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10th Standard Maths 1 Practice Set 3.1 Chapter 3 Arithmetic Progression Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 3.1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 3 Arithmetic Progression.

Class 10 Maths Part 1 Practice Set 3.1 Chapter 3 Arithmetic Progression Questions With Answers Maharashtra Board

Question 1.
Which of the following sequences are A.P.? If they are A.P. find the common difference.

Solution:
i. The given sequence is 2, 4, 6, 8,…
Here, t₁ = 2, t₂ = 4, t₃ = 6, t₄ = 8
∴ t₂ – t₁ = 4 – 2 = 2
t₃ – t₂ = 6 – 4 = 2
t₄ – t₃ = 8 – 6 = 2
∴ t₂ – t₁ =  t₃ – t₂ = … = 2 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = 2.

The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = \(\frac { 1 }{ 2 } \).

iii. The given sequence is -10, -6, -2, 2,…
Here, t₁ = -10, t₂ = – 6, t₃ = -2, t₄ = 2
∴ t₂ – t₁ = -6 – (-10) = -6 + 10 = 4
t₃ – t₂ = -2 -(-6) = -2 + 6 = 4
t₄ – t₃ = 2 – (-2) = 2 + 2 = 4
∴ t₂ – t₁ = t₃ – t₂ = … = 4 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = 4.

iv. The given sequence is 0.3, 0.33, 0.333,…
Here, t₁ = 0.3, t₂ = 0.33, t₃ = 0.333
∴ t₂ -t₁ = 0.33 – 0.3 = 0.03
t₃ – t₂ = 0.333 – 0.33 = 0.003
∴ t₂ – t₁ ≠ t₃ – t₂
The difference between two consecutive terms is not constant.
∴ The given sequence is not an A.P.

v. The given sequence is 0, -4, -8, -12,…
Here, t₁ = 0, t₂ = -4, t₃ = -8, t₄ = -12
∴ t₂ – t₁ = -4 – 0 = -4
t₃ – t₂ = -8 – (-4) = -8 + 4 = -4
t₄ – t₃ = -12 – (-8) = -12 + 8 = -4
∴ t₂ – t₁ = t₃ – t₂ = … = —4 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = -4.

The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = 0.

The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = √2.

viii. The given sequence is 127, 132, 137,…
Here, t₁ = 127, t₂ = 132, t₃ = 137
∴ t₂ – t₁ = 132 – 127 = 5
t₃ – t₂ = 137 – 132 = 5
∴ t₂ – t₁ = t₃ – t₂ = … = 5 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = 5.

Question 2.
Write an A.P. whose first term is a and common difference is d in each of the following.
i. a = 10, d = 5
ii. a = -3, d = 0
iii. a = -7, d = \(\frac { 1 }{ 2 } \)
iv. a = -1.25, d = 3
v. a = 6, d = -3
vi. a = -19, d = -4
Solution:
i. a = 10, d = 5 …[Given]
∴ t₁ = a = 10
t₂ = t₁ + d = 10 + 5 = 15
t₃ = t₂ + d = 15 + 5 = 20
t₄ = t₃ + d = 20 + 5 = 25
∴ The required A.P. is 10,15, 20, 25,…

ii. a = -3, d = 0 …[Given]
∴ t₁ = a = -3
t₂ = t₁ + d = -3 + 0 = -3
t₃ = t₂ + d = -3 + 0 = -3
t₄ = t₃ + d = -3 + 0 = -3
∴ The required A.P. is -3, -3, -3, -3,…

∴ The required A.P. is -7, – 6.5, – 6, – 5.5,

iv. a = -1.25, d = 3 …[Given]
t₁ = a = -1.25
t₂ = t₁ + d = – 1.25 + 3 = 1.75
t₃ = t₂ + d = 1.75 + 3 = .4.75
t₄ = t₃ + d = 4.75 + 3 = 7.75
∴ The required A.P. is -1.25, 1.75, 4.75, 7.75,…

v. a = 6, d = -3 …[Given]
∴ t₁ = a = 6
t₂ = t₁ + d = 6 – 3 = 3
t₃ = t₂ + d = 3 – 3 = 0
t₄ = t₃ + d = 0- 3 = -3
∴ The required A.P. is 6, 3, 0, -3,…

vi. a = -19, d = -4 …[Given]
t₁ = a = -19
t₂ = t₁ + d = -19 – 4 = -23
t₃ = t₂ + d = -23 – 4 = -27
t₄ = t₃ + d = -27 – 4 = -31
∴ The required A.P. is -19, -23, -27, -31,…

Question 3.
Find the first term and common difference for each of the A.P.

Solution:
i. The given A.P. is 5, 1,-3,-7,…
Here, t₁ = 5, t₂ = 1
∴ a = t₁ = 5 and
d = t₂ – t₁ = 1 – 5 = -4
∴ first term (a) = 5,
common difference (d) = -4

ii. The given A.P. is 0.6, 0.9, 1.2, 1.5,…
Here, t₁ = 0.6, t₂ = 0.9
∴ a = t₁ = 0.6 and
d = t₂ – t₁ = 0.9 – 0.6 = 0.3
∴ first term (a) = 0.6,
common difference (d) = 0.3

iii. The given A.P. is 127, 135, 143, 151,…
Here, t₁ = 127, t₂ = 135
∴ a = t₁ = 127 and
d = t₂ – t₁ = 135 – 127 = 8
∴ first term (a) = 127,
common difference (d) = 8

Question 1.
Complete the given pattern. Look at the pattern of the numbers. Try to find a rule to obtain the next number from its preceding number. Write the next numbers. (Textbook pg, no. 55 and 56)
Answer:

Every pattern is formed by adding a circle in horizontal and vertical rows to the preceding pattern.
∴ The sequence for the above pattern is 1,3, 5, 7, 9,11,13,15,17,….

Every pattern is formed by adding 2 triangles horizontally and 1 triangle vertically to the preceding pattern.
∴ The sequence for the above pattern is 5,8,11,14,17,20,23,…

Question 2.
Some sequences are given below. Show the positions of the terms by t₁, t₂, t₃,…
Answer:

Question 3.
Some sequences are given below. Check whether there is any rule among the terms. Find the similarity between two sequences. To check the rule for the terms of the sequence look at the arrangements and fill the empty boxes suitably. (Textbook pg. no. 56 and 57)
i. .1,4,7,10,13,…
ii. 6,12,18,24,…
iii. 3,3,3,3,…
iv. 4, 16, 64,…
v. -1, -1.5, -2, -2.5,…
vi. 1³, 2³, 3³, 4³
Answer:


The similarity in the sequences i., ii., iii. and v. is that the next term is obtained by adding a particular fixed number to the previous term.

Note : A Geometric Progression is a sequence in which the ratio of any two consecutive terms is a constant,

Sequence iv. is a geometric progression.

Question 4.
Write one example of finite and infinite A.P. each. (Textbook pg. no. 59)
Answer:
Finite A.P.:
Even natural numbers from 4 to 50:
4, 6, 8, ………………. 50.
Infinite A. P.:
Positive multiples of 5:
5, 10, 15, ……………..

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

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

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

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

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

i. The sequence – 10,- 6,- 2, 2, …
(A) is an A.P. Reason d = – 16
(B) is an A.P. Reason d = 4
(C) is an A.P. Reason d = – 4
(D) is not an A.P.
Answer:
(B)

ii. First four terms of an A.P. are …, whose first term is -2 and common difference is -2.
(A) -2, 0, 2, 4
(B) -2, 4,- 8, 16
(C) -2, -4, -6, -8
(D) -2,-4, -8, -16
Answer:
(C)

iii. What is the sum of the first 30 natural numbers?
(A) 464
(B) 465
(C) 462
(D) 461
Answer:
(B)

iv. For an given A.P. t₇ = 4, d = – 4, then a = ………
(A) 6
(B) 7
(C) 20
(D) 28
Answer:
(D)

v. For an given A.P. a = 3.5, d = 0, n = 101, then t_n = ….
(A) 0
(B) 3.5
(c) 103.5
(D) 104.5
Answer:
(B)

vi. In an A.P. first two terms are – 3, 4, then 21st term is ….
(A) -143
(B) 143
(C) 137
(D) 17
Answer:
(C)

vii. If for any A.P. d = 5, then t₁₈ – t₁₃ = ….
(A) 5
(B) 20
(C) 25
(D) 30
Answer:
(C)

viii. Sum of first five multiples of 3 is …
(A) 45
(B) 55
(C) 15
(D) 75
Answer:
(A)

ix. 15, 10, 5, … In this A.P. sum of first 10 terms is…
(A) -75
(B) -125
(C) 75
(D) 125
Answer:
(A)

x. In an A.P. 1^st term is 1 and the last term is 20. The sum of all terms is 399, then n = ….
(A) 42
(B) 38
(C) 21
(D) 19
Answer:
(B)

Hints:

Question 2.
Find the fourth term from the end in an
A.P.: -11, -8, -5, …, 49.
Solution:
The given A.P. is
-11,-8,-5, ……. 49
Reversing the A.P., we get 49, …, -5, -8, -11
Here, a = 49, d = -11 -(-8) = -11 + 8 = -3
Since, t_n = a + (n – 1)d
∴ t₄ = 49 + (4 – 1)(-3)
= 49 + (3) (-3)
= 49 – 9
= 40
∴ Fourth term from the end in the given A.P. is 40.
[Note: If an AY. is reversed, then the resulting sequence is also an A.P.]

Question 3.
In an A.P. the 10^th term is 46, sum of the 5^th and 7^th term is 52. Find the A.P.
Solution:
For an A.P., let a be the first term and d be the common difference.
t₁₀ = 46, t₅ + t₇ = 52 …[Given]
Since, t_n = a + (n – 1)d
∴ t₁₀ = a + (10 – 1)d
∴ 46 = a + 9d
i. e. a + 9d = 46 …(i)
Also, t₅ + t₇ = 52
∴ a + (5 – 1)d + a + (7 – 1)d = 52
∴ a + 4d + a + 6d = 52
∴ 2a + 10d = 52
∴ 2 (a + 5d) = 52
∴ a + 5d = \(\frac { 52 }{ 2 } \)
∴ a + 5d = 26 …(ii)
Subtracting equation (ii) from (i), we get

Substituting d = 5 in equation (ii), we get
a + 5(5) = 26
∴ a + 25 = 26
∴ a = 26 – 25 = 1
t₁ = a = 1
t₂ = t₁ + d = 1 + 5 = 6
t₃ = t₂ + d = 6 + 5 = 11
t₄ = t₃ + d = 11 + 5 = 16
The required A.P. is 1,6,11,16,….

Question 4.
The A.P. in which 4^th term is -15 and 9^th term is -30. Find the sum of the first 10 numbers.
Solution:
t₄ = -15, t9 = – 30 …[Given]
Since, t_n = a + (n – 1)d
∴ t₄ = a + (4 – 1)d
∴ – 15 = a + 3d
i. e. a + 3d = -15 …(i)
Also, t₉ = a + (9 – 1)d
∴ -30 = a + 8d
i.e. a + 8d = -30 …(ii)

∴ The sum of the first 10 numbers is -195.

Question 5.
Two given A.P.’s are 9, 7, 5, … and 24, 21, 18, … If nth term of both the .progressions are equal then find the value of n and n,h term.
Solution:
The first A.P. is 9, 7, 5,…
Here, a = 9, d = 7- 9 = -2
∴n^th term = a + (n – 1)d
= 9 + (n – 1) (-2)
= 9 – 2n + 2
= 11 – 2n
The second A.P. is 24, 21, 18, …
Here, a = 24, d = 21 – 24 = – 3
∴ n^th term = a + (n – 1)d
= 24 + (n – 1) (-3)
= 24 – 3n + 3
= 27 – 3n
Since, the nth terms of the two A.P.’s are equal.
∴ 11 – 2n = 27 – 3n
∴ 3n – 2n = 27 – 11
∴ n = 16
∴ t₁₆ = 9 + (16 – 1) (-2)
= 9 + 15 × (-2)
= 9 – 30
∴ t₁₆ = -21
∴ The values of n and n^th term are 16 and -21 respectively.

Question 6.
If sum of 3^rd and 8^th terms of an A.P. is 7 and sum of 7th and 14th terms is -3, then find the 10th term.
Solution:
for an A.P., let a be the first term and d be the common difference.
According to the first condition,
t₃ + tg = 7
∴ a + (3 – 1) d + a + (8 – 1)d = 7 …[∵ t_n = a + (n – 1)d]
∴ a + 2d + a + 7d = 7
∴ 2a + 9d = 7 …(i)
According to the second condition,
t₇ + t₁₄ = -3
∴ a + (7 – 1)d + a + (14 – 1 )d = -3
∴ a + 6d + a + 13d = -3
∴ 2a + 19 d = – 3 …(ii)
Subtracting equation (i) from (ii), we get

Question 7.
In an A.P. the first term is -5 and last term is 45. If sum of all numbers in the A.P. is 120, then how many terms are there? What is the common difference?
Solution:
Let the number of terms in the A.P. be n and the common difference be d.
Then, a = -5, t_n = 45, S_n = 120
Since, t_n = a + (n – 1)d
∴ 45 = -5 + (n – 1)d
∴ 45 + 5 = (n – 1)d
∴ (n – 1)d = 50 …(i)

Substituting n = 6 in equation (i), we get
(6 – 1)d = 50
∴ 5d = 50
∴ d = \(\frac { 50 }{ 5 } \) = 10
∴ There are 6 terms in the A.P. and the common difference is 10.

Alternate Method:
Let the number of terms in the A.P. be n.
Then, t₁ = a = -5, t_n = 45, S_n = 120

∴ There are 6 terms in the A.P. and the common difference is 10.

Question 8.
Sum of 1 to n natural numbers is 36, then find the value of n.
Solution:
The natural numbers from 1 to n are
1,2, 3, ……, n.
The above sequence is an A.P.
∴ a = 1, d = 2 – 1 = 1
S_n = 36 …[Given]
Now, S_n = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
∴ 36 = \(\frac { n }{ 2 } \) [2(1) + (n – 1)(1)]
∴ 36 = \(\frac { n }{ 2 } \) (2 + n – 1)
∴ 36 × 2 = n (n + 1)
∴ 72 = n (n + 1)
∴ 72 = n² + n
∴ n² + n – 72 = 0
∴ n² + 9_n – 8n – 72 = 0
∴ n(n + 9) – 8 (n + 9) = 0
∴ (n + 9) (n – 8) = 0
∴ n + 9 = 0 or n – 8 = 0
∴ n = -9 or n = 8
But, n cannot be negative.
∴ n = 8
∴ The value of n is 8.

Question 9.
Divide 207 in three parts, such that all parts are in A.P. and product of two smaller parts will be 4623.
Solution:
Let the three parts of 207 that are in A.P. be
a – d, a, a + d
According to the first condition,
(a – d) + a + (a + d) = 207
∴ 3a = 207
∴ a = \(\frac { 207 }{ 3 } \)
∴ a = 69 …(i)
According to the second condition,
(a – d) × a = 4623
∴ (69 – d) × 69 = 4623 …[From (i)]
∴ 69 – d = \(\frac { 4623 }{ 69 } \)
∴ d = 69 – 67
∴ d = 2
∴ a – d = 69 – 2 = 67
a = 69
a + d = 69 + 2 = 71
∴ The three parts of 207 that are in A.P. are 67, 69 and 71.

Question 10.
There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429. Write the A.P.
Solution:
Since, there are 37 terms in the A.P.

Substituting d = 4 in equation (i), we get
3a + 54(4) = 225
∴ 3a + 216 = 225
∴ 3a = 225 – 216
∴ 3a = 9
∴ a = \(\frac { 9 }{ 3 } \) = 3
∴The required A. P. is
a, a + d, a + 2d, a + 3d, …., a + (n – 1)d
i.e. 3, 3 + 4,3 + 2 × 4, 3 + 3 × 4,…, 3 + (37 – 1)4
i.e. 3, 7,11,15, …,147

Question 11.
If first term of an A.P. is a, second term is b and last term is c, then show that sum of all

Solution:

Question 12.
If the sum of first p terms of an A.P. is equal to the sum of first q terms then show that the sum of its first (p + q) terms is zero, (p ≠ q)
Solution:
For an A.P., let a be the first term and d be the common difference.
The sum of first n terms of an A.P. is given by
Sn = [2a + (n – 1)d]
According to the given condition,
Sp = Sq


∴ The sum of the first (p + q) terms is zero

Question 13.
If m times the m^th term of an A.P. is equal to n times n^th term, then show that the (m + n)^th term of the A.P. is zero.
Solution:
According to the given condition,
mt_m = nt_n
∴ m[a + (m – 1)d] = n[a + (n – 1)d]
∴ ma + md(m – 1) = na + nd(n- 1)
∴ ma + m²d – md = na + n²d – nd
∴ ma + m²d – md – na – n²d + nd = 0
∴ (ma – na) + (m²d – n²d) – (md – nd) = 0
∴ a(m – n) + d(m² – n²) – d(m – n) = 0
∴ a(m – n) + d(m + n) (m – n) – d(m – n) = 0
∴ (m – n)[a + (m + n – 1) d] = 0
∴ [a+ (m + n – 1)d] = 0 …[Dividing both sides by (m – n)]
∴ t_(m+n) = 0
∴ The (m + n)^th term of the A.P. is zero.

Question 14.
₹ 1000 is invested at 10 percent simple interest. Check at the end of every year if the total interest amount is in A.P. If this is an A.P. then find interest amount after 20 years. For this complete the following activity.
Solution:

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

9th Standard Maths 1 Practice Set 7.4 Chapter 7 Statistics Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.4 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.

Class 9 Maths Part 1 Practice Set 7.4 Chapter 7 Statistics Questions With Answers Maharashtra Board

Question 1.
Complete the following cumulative frequency table:

Solution:

Question 2.
Complete the following Cumulative Frequency Table:

Solution:

Question 3.
The data is given for 62 students in a certain class regarding their mathematics marks out of 100. Take the classes 0 – 10, 10 – 20,… and prepare frequency distribution
table and cumulative frequency table more than or equal to type.
55. 60, 81, 90, 45, 65, 45, 52, 30, 85, 20, 10,
75, 95, 09, 20, 25, 39, 45, 50, 78, 70, 46, 64,
42. 58. 31, 82, 27, 11, 78, 97, 07, 22, 27, 36,
35, 40, 75, 80, 47, 69, 48, 59, 32, 83, 23, 17,
77, 45, 05, 23, 37, 38, 35, 25, 46, 57, 68, 45.
47,49
From the prcparcd table, answer the following questions :
i. How many students obtained marks 40 or above 40?
ii. How many students obtained marks 90 or above 90?
iii. How many students obtained marks 60 or above 60?
iv. What is the cumulative frequency of equal to or more than type of the class 0 – 10?
Solution:

i. 38 students obtained marks 40 or above 40.
ii. 3 students obtained marks 90 or above 90.
iii. 19 students obtained marks 60 or above 60.
iv. Cumulative frequency of equal to or more than type of the class 0 – 10 is 62.

Question 4.
Using the data In example (3) above, prepare less than type cumulative frequency table and answer the following questions.
i. How many students obtained less than 40 marks?
ii. How many students obtained less than 10 marks?
iii. How many students obtained less than 60 marks?
iv. Find the cumulative frequency of the class 50 – 60.
Solution:

i. 24 students obtained less than 40 marks.
ii. 3 students obtained less than 10 marks.
iii. 43 students obtained less than 60 marks.
iv. Cumulative frequency of the class 50 – 60 is 43.

Maharashtra Board Class 9 Maths Chapter 7 Statistics Practice Set 7.4 Intext Questions and Activities

Question 1.
The following information is regarding marks in mathematics, obtained out of 40, scored by 50 students of 9th std. ¡n the first unit test. (Textbook pg. no. 120)

From the table, fill in the blanks in the following statements.
i. For class interval 10 – 20 the lower class limit is _____ and upper class limit is _____
ii. How many students obtained marks less than 10? 2
iii. How many students obtained marks less than 20? 2 + ____ = 14
iv. How many students obtained marks less than 30? ______ + _____ = 34
v. How many students obtained marks less than 40? ______ + ______ =50
Solution:
i. 10, 20
iii. 12
iv. 14 + 20
v. 34 + 16

Question 2.
A sports club has organised a table-tennis tournaments. The following table gives the distribution of players ages. Find the cumulative frequencies equal to or more than the lower class limit and complete the table (Textbook pg. no. 121)
Solution:
Equal to lower limit or more than lower limit type of cumulative table.

Maharashtra Board Class 9 Maths Solutions

• Statistics Practice Set 7.1 Class 9 Maths Solutions
• Statistics Practice Set 7.2 Class 9 Maths Solutions
• Statistics Practice Set 7.3 Class 9 Maths Solutions
• Statistics Practice Set 7.4 Class 9 Maths Solutions
• Statistics Practice Set 7.5 Class 9 Maths Solutions
• Statistics Problem Set 7 Class 9 Maths Solutions

9th Standard Maths 1 Practice Set 7.3 Chapter 7 Statistics Textbook Answers Maharashtra Board

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

Class 9 Maths Part 1 Practice Set 7.3 Chapter 7 Statistics Questions With Answers Maharashtra Board

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

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

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

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

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

Question 4.
Complete the following table.

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

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

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

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

ii.

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

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

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

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

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

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

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

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

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

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

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

Maharashtra Board Class 9 Maths Solutions

• Statistics Practice Set 7.1 Class 9 Maths Solutions
• Statistics Practice Set 7.2 Class 9 Maths Solutions
• Statistics Practice Set 7.3 Class 9 Maths Solutions
• Statistics Practice Set 7.4 Class 9 Maths Solutions
• Statistics Practice Set 7.5 Class 9 Maths Solutions
• Statistics Problem Set 7 Class 9 Maths Solutions

10th Standard Maths 1 Practice Set 2.2 Chapter 2 Quadratic Equations Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.2 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 2 Quadratic Equations.

Class 10 Maths Part 1 Practice Set 2.2 Chapter 2 Quadratic Equations Questions With Answers Maharashtra Board

Question 1.
Solve the following quadratic equations by factorisation.
i. x² – 15x + 54 = 0
ii. x² + x – 20 = 0
iii. 2y² + 27y + 13 = 0
iv. 5m² = 22m + 15
v. 2x² – 2x + \(\frac { 1 }{ 2 } \) = 0
vi. 6x – \(\frac { 2 }{ x } \) = 1
vii. √2x² + 7x + 5√2 = 0 to solve this quadratic equation by factorisation complete the following activity
viii. 3x² – 2√6x + 2 = 0
ix. 2m(m – 24) = 50
x. 252 = 9
xi. 7m² = 21 m
xii. m² – 11 = 0
Solution:

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x – 9 = 0 or x – 6 = 0
∴ x = 9 or x = 6
∴ The roots of the given quadratic equation are 9 and 6.

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 5 = 0 or x – 4 = 0
∴ x = -5 or x = 4
∴ The roots of the given quadratic equation are -5 and 4.

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ y + 13 = 0 or 2y + 1 = 0
∴ y = – 13 or 2y = -1
∴ y = -13 or y = –\(\frac { 1 }{ 2 } \)
∴ The roots of the given quadratic equation are -13 and – \(\frac { 1 }{ 2 } \)

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ m – 5 = 0 or 5m + 3 = 0
∴ m = 5 or 5m = -3
∴ m = 5 or m = \(\frac { -3 }{ 5 } \)
∴ The roots of the given quadratic equation are 5 and – \(\frac { 3 }{ 5 } \)

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ 3x – 2 = 0 or 2x + 1 = 0
∴ 3x = 2 or 2x = -1
∴ x = \(\frac { 2 }{ 3 } \) or 2x = -1
∴ The roots of the given quadratic equation are \(\frac { 2 }{ 3 } \) and \(\frac { -1 }{ 2 } \).

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get

ix. 2m (m – 24) = 50
∴ 2m² – 48m = 50
∴ 2m² – 48m – 50 = 0
∴m² – 24m – 25 = 0 …[Dividing both sides by 2]

∴ m – 25 = 0 or m + 1 = 0
∴ m = 25 or m = -1
∴ The roots of thes given quadratic equation are 25 and -1.

x. 25m² = 9
∴ 25m² – 9 = 0
∴ (5m)² – (3)² = 0
∴ (5m + 3) (5m – 3) = 0
…. [∵a² – b² = (a + b) (a – b)]
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ 5m + 3 = 0 or 5m – 3 = 0
∴ 5m = -3 or 5m = 3
∴ m = \(\frac { -3 }{ 5 } \) or m = \(\frac { 3 }{ 5 } \)
∴ The roots of the given quadratic equation are \(\frac { -3 }{ 5 } \) and \(\frac { 3 }{ 5 } \).

xi. 7m² = 21m
∴ 7m² – 21m = 0
∴ m² – 3m = 0 …[Dividing both sides by 7]
∴ m(m – 3) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ m = 0 or m – 3 = 0
∴ m = 0 or m = 3
∴ The roots of the given quadratic equation are 0 and 3.

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ m + √11 = 0 or m – √11 = 0
∴ m = -√11 or m = √11
∴ The roots of the given quadratic equation are – √11 and √11

Maharashtra State Board Class 10 Maths Solutions Part 1

• Linear Equations in Two Variables Practice Set 1.1 Class 10 Maths Solutions
• Linear Equations in Two Variables Practice Set 1.2 Class 10 Maths Solutions
• Linear Equations in Two Variables Practice Set 1.3 Class 10 Maths Solutions
• Linear Equations in Two Variables Practice Set 1.4 Class 10 Maths Solutions
• Linear Equations in Two Variables Practice Set 1.5 Class 10 Maths Solutions
• Linear Equations in Two Variables Problem Set 1 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.1 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.2 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.3 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.4 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.5 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.6 Class 10 Maths Solutions
• Quadratic Equations Problem Set 2 Class 10 Maths Solutions

9th Standard Maths 1 Practice Set 7.2 Chapter 7 Statistics Textbook Answers Maharashtra Board

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

Class 9 Maths Part 1 Practice Set 7.2 Chapter 7 Statistics Questions With Answers Maharashtra Board

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

Maharashtra Board Class 9 Maths Solutions

• Statistics Practice Set 7.1 Class 9 Maths Solutions
• Statistics Practice Set 7.2 Class 9 Maths Solutions
• Statistics Practice Set 7.3 Class 9 Maths Solutions
• Statistics Practice Set 7.4 Class 9 Maths Solutions
• Statistics Practice Set 7.5 Class 9 Maths Solutions
• Statistics Problem Set 7 Class 9 Maths Solutions

9th Standard Maths 1 Practice Set 7.1 Chapter 7 Statistics Textbook Answers Maharashtra Board

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

Class 9 Maths Part 1 Practice Set 7.1 Chapter 7 Statistics Questions With Answers Maharashtra Board

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

Solution:


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

Solution:
i. Sub-divided bar diagram:

ii. Percentage bar diagram:

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

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


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

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

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

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

Maharashtra Board Class 9 Maths Solutions

• Statistics Practice Set 7.1 Class 9 Maths Solutions
• Statistics Practice Set 7.2 Class 9 Maths Solutions
• Statistics Practice Set 7.3 Class 9 Maths Solutions
• Statistics Practice Set 7.4 Class 9 Maths Solutions
• Statistics Practice Set 7.5 Class 9 Maths Solutions
• Statistics Problem Set 7 Class 9 Maths Solutions

9th Standard Maths 1 Problem Set 5 Chapter 5 Linear Equations in Two Variables Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 5 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 5 Linear Equations in Two Variables.

Class 9 Maths Part 1 Problem Set 5 Chapter 5 Linear Equations in Two Variables Questions With Answers Maharashtra Board

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

i. If 3x + 5y = 9 and 5x + 3y = 7, then what is the value of x + y ?
(A) 2
(B) 16
(C) 9
(D) 7
Answer:
(A) 2

ii. ‘When 5 is subtracted from length and breadth of the rectangle, the perimeter becomes 26’. What is the mathematical form of the statement ?
(A) x – y = 8
(B) x + y = 8
(C) x + y = 23
(D) 2x + y = 21
Answer:
(C) x + y = 23

iii. Ajay is younger than Vijay by 5 years. Sum of their ages is 25 years. What is Ajay’s age?
(A) 20 years
(B) 15 years
(C) 10 years
(D) 5 years
Answer:
(C) 10 years

Hints:
i. Adding the given equations,
3x+ 5y = 9
5x + 3y = 7
8x + 8y = 16
∴ x + y = 2 .. [Dividing both sides by 8]

ii. Let the length of the rectangle be ‘x’ and that of breadth be ‘y’.
Perimeter of rectangle = 2[(x – 5) + (y – 5)]
∴ 26 = 2(x + y – 10)
∴ x + y – 10 = 13
∴ x + y = 23

iii. Let the age of Ajay bex years.
∴ x + (x + 5) = 25
∴ 2x = 20
∴ x = 10 years

Question 2.
Solve the following simultaneous equations.
i. 2x + y = 5 ; 3x – y = 5
ii. x – 2y = -1 ; 2x – y = 7
iii. x + y = 11 ; 2x – 3y = 7
iv. 2x + y = -2 ; 3x – y = 7
V. 2x – y = 5 ; 3x + 2y = 11
vi. x – 2y – 2 ; x + 2y = 10
Solution:
ii. 2x + y = 5 …(i)
3x – y = 5 …(ii)
Adding equations (i) and (ii),
2x + y = 5
+ 3x – y = 5
5x = 10
∴ x = 10/5
∴ x = 2
Substituting x = 2 in equation (i),
2(2) + y = 5
4 + y = 5
∴ y = 5 – 4 = 1
∴ (2, 1) is the solution of the given equations.

ii. x – 2y = -1
∴x = 2y – 1 … .(i)
∴ 2x – y = 7 ….(ii)
Substituting x = 2y – 1 in equation (ii),
2(2y – 1) – y = 7
∴ 4y – 2 – y = 7
∴ 3y = 7 + 2
∴ 3y = 9
∴ y = 9/3
∴ y = 3
Substituting y = 3 in equation (i),
x = 2y – 1
∴ x = 2(3) – 1
∴ x = 6 – 1 = 5
∴ (5, 3) is the solution of the given equations.

iii. x + y = 11
∴ x = 11 – y …(i)
2x – 3y = 7 …….(ii)
Substituting x = 11 -y in equation (ii),
2(11 – y) – 3y = 7
∴ 22 – 2y – 3y = 1
∴ 22 – 5y = 7
∴ 22 – 7 = 5y
∴ 15 = 5y
∴ y = \(\frac { 15 }{ 5 }\)
∴ y = 3
Substituting y = 3 in equation (i),
x = 11 – y
∴ x = 11 – 3 = 8
∴ (8, 3) is the solution of the given equations.

iv. 2x + y = -2 …(i)
3x – y = 7 …(ii)
Adding equations (i) and (ii),
2x + y = -2
+ 3x – y = l
5x = 5
∴ x = \(\frac { 5 }{ 5 }\)
∴ x = 1
Substituting x = 1 in equation (i),
2x + y = -2
∴ 2(1) +y = -2
2 + y = -2
∴ y = – 2 – 2
∴ y = -4
∴ (1, -4) is the solution of the given equations.

v. 2x – y = 5
∴ -y = 5 – 2x
∴ y = 2x – 5 …(i)
3x + 2y = 11 ……(ii)
Substituting y = 2x – 5 in equation (ii),
3x + 2(2x – 5) = 11
∴ 3x + 4x- 10= 11
∴ 7x = 11 + 10
∴ 7x = 21
∴ x = \(\frac { 21 }{ 7 }\)
∴ x = 3
Substituting x = 3 in equation (i),
y = 2x – 5
∴ y = 2(3) – 5
∴ y = 6 – 5 = 1
∴(3,1) is the solution of the given equations.

vi. x – 2y = -2
∴ x = 2y – 2 …(i)
x + 2y = 10 …..(ii)
Substituting x = 2y – 2 in equation (ii),
2y – 2 + 2y = 10
∴ 4y = 10 + 2
∴ 4y= 12
∴ y = \(\frac { 12 }{ 7 }\)
∴ y = 3
Substituting y = 3 in equation (i),
x = 2y – 2
∴ x = 2(3) – 2
∴ x = 6 – 2 = 4
∴ (4, 3) is the solution of the given equations.

Question 3.
By equating coefficients of variables, solve the following equations. [3 Marks each]
i. 3x – 4y = 7 ; 5x + 2y = 3
ii. 5x + ly= 17 ; 3x – 2y = 4
iii. x – 2y = -10 ; 3x – 3y = -12
iv. 4x+y = 34 ; x + 4y = 16
Solution:
i. 3x – 4y = 7 …(i)
5x + 2y = 3 ….(ii)
Multiplying equation (ii) by 2,
10x + 4y = 6 …(iii)
Adding equations (i) and (iii),

∴ x = 1
Substituting x = 1 in equation (i),
3x – 4y = 7
∴ 3(1) – 4y = 7
∴ 3 – 4y = 7
∴ 3 – 7 = 4y
∴ -4 = 4y
∴ y = \(\frac { -4 }{ 4 }\)
∴ y = -1
∴ (1, -1) is the solution of the given equations.

ii. 5x + 7y = 17 …(i)
3x – 2y = 4 ….(ii)
Multiplying equation (i) by 2,
10x + 14y = 34 …(iii)
Multiplying equation (ii) by 7,
21x – 14y = 28 …..(iv)
Adding equations (iii) and (iv),

∴ x = 2
Substituting x = 2 in equation (ii),
3x – 2y = 4
∴ 3(2) – 2y = 4
∴ 6 – 2y = 4
∴ 6 – 4 = 2y
∴ 2 = 2y
∴ y = \(\frac { 2 }{ 2 }\)
∴ y = 1
∴ (2,1) is the solution of the given equations.

iii. x – 2y = -10 ….(i)
3x – 5y = -12 …….(ii)
Multiplying equation (i) by 3,
3x – 6y = -30 …(iii)
Subtracting equation (ii) from (iii),

∴ y = 18
Substituting y = 18 in equation (i),
x – 2y = -10
∴ x – 2(18) = -10
∴ x – 36 = -10
∴ x = -10 + 36 = 26
∴ (26, 18) is the solution of the given equations.

iv. 4x + y = 34 …(i)
x + 4y = 16 …… (ii)
Multiplying equation (i) by 4,
16x + 4y = 136 …(iii)
Subtracting equation (ii) from (iii),

x = 8
Substituting x = 8 in equation (i),
4x + y = 34
∴ 4(8) + y = 34
∴ 32 + y = 34
∴ y = 34 – 32 = 2
∴ (8, 2) is the solution of the given equations.

Question 4.
Solve the following simultaneous equations.

Solution:
i. \(\frac{x}{3}+\frac{y}{4}=4\)
Multiplying both sides by 12,
4x + 3y = 48 …(i)
\(\frac{x}{2}-\frac{y}{4}=1\)
Multiplying both sides by 8,
4x – 2y = 8 …..(ii)
Subtracting equation (ii) from (i),

∴ y = 8
Substituting y = 8 in equation (ii),
4x – 2y = 8
∴ 4x – 2(8) = 8
∴ 4x – 16 = 8
∴ 4x = 8+ 16
∴ 4x = 24
∴ x = \(\frac { 24 }{ 4 }\)
∴ x = 6
∴ (6, 8) is the solution of the given equations.

ii. \(\frac { x }{ 3 }\) + 5y = 13
Multiplying both sides by 3,
x + 15y = 39 …(i)
2x + \(\frac { y }{ 2 }\) =19
Multiplying both sides by 2,
4x + y = 38 …….(ii)
Multiplying equation (i) by 4,
4x + 60y = 156 …(iii)
Subtracting equation (ii) from (iii),
4x + 60y =156 4x + y= 38

∴ y = 2
Substituting y = 2 in equation (i),
x + 15y = 39
∴ x+ 15(2) = 39
∴ x + 30 = 39
∴ x = 39 – 30 = 9
∴ (9,2) is the solution of the given equations.

iii. \(\frac { 2 }{ x }\) + \(\frac { 3 }{ y }\) = 13
Multiplying both sides by 5,

Question 5.
A two digit number is 3 more than 4 times the sum of its digits. If 18 is added to this number, the sum is equal to the number obtained by interchanging the digits. Find the number.
Solution:
Let the digit in unit’s place be ‘x’ and the digit in ten’s place be ‘y’.

According to the first condition,
a two digit number is 3 more than 4 times the sum of its digits.
10y + x = 4(x + y) + 3
∴ 10y + x = 4x + 4y + 3
∴ x – 4x + 10y – 4y = 3
∴ – 3x + 6y = 3
Dividing both sides by -3,
x – 2y = -1 …(i)
According to the second condition,
if 18 is added to the number, the sum is equal to the number obtained by interchanging the digits.
10y + x + 18= 10x + y
∴ x – 10x + 10y – y = -18
∴ – 9x + 9y = -18
Dividing both sides by – 9,
x – y = 2 ……(ii)
Subtracting equation (ii) from (i),

∴ y = 3
Substituting y = 3 in equation (ii),
x – y = 2
∴ x – 3 = 2
∴ x = 2 + 3 = 5
∴ Original number = 10y + x
= 10(3) + 5
= 30 + 5
= 35
The required number is 35.

Question 6.
The total cost of 8 books and 5 pens is ₹ 420 and the total cost of 5 books and 8 pens is ₹321. Find the cost of 1 book and 2 pens.
Solution:
Let the cost of one book be ₹ x and the cost of one pen be ₹ y.
According to the first condition,
the total cost of 8 books and 5 pens is ₹ 420.
∴ 8x + 5y = 420 …(i)
According to the second condition, the total cost of 5 books and 8 pens is ₹ 321.
5x + 8y = 321 ….(ii)
Multiplying equation (i) by 5,
40x + 25y = 2100 …(iii)
Multiplying equation (ii) by 8,
40x + 64y = 2568 … (iv)
Subtracting equation (iii) from (iv),

∴ y = 12
Substituting y = 12 in equation (i),
8x + 5y = 420
∴ 8x + 5(12) = 420
∴ 8x + 60 = 420
∴ 8x = 420 – 60
∴ 8x = 360
∴ x = \(\frac { 360 }{ 8 }\)
∴ x = 45
Cost of 1 book and 2 pens = x + 2y
= 45 + 2(12)
= 45 + 24
= ₹69
∴ The cost of 1 book and 2 pens is ₹69.

Question 7.
The ratio of incomes of two persons is 9 : 7. The ratio of their expenses is 4 : 3. Every person saves ₹ 200, find the income of each.
Solution:
Let the income of first person be ₹ x and that of second person be ₹ y.
According to the first condition,
the ratio of their incomes is 9 : 7.
∴ \(\frac { x }{ y }\) = \(\frac { 9 }{ 7 }\)
∴ 7x = 9y
∴ 7x – 9y = 0 …….(i)
Each person saves ₹ 200.
Expenses of first person = Income – Saving = x – 200
Expenses of second person = y – 200
According to the second condition,
the ratio of their expenses is 4 : 3
∴ \(\frac { x – 200 }{ y – 200 }\) = \(\frac { 4 }{ 3 }\)
∴ 3(x – 200) = 4(y – 200)
∴ 3x – 600 = 4y – 800
∴ 3x – 4y = – 800 + 600
∴ 3x – 4y = -200 …(ii)
Multiplying equation (i) by 4,
28x-36y =0 …(iii)
Multiplying equation (ii) by 9,
27x-36y = -1800 …(iv)
Subtracting equation (iv) from (iii),

Substituting x = 1800 in equation (i),
7x – 9y = 0
∴ 7(1800) – 9y = 0
∴ 9y = 7 x 1800
∴ y = \(\frac { 7 \times 1800 }{ 9 }\)
y = 7 x 200
∴ y = 1400
∴ The income of first person is ₹ 1800 and that of second person is ₹ 1400.

Question 8.
If the length of a rectangle is reduced by 5 units and its breadth is increased by 3 units, then the area of the rectangle is reduced by 9 square units. If length is reduced by 3 units and breadth is increased by 2 units, then the area of rectangle will increase by 67 square units. Then find the length and breadth of the rectangle.
Solution:
Let the length of the rectangle be ‘x’ units and the breadth of the rectangle be ‘y’ units.
Area of the rectangle = xy sq. units
length of the rectangle is reduced by 5 units
∴ length = x – 5
breadth of the rectangle is increased by 3 units
∴ breadth = y + 3
area of the rectangle is reduced by 9 square units
∴ area of the rectangle = xy – 9
According to the first condition,
(x – 5)(y + 3) = xy – 9
∴ xy + 3x – 5y – 15 = xy – 9
∴ 3x – 5y = -9 + 15
∴ 3x – 5y = 6 …(i)
length of the rectangle is reduced by 3 units
∴ length = x – 3
breadth of the rectangle is increased by 2 units
∴ breadth = y + 2
area of the rectangle is increased by 67 square units
∴ area of the rectangle = xy + 61
According to the second condition,
(x – 3)(y + 2) = xy + 67
∴ xy + 2x – 3y – 6 = xy + 67
∴ 2x – 3y = 67 + 6
∴ 2x – 3y = 73 …(ii)
Multiplying equation (i) by 3,
9x – 15y = 18 . ..(iii)
Multiplying equation (ii) by 5,
10x – 15y = 365 …(iv)
Subtracting equation (iii) from (iv), 10x- 15y= 365 9x-15y= 18

Substituting x = 347 in equation (ii),
2x – 3y = 73
∴ 2(347) – 3y = 73
∴ 694 – 73 = 3y
∴ 621 = 3y
∴ y = \(\frac { 621 }{ 3 }\)
∴ y = 207
∴ The length and breadth of rectangle are 347 units and 207 units respectively.

Question 9.
The distance between two places A and B on a road is 70 kilometres. A car starts from A and the other from B. If they travel in the same direction, they will meet in 7 hours. If they travel towards each other they will meet in 1 hour, then find their speeds.
Solution:
Let the speed of the car starting from A (first car) be ‘x’ km/hr and that starting from B (second car) be ‘y’ km/hr. (x > y)
According to the first condition,
Distance covered by the first car in 7 hours = 7x km
Distance covered by the second car in 7 hours = 7y km

If the cars are travelling in the same direction, 7x – 7y = 70
Dividing both sides by 7,
x – y = 10 …(i)
According to the second condition,
Distance covered by the first car in
1 hour = x km
Distance covered by the second car in 1 hour = y km

If the cars are travelling in the opposite direction
x + y = 70 …(ii)
Adding equations (i) and (ii),

∴ x = 40
Substituting x = 40 in equation (ii), x + y = 70
∴ 40 +y = 70
∴ y = 70 – 40 = 30
∴ The speed of the cars starting from places A and B are 40 km/hr and 30 km/hr respectively.

Question 10.
The sum of a two digit number and the number obtained by interchanging its digits is 99. Find the number.
Solution:
Let the digit in unit’s place be ‘x’ and the digit in ten’s place be ‘y’.

According to the given condition,
the sum of a two digit number and the number
obtained by interchanging its digits is 99.
∴ 10y + x + 10x +y = 99
∴ 11x + 11y = 99
Dividing both sides by 11,
x + y = 9
If y = 1, then x = 8
If y = 2, then x = 7
If y = 3, then x = 6 and so on.
∴ The number can be 18, 27, 36, … etc.

Maharashtra Board Class 9 Maths Chapter 5 Linear Equations in Two Variables Practice Set 5 Intext Questions and Activities

Question 1.
On the glasses of following spectacles, write numbers such that (Textbook pg. no. 82)
i. Their sum is 42 and difference is 16.

ii. Their sum is 37 and difference is 11.

iii. Their sum is 54 and difference is 20.

iv. Their sum is … and difference is … .

Answer:
ii. x + y = 37 and x – y = 11
∴ x = 24, y = 13
iii. x +y = 54 and x – y = 20
∴ x = 37, y =17

Question 2.
There are instructions written near the arrows in the following diagram. From this information form suitable equations and write in the boxes indicated by arrows. Select any two equations from these boxes and find their solutions. Also verify the solutions. By taking one pair of equations at a time, how many pairs can be formed ? Discuss the solutions for these pairs. (Textbook pg. no. 92)

Answer:
Here, if we take a pair of any two equations, we get following 6 pairs.

1. equation (i) and (ii)
2. equation (i) and (iii)
3. equation (i) and (iv)
4. equation (ii) and (iii)
5. equation (ii) and (iv)
6. equation (iii) and (iv)

Solution of each pair given above is (21, 15).
Here, all four equations are of same rectangle. By solving any two equations simultaneously, we get length and breadth of the rectangle.

Question 3.
Find the function.

∴ Given function = \(\frac { 6 }{ 14 }\)
Verify the answer obtained. (Textbook pg. no. 92)
Answer:
For the fraction\(\frac { 6 }{ 14 }\), if the numerator is multiplied by 3 and 3 is subtracted from the denominator, we get fraction \(\frac { 18 }{ 11 }\).
Similarly, for the fraction \(\frac { 6 }{ 14 }\), if the numerator is increased by 8 and the denominator is doubled, we get fraction \(\frac { 1 }{ 2 }\).

9th Class Maths Solution Maharashtra Board 

• Linear Equations in Two Variables Practice Set 5.1 Class 9 Maths Solutions
• Linear Equations in Two Variables Practice Set 5.2 Class 9 Maths Solutions
• Linear Equations in Two Variables Problem Set 5 Class 9 Maths Solutions

9th Standard Maths 1 Practice Set 6.1 Chapter 6 Financial Planning Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 6.1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 6 Financial Planning.

Class 9 Maths Part 1 Practice Set 6.1 Chapter 6 Financial Planning Questions With Answers Maharashtra Board

Question 1.
Alka spends 90% of the money that she receives every month, and saves ₹ 120. How much money does she get monthly?
Solution:
Let Alka’s monthly income be ₹ x.
Alka spends 90% of the money that she receives every month.
∴ Amount spent by Alka = 90% of x
= \(\frac { 90 }{ 100 }\) × x = 0.9x 100
Now, Savings = Income – Expenditure
∴ 120 = x – 0.9x
∴120 = 0.1 x
∴ \(x=\frac{120}{0.1}=\frac{120 \times 10}{0.1 \times 10}\)
∴ x = 1200
Alka gets ₹ 1200 monthly.

Question 2.
Sumit borrowed a capital of ₹ 50,000 to start his food products business. In the first year he suffered a loss of 20%. He invested the remaining capital in a new sweets business and made a profit of 5%. How much was his profit or loss computed on his original capital ?
Solution:
Original capital borrowed by Sumit = ₹ 50000
Sumit suffered a loss of 20% in his food products business.
∴ Loss suffered in the first year = 20% of 50000
= \(\frac { 20 }{ 100 }\) × 50000
= ₹10000
Remaining capital = Original capital – loss suffered = 50000- 10000
= ₹ 40000
Sumit invested the remaining capital i.e. ₹ 40,000 in a new sweets business. He made a profit of 5%.
Profit in sweets business = 5% of 40000
= \(\frac { 5 }{ 100 }\) x 40000 100
= ₹ 2000
New capital with Sumit after the profit in new sweets business = 40000 + 2000 = ₹42000
Since, the new capital is less than the original capital, we can conclude that Sumit suffered a loss.
Total loss on original capital = Original capital – New capital
= 50000 – 42000 = ₹ 8000

∴ Sumit suffered a loss of 16% on the original capital.

Question 3.
Nikhil spent 5% of his monthly income on his children’s education, invested 14% in shares, deposited 3% in a bank and used 40% for his daily expenses. He was left with a balance of ₹ 19,000. What was his income that month?
Solution:
Let the monthly income of Nikhil be ₹ x.
Nikhil invested 14% in shares and deposited 3% in a bank.
∴ Total investment = (14% + 3%) of x
= 17% of x
= \(\frac { 17 }{ 100 }\) × x
= 0.1 7 x
Nikhil spent 5% on his children’s education and used 40% for his daily expenses.

∴ Total expenditure = (5% + 40%) of x
= 45% of x
= \(\frac { 45 }{ 100 }\) × x
= 0.45x
Amount left with Nikhil = 19,000
Amount left with Nikhil = Income – (Total investment + Total expenditure)
∴ 19000 = x – (0.17x + 0.45x)
∴ 19000 = x – 0.62x ,
∴ 19000 = 0.38x
∴ \(x=\frac{19000}{0.38}=\frac{19000 \times 100}{0.38 \times 100}=\frac{1900000}{38}\)
= 50000
∴ The monthly income of Nikhil is ₹ 50000.

Question 4.
Mr. Sayyad kept ₹ 40,000 in a bank at 8% compound interest for 2 years. Mr. Fernandes invested ₹ 1,20,000 in a mutual fund for 2 years. After 2 years, Mr. Fernandes got ₹ 1,92,000. Whose investment turned out to be more profitable?
Solution:
Mr. Sayyad:
Mr. Sayyad kept ₹ 40,000 in a bank at 8% compound interest for 2 years P = ₹ 40000, R = 8%, n = 2 years
∴ Compound interest (I)
= Amount (A) – Principal (P)

= 40000 [(1 +0.08)² – 1]
= 40000 [(1.08)² – 1]
= 40000(1.1664 – 1)
= 40000 (0.1664)
= ₹ 6656
∴ Mr. Sayyad’s percentage of profit Interest

Mr. Fernandes:
Amount invested by Mr. Fernandes in mutual fund = ₹ 120000
Amount received by Mr. Fernandes after 2 years = ₹ 192000
∴ Profit earned by Mr. Fernandes
= Amount received – Amount invested
= 192000- 120000
= ₹72000
∴ Mr. Fernandes percentage of profit Profit earned

= 60%
From (i) and (ii),
Investment of Mr. Fernandes turned out to be more profitable.

Question 5.
Sameera spent 90% of her income and donated 3% for socially useful causes. If she was left with ₹ 1750 at the end of the month, what was her actual income ?
Solution:
Let the actual income of Sameera be ₹ x.
Sameera spent 90% of her income and donated 3%.
∴ Sameera’s total expenditure
= (3% + 90%) of x
= 93% of x
= \(\frac { 93 }{ 100 }\) × x
= 0.93x
Now, Savings = Income – Expenditure
∴ 1750 = x-0.93x
∴ 1750 = 0.07x
\(x=\frac{1750}{0.07}=\frac{1750 \times 100}{0.07 \times 100}=\frac{175000}{7}=25000\)
∴ The actual income of Sameera is ₹ 25000.

Maharashtra Board Class 9 Maths Chapter 6 Financial Planning Practice Set 6.1 Intext Questions and Activities

Question 1.
Amita invested some part of ₹ 35000 at 4% and the rest at 5% interest for one year. Altogether her gain was ₹ 1530. Find out the amounts she had invested at the two different rates. Write your answer in words. (Textbook pg. no. 97)
Solution:
Let the amount invested at the rate of 4% and 5% be ₹ x and ₹ y respectively.
According to the first condition, total amount invested = ₹ 35000
∴ x + y = 35000 …(i)
According to the second condition,
total interest received at 4% and 5% is ₹ 1530.
∴ 4 % of x + 5 % of y = 1530
∴ \(\frac { 4 }{ 100 }\) x x + \(\frac { 5 }{ 100 }\) x y = 1530
∴ 4x + 5y = 153000 …(ii)
Multiplying equation (i) by 4, we get
4x + 4y = 140000 …(iii)
Subtracting equation (iii) from (ii),

Substituting y = 13000 in equation (i),
x + 13000 = 35000
∴ x = 35000 – 13000 = 22000

∴ Amita invested ₹ 22000 at the rate of 4% and ₹ 13000 at the rate of 5%.

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 1 Practice Set 2.4 Chapter 2 Quadratic Equations Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.4 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 2 Quadratic Equations.

Class 10 Maths Part 1 Practice Set 2.4 Chapter 2 Quadratic Equations Questions With Answers Maharashtra Board

Question 1.
Compare the given quadratic equations to the general form and write values of a, b, c.
i. x² – 7x + 5 = 0
ii. 2m² = 5m – 5
iii. y² = 7y
Solution:
i. x² – 7x + 5 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = -7, c = 5

ii. 2m² = 5m – 5
∴ 2m² – 5m + 5 = 0
Comparing the above equation with
am² + bm + c = 0, we get
a = 2, b = -5, c = 5

iii. y² = 7y
∴ y² – 7y + 0 = 0
Comparing the above equation with
ay² + by + c = 0, we get
a = 1, b = -7, c = 0

Question 2.
Solve using formula.
i. x² + 6x + 5 = 0
ii. x² – 3x – 2 = 0
iii. 3m² + 2m – 7 = 0
iv. 5m² – 4m – 2 = 0
v. y² + \(\frac { 1 }{ 3 } \) y = 2
vi. 5x² + 13x + 8 = 0
Solution:
i. x² + 6x + 5 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = 6, c = 5
∴ b² – 4ac = (6)2 – 4 × 1 × 5
= 36 – 20 = 16

∴ x = -3 + 2 or x = -3 -2
∴ x = -1 or x = -5
∴ The roots of the given quadratic equation are -1 and -5.

ii. x² – 3x – 2 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = -3, c = -2
∴ b² – 4ac = (-3)2 – 4 × 1 × (-2)
= 9 + 8 = 17

iii. 3m² + 2m – 7 = 0
Comparing the above equation with
am² + bm + c = 0, we get
a = 3, b = 2, c = -7
∴ b² – 4ac = (2)² – 4 × 3 × ( -7)
= 4 + 84 = 88

iv. 5m² – 4m – 2 = 0
Comparing the above equation with
am² + bm + c = 0, we get
a = 5, b = -4, c = -2
∴ b² – 4ac = (-4)² – 4 × 5 × (-2)
= 16 + 40 = 56

v. y² + \(\frac { 1 }{ 3 } \)y = 2
∴ 3y² + y = 6 …(Multiplying both sides by 3]
∴ 3y² + y – 6 = 0
Comparing the above equation with
ay² + by + c = 0, we get
a = 3, b = 1, c = -6
∴ b² – 4ac = (1)² – 4 × 3 × (-6)
= 1 + 72 = 73

vi. 5x² + 13x + 8 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 5, b = 13, c = 8
∴ b² – 4ac = (13)2 – 4 × 5 × 8
= 169 – 160 = 9


The roots of the given quadratic equation are -1 and \(\frac { -8 }{ 5 } \).

Question 3.
With the help of the flow chart given below solve the equation x² + 2√3 x + 3 = 0 using the formula.

Solution:
i. x² + 2√3 x + 3 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = 2√3 ,c = 3

ii. b² – 4ac = (2√3)2 -4 × 1 × 3
= 12 – 12
= 0

Question 1.
Solve the equation 2x² + 13x + 15 = 0 by factorisation method, by completing the square method and by using the formula. Verify that you will get the same roots every time. (Textbook pg. no. 43)
Solution:

By using the property, if the product of two numbers is zero, then at least zero, we get
∴ x + 5 = 0 or 2x + 3 = 0
∴ x + -5 = 0 or 2x = -3 = 0
∴ x + -5 = or x = \(\frac { -3 }{ 2 } \)
∴ The roots of the given quadratic equation are \(\frac { -3 }{ 2 } \) and -5.

ii. Completing the square method:
2x² + 13x + 15 = 0


∴ The roots of the given quadratic equation are \(\frac { -3 }{ 2 } \) and -5.

iii. Formula method:
2x² + 13x + 15 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 2, b = 13, c = 15
∴ b² – 4ac = (13)2 – 4 × 2 × 15
= 169 – 120 = 49

∴ The roots of the given quadratic equation are \(\frac { -3 }{ 2 } \) and -5.
∴ By all the above three methods, we get the same roots of the given quadratic equation.

Maharashtra State Board Class 10 Maths Solutions Part 1

• Linear Equations in Two Variables Practice Set 1.1 Class 10 Maths Solutions
• Linear Equations in Two Variables Practice Set 1.2 Class 10 Maths Solutions
• Linear Equations in Two Variables Practice Set 1.3 Class 10 Maths Solutions
• Linear Equations in Two Variables Practice Set 1.4 Class 10 Maths Solutions
• Linear Equations in Two Variables Practice Set 1.5 Class 10 Maths Solutions
• Linear Equations in Two Variables Problem Set 1 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.1 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.2 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.3 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.4 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.5 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.6 Class 10 Maths Solutions
• Quadratic Equations Problem Set 2 Class 10 Maths Solutions