Category: Class 10

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

10th Standard Maths 1 Practice Set 3.4 Chapter 3 Arithmetic Progression Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 3.4 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 3 Arithmetic Progression.

Class 10 Maths Part 1 Practice Set 3.4 Chapter 3 Arithmetic Progression Questions With Answers Maharashtra Board

Question 1.
On 1^st Jan 2016, Sanika decides to save ₹ 10, ₹ 11 on second day, ₹ 12 on third day. If she decides to save like this, then on 31st Dec 2016 what would be her total saving?
Solution:

∴ Sanika’s total saving on 31^st December 2016 would be ₹ 70455.

Question 2.
A man borrows ₹ 8000 and agrees to repay with a total interest of ₹ 1360 in 12 monthly instalments. Each instalment being less than the preceding one by ₹ 40. Find the amount of the first and last instalment.
Solution:
i. The instalments are in A.P.
Amount repaid in 12 instalments (S12)
= Amount borrowed + total interest
= 8000 + 1360
∴ S₁₂ = 9360
Number of instalments (n) = 12
Each instalment is less than the preceding one by ₹ 40.
∴ d = -40

∴ Amount of the first instalment is ₹ 1000 and that of the last instalment is ₹ 560.

Question 3.
Sachin invested in a national saving certificate scheme. In the first year he invested ₹ 5000, in the second year ₹ 7000, in the third year ₹ 9000 and so on. Find the total amount that he invested in 12 years.
Solution:
i. Amount invested by Sachin in each year are as follows:
5000, 7000, 9000, …
The above sequence is an A.P.
∴ a = 5000, d = 7000 – 5000 = 2000, n = 12

∴ The total amount invested by Sachin in 12 years is ₹ 1,92,000.

Question 4.
There is an auditorium with 27 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. Find the number of seats in the 15th row and also find how many total seats are there in the auditorium?
Solution:
i. The number of seats arranged row-wise are as follows:
20, 22, 24,
The above sequence is an A.P.
∴ a = 20, d = 22 – 20 = 2, n = 27

ii. t_n = a + (n – 1)d
∴ t₁₅ = 20 + (15 – 1)2
= 20 + 14 × 2
= 20 + 28
∴ t₁₅ = 48
∴ The number of seats in the 15^th row is 48.

∴ Total seats in the auditorium are 1242.

Question 5.
Kargil’s temperature was recorded in a week from Monday to Saturday. All readings were in A.P. The sum of temperatures of Monday and Saturday was 5°C more than sum of temperatures of Tuesday and Saturday. If temperature of Wednesday was -30° Celsius then find the temperature on the other five days.
Solution:
Let the temperatures from Monday to Saturday in A.P. be
a, a + d, a + 2d, a + 3d, a + 4d, a + 5d.
According to the first condition,
(a) + (a + 5d) = (a + d) + (a + 5d) + 5°
∴ d = -5°
According to the second condition,
a + 2d = -30°
∴ a + 2(-5°) = -30°
∴ a – 10° = -30°
∴ a = -30° + 10° = -20°
∴ a + d = -20° – 5° = – 25°
a + 3d = -20° + 3(- 5°) = -20° – 15° = -35°
a + 4d = -20° + 4(-5°) = -20° – 20° = -40°
a + 5d = -20° + 5(-5°) = -20° – 25° = -45°
∴ The temperatures on the other five days are
-20°C, -25° C, -35° C, -40° C and -45° C.

Question 6.
On the world environment day tree plantation programme was arranged on a land which is triangular in shape. Trees are planted such that in the first row there is one tree, in the second row there are two trees, in the third row three trees and so on. Find the total number of trees in the 25 rows.
Solution:
i. The number of frees planted row-wise are as follows:
1,2,3,…
The above sequence is an A.P.
∴ a = 1, d = 2 – 1 = 1,n = 25

∴ The total number of trees in 25 rows are 325.

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 Practice Set 2.3 Chapter 2 Quadratic Equations Textbook Answers Maharashtra Board

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.

Class 10 Maths Part 1 Practice Set 2.3 Chapter 2 Quadratic Equations Questions With Answers Maharashtra Board

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

∴ 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

4. 9y² – 12y + 2 = 0

5. 2y² + 9y + 10 = 0

Taking square root of both sides, we get

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

6. 5x² = 4x + 7
∴ 5x² – 4x – 7 = 0

Comparing the coefficients, we get

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

10th Standard Maths 1 Practice Set 2.6 Chapter 2 Quadratic Equations Textbook Answers Maharashtra Board

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.

Class 10 Maths Part 1 Practice Set 2.6 Chapter 2 Quadratic Equations Questions With Answers Maharashtra Board

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.

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,

∴ 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.

∴ 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,

∴ 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

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.

Solution:

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

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 2.5 Chapter 2 Real Numbers Textbook Answers Maharashtra Board

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.

Class 9 Maths Part 1 Practice Set 2.5 Chapter 2 Real Numbers Questions With Answers Maharashtra Board

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.

Solution:
i. |3x – 5| = 1
∴ 3x – 5 = 1 or 3x – 5 = -1
∴ 3x = 1 + 5 or 3x = -1 + 5
∴ 3x = 6 or 3x = 4

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

∴ 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

Maths Solution Class 9 Maharashtra Board 

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

10th Standard Maths 1 Practice Set 2.1 Chapter 2 Quadratic Equations Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 2 Quadratic Equations.

Class 10 Maths Part 1 Practice Set 2.1 Chapter 2 Quadratic Equations Questions With Answers Maharashtra Board

Question 1.
Write any two quadratic equations.
Solution:
i. y² – 7y + 12 = 0
ii. x² – 8 = 0

Question 2.
Decide which of the following are quadratic
i. x² – 7y + 2 = 0
ii. y² = 5y – 10
iii. y² + \(\frac { 1 }{ y } \) = 2
iv. x + \(\frac { 1 }{ x } \) = -2
v. (m + 2) (m – 5) = 03
vi. m³ + 3m² – 2 = 3m³
Solution:
i. The given equation is x² + 5x – 2 = 0
Here, x is the only variable and maximum index of the variable is 2.
a = 1, b = 5, c = -2 are real numbers and a ≠ 0.
∴ The given equation is a quadratic equation.

ii. The given equation is
y² = 5y – 10
∴ y² – 5y + 10 = 0
Here, y is the only variable and maximum index of the variable is 2.
a = 1, b = -5, c = 10 are real numbers and a ≠ 0.
∴ The given equation is a quadratic equation.

iii. The given equation is
y² + \(\frac { 1 }{ y } \) = 2
∴ y³ + 1 = 2y …[Multiplying both sides by y]
∴ y³ – 2y + 1 = 0
Here, y is the only variable and maximum index of the variable is not 2.
∴ The given equation is not a quadratic equation.

iv. The given equation is
x + \(\frac { 1 }{ x } \) = -2
∴ x² + 1 = -2x …[Multiplying both sides by x]
∴ x² + 2x+ 1 = 0
Here, x is the only variable and maximum index of the variable is 2.
a = 1, b = 2, c = 1 are real numbers and a ≠ 0.
∴ The given equation is a quadratic equation.

v. The given equation is
(m + 2) (m – 5) = 0
∴ m(m – 5) + 2(m – 5) = 0
∴ m² – 5m + 2m – 10 = 0
∴ m² – 3m – 10 = 0
Here, m is the only variable and maximum index of the variable is 2.
a = 1, b = -3, c = -10 are real numbers and a ≠ 0.
∴ The given equation is a quadratic equation.

vi. The given equation is
m³ + 3m² – 2 = 3m³
∴ 3m³ – m³ – 3m² + 2 = 0
∴ 2m³ – 3m² + 2 = 0
Here, m is the only variable and maximum
index of the variable is not 2.
∴ The given equation is not a quadratic equation.

Question 3.
Write the following equations in the form ax² + bx + c = 0, then write the values of a, b, c for each equation.
i. 2y = 10 – y²
ii. (x – 1)² = 2x + 3
iii. x² + 5x = – (3 – x)
iv. 3m² = 2m² – 9
v. P (3 + 6p) = – 5
vi. x² – 9 = 13
Solution:
i. 2y – 10 – y²
∴ y² + 2y – 10 = 0
Comparing the above equation with
ay² + by + c = 0, we get
a = 1, b = 2, c = -10

ii. (x – 1)² = 2x + 3
∴ x² – 2x + 12x + 3
x² – 2x + 1 – 2x – 30
∴ x² – 4x – 2 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = -4, c = -2

iii. x² + 5x = – (3 – x)
∴ x² + 5x = -3 + x
∴ x² + 5x – x + 3 = 0
∴ x² + 4x + 3 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = 4, c = 3

iv. 3m² = 2m² – 9
∴ 3m² – 2m² + 9 = 0
∴ m² + 9 = 0
∴ m² + 0m + 9 = 0
Comparing the above equation with
am² + bm + c = 0, we get
a = 1, b = 0, c = 9

v. p (3 + 6p) = – 5
∴ 3p + 6p² = -5
∴ 6p² + 3p + 5 = 0
Comparing the above equation with
ap² + bp + c = 0, we get
a = 6, b = 3, c = 5

vi. x² – 9 = 13
∴ x² – 9 – 13 = 0
∴ x² – 22 = 0
∴ x² + 0x – 22 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = 0, c = -22

Question 4.
Determine whether the values given against each of the quadratic equation are the roots of the equation.
i. x² + 4x – 5 = 0; x = 1,-1
ii. 2m² – 5m = 0; m = 2, \(\frac { 5 }{ 2 } \)
Solution:
i. The given equation is
x² + 4x – 5 = 0 …(i)
Putting x = 1 in L.H.S. of equation (i), we get
L.H.S. = (1)² + 4(1) – 5 = 1 + 4 – 5 = 0
∴ L.H.S. = R.H.S.
∴ x = 1 is the root of the given quadratic equation.
Putting x = -1 in L.H.S. of equation (i), we get
L.H.S. = (-1)² + 4(-1) – 5 = 1 – 4 – 5 = -8
∴ LH.S. ≠ R.H.S.
∴ x = -1 ¡s not the root of the given quadratic equation.

ii. The given equation is
2m² – 5m = 0 …(i)
Putting m = 2 in L.H.S. of equation (i), we get
L.H.S. = 2(2)² – 5(2) = 2(4) -10 = 8 – 10 = -2
∴ L.H.S. ≠ R.H.S.
∴ m = 2 is not the root of the given quadratic equation.
Putting m = \(\frac { 5 }{ 2 } \) in L.H.S. of equation (i), we get

Question 5.
Find k if x = 3 is a root of equation kx² – 10x + 3 = 0.
Solution:
x = 3 is the root of the equation kx² – 10x + 3 = 0.
Putting x = 3 in the given equation, we get
k(3)² – 10(3) + 3 = 0
∴ 9k – 30 +3 = 0
∴ 9k – 27 = 0
∴ 9k = 27
∴ k = \(\frac { 27 }{ 9 } \)
∴ k = 3

Question 6.
One of the roots of equation 5m² + 2m + k = 0 is \(\frac { -7 }{ 5 } \) Complete the following activity to find the value of ‘k’.
Solution:

Question 1.
x² + 3x – 5, 3x² – 5x, 5x²; Write the polynomials In the index form. Observe the coefficients and fill in the boxes. (Textbook p. no. 31)
Answer:
Index form of the given polynomials:
x² + 3x – 5, 3x² – 5x + 0, 5x² + 0x + 0
i. Coefficients of x2 are [1], [3] and [5] respectively, and these coefficients are non zero.
ii. Coefficients of x are 3, [-5] and [0] respectively.
iii. Constant terms are [-5], [0] and [0] respectively.
Here, constant terms of second and third polynomial is zero.

Question 2.
Complete the following table (Textbook p. no. 31)

Answer:

Question 3.
Decide which of the following are quadratic equations? (Textbook pg. no. 31)
i. 9y² + 5 = 0
ii. m³ – 5m² + 4 = 0
iii. (l + 2)(l – 5) = 0
Solution:
i. In the equation 9y² + 5 = 0, [y] is the only variable and maximum index of the variable is [2].
∴ It [is] a quadratic equation.

ii. In the equation m³ – 5m² + 4 = 0, [m] is the only variable and maximum index of the variable is not 2.
∴ It [is not] a quadratic equation.

iii. (l + 2)(l – 5) = 0
∴ l(l – 5) + 2(l – 5) = 0
∴ l² – 5l + 2l – 10 = 0
∴ l² – 3l – 10 = 0.
In this equation [l] is the only variable and maximum index of the variable is [2]
∴ it [is] a quadratic equation.

Question 4.
If x = 5 is a root of equation kx² – 14x – 5 = 0, then find the value of k by completing the following activity. (Textbook pg, no. 33)
Solution:

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

10th Standard Maths 1 Practice Set 1.5 Chapter 1 Linear Equations in Two Variables Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.5 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 1 Linear Equations in Two Variables.

Class 10 Maths Part 1 Practice Set 1.5 Chapter 1 Linear Equations in Two Variables Questions With Answers Maharashtra Board

Question 1.
Two numbers differ by 3. The sum of twice the smaller number and thrice the greater number is 19. Find the numbers.
Solution:
Let the greater number be x and the smaller number be y.
According to the first condition, x – y = 3 …(i)
According to the second condition,
3x + 2y = 19 …(ii)
Multiplying equation (i) by 2, we get
2x – 2y = 6 …(iii)
Adding equations (ii) and (iii), we get

Substituting x = 5 in equation (i), we get
5 – y = 3
∴ 5 – 3 = y
∴ y = 2
∴ The required numbers are 5 and 2.

Question 2.
Complete the following.

Solution:
Opposite sides of a rectangle are equal.
∴ 2x + y + 8 = 4x – y
∴ 8 = 4x – 2x – y – y
∴ 2x – 2y = 8
∴ x – y = 4 …(i)[Dividingboth sides by 2]
Also, x + 4= 2y
∴ x – 2y = -4 …(ii)
Subtracting equation (ii) from (i), we get

Substituting y = 8 in equation (i), we get
x – 8 = 4
∴ x = 4 + 8
∴ x = 12
Now, length of rectangle = 4x – y
= 4(12) – 8
= 48 – 8
∴ Length of rectangle = 40
Breadth of rectangle = 2y = 2(8) = 16
Perimeter of rectangle = 2(length + breadth)
= 2(40 + 16)
= 2(56)
∴ Perimeter of rectangle =112 units
Area of rectangle = length × breadth
= 40 × 16
∴ Area of rectangle = 640 sq. units
∴ x = 12 and y = 8, Perimeter of rectangle is 112 units and area of rectangle is 640 sq. units.

Question 3.
The sum of father’s age and twice the age of his son is 70. If we double the age of the father and add it to the age of his son the sum is 95. Find their present ages.
Solution:
Let the present ages of father and son be x years and y years respectively.
According to the first condition,
x + 2y = 70 …(i)
According to the second condition,
2x + y = 95 …(ii)
Multiplying equation (i) by 2, we get
2x + 4y = 140 …(iii)
Subtracting equation (ii) from (iii), we get

Substituting y = 15 in equation (i), we get
x + 2(15) = 7O
⇒ x + 30 = 70
⇒ x = 70 – 30
∴ x = 40
∴ The present ages of father and son are 40 years and 15 years respectively.

Question 4.
The denominator of a fraction is 4 more than twice its numerator. Denominator becomes 12 times the numerator, if both the numerator and the denominator are reduced by 6. Find the fraction.
Solution:
Let the numerator of the fraction be x and the denominator be y.
∴ Fraction = \(\frac { x }{ y } \)
According to the first condition,
y = 2x + 4
∴ 2x – y = -4 …(i)
According to the second condition,
(y – 6)= 12(x – 6)
∴ y – 6 = 12x – 72
∴ 12x – y = 72 – 6
∴ 12x – y = 66 …(ii)
Subtracting equation (i) from (ii), we get

Question 5.
Two types of boxes A, B ,are to be placed in a truck having capacity of 10 tons. When 150 boxes of type A and 100 boxes of type B are loaded in the truck, it weights 10 tons. But when 260 boxes of type A are loaded in the truck, it can still accommodate 40 boxes of type B, so that it is fully loaded. Find the weight of each type of box.
Solution:
Let the weights of box of type A be x kg and that of box of type B be y kg.
1 ton = 1000 kg
∴ 10 tons = 10000 kg
According to the first condition,
150x + 100y = 10000
∴ 3x + 2y = 200 …(i) [Dividing both sides by 50]
According to the second condition,
260x + 40y = 10000
∴ 13x + 2y = 500 …(ii) [Dividing both sides by 20]
Subtracting equation (i) from (ii), we get

∴ The weights of box of type A is 30 kg and that of box of type B is 55 kg.

Question 6.
Out of 1900 km, Vishal travelled some distance by bus and some by aeroplane. Bus travels with average speed 60 km/hr and the average speed of aeroplane is 700 km/hr. It takes 5 hours to complete the journey. Find the distance Vishal travelled by bus.
Solution:
Let the distance Vishal travelled by bus be x km and by aeroplane be y km.
According to the first condition,
x + y = 1900 …(i)
\(\text { Time }=\frac{\text { Distance }}{\text { Speed }} \)
∴ Time required to cover x km by bus = \(\frac { x }{ 60 } \) hr
Time required to cover y km by aeroplane
= \(\frac { y }{ 700 } \) hr
According to the second condition,

Multiplying equation (i) by 6, we get
6x + 6y= 11400 …(iii)
Subtracting equation (iii) from (ii), we get

∴ The distance Vishal travelled by bus is 150 km.

Question 1.
There are some instructions given below. Frame the equations from the information and write them in the blank boxes shown by arrows. (Textbook pg. no. 20)
Answer:

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

10th Standard Maths 1 Problem Set 1 Chapter 1 Linear Equations in Two Variables Textbook Answers Maharashtra Board

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

Class 10 Maths Part 1 Problem Set 1 Chapter 1 Linear Equations in Two Variables Questions With Answers Maharashtra Board

Choose correct alternative for each of the following questions.

Question 1.
To draw graph of 4x + 5y = 19, find y when x = 1.
(a) 4
(b) 3
(c) 2
(d) -3
Answer:
(b)

Question 2.
For simultaneous equations in variables x and y, D_x = 49, Dy = – 63, D = 7 then what is x?
(a) 7
(b) -7
(c) \(\frac { 1 }{ 7 } \)
(d) \(\frac { -1 }{ 7 } \)
Answer:
(a)

Question 3.
Find the value of

(a) -1
(b) -41
(c) 41
(d) 1
Answer:
(d)

Question 4.
To solvex + y = 3; 3x – 2y – 4 = 0 by determinant method find D.
(a) 5
(b) 1
(c) -5
(d) -1
Answer:
(c)

Question 5.
ax + by = c and mx + n y = d and an ≠ bm then these simultaneous equations have-
(a) Only one common solution
(b) No solution
(c) Infinite number of solutions
(d) Only two solutions.
Answer:
(a)

Question 2.
Complete the following table to draw the graph of 2x – 6y = 3.
Answer:

Question 3.
Solve the following simultaneous equations graphically.
i. 2x + 3y = 12 ; x – y = 1
ii. x – 3y = 1 ; 3x – 2y + 4 = 0
iii. 5x – 6y + 30 = 0; 5x + 4y – 20 = 0
iv. 3x – y – 2 = 0 ; 2x + y = 8
v. 3x + y= 10 ; x – y = 2
Answer:
i. The given simultaneous equations are

The two lines interest at point (3,2).
∴ x = 3 and y = 2 is the solution of the simultaneous equations 2x + 3y = 12 and x – y = 1.

ii. The given simultaneous equations are

The two lines intersect at point (-2, -1).
∴ x = -2 and y = -1 is the solution of the simultaneous equations x – 3y = 1 and 3x – 2p + 4 = 0.

iii. The given simultaneous equations are

The two lines intersect at point (0, 5).
∴ x = 0 and y = 5 is the solution of the simultaneous equations 5x – 6y + 30 = 0 and 5x + 4y – 20 = 0.

iv. The given simultaneous equations are


The two lines intersect at point (2, 4).
∴ x = 2 and y = 4 is the solution of the simultaneous equations 3x – y – 2 = 0 and 2x + y = 8.

v. The given simultaneous equations are

The two lines intersect at point (3, 1).
∴ x = 3 and y = 1 is the solution of the simultaneous equations 3x + y = 10 and x – y = 2.

Question 4.
Find the values of each of the following determinants.

Solution:

Question 5.
Solve the following equations by Cramer’s method.

Solution:
i. The given simultaneous equations are
6x – 3y = -10 …(i)
3x + 5y – 8 = 0
∴ 3x + 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 = 6, b₁ = -3, c₁ = 10 and
a₂ = 3, b₂ = 5, c₂ = 8

ii. The given simultaneous equations are
4m – 2n = -4 …(i)
4m + 3n = 16 …(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 = 4, b₁ = -2, c₁ = -4 and
a₂ = 4, b₂ = 3, c₂ = 16


∴ (m, n) = (1, 4) is the solution of the given simultaneous equations.

iii. The given simultaneous equations are

iv. The given simultaneous equations are
7x + 3y = 15 …(i)
12y – 5x = 39
i.e. -5x + 12y = 39 …(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 = 7, b₁ = 3, c₁ = 15 and
a₂ = -5, b₂ = 12, c₂ = 39

v. The given simultaneous equations are

∴ 4(x + y – 8) = 2(3x – y)
∴ 4x + 4y – 32 = 6x – 2y
∴ 6x – 4x – 2y – 4y = -32
∴ 2x – 6y = -32
∴ x – 3y = -16 …(ii)[Dividing 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 = 1, b₁ = -1, c₁ = -4 and
a₂ = 1, b₂ = -3, c₂ = -16

∴ (x, y) = (2, 6) is the solution of the given simultaneous equations.

Question 6.
Solve the following simultaneous equations:

Answer:
i. The given simultaneous equations are

Subtracting equation (iv) from (iii), we get

∴ (x, y) = (6, – 4) is the solution of the given simultaneous equations.

ii. The given simultaneous equations are

Adding equations (v) and (vi), we get

iii. The given simultaneous equations are


∴ (x, y) = (1, 2) is the solution of the given simultaneous equations.

iv. The given simultaneous equations are

∴ Equations (i) and (ii) become 7q – 2p = 5 …(iii)
8q + 7p = 15 …(iv)
Multiplying equation (iii) by 7, we get
49q – 14p = 35 …(v)
Multiplying equation (iv) by 2, we get
16q + 14p = 30 …(vi)
Adding equations (v) and (vi), we get

Substituting q = 1 in equation (iv), we get
8(1) + 7p = 15
∴ 8 + 7p = 15
∴ 7p = 15 – 8
∴ 7p = 7
∴ p = \(\frac { 7 }{ 7 } \) = 1
∴ (P, q) = (1,1)
Resubstituting the values of p and q, we get
1 = \(\frac { 1 }{ x } \) and 1 = \(\frac { 1 }{ y } \)
∴ x = 1 and y = 1
∴ (x, y) = (1, 1) is the solution of the given simultaneous equations.

v. The given simultaneous equations are

Resubstituting the values of p and q, we get

∴ 3x + 4y = 10 …(v)
and 2x – 3y = 1 …(vi)
Multiplying equation (v) by 3, we get
9x + 12y = 30 …(vii)
Multiplying equation (vi) by 4, we get
8x – 12y = 4 …(viii)
Adding equations (vii) and (viii), we get

Substituting x = 2 in equation (v), we get
3(2) + 4y = 10
⇒ 6 + 4y = 10
⇒ 4y = 10 – 6
⇒ y = 4/4 = 1
∴ y = 1
∴ (x, y) = (2, 1) is the solution of the given simultaneous equations.

Question 7.
Solve the following word problems, i. A two digit number and the number with digits interchanged add up to 143. In the given number the digit in unit’s place is 3 more than the digit in the ten’s place. Find the original number.
Solution:
Let the digit in unit’s place be x
and that in the ten’s place be y.

ii. Kantabai bought 1 \(\frac { 1 }{ 2 } \) kg tea and 5 kg sugar from a shop. She paid ₹ 50 as return fare for rickshaw. Total expense was ₹ 700. Then she realised that by ordering online the goods can be bought with free home delivery at the same price. So, next month she placed the order online for 2 kg tea and 7 kg sugar. She paid ₹ 880 for that. Find the rate of sugar and tea per kg.
Solution:
Let the rate of tea be ₹ x per kg and that of sugar be ₹ y per kg.
According to the first condition,
cost of 1 \(\frac { 1 }{ 2 } \) kg tea + cost of 5 kg sugar + fare for rickshaw = total expense

According to the second condition,
cost of 2 kg tea + cost of 7 kg sugar = total expense
2x + 7y = 880 …(ii)
Multiplying equation (i) by 2, we get
6x + 20y = 2600 …(iii)
Multiplying equation (ii) by 3, we get
6x + 21y = 2640 …(iv)
Subtracting equation (iii) from (iv), we get

∴ The rate of tea is ₹ 300 per kg and that of sugar is ₹ 40 per kg.

iii. To find number of notes that Anushka had, complete the following activity.

Solution:
Anushka had x notes of ₹ 100 and y notes of ₹ 50.
According to the first condition,
100x + 50y = 2500
∴ 2x + y = 50 …(i) [Dividing both sides by 50]
According to the second condition,
100y + 50x = 2000
∴ 2y + x = 40 … [Dividing both sides by 50]
i.e. x + 2y = 40
∴ 2x + 4y = 80 …(ii) [Multiplying both sides by 2]
Subtracting equation (i) from (ii), we get

∴ Anushka had 20 notes of ₹ 100 and 10 notes of ₹ 50.

iv. Sum of the present ages of Manish and Savita is 31, Manish’s age 3 years ago was 4 times the age of Savita. Find their present ages.
Solution:
Let the present ages of Manish and Savita be x years and y years respectively.
According to the first condition,
x + y = 31 …(i)
3 years ago,
Manish’s age = (x – 3) years
Savita’s age = (y – 3) years
According to the second condition,
(x – 3) = 4 (y – 3)
∴ x – 3 = 4y – 12
∴ x – 4y = -12 + 3
∴ x – 4y = -9 …(ii)
Subtracting equation (ii) from (i), we get

∴ x = 31 – 8
∴ x = 23
The present ages of Manish and Savita are 23 years and 8 years respectively.

v. In a factory the ratio of salary of skilled and unskilled workers is 5 : 3. Total salary of one day of both of them is ₹ 720. Find daily wages of skilled and unskilled workers.
Solution:
Let the daily wages of skilled workers be ₹ x
that of unskilled workers be ₹ y.
According to the first condition,

∴ The daily wages of skilled workers is ₹ 450 and that of unskilled workers is ₹ 270.

vi. Places A and B are 30 km apart and they are on a straight road. Hamid travels from A to B on bike. At the same time Joseph starts from B on bike, travels towards A. They meet each other after 20 minutes. If Joseph would have started from B at the same time but in the opposite direction (instead of towards A), Hamid would have caught him after 3 hours. Find the speed of Hamid and Joseph.
Solution:
Let the speeds of Hamid and Joseph be x km/hr andy km/hr respectively.
Distance travelled by Hamid in 20 minutes

According to the first condition,

∴ The speeds of Hamid and Joseph 50 km/hr and 40 km/hr respectively.

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

10th Standard Maths 1 Practice Set 1.4 Chapter 1 Linear Equations in Two Variables Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.4 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 1 Linear Equations in Two Variables.

Class 10 Maths Part 1 Practice Set 1.4 Chapter 1 Linear Equations in Two Variables Questions With Answers Maharashtra Board

Question 1.
Solve the following simultaneous equations.

Solution:
i. The given simultaneous equations are

∴ Equations (i) and (ii) become
2p – 3q = 15 …(iii)
8p + 5q = 77 …(iv)
Multiplying equation (iii) by 4, we get
8p – 12q = 60 …(v)
Subtracting equation (v) from (iv), we get

ii. The given simultaneous equations are



Substituting x = 3 in equation (vi), we get
3 + y = 5
∴ y = 5 – 3 = 2
∴ (x, y) = (3, 2) is the solution of the given simultaneous equations.

iii. The given simultaneous equations are

∴ Equations (i) and (ii) become
27p + 31q = 85 …(iii)
31p + 27q = 89 …(iv)
Adding equations (iii) and (iv), we get

iv. The given simultaneous equations are


Substituting x = 1 in equation (vi), we get
3(1) + y = 4
∴ 3 + y = 4
∴ y = 4 – 3 = 1
∴ (x, y) = (1, 1) is the solution of the given simultaneous equations.

Question 1.
Complete the following table. (Textbook pg. no. 16)
Solution:

Question 2.
In the above table the equations are not linear. Can you convert the equations into linear equations? (Textbook pg. no. 17)
Answer:
Yes, the above given simultaneous equations can be converted to a pair of linear equations by making suitable substitutions.

Steps for solving equations reducible to a pair of linear equations.

• Step 1: Select suitable variables other than those which are in the equations.
• Step 2: Replace the given variables with new variables such that the given equations become linear equations in two variables.
• Step 3: Solve the new simultaneous equations and find the values of the new variables.
• Step 4: By resubstituting the value(s) of the new variables, find the replaced variables which are to be determined.

Question 3.
To solve given equations fill the below boxes suitably. (Text book pg.no. 19)
Answer:

Question 4.
The examples on textbook pg. no. 17 and 18 obtained by transformation are solved by elimination method. If you solve these equations by graphical method and by Cramer’s rule will you get the same answers? Solve and check it. (Textbook pg. no. 18)

Solution:

The two lines intersect at point (1,-1).
∴ p = 1 and q = -1 is the solution of the simultaneous equations 4p + q = 3 and 2p – 3q = 5.
Re substituting the values of p and q, we get

The two lines intersect at point (0, -1).
∴ x = 0 and y = -1 is the solution of the simultaneous equations x – y = 1 and x + y = -1.
∴ (x, y) = (0, -1) is the solution of the given simultaneous equations.

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

10th Standard Maths 1 Practice Set 1.3 Chapter 1 Linear Equations in Two Variables Textbook Answers Maharashtra Board

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.

Class 10 Maths Part 1 Practice Set 1.3 Chapter 1 Linear Equations in Two Variables Questions With Answers Maharashtra Board

Question 1.
Fill in the blanks with correct number.
Solution:

Question 2.
Find the values of following determinants.

Solution:

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

∴ (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

∴ (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

∴ (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

∴ (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


∴ (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

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

Question 2.
Complete the following activity. (Textbook pg. no. 15)
Solution:

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 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