Important Questions Class 10 Maths Chapter 3 Linear Equations in Two Variables

# Important Questions Class 10 Maths Chapter 3 Linear Equations in Two Variables

The mathematical equation which can be shown as ‘ax by c = 0’ where a, b, and c are real numbers and both ‘a’ and ‘b’ are not ‘0’ are called linear equations in two variables ‘x’ and ‘y’. The solution of such an equation is a pair of values, one value for ‘x’ and one value for ‘y’ which causes the equality of the equation on both sides (LHS = RHS). Such an equation can also be plotted geometrically over a sheet of paper or graph. in this article explore, Important Questions Class 10 Maths Chapter 3 Linear Equations in Two Variables.

Students looking for  important questions in class 10 maths Chapter 3 Linear Equations in Two Variables should focus on the following topics:

• Consistency and inconsistency of linear equations
• Infinitely many solutions and unique solution of linear equations
• Algebraic methods of linear equation
• Substitution method
• Elimination method
• Cross multiplication method
• Geometrical/graphical solution of linear equations

## Important Questions Chapter 3 – Linear Equations in Two Variables

Q1. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Solution:

Given, half the perimeter of a rectangular garden = 36 m

so, 2(l + b)/2 = 36

(l + b) = 36 ……….(1)

Given, the length is 4 m more than its width.

Let width = x

And length = x + 4

Substituting this in eq(1), we get;

x + x + 4 = 36

2x + 4 = 36

2x = 32

x = 16

Therefore, the width is 16 m and the length is 16 + 4 = 20 m.

Q2. Use elimination method to find all possible solutions of the following pair of linear equation:

2x + 3y = 8

4x + 6y = 7

Solution:

Given,

2x + 3y = 8….(i)

4x + 6y = 7….(ii)

Multiply Equation (1) by 2 and Equation (2) by 1 to make the coefficients of x equal.

4x + 6y = 16….(iii)

4x + 6y = 7….(iv)

Subtracting (iv) from (iii),

4x + 6y – 4x – 6y = 16 – 7

0 = 9, it is not possible

Therefore, the pair of equations has no solution.

Q3. Five years ago, A was thrice as old as B and ten years later, A shall be twice as old as B. What are the present ages of A and B?

Solution:

Let the present ages of B and A be x years and y years respectively. Then

B’s age 5 years ago = (x – 5) years

and A’s age 5 years ago = (- 5) years

(-5) = 3 (x – 5) = 3x – y = 10 …….(i)

B’s age 10 years hence = (x + 10) years

A’s age 10 years hence = (y + 10) years

y + 10 = 2 (x + 10) = 2x – y = -10 …….. (ii)

On subtracting (ii) from (i) we get x = 20

Putting x = 20 in (i) we get

(3 × 20) – y = 10 ⇒ y = 50

∴ x = 20 and y = 50

Hence, B’s present age = 20 years and A’s present age = 50 years.

Q4. Find the value(s) of k so that the pair of equations x + 2y = 5 and 3x + ky + 15 = 0 has a unique solution.

Solution:

Given,

x + 2y = 5

3x + ky + 15 = 0

Also, given that the pair of equations has a unique solution.

Comparing the given equations with standard form,

a1 = 1, b1 = 2, c1 = -5

a2 = 3, b2 = k, c2 = 15

Condition for unique solution is:

a1/a2 ≠ b1/b2

1/3 ≠ 2/k

k ≠ (2)(3)

k ≠ 6

Thus, for all real values of k except 6, the given pair of equations has a unique solution.

Q5. Solve the following pair of linear equations by the substitution and cross-multiplication methods:

8x + 5y = 9

3x + 2y = 4

Solution:

8x + 5y = 9 …………………..(1)

3x + 2y = 4 ……………….….(2)

From equation (2) we get;

x = (4 – 2y) / 3 ……………………. (3)

Substituting this value in equation 1, we get

8[(4 – 2y)/3] + 5y = 9

32 – 16y + 15y = 27

-y = -5

y = 5 ……………………………….(4)

Substituting this value in equation (2), we get

3x + 10 = 4

3x = -6

x = -2

Thus, x = -2 and y = 5.

Now, Using Cross Multiplication method:

8x +5y – 9 = 0

3x + 2y – 4 = 0

x/(-20 + 18) = y/(-27 + 32 ) = 1/(16 – 15)

-x/2 = y/5 = 1/1

∴ x = -2 and y =5.

Q6. The sum of the digits of a two digit number is 8 and the difference between the number and that formed by reversing the digits is 18. Find the number.

Solution:

Let unit and tens digit be x and y.

∴ Original number = 1x + 10y …(i)

Reversed number = 10x + 1y

According to question,

x + y = 8

⇒ y = 8 – x …(ii)

Also, 1x + 10Oy – (10x + y) = 18

⇒ x + 10y – 10x – y = 18

⇒ 9y – 9x = 18

⇒ y – x = 2 …[Dividing both sides by 9

⇒ 8 – x – x = 2 …[From (it)

⇒ 8 – 2 = 2x

⇒ 2x = 6

From (it), y = 8 – 3 = 5

From (i), Original number = 3 + 10(5) = 53

Q7. Solve the following pairs of equations by reducing them to a pair of linear equations:

1/2x + 1/3y = 2

1/3x + 1/2y = 13/6

Solution:

Given,

1/2x + 1/3y = 2

1/3x + 1/2y = 13/6

Let us assume 1/x = m and 1/y = n , then the equations will change as follows.

m/2 + n/3 = 2

⇒ 3m+2n-12 = 0….(1)

m/3 + n/2 = 13/6

⇒ 2m+3n-13 = 0….(2)

Now, using cross-multiplication method, we get,

m/(-26-(-36) ) = n/(-24-(-39)) = 1/(9-4)

m/10 = n/15 = 1/5

m/10 = 1/5 and n/15 = 1/5

So, m = 2 and n = 3

1/x = 2 and 1/y = 3

x = 1/2 and y = 1/3

Q8. The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for Rs.1750. Find the cost of each bat and each ball.

Solution:

Let the cost of a bat be x and the cost of a ball be y.

According to the question,

7x + 6y = 3800 ………………. (i)

3x + 5y = 1750 ………………. (ii)

From (i), we get;

y = (3800 – 7x)/6 …………………… (iii)

Substituting (iii) in (ii). we get,

3x + 5[(3800 – 7x)/6] = 1750

⇒3x + (9500/3) – (35x/6) = 1750

3x – (35x/6) = 1750 – (9500/3)

(18x – 35x)/6 = (5250 – 9500)/3

⇒-17x/6 = -4250/3

⇒-17x = -8500

x = 500

Putting the value of x in (iii), we get;

y = (3800 – 7 × 500)/6 = 300/6 = 50

Hence, the cost of a bat is Rs 500 and the cost of a ball is Rs 50.

Q9. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

Solution:

2x + 3y = 11…………………………..(i)

2x – 4y = -24………………………… (ii)

From equation (i), we get;

x = (11 – 3y)/2 ……….…………………………..(iii)

Putting the value of x in equation (ii), we get

2[(11 – 3y)/2] – 4y = −24

11 – 3y – 4y = -24

-7y = -35

y = 5……………………………………..(iv)

Putting the value of y in equation (iii), we get;

x = (11 – 15)/2 = -4/2 = −2

Hence, x = -2, y = 5

Also,

y = mx + 3

5 = -2m +3

-2m = 2

m = -1

Therefore, the value of m is -1.

Q10. On reversing the digits of a two digit number, number obtained is 9 less than three times the original number. If difference of these two numbers is 45, find the original number.

Solution:

Let unit’s place digit be x and ten’s place digit bey.

∴ Original number = x + 10y Reversed number = 10x + y

According to the Question,

10x + y = 3(x + 10y) – 9

⇒ 10x + y = 3x + 30y – 9

⇒ 10x + y – 3x – 30y = -9

⇒ 7x – 29y = -9 …(i)

10x + y – (x + 10y) = 45

⇒ 9x – 9y = 45

⇒ x – y = 5 …[Dividing both sides by 9

⇒ x – 5 + y …(ii)

Solving (i),

7x – 29y = -9

7(5 + y) – 29y = -9 …[From (ii)

35+ 7y – 29y = -9

-22y = -9 – 35

-22y = -44 ⇒ y = 44/22 = 2

Putting the value of y in (ii),

x = 5 + 2 = 7

∴ Original number = x + 10y

= 7 + 10(2) = 27