Important questions of Chapter 3 Linear Equations In Two Variables with solutions for class 10 maths are available here for students who are preparing for CBSE board exam. By practising these questions, students will be able to score good marks, as it covers the major topics included in the NCERT syllabus.

Chapter 3 linear equations In two variablesÂ deals with finding the solutions for pairs of such equations which has two variables present in it. These solutions can also be represented in graphs. Let us solve the questions which are important for the final exams of 10th standard.

**Also Check:**

- Important 2 Marks Questions for CBSE 10th Maths
- Important 3 Marks Questions for CBSE 10th Maths
- Important 4 Marks Questions for CBSE 10th Maths

## Class 10 Maths Chapter 3 Important Questions With Answers

**Q.1: The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs.160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs.300. Represent the situation algebraically.**

Solution: Let the cost of 1 kg of apples be â€˜Rs. xâ€™.

And, let the cost of 1 kg of grapes be â€˜Rs. yâ€™.

According to the question, the algebraic representation is

2x + y = 160

And 4x + 2y = 300

For, 2x + y = 160 or y = 160 âˆ’ 2x, the solution table is;

x | 50 | 60 | 70 |

y | 60 | 40 | 20 |

For 4x + 2y = 300 or y = (300 – 4x)/ 2, the solution table is;

x | 70 | 80 | 75 |

y | 10 | -10 | 0 |

Note: Students can also represent these two equations graphically, by using the given points of x-coordinate and y-coordinate.

**Q.2:Â 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 itsÂ 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.

**Q.3:Â On comparing the ratios a _{1}/a_{2}, b_{1}/b_{2}, and c_{1}/c_{2}, find out whether the following pair of linear equations are consistent, or inconsistent.**

**(i) 3x + 2y = 5 ; 2x â€“ 3y = 7 **

**(ii) 2x â€“ 3y = 8 ; 4x â€“ 6y = 9**

Solution:

(i) Given : 3x + 2y = 5 or 3x + 2y – 5 = 0

and 2x â€“ 3y = 7 or 2x â€“ 3y – 7 = 0

Comparing the above equations with a_{1}x + b_{1}y + c_{1}=0

And a_{2}x + b_{2}y + c_{2} = 0

We get,

a_{1Â }= 3, b_{1Â }= 2, c_{1Â }= -5

a_{2Â }= 2, b_{2Â }= -3, c_{2Â }= -7

a_{1/}a_{2Â }= 3/2, b_{1}/b_{2Â }= 2/-3, c_{1}/c_{2Â }= -5/-7 = 5/7

Since, a_{1}/a_{2}â‰ b_{1}/b_{2 }the lines intersect each other at a point and have only one possible solution.

Hence, the equations are consistent.

(ii)Â Given 2x â€“ 3y = 8 and 4x â€“ 6y = 9

Therefore,

a_{1Â }= 2, b_{1Â }= -3, c_{1Â }= -8

a_{2Â }= 4, b_{2Â }= -6, c_{2Â }= -9

a_{1}/a_{2Â }= 2/4 = 1/2, b_{1}/b_{2Â }= -3/-6 = 1/2, c_{1}/c_{2Â }= -8/-9 = 8/9

Since, a_{1}/a_{2}=b_{1}/b_{2}â‰ c_{1}/c_{2}

Therefore, the lines are parallel to each other and they have no possible solution. Hence, the equations are inconsistent.

**Q.4:Â Solve the following pair of linear equations by the substitution method.**

**(i) x + y = 14**

**x â€“ y = 4**

**(ii) 3x â€“ y = 3**

**9x â€“ 3y = 9**

Solution:

(i)Â Given,

x + y = 14 and x â€“ y = 4 are the two equations.

From 1st equation, we get,

x = 14 â€“ y

Now, put the value of x in second equation to get,

(14 â€“ y) â€“ y = 4

14 â€“ 2y = 4

2y = 10

Or y = 5

By the value of y, we can now find the value of x;

âˆµ x = 14 â€“ y

âˆ´ x = 14 – 5

Or x = 9

Hence, x = 9 and y = 5.

(ii) Given,

3x â€“ y = 3 and 9x â€“ 3y = 9 are the two equations.

From 1st equation, we get,

x = (3 + y)/3

Now, substitute the value of x in the given second equation to get,

9[(3 + y)/3] â€“ 3y = 9

â‡’ 3(3+y) â€“ 3y = 9

â‡’ 9 + 3y – 3y = 9

â‡’ 9 = 9

Therefore, y has infinite values and since, x = (3 + y)/3, so x also has infinite values.

**Q.5:Â Solve 2x + 3y = 11 and 2x â€“ 4y = â€“ 24 and hence find the value of â€˜mâ€™ for which y = mx + 3.**

Solution:

2x + 3y = 11â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(i)

2x â€“ 4y = -24â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (ii)

From equation (ii), we get;

x = (11 – 3y)/2 â€¦â€¦â€¦.â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(iii)

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

2[(11 – 3y)/2] â€“ 4y = âˆ’24

11 â€“ 3y â€“ 4y = -24

-7y = -35

y = 5â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(iv)

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

x = (11 – 15)/2 = -4/2 = âˆ’2

Hence, x = -2, y = 5

Also,

y = mx + 3

5 = -2m +3

-2m = 2

m = -1

Therefore, the value of m is -1.

**Q.6: 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 â€¦â€¦â€¦â€¦â€¦â€¦. (i)

3x + 5y = 1750 â€¦â€¦â€¦â€¦â€¦â€¦. (ii)

From (i), we get;

y = (3800 – 7x)/6 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (iii)

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

3x + 5[(3800 – 7x)/6] = 1750

â‡’3x + (9500/3) – (35x/6) = 1750

3x – (35x/6) = 1750 – (9500/3)

(18x – 35x)/6 = (5250 – 9500)/3

â‡’-17x/6 = -4250/3

â‡’-17x = -8500

x = 500

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

y = (3800 – 7 Ã— 500)/6 = 300/6 = 50

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

**Q.7:Â A fraction becomes 9/11 if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.**

Solution:

Let the fraction be x/y.

According to the question,

(x + 2)/(y + 2) = 9/11

11x + 22 = 9y + 18

11x â€“ 9y = -4 â€¦â€¦â€¦â€¦â€¦.. (1)

(x + 3)/(y + 3) = 5/6

6x + 18 = 5y +15

6x â€“ 5y = -3 â€¦â€¦â€¦â€¦â€¦â€¦. (2)

From (1), we get

x = (-4 + 9y)/11 â€¦â€¦â€¦â€¦â€¦.. (3)

Substituting the value of x in (2), we get

6[(-4 + 9y)/11] – 5y = -3

-24 + 54y â€“ 55y = -33

-y = -9

y = 9 â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (4)

Substituting the value of y in (3), we get

x = (-4 + 81)/11 = 77/11 = 7

Hence, the fraction is 7/9.

**Q.8Â Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:**

**(i) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?**

Solution:

Let us assume, the present age of Nuri is x.

And the present age of Sonu is y.

According to the given condition, we can write as;

x â€“ 5 = 3(y â€“ 5)

x â€“ 3y = -10â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(1)

Now,

x + 10 = 2(y +10)

x â€“ 2y = 10â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(2)

Subtract eq. 1 from 2, to get,

y = 20 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(3)

Substituting the value of y in eq.1, we get,

x â€“ 3(20) = -10

x â€“ 60 = -10

x = 50

Therefore,

Age of Nuri is 50 years

Age of Sonu is 20 years.

**(ii) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs.27 for a book kept for seven days, while Susy paid Rs.21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.**

Solution:

Let the fixed charge for the first three days be Rs. A and the charge for each day extra be Rs. B.

According to the information given,

A + 4B = 27 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (i)

A + 2B = 21 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (ii)

When equation (ii) is subtracted from equation (i) we get,

2B = 6

B = 3 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(iii)

Substituting B = 3 in equation (i) we get,

A + 12 = 27

A = 15

Hence, the fixed charge is Rs. 15.

And the Additional charge per day is Rs. 3.

**Q.9:Â Solve the following pair of linear equations by the substitution and cross-multiplication methods: **

**8x + 5y = 9 **

**3x + 2y = 4**

Solution:

8x + 5y = 9 â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(1)

3x + 2y = 4 â€¦â€¦â€¦â€¦â€¦â€¦.â€¦.(2)

From equation (2) we get;

x = (4 â€“ 2y) / 3 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (3)

Substituting this value in equation 1, we get

8[(4 – 2y)/3] + 5y = 9

32 â€“ 16y + 15y = 27

-y = -5

y = 5 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(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 â€“ 9 = 0

3x + 2y â€“ 4 = 0

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

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

âˆ´ x = -2 and y =5.

**Q.10:Â Formulate the following problems as a pair of equations, and hence find their solutions:**

**(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.**

Solution:

(i) Let us consider,

Speed of Ritu is still water = x km/hr

Speed of Stream = y km/hr

Now, speed of Ritu, during,

Downstream = x + y km/hr

Upstream = x â€“ y km/hr

As per the question given,

2(x + y) = 20

Or x + y = 10â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(1)

And, 2(x – y) = 4

Or x â€“ y = 2â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(2)

Adding both the eq.1 and 2, we get,

2x = 12

x = 6

Putting the value of x in eq.1, we get,

y = 4

Therefore,

Speed of Ritu is still water = 6 km/hr

Speed of Stream = 4 km/hr

Your extra questions really helped me 4 d preparation for mid-term examination. ………..Tysm

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