Maharashtra Board 10th Class Maths Part 1 Problem Set 3 Solutions Chapter 3 Arithmetic Progression

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.

Problem Set 3 Algebra 10th Std Maths Part 1 Answers Chapter 3 Arithmetic Progression

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 tn = ….
(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. 1st 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:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 1

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, tn = 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 10th term is 46, sum of the 5th and 7th 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, tn = 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
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 2
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 4th term is -15 and 9th term is -30. Find the sum of the first 10 numbers.
Solution:
t4 = -15, t9 = – 30 …[Given]
Since, tn = 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)
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 3
∴ 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
∴nth 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
∴ nth 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 nth term are 16 and -21 respectively.

Question 6.
If sum of 3rd and 8th 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 …[∵ tn = 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
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 4

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, tn = 45, Sn = 120
Since, tn = a + (n – 1)d
∴ 45 = -5 + (n – 1)d
∴ 45 + 5 = (n – 1)d
∴ (n – 1)d = 50 …(i)
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 5
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, tn = 45, Sn = 120
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 6
∴ 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
Sn = 36 …[Given]
Now, Sn = \(\frac { n }{ 2 } \) [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 + 9n – 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.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 7
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
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 8
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 9

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
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 10
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 11
∴ The sum of the first (p + q) terms is zero

Question 13.
If m times the mth term of an A.P. is equal to n times nth term, then show that the (m + n)th term of the A.P. is zero.
Solution:
According to the given condition,
mtm = ntn
∴ 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 Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 12
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 13

Maharashtra Board 9th Class Maths Part 1 Practice Set 2.5 Solutions Chapter 2 Real Numbers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.5 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 2 Real Numbers.

Practice Set 2.5 Algebra 9th Std Maths Part 1 Answers Chapter 2 Real Numbers

Question 1.
Find the value.
i. | 15 – 2|
ii. | 4 – 9|
iii. | 7| x | -4|
Solution:
i. |15 – 2| = |13| = 13
ii. |4 – 9| = |-5| = 5
iii. |7| x |- 4| = 7 x 4 = 28

Question 2.
Solve.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.5 1
Solution:
i. |3x – 5| = 1
∴ 3x – 5 = 1 or 3x – 5 = -1
∴ 3x = 1 + 5 or 3x = -1 + 5
∴ 3x = 6 or 3x = 4
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.5 2

ii. |7 – 2x| = 5
∴ 7 – 2x = 5 or 7 – 2x = -5
∴ 7 – 5 = 2x or 7 + 5 = 2x
∴ 2x = 2 or 2x = 12
∴ x = \(\frac { 2 }{ 2 }\) or x = \(\frac { 12 }{ 2 }\)
∴ x = 1 or x = 6

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.5 3
∴ 8 – x = 10 or 8 – x = -10 .. [Multiplying both the sides by 2]
∴ 8 – 10 = x or 8 + 10 = x
∴ x = -2 or x = 18

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.5 4

Maharashtra Board 10th Class Maths Part 1 Practice Set 2.3 Solutions Chapter 2 Quadratic Equations

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.3 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 2 Quadratic Equations.

Practice Set 2.3 Algebra 10th Std Maths Part 1 Answers Chapter 2 Quadratic Equations

Question 1.
Solve the following quadratic equations by completing the square method.
1. x2 + x – 20 = 0
2. x2 + 2x – 5 = 0
3. m2 – 5m = -3
4. 9y2 – 12y + 2 = 0
5. 2y2 + 9y + 10 = 0
6. 5x2 = 4x + 7
Solution:
1. x2 + x – 20 = 0
If x2 + x + k = (x + a)2, then
x2 + x + k = x2 + 2ax + a2
Comparing the coefficients, we get
1 = 2a and k = a2
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.3 1
∴ The roots of the given quadratic equation are 4 and -5.

2. x2 + 2x – 5 = 0
If x2 + 2x + k = (x + a)2, then
x2 + 2x + k = x2 + 2ax + a2
Comparing the coefficients, we get
2 = 2a and k = a2
∴ a = 1 and k = (1)2 = 1
Now, x2 + 2x – 5 = 0
∴ x2 + 2x + 1 – 1 – 5 = 0
∴ (x + 1)2 – 6 = 0
∴ (x + 1)2 = 6
Taking square root of both sides, we get
x + 1 = ± √6
∴ x + 1 √6 or x + 1 = √6
∴ x = √6 – 1 or x = -√6 – 1
∴ The roots of the given quadratic equation are √6 -1 and – √6 -1.

3. m2 – 5m = -3
∴ m2 – 5m + 3 = 0
If m2 – 5m + k = (m + a)2, then
m2 – 5m + k = m2 + 2am + a2
Comparing the coefficients, we get
-5 = 2a and k = a2
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.3 2

4. 9y2 – 12y + 2 = 0
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.3 3
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.3 4

5. 2y2 + 9y + 10 = 0
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.3 5
Taking square root of both sides, we get
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.3 6
∴ The roots of the given quadratic equation are -2 and \(\frac { -5 }{ 2 } \).

6. 5x2 = 4x + 7
∴ 5x2 – 4x – 7 = 0
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.3 7
Comparing the coefficients, we get
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.3 8

Maharashtra Board 10th Class Maths Part 1 Practice Set 2.6 Solutions Chapter 2 Quadratic Equations

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.6 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 2 Quadratic Equations.

Practice Set 2.6 Algebra 10th Std Maths Part 1 Answers Chapter 2 Quadratic Equations

Question 1.
Product of Pragati’s age 2 years ago and years hence is 84. Find her present age.
Solution:
Let the present age of Pragati be x years.
∴ 2 years ago,
Age of Pragati = (x – 2) years
After 3 years,
Age of Pragati = (x + 3) years
According to the given condition,
(x – 2) (x + 3) = 84
∴ x(x + 3) – 2(x + 3) = 84
∴ x2 + 3x – 2x – 6 = 84
∴ x2 + x – 6 – 84 = 0
∴ x2 + x – 90 = 0
x2 + 10x – 9x – 90 = 0
∴ x(x + 10) – 9(x + 10) = 0
∴ (x + 10)(x – 9) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 10 = 0 or x – 9 = 0
∴ x = -10 or x = 9
But, age cannot be negative.
∴ x = 9
∴ Present age of Pragati is 9 years.

Question 2.
The sum of squares of two consecutive even natural numbers is 244; find the numbers.
Solution:
Let the first even natural number be x.
∴ the next consecutive even natural number will be (x + 2).
According to the given condition,
x2 + (x + 2)2 = 244
∴ x2 + x2 + 4x + 4 = 244
∴ 2x2 + 4x + 4 – 244 = 0
∴ 2x2 + 4x – 240 = 0
∴ x2 + 2x – 120 = 0 …[Dividing both sides by 2]
∴ x2 + 12x – 10x – 120 = 0
∴ x(x + 12) – 10 (x + 12) = 0
∴ (x + 12) (x – 10) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 12 = 0 or x – 10 = 0
∴ x = -12 or x = 10
But, natural number cannot be negative.
∴ x = 10 and x + 2 = 10 + 2 = 12
∴ The two consecutive even natural numbers are 10 and 12.

Question 3.
In the orange garden of Mr. Madhusudan there are 150 orange trees. The number of trees in each row is 5 more than that in each column. Find the number of trees in each row and each column with the help of following flow chart.
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.6 1
Solution:
i. Number of trees in a column is x.
ii. Number of trees in a row = x + 5
iii. Total number of trees = x x (x + 5)
iv. According to the given condition,
x(x + 5) = 150
∴ x2 + 5x = 150
∴ x2 + 5x – 150 = 0
v. x2 + 15x – 10x – 150 = 0
∴ x(x+ 15) – 10(x + 15) = 0
∴ (x + 15)(x – 10) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 15 = 0 or x – 10 = 0
∴ x = -15 or x = 10
But, number of trees cannot be negative.
∴ x = 10
vi. Number of trees in a column is 10.
vii. Number of trees in a row = x + 5 = 10 + 5 = 15
∴ Number of trees in a row is 15.

Question 4.
Vivek is older than Kishor by 5 years. The Find their present ages is \(\frac { 1 }{ 6 } \) Find their Present ages
Solution:
Let the present age of Kishor be x.
∴ Present age of Vivek = (x + 5) years
According to the given condition,
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.6 2
∴ 6(2x + 5) = x(x + 5)
∴ 12x + 30 = x2 + 5x
∴ x2 + 5x – 12x – 30 = 0
∴ x2 – 7x – 30 = 0
∴ x2 – 10x + 3x – 30 = 0
∴ x(x – 10) + 3(x – 10) = 0
∴ (x – 10)(x + 3) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x – 10 = 0 or x + 3 = 0
∴ x = 10 or x = – 3
But, age cannot be negative.
∴ x = 10 andx + 5 = 10 + 5 = 15
∴ Present ages of Kishor and Vivek are 10 years and 15 years respectively.

Question 5.
Suyash scored 10 marks more in second test than that in the first. 5 times the score of the second test is the same as square of the score in the first test. Find his score in the first test.
Solution:
Let the score of Suyash in the first test be x.
∴ Score in the second test = x + 10 According to the given condition,
5(x + 10) = x2
∴ 5x + 50 = x2
∴ x2 – 5x – 50 = 0
∴ x2 – 10x + 5x – 50 = 0
∴ x(x – 10) + 5(x – 10) = 0
∴ (x – 10) (x + 5) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x – 10 = 0 or x + 5 = 0
∴ x = 10 or x = – 5
But, score cannot be negative.
∴ x = 10
∴ The score of Suyash in the first test is 10.

Question 6.
‘Mr. Kasam runs a small business of making earthen pots. He makes certain number of pots on daily basis. Production cost of each pot is ₹ 40 more than 10 times total number of pots, he makes in one day. If production cost of all pots per day is ₹ 600, find production cost of one pot and number of pots he makes per day.
Solution:
Let Mr. Kasam make x number of pots on daily basis.
Production cost of each pot = ₹ (10x + 40)
According to the given condition,
x(10x + 40) = 600
∴ 10x2 + 40x = 600
∴ 10x2 + 40x- 600 = 0
∴ x2 + 4x – 60 = 0 …[Dividing both sides by 10]
∴ x2 + 10x – 6x – 60 = 0
∴ x(x + 10) – 6(x + 10) = 0
∴ (x + 10) (x – 6) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 10 = 0 or x – 6 = 0
∴ x = – 10 or x = 6
But, number of pots cannot be negative.
∴ x = 6
∴ Production cost of each pot = 7(10 x + 40)
= ₹ [(10×6)+ 40]
= ₹(60 + 40) = ₹ 100
Production cost of one pot is ₹ 100 and the number of pots Mr. Kasam makes per day is 6.

Question 7.
Pratik takes 8 hours to travel 36 km downstream and return to the same spot. The speed of boat in still water is 12 km. per hour. Find the speed of water current.
Solution:
Let the speed of water current be x km/hr. Speed of boat is 12 km/hr. (x < 12)
In upstream, speed of the water current decreases the speed of the boat and it is the opposite in downstream.
∴ speed of the boat in upstream = (12 – x) km/hr and speed of the boat in downstream = (12 + x) km/hr.
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.6 3
∴ The speed of water current is 6 km/hr.

Question 8.
Pintu takes 6 days more than those of Nishu to complete certain work. If they work together they finish it in 4 days. How many days would it take to complete the work if they work alone.
Solution:
Let Nishu take x days to complete the work alone.
∴ Total work done by Nishu in 1 day = \(\frac { 1 }{ x } \)
Also, Pintu takes (x + 6) days to complete the work alone.
∴ Total work done by Pintu in 1 day = \(\frac { 1 }{ x+6 } \)
∴ Total work done by both in 1 day = (\(\frac { 1 }{ x } \) + \(\frac { 1 }{ x+6 } \))
But, both take 4 days to complete the work together.
∴ Total work done by both in 1 day = \(\frac { 1 }{ 4 } \)
According to the given condition,
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.6 4
∴ 4(2x + 6) = x(x + 6)
∴ 8x + 24 = x2 + 6x
∴ x2 + 6x – 8x – 24 = 0
∴ x2 – 2x – 24 = 0
∴ x2 – 6x + 4x – 24 = 0
∴ x(x – 6)+ 4(x – 6) = 0
∴ (x – 6) (x + 4) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x – 6 = 0 or x + 4 = 0
∴ x = 6 or x = -4
But, number of days cannot be negative,
∴ x = 6 and x + 6 = 6 + 6 = 12
∴ Number of days taken by Nishu and Pintu to complete the work alone is 6 days and 12 days respectively.

Question 9.
If 460 is divided by a natural number, quotient is 6 more than five times the divisor and remainder is 1. Find quotient and divisor.
Solution:
Let the natural number be x.
∴ Divisor = x, Quotient = 5x + 6
Also, Dividend = 460 and Remainder = 1
Dividend = Divisor × Quotient + Remainder
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.6 5
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 5x + 51 = 0
∴ x = 9 or x = \(\frac { -51 }{ 5 } \)
But, natural number cannot be negative,
∴ x = 9
∴ Quotient = 5x + 6 = 5(9) + 6 = 45 + 6 = 51
∴ Quotient is 51 and Divisor is 9.

Question 10.
In the given fig. []ABCD is a trapezium, AB || CD and its area is 33 cm2. From the information given in the figure find the lengths of all sides of the []ABCD. Fill in the empty boxes to get the solution.
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.6 6
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.6 7
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.6 8

Maharashtra Board 10th Class Maths Part 1 Practice Set 2.5 Solutions Chapter 2 Quadratic Equations

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.5 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 2 Quadratic Equations.

Practice Set 2.5 Algebra 10th Std Maths Part 1 Answers Chapter 2 Quadratic Equations

Question 1.
Fill in the gaps and complete.
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.5 1
Answer:
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.5 2

Question 2.
Find the value of discriminant.
i. x2 + 7x – 1 = 0
ii. 2y2 – 5y + 10 = 0
iii. √2 x2 + 4x + 2√2 = 0
Solution:
i. x2 +7 x – 1 = 0
Comparing the above equation with
ax2 + bx + c = 0, we get
a = 1, b = 7, c = -1
∴ b2– 4ac = (7)2 – 4 × 1 × (-1)
= 49 + 4
∴ b2 – 4ac = 53

ii. 2y2 – 5y + 10 = 0
Comparing the above equation with
ay2 + by + c = 0, we get
a = 2, b = -5, c = 10
∴ b2 – 4ac = (-5)2 -4 × 2 × 10
= 25 – 80
∴ b2 – 4ac = -55

iii. √2 x2 + 4x + 2√2 = 0
Comparing the above equation with
ax + bx + c = 0, we get
a = √2,b = 4, c = 2√2
∴ b2 – 4ac = (4)2 – 4 × √2 × 2√2
= 16 – 16
∴ b2 – 4ac =0

Question 3.
Determine the nature of roots of the following quadratic equations.
i. x2 – 4x + 4 = 0
ii. 2y2 – 7y + 2 = 0
iii. m2 + 2m + 9 = 0
Solution:
i. x2 – 4x + 4= 0
Comparing the above equation with
ax2 + bx + c = 0, we get
a = 1,b = -4, c = 4
∴ ∆ = b2 – 4ac
= (- 4)2 – 4 × 1 × 4
= 16 – 16
∴ ∆ = 0
∴ Roots of the given quadratic equation are real and equal.

ii. 2y2 – 7y + 2 = 0
Comparing the above equation with
ay2 + by + c = 0, we get
a = 2, b = -7, c = 2
∴ ∆ = b2 – 4ac
= (- 7)2 – 4 × 2 × 2
= 49 – 16
∴ ∆ = 33
∴ ∆ > 0
∴ Roots of the given quadratic equation are real and unequal.

iii. m2 + 2m + 9 = 0
Comparing the above equation with
am2 + bm + c = 0, we get
a = 1,b = 2, c = 9
∴ ∆ = b2 – 4ac
= (2)2 – 4 × 1 × 9
= 4 – 36
∴ ∆ = -32
∴ ∆ < 0
∴ Roots of the given quadratic equation are not real.

Question 4.
Form the quadratic equation from the roots given below.
i. 0 and 4
ii. 3 and -10
iii. \(\frac { 1 }{ 2 } \) , \(\frac { 1 }{ 2 } \)
iv. 2 – √5, 2 + √5
Solution:
i. Let a = 0 and β = 4
∴ α + β = 0 + 4 = 4
and α × β = 0 × 4 = 0
∴ The required quadratic equation is
x2 – (α + β) x + αβ = 0
∴ x2 – 4x + 0 = 0
∴ x2 – 4x = 0

ii. Let α = 3 and β = -10
∴ α + β = 3 – 10 = -7
and α × β = 3 × -10 = -30
∴ The required quadratic equation is
x2 – (α + β)x + αβ = 0
∴ x2 – (-7) x + (-30) = 0

Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.5 3

Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.5 4
∴ The required quadratic equation is
x2 – (α + β)x + αβ = 0
∴ x2 – 4x – 1 = 0

Question 5.
Sum of the roots of a quadratic equation is double their product. Find k if equation is x2 – 4kx + k + 3 = 0.
Solution:
x2 – 4kx + k + 3 = 0
Comparing the above equation with
ax2 + bx + c = 0, we get
a = 1, b = – 4k, c = k + 3
Let α and β be the roots of the given quadratic equation.
Then, α + β  = \(\frac { -b }{ a } \) and αβ = \(\frac { c }{ a } \)
According to the given condition,
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.5 5

Question 6.
α, β are roots of y2 – 2y – 7 = 0 find,
i. α2 + β2
ii. α3 + β3
Solution:
y2 – 2y – 7 = 0
Comparing the above equation with
ay2 + by + c = 0, we get
a = 1, b = -2, c = -7
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.5 6

Question 7.
The roots of each of the following quadratic equations are real and equal, find k.
i. 3y2 + ky + 12 = 0
ii. kx (x-2) + 6 = 0
Solution:
i. 3y2 + kg + 12 = 0
Comparing the above equation with
ay2 + by + c = 0, we get
a = 3, b = k, c = 12
∴ ∆ = b2 – 4ac
= (k)2 – 4 × 3 × 12
= k2 – 144 = k2 – (12)2
∴ ∆ = (k + 12) (k – 12) …[∵ a2 – b2 = (a + b) (a – b)]
Since, the roots are real and equal.
∴ ∆ = 0
∴ (k + 12) (k – 12) = 0
∴ k + 12 = 0 or k – 12 = 0
∴ k = -12 or k = 12

ii. kx (x – 2) + 6 = 0
∴ kx2 – 2kx + 6 = 0
Comparing the above equation with
ax2 + bx + c = 0, we get
a = k, b = -2k, c = 6
∴ ∆ = b2 – 4ac
= (-2k)2 – 4 × k × 6
= 4k2 – 24k
∴ ∆ = 4k (k – 6)
Since, the roots are real and equal.
∴ ∆ = 0
∴ 4k (k – 6) = 0
∴ k(k – 6) = 0
∴ k = 0 or k – 6 = 0
But, if k = 0 then quadratic coefficient becomes zero.
∴ k ≠ 0
∴ k = 6

Question 1.
Fill in the blanks. (Textbook pg. no. 44)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.5 7

Question 2.
Determine nature of roots of the quadratic equation: x2 + 2x – 9 = 0 (Textbook pg. no. 45)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.5 8
∴ The roots of the given equation are real and unequal.

Question 3.
Fill in the empty boxes properly. (Textbook pg. no. 46)
Solution:
10x2 + 10x + 1 = 0
Comparing the above equation with
ax2 + bx + c = 0, we get
a = 10, b = 10, c = 1
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.5 9

Question 4.
Write the quadratic equation if addition of the roots is 10 and product of the roots is 9. (Textbook pg, no. 48)
Answer:
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.5 10

Question 5.
What will be the quadratic equation if α = 2, β = 5. (Textbook pg. no, 48)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.5 11

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Maharashtra Board 10th Class Maths Part 1 Practice Set 3.1 Solutions Chapter 3 Arithmetic Progression

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.

Practice Set 3.1 Algebra 10th Std Maths Part 1 Answers Chapter 3 Arithmetic Progression

Question 1.
Which of the following sequences are A.P.? If they are A.P. find the common difference.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 1
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.

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 2
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 3
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.

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 4
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = 0.

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 5
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,…

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 6
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 7
∴ 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.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 8
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

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 9

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:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 10
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,….
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 11
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:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 12

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:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 13
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 14
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,
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 15
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 Board 10th Class Maths Part 1 Practice Set 1.3 Solutions Chapter 1 Linear Equations in Two Variables

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.3 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 1 Linear Equations in Two Variables.

Practice Set 1.3 Algebra 10th Std Maths Part 1 Answers Chapter 1 Linear Equations in Two Variables

Question 1.
Fill in the blanks with correct number.
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.3 1

Question 2.
Find the values of following determinants.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.3 2
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.3 3
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.3 4

Question 3.
Solve the following simultaneous equations using Cramer’s rule.
i. 3x – 4y = 10 ; 4x + 3y = 5
ii. 4x + 3y – 4 = 0 ; 6x = 8 – 5y
iii. x + 2y = -1 ; 2x – 3y = 12
iv. 6x – 4y = -12 ; 8x – 3y = -2
v. 4m + 6n = 54 ; 3m + 2n = 28
vi. 2x + 3y = 2 ; x – \(\frac { y }{ 2 } \) = \(\frac { 1 }{ 2 } \)
Solution:
i. The given simultaneous equations are 3x – 4y = 10 …(i)
4x + 3y = 5 …(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a1x + b1y = c1 and a2x + b2y = c2, we get
a1 = 3, b1 = -4, c1 = 10 and
a2 = 4, b2 = 3, c2 = 5
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.3 5
∴ (x, y) = (2, -1) is the solution of the given simultaneous equations.

ii. The given simultaneous equations are
4x + 3y – 4 = 0
∴ 4x + 3y = 4 …(i)
6x = 8 – 5y
∴ 6x + 5y = 8 …(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a1x + b1y = c1 and a2x + b2y = c2, we get
a1 = 4, b1 = 3, c1 = 4 and
a2 = 6, b2 = 5, c2 = 8
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.3 6
∴ (x, y) = (-2, 4) is the solution of the given simultaneous equations.

iii. The given simultaneous equations are
x + 2y = -1 …(i)
2x – 3y = 12 …(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a1x + b1y = C1 and a2x + b2y = c2, we get
a1 = 1, b1 = 2, c1 = -1 and
a2 = 2, b2 = -3, c2 = 12
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.3 7
∴ (x, y) = (3, -2) is the solution of the given simultaneous equations.

iv. The given simultaneous equations are
6x – 4y = -12
∴ 3x – 2y = -6 …(i) [Dividing both sides by 2]
8x – 3y = -2 …(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a1x + b1y = c1 and a2x + b2y = c2, we get
a1 = 3, b1 = -2, c1 = -6 and
a2 = 8, b2 = -3, c2 = -2
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.3 8
∴ (x, y) = (2, 6) is the solution of the given simultaneous equations.

v. The given simultaneous equations are
4m + 6n = 54
2m + 3n = 27 …(i) [Dividing both sides by 2]
3m + 2n = 28 …(ii)
Equations (i) and (ii) are in am + bn = c form.
Comparing the given equations with
a1m + b1n = c1 and a2m + b2n = c2, we get
a1 = 2, b1 = 3, c1 = 27 and
a2 = 3, b2 = 2, c2 = 28
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.3 9
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.3 10
∴ (m, n) = (6, 5) is the solution of the given simultaneous equations.

vi. The given simultaneous equations are
2x + 3y = 2 …(i)
x = \(\frac { y }{ 2 } \) = \(\frac { 1 }{ 2 } \)
∴ 2x – y = 1 …(ii) [Multiplying both sides by 2]
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a1x + b1y = c1 and a2x + b2y = c2, we get
a1 = 2, b1 = 3, c1 = 2 and
a2 = 2, b2 = -1, c2 = 1
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.3 11

Question 1.
To solve the simultaneous equations by determinant method, fill in the blanks,
y + 2x – 19 = 0; 2x – 3y + 3 = 0 (Textbookpg.no. 14)
Solution:
Write the given equations in the form
ax + by = c.
2x + y = 19
2x – 3y = -3
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.3 12

Question 2.
Complete the following activity. (Textbook pg. no. 15)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.3 13

Question 3.
What is the nature of solution if D = 0? (Textbook pg. no. 16)
Solution:
If D = 0, i.e. a1b2 – b1a2 = 0, then the two simultaneous equations do not have a unique solution.
Examples:
i. 2x – 4y = 8 and x – 2y = 4
Here, a1b2 – b1a2 = (2)(-2) – (-4) (1)
= -4 + 4 = 0
Graphically, we can check that these two lines coincide and hence will have infinite solutions.

ii. 2x – y = -1 and 2x – y = -4
Here, a1 b2 – b1 a2 = (2)(-1) – (-1) (2)
= -2 + 2 = 0
Graphically, we can check that these two lines are parallel and hence they do not have a solution.

Question 4.
What can you say about lines if common solution is not possible? (Textbook pg. no. 16)
Answer:
If the common solution is not possible, then the lines will either coincide or will be parallel to each other.

Maharashtra Board 10th Class Maths Part 1 Practice Set 1.2 Solutions Chapter 1 Linear Equations in Two Variables

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.2 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 1 Linear Equations in Two Variables.

Practice Set 1.2 Algebra 10th Std Maths Part 1 Answers Chapter 1 Linear Equations in Two Variables

10th Maths 2 Practice Set 1.2 Question 1.
Complete the following table to draw graph of the equations.
i. x + y = 3
ii. x – y = 4
Answer:
i. x + y = 3
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 1
ii. x – y = 4
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 2

Linear Equations In Two Variables Practice Set 1.2  Question 2.
Solve the following simultaneous equations graphically.
i. x + y = 6 ; x – y = 4
ii. x + y = 5 ; x – y = 3
iii. x + y = 0 ; 2x – y = 9
iv. 3x – y = 2 ; 2x – y = 3
v. 3x – 4y = -7 ; 5x – 2y = 0
vi. 2x – 3y = 4 ; 3y – x = 4
Solution:
i. The given simultaneous equations are
x + y = 6                                                                                                        x – y = 4
∴ y = 6 – x                                                                                                     ∴ y = x – 4Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 3
The two lines intersect at point (5, 1).
∴ x = 5 and y = 1 is the solution of the simultaneous equations x + y = 6 and x – y = 4.

ii. The given simultaneous equations are
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 6
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 7
The two lines intersect at point (4, 1).
∴ x = 4 and y = 1 is the solution of the simultaneous equations x+y = 5 and x – y = 3.

iii. The given simultaneous equations are
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 4
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 5
The two lines intersect at point (3, -3).
∴ x = 3 and y = -3 is the solution of the simultaneous equations x + y = 0 and 2x – y = 9.

iv. The given simultaneous equations are
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 8
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 9
The two lines intersect at point (-1, -5).
∴ x = -1 and y = -5 is the solution of the simultaneous equations 3x- y = 2 and 2x- y = 3.

v. The given simultaneous equations are
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 10
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 11
The two lines intersect at point (1, 2.5).
∴ x = 1 and y = 2.5 is the solution of the simultaneous equations 3x – 4y = -7 and 5x – 2y = 0.

vi. The given simultaneous equations are
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 12
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 13
The two lines intersect at point (8, 4).
∴ x = 8 and y = 4 is the solution of the simultaneous equations 2x – 3y = 4 and 3y – x = 4.

10th Math Part 2 Practice Set 1.2  Question 1.
Solve the following simultaneous equations by graphical method. Complete the following tables to get ordered pairs.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 14
i. Plot the above ordered pairs on the same co-ordinate plane.
ii. Draw graphs of the equations.
iii. Note the co-ordinates of the point of intersection of the two graphs. Write solution of these equations. (Textbook pg. no. 8)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 15 Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 16
The two lines intersect at point (-1, -2).
∴ (x , y) = (-1, -2) is the solution of the given simultaneous equations.

Mathematics Part 1 Standard 9 Practice Set 1.2 Answer  Question 1.
Solve the above equations by method of elimination. Check your solution with the solution obtained by graphical method. (Textbook pg. no. 8)
Solution:
The given simultaneous equations are
x – y = 1 …(i)
5x – 3y = 1 …(ii)
Multiplying equation (i) by 3, we get
3x – 3y = 3 …(iii)
Subtracting equation (iii) from (ii), we get
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 17
Substituting x = -1 in equation (i), we get
-1 -y= 1
∴ -y = 1 + 1
∴ -y = 2
∴ y = -2
∴ (x,y) = (-1, -2) is the solution of the given simultaneous equations.
∴ The solution obtained by elimination method and by graphical method is the same.

1.2 Maths Class 10 Question 2.
The following table contains the values of x and y co-ordinates for ordered pairs to draw the graph of 5x – 3y = 1.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 18
i. Is it easy to plot these points?
ii. Which precaution is to be taken to find ordered pairs so that plotting of points becomes easy? (Textbook pg. no. 8)
Solution:
i. No
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 19
The above numbers are non-terminating and recurring decimals.
∴ It is not easy to plot the given points.

ii. While finding ordered pairs, numbers should be selected in such a way that the co-ordinates obtained will be integers.

Linear Equations ¡n Two Variables Class 10 Maths Question 3.
To solve simultaneous equations x + 2y = 4; 3x + 6y = 12 graphically, following are the ordered pairs.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 20
Plotting the above ordered pairs, graph is drawn. Observe it and find answers of the following questions.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 21
i. Are the graphs of both the equations different or same?
ii. What are the solutions of the two equations x + 2y = 4 and 3x + 6y = 12? How many solutions are possible?
iii. What are the relations between coefficients of x, coefficients of y and constant terms in both the equations?
iv. What conclusion can you draw when two equations are given but the graph is only one line? (Textbook pg. no. 9)
Solution:
i. The graphs of both the equations are same.
ii. The solutions of the given equations are (-2, 3), (0, 2), (1, 1.5), etc.
∴ Infinite solutions are possible.
iii. Ratio of coefficients of x = \(\frac { 1 }{ 3 } \)
Ratio of coefficients of y = \(\frac { 2 }{ 6 } \) = \(\frac { 1 }{ 3 } \)
Ratio of constant terms = \(\frac { 4 }{ 12 } \) = \(\frac { 1 }{ 3 } \)
∴ Ratios of coefficients of x = ratio of coefficients of y = ratio of the constant terms
iv. When two equations are given but the graph is only one line, the equations will have infinite solutions.

Class 10 Maths Part 1 Practice Set 1.2 Question 4.
Draw graphs of x- 2y = 4, 2x – 4y = 12 on the same co-ordinate plane. Observe it. Think of the relation between the coefficients of x, coefficients ofy and the constant terms and draw the inference. (Textbook pg. no. 10)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 22 Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 23
ii. Ratio of coefficients of x =\(\frac { 1 }{ 2 } \)
Ratio of coefficients of y = \(\frac { -2 }{ -4 } \) = \(\frac { 1 }{ 2 } \)
Ratio of constant terms = \(\frac { 4 }{ 12 } \) = \(\frac { 1 }{ 3 } \)
∴ Ratio of coefficients of x = ratio of coefficients of y ratio of constant terms
iii. If ratio of coefficients of x = ratio of coefficients of y ≠ ratio of constant terms, then the graphs of the two equations will be parallel to each other.

Condition of consistency in Equations:
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.2 24

Maharashtra Board 10th Class Maths Part 1 Practice Set 5.2 Solutions Chapter 5 Probability

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.2 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 5 Probability.

Practice Set 5.2 Algebra 10th Std Maths Part 1 Answers Chapter 5 Probability

Question 1.
For each of the following experiments write sample space ‘S’ and number of sample Point n(S)
i. One coin and one die are thrown simultaneously.
ii. Two digit numbers are formed using digits 2,3 and 5 without repeating a digit.
Solution:
i. Sample space,
S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
∴ n(S) =12
ii. Sample space,
S = {23,25,32, 35, 52, 53}
∴ n(S) = 6

Question 2.
The arrow is rotated and it stops randomly on the disc. Find out on which colour it may stop.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 1
Solution:
There are total six colours on the disc.
Sample space,
S = {Red, Orange, Yellow, Blue, Green, Purple}
∴ n(S) = 6
∴ Arrow may stop on any one of the six colours.

Question 3.
In the month of March 2019, find the days on which the date is a multiple of 5. (see the given page of the calendar).
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 2
Solution:
Dates which are multiple of 5:
5,10, 15,20,25,30
∴ S = {Tuesday, Sunday, Friday, Wednesday, Monday, Saturday}
∴ n(S) = 6
∴ The days on which the date will be a multiple of 5 are Tuesday, Sunday, Friday, Wednesday, Monday and Saturday.

Question 4.
Form a ‘Road safety committee’ of two, from 2 boys (B1 B2) and 2 girls (G1, G2). Complete the following activity to write the sample space.
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 3

Question 1.
Sample Space

  • The set of all possible outcomes of a random experiment is called sample space.
  • It is denoted by ‘S’ or ‘Ω’ (omega).
  • Each element of a sample space is called a sample point.
  • The number of elements in the set S is denoted by n(S).
  • If n(S) is finite, then the sample space is called a finite sample space.

Some examples of finite sample space. (Textbook pg. no, 117)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 4 Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 5

Maharashtra Board 10th Class Maths Part 1 Practice Set 5.1 Solutions Chapter 5 Probability

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 5 Probability.

Practice Set 5.1 Algebra 10th Std Maths Part 1 Answers Chapter 5 Probability

Question 1.
How many possibilities are there in each of the following?
i. Vanita knows the following sites in Maharashtra. She is planning to visit one of them in her summer vacation. Ajintha, Mahabaleshwar, Lonar Sarovar, Tadoba wild life sanctuary, Amboli, Raigad, Matheran, Anandavan.
ii. Any day of a week is to be selected randomly.
iii. Select one card from the pack of 52 cards.
iv. One number from 10 to 20 is written on each card. Select one card randomly.
Solution:
i. Here, 8 sites of Maharashtra are given.
∴ There are 8 possibilities in a random experiment of visiting a site out of 8 sites in Maharashtra.

ii. There are 7 days in a week.
∴ There are 7 possibilities in a random experiment of selecting a day of the week.

iii. There are 52 cards in a pack of cards.
∴ There are 52 possibilities in a random experiment of selecting one card from the pack of 52 cards.

iv. There are 11 cards numbered from 10 to 20.
∴ There are 11 possibilities in a random experiment of selecting one card from the given set of cards.

Question 1.
In which of the following experiments possibility of expected outcome is more? (Textbook pg, no. 116)
i. Getting 1 on the upper face when a die is thrown.
ii. Getting head by tossing a coin.
Solution:
i. On a die there are 6 numbers.
∴ There are 6 possibilities of getting any one number from 1 to 6 on the upper face i.e. \(\frac { 1 }{ 6 } \) is the possibility.

ii. There are two possibilities (H or T) on tossing a coin i.e. \(\frac { 1 }{ 2 } \) possibility.
∴ In the second experiment, the possibility of expected outcome is more.

Question 2.
Throw a die, once. What are the different possibilities of getting dots on the upper face? (Textbook pg. no. 114)
Answer:
There are six different possibilities of getting dots on the upper face. They are
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.1