁴ θ – sec² θ

vii. L.H.S. = sec⁶ x – tan⁶ x
= (sec² x)³ – tan⁶ x
= (1 + tan² x)³ – tan⁶ x …[∵ 1 + tan² θ = sec² θ]
= 1 + 3tan² x + 3(tan² x)² + (tan² x)³ – tan⁶ x …[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= 1 + 3 tan² x (1 + tan² x) + tan⁶ x – tan⁶ x
= 1 + 3 tan² x sec² x …[∵ 1 + tan² θ = sec² θ]
= R.H.S.
∴ sec³x – tan⁶x = 1 + 3sec²x.tan²x


x. We know that,
sin² θ + cos² θ = 1
∴ 1 – sin² θ = cos² θ
∴ (1 – sin θ) (1 + sin θ) = cos θ. cos θ

Question 6.
A boy standing at a distance of 48 metres from a building observes the top of the building and makes an angle of elevation of 30°. Find the height of the building.
Solution:
Let AB represent the height of the building and point C represent the position of the boy.
Angle of elevation = ∠ACB = 30°
BC = 48 m

In right angled ∆ABC,
tan 30° = \(\frac { AB }{ BC } \) … [By definition]

∴ The height of the building is 16\(\sqrt { 3 }\) m.

Question 7.
From the top of the lighthouse, an observer looks at a ship and finds the angle of depression to be 30°. If the height of the lighthouse is 100 metres, then find how far the ship is from the lighthouse.
Solution:
Let AB represent the height of lighthouse and point C represent the position of the ship.

Angle of depression ∠PAC 30°
AB = 100m.
Now, ray AP || seg BC
∴ ∠ACB = ∠PAC … [Alternate angles]
∴ ∠ACB = 30°
AB = 100m
In right angled ∆ABC,
tan 30° = \(\frac { AB }{ BC } \) …[By definition]
∴ \(\frac{1}{\sqrt{3}}=\frac{100}{\mathrm{BC}}\)
∴ BC = 100\(\sqrt { 3 }\)m
∴ The ship is 1oo\(\sqrt { 3 }\)m far from the lighthouse.

Question 8.
Two buildings are in front of each other on a road of width 15 metres. From the top of the first building, having a height of 12 metre, the angle of elevation of the top of the second building is 30°. What is the height of the second building?
Solution:
Let AB and CD represent the heights of the two buildings, and BD represent the width of the road.

Draw seg AM ⊥ seg CD
Angle of elevation = ∠CAM = 30°
AB = 12m
BD = 15m
In ꠸ ABDM,
∠B = ∠D = 90°
∠M 90° …[segAM ⊥ segCD]
∠A 90° …[Remaining angle of ꠸ABDM]
꠸ABDM is a rectangle …[Each angle is 90°]

∴ The height of the second building is 20.65 m.

Question 9.
A ladder on the platform of a fire brigade van can be elevated at an angle of 70° to the maximum. The length of the ladder can be extended upto 20 m. If the platform is 2 m above the ground, find the maximum height from the ground upto which the ladder can reach. (sin 70° = 0.94)
Solution:
Let AB represent the length of the ladder and AE represent the height of the platform.

Draw seg AC ⊥ seg BD.
Angle of elevation = ∠BAC = 70°
AB = 20 m
AE = 2m
In right angled ∆ABC,
sin 70° = \(\frac { BC }{ AB } \) …..[By definition]
∴ 0.94 = \(\frac { BC }{ 20 } \)
∴ BC = 0.94 × 20
= 18.80 m
In ꠸ACDE,
∠E = ∠D = 90°
∠C = 90° … [seg AC ⊥ seg BD]
∴ ∠A = 90° … [Remaining angle of ꠸ACDE]
∴ ꠸ACDE is a rectangle. … [Each angle is 90°]
∴ CD = AE = 2 m … [Opposite sides of a rectangle]
Now, BD = BC + CD … [B – C – D]
= 18.80 + 2
= 20.80 m
∴ The maximum height from the ground upto which the ladder can reach is 20.80 metres.

Question 10.
While landing at an airport, a pilot made an angle of depression of 20°. Average speed of the plane was 200 km/hr. The plane reached the ground after 54 seconds. Find the height at which the plane was when it started landing, (sin 20° = 0.342)
Solution:
Let AC represent the initial height and point A represent the initial position of the plane.
Let point B represent the position where plane lands.
Angle of depression = ∠EAB = 20°

Now, seg AE || seg BC
∴ ∠ABC = ∠EAB … [Alternate angles]
∴ ∠ABC = 20°
Speed of the plane = 200 km/hr
= 200 × \(\frac { 1000 }{ 3600 } \) m/sec
= \(\frac { 500 }{ 9 } \) m/sec
∴ Distance travelled in 54 sec = speed × time
= \(\frac { 500 }{ 9 } \) × 54
= 3000 m
∴ AB = 3000 m
In right angled ∆ABC,
sin 20° = \(\frac { AC }{ AB } \) ….[By definition]
∴ 0.342 = \(\frac { AC }{ 3000 } \)
∴ AC = 0.342 × 3000
= 1026 m
∴ The plane was at a height of 1026 m when it started landing.

Maharashtra State Board Class 10 Maths Solutions Part 2

• Co-ordinate Geometry Practice Set 5.1 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.2 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.3 Class 10 Maths Solutions
• Co-ordinate Geometry Problem Set 5 Class 10 Maths Solutions
• Trigonometry Practice Set 6.1 Class 10 Maths Solutions
• Trigonometry Practice Set 6.2 Class 10 Maths Solutions
• Trigonometry Problem Set 6 Class 10 Maths Solutions
• Mensuration Practice Set 7.1 Class 10 Maths Solutions
• Mensuration Practice Set 7.2 Class 10 Maths Solutions
• Mensuration Practice Set 7.3 Class 10 Maths Solutions
• Mensuration Practice Set 7.4 Class 10 Maths Solutions
• Mensuration Problem Set 7 Class 10 Maths Solutions

10th Standard Maths 1 Practice Set 3.2 Chapter 3 Arithmetic Progression Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 3.2 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 3 Arithmetic Progression.

Class 10 Maths Part 1 Practice Set 3.2 Chapter 3 Arithmetic Progression Questions With Answers Maharashtra Board

Question 1.
Write the correct number in the given boxes from the following A.P.

Solution:

Question 2.
Decide whether following sequence is an A.P., if so find the 20^th term of the progression.
-12, -5, 2, 9,16, 23,30,…
Solution:
i. The given sequence is
-12, -5,2, 9, 16, 23,30,…
Here, t₁ = -1₂, t₂ = -5, t₃ = 2, t₄ = 9
∴ t₂ – t₁ – 5 – (-12) – 5 + 12 = 7
t₃ – t₂ = 2 – (-5) = 2 + 5 = 7
∴ t₄ – t₃ – 9 – 2 = 7
∴ t₂ – t₁ = t₃ – t₂ = … = 7 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P.

ii. t_n = a + (n – 1)d
∴ t₂₀ = -12 + (20 – 1)7 …[∵a = -12, d = 7]
= -12 + 19 × 7
= -12 + 133
∴ t₂₀ = 121
∴ 20^th term of the given A.P. is 121.

Question 3.
Given Arithmetic Progression is 12, 16, 20, 24, … Find the 24^th term of this progression.
Solution:
The given A.P. is 12, 16, 20, 24,…
Here, a = 12, d = 16 – 12 = 4 Since,
t_n = a + (n – 1)d
∴ t₂₄ = 12 + (24 – 1)4
= 12 + 23 × 4
= 12 + 92
∴ t₂₄ = 104
∴ 24th term of the given A.P. is 104.

Question 4.
Find the 19th term of the following A.P. 7,13,19,25…..
Solution:
The given A.P. is 7, 13, 19, 25,…
Here, a = 7, d = 13 – 7 = 6
Since, t_n = a + (n – 1)d
∴ t₁₉ = 7 + (19 – 1)6
= 7 + 18 × 6
= 7 + 108
∴ t₁₉ = 115
∴ 19^th term of the given A.P. is 115.

Question 5.
Find the 27^th term of the following A.P. 9,4,-1,-6,-11,…
Solution:
The given A.P. is 9, 4, -1, -6, -11,…
Here, a = 9, d = 4- 9 = -5
Since, t_n = a + (n – 1)d
∴ t₂₇ = 9 + (27 – 1)(-5)
= 9 + 26 × (-5)
= 9 – 130
∴ t₂₇ = -121
∴ 27^th term of the given A.P. is -121.

Question 6.
Find how many three digit natural numbers are divisible by 5.
Solution:
The three digit natural numbers divisible by
5 are 100, 105, 110, …,995
The above sequence is an A.P.
∴ a = 100, d = 105 – 100 = 5
Let the number of terms in the A.P. be n.
Then, t_n = 995
Since, t_n = a + (n – 1)d
∴ 995 = 100 +(n – 1)5
∴ 995 – 100 = (n – 1)5
∴ 895 = (n – 1)5
∴ n – 1 = \(\frac { 895 }{ 5 } \)
∴ n – 1 = 179
∴ n = 179 + 1 = 180
∴ There are 180 three digit natural numbers which are divisible by 5.

Question 7.
The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41st term of that A.P.
Solution:
Bor an A.P., let a be the first term and d be the common difference,
t₁₁ = 16, t₂₁ = 29 …[Given]
t_n = a + (n – 1)d
∴ t₁₁, = a + (11 – 1)d
∴ 16 = a + 10d
i.e. a + 10d = 16 …(i)
Also, t₂₁ = a + (21 – 1)d
∴ 29 = a + 20d
i.e. a + 20d = 29 …(ii)
Subtracting equation (i) from (ii), we get a

Question 8.
8. 11, 8, 5, 2, … In this A.P. which term is number-151?
Solution:
The given A.P. is 11, 8, 5, 2,…
Here, a = 11, d = 8 – 11 = -3
Let the n^th term of the given A.P. be -151.
Then, t_n = – 151
Since, t_n = a + (n – 1)d
∴ -151= 11 + (n – 1)(-3)
∴ -151 – 11 =(n – 1)(-3)
∴ -162 = (n – 1)(-3)
∴ n – 1 = \(\frac { -162 }{ -3 } \)
∴ n – 1 = 54
∴ n = 54 + 1 = 55
∴ 55th term of the given A.P. is -151.

Question 9.
In the natural numbers from 10 to 250, how many are divisible by 4?
Solution:
The natural numbers from 10 to 250 divisible
by 4 are 12, 16, 20, …,248
The above sequence is an A.P.
∴ a = 12, d = 16 – 12 = 4
Let the number of terms in the A.P. be n.
Then, t_n = 248
Since, t_n = a + (n – 1)d
∴ 248 = 12 + (n – 1)4
∴ 248 – 12 = (n – 1)4
∴ 236 = (n – 1)4
∴ n – 1 = \(\frac { 236 }{ 4 } \)
∴ n – 1 = 59
∴ n = 59 + 1 = 60
∴ There are 60 natural numbers from 10 to 250 which are divisible by 4.

Question 10.
In an A.P. 17^th term is 7 more than its 10th term. Find the common difference.
Solution:
For an A.P., let a be the first term and d be the common difference.
According to the given condition,
t₁₇ = t₁₀ + 7
∴ a + (17 – 1)d = a + (10 – 1)d + 7 …[∵ t_n = a + (n – 1)d]
∴ a + 16d = a + 9d + 7
∴ a + 16d – a – 9d = 7
∴ 7d = 7
∴ d = \(\frac { 7 }{ 7 } \) = 1
∴ The common difference is 1.

Question 1.
Kabir’s mother keeps a record of his height on each birthday. When he was one year old, his height was 70 cm, at 2 years he was 80 cm tall and 3 years he was 90 cm tall. His aunt Meera was studying in the 10th class. She said, “it seems like Kabir’s height grows in Arithmetic Progression”. Assuming this, she calculated how tall Kabir will be at the age of 15 years when he is in 10th! She was shocked to find it. You too assume that Kabir grows in A.P. and find out his height at the age of 15 years. (Textbook pg. no. 63)
Solution:
Height of Kabir when he was 1 year old = 70 cm Height of Kabir when he was 2 years old = 80 cm
Height of Kabir when he was 3 years old = 90 cm The heights of Kabir form an A.P.
Here, a = 70, d = 80 – 70 = 10
We have to find height of Kabir at the age of 15years i.e. t₁₅.
Now, tn = a + (n – 1)d
∴ t₁₅ = 70 + (15 – 1)10
= 70 + 14 × 10 = 70 + 140
∴ t₁₅ = 210
∴ The height of Kabir at the age of 15 years will be 210 cm.

Question 2.
Is 5, 8, 11, 14, …. an A.P.? If so then what will be the 100^th term? Check whether 92 is in this A.P.? Is number 61 in this A.P.? (Textbook pg. no, 62)
Solution:
i. The given sequence is
5, 8,11,14,…
Here, t₁ = 5, t₂ = 8, t₃ = 11, t₄ = 14
∴ t₂ – t₁ = 8 – 5 = 3
t₃ – t₂ = 11 – 8 = 3
t₄ – t₃ = 14 – 11 = 3
∴ t₂ – t₁ = t₃ – t₂ = t₄ – t₃ = 3 = d = constant
The difference between two consecutive terms is constant
∴ The given sequence is an A.P.

ii. t_n = a + (n – 1)d
∴ t₁₀₀ = 5 + (100 – 1)3 …[∵ a = 5, d = 3]
= 5 + 99 × 3
= 5 + 297
∴ t₁₀₀ = 302
∴ 100^th term of the given A.P. is 302.

iii. To check whether 92 is in given A.P., let t_n = 92
∴ t_n = a + (n – 1)d
∴ 92 = 5 + (n – 1)3
∴ 92 = 5 + 3n – 3
∴ 92 = 2 + 3n
∴ 90 = 3n
∴ n = \(\frac { 90 }{ 3 } \) = 30
∴ 92 is the 30th term of given A.P.

iv. To check whether 61 is in given A.P., let t_n = 61
61 = 5 + (n – 1)3
∴ 61 = 5 + 3n – 3
∴ 61 = 2 + 3n
∴ 61 – 2 = 3n
∴ 59 = 3n
∴ n = \(\frac { 59 }{ 3 } \)
But, n is natural number 59
∴ n ≠ \(\frac { 59 }{ 3 } \)
∴ 61 is not in given A.P.

Maharashtra State Board Class 10 Maths Solutions Part 1

• Arithmetic Progression Practice Set 3.1 Class 10 Maths Solutions
• Arithmetic Progression Practice Set 3.2 Class 10 Maths Solutions
• Arithmetic Progression Practice Set 3.3 Class 10 Maths Solutions
• Arithmetic Progression Practice Set 3.4 Class 10 Maths Solutions
• Arithmetic Progression Problem Set 3 Class 10 Maths Solutions
• Financial Planning Practice Set 4.1 Class 10 Maths Solutions
• Financial Planning Practice Set 4.2 Class 10 Maths Solutions
• Financial Planning Practice Set 4.3 Class 10 Maths Solutions
• Financial Planning Practice Set 4.4 Class 10 Maths Solutions
• Financial Planning Problem Set 4A Class 10 Maths Solutions
• Financial Planning Problem Set 4B Class 10 Maths Solutions

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