**Download NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.5**

**1. Which of the following pairs of linear equations has a equation, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.**

**Solution:** We know that If

a_{1}x+b_{1}y +c_{1 }= 0………(1)

a_{2}x+b_{2}y +c_{2 }= 0………(2)

be two pair of liner equation then

When , we get a unique solution.

When , there are infinitely many solutions.

When , there is no solution.

And if there is a unique solution, then the solution is given by,

**(i) x-3y-3 = 0**

**3x-9y-2=0**

**Solution:** Comparing the given pair of equations by the equations

a_{1}x+b_{1}y +c_{1 }= 0

a_{2}x+b_{2}y +c_{2 }= 0 we have

a_{1 }= 1, b_{1} =-3 c_{1 }= -3

a_{2}= 3, b_{2 }= -9, c_{2 }= -2

Now, ,

,

Therefore we get

Hence the given equations has no solution.

**(ii) 2x+y = 5**

**3x+2y = 8**

**Solution:** The given equations can be written as

2x+y-5 = 0……..(1)

3x+2y-8= 0 ………(2)

Comparing the above pair of equations by the equations

a_{1}x+b_{1}y +c_{1 }= 0

a_{2}x+b_{2}y +c_{2 }= 0 we have

a_{1 }= 2, b_{1} =1 c_{1 }= -5

a_{2}= 3, b_{2 }= 2, c_{2 }= -8

Now, ,

,

Since here , then we get a unique solution

And the solution is given by,

and

x = 2 and y = 1

Hence the solution is x = 2and y = 1

**(iii) 3x-5y = 20**

**6x-10y = 40**

**Solution:** The given equations can be written as

3x-5y-20 = 0……..(1)

6x-10y-40= 0 ………(2)

Comparing the above pair of equations by the equations

a_{1}x+b_{1}y +c_{1 }= 0

a_{2}x+b_{2}y +c_{2 }= 0 we have

a_{1 }= 3, b_{1} = -5 c_{1 }= -20

a_{2}= 6, b_{2 }= -10, c_{2 }= -40

Now, ,

,

Since,

Therefore the given pair of equations has infinitely many solutions.

**(iv) x-3y-7 = 0**

**3x-3y-15 = 0**

**Solution:** Comparing the given pair of equations by the equations

a_{1}x+b_{1}y +c_{1 }= 0

a_{2}x+b_{2}y +c_{2 }= 0 we have

a_{1 }= 1, b_{1} =-3 c_{1 }= -7

a_{2}= 3, b_{2 }= -3, c_{2 }= -15

Now, ,

= 1

Since , then there is a unique solution

And the solution is given by,

and

x==4

y = =-1

Hence the solution is x = 4 and y = -1

**2. For which values of a and b does the following pair of linear equations have been an infinite number of solutions?**

**(i) 2x+3y = 7**

**(a-b)x+(a+b)y = 3a+b-2**

**Solution:** We know that the pair of equations

a_{1}x+b_{1}y +c_{1 }= 0……(1)

a_{2}x+b_{2}y +c_{2 }= 0……..(2) have infinitely many solutions if

The given pair of equations can be written as,

2x+3y-7 = 0……..(3)

(a-b)x+(a+b)y -(3a+b-2) = 0……..(4)

Comparing the equations (3) and (4) with the equations (1) and (2)

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

a_{2}= a-b, b_{2 }= a+b, c_{2 }= -(3a+b-2)

Therefore the given pair of equations have infinitely many solutions if

or,

2a+2b = 3a-3b

a = 5b…….(5)

Also,

9a+3b-6 = 7a+7b

2a-4b = 6……..(6)

Substituting a = 5b from equation (5) in equation (6)

2×5b-4b = 6

6b = 6

b = 1

From equation (5), we have

a = 5b = 5×1 = 5

Hence the required values are a = 5 and b = 1.

**(ii) For which value of k will the following pair of liner equations have no solutions?**

**3x+y = 1**

**(2k-1)x+(k-1)y = 2k+1**

**Solution:** We know that the pair of equations

a_{1}x+b_{1}y +c_{1 }= 0……(1)

a_{2}x+b_{2}y +c_{2 }= 0……..(2) have no solutions if ,

The given pair of equations can be written as,

3x+y-1 = 0……..(3)

(2k-1)x+(k-1)y -(2k+1) = 0……..(4)

Comparing the equations (3) and (4) with the equations (1) and (2)

a_{1 }= 3, b_{1} = 1 c_{1 }= -1

a_{2}= 2k-1, b_{2 }= k-1, c_{2 }= -(2k+1)

Therefore the given pair of equations have no solutions if

or,

3k-3 = 2k-1

k = 3-1

k = 2

Substituting k = 2 in the last two equations

Hence the value k =2 satisfy the condition for having no solution.

Therefore the required value of k is 2.

**3. Solve the following pair of equations by substitution and cross multiplications:**

**8x+5y = 9**

**3x+2y = 4**

**Solution:**

**Substitution method:**

The given equations are

8x+5y = 9……..(1)

3x+2y = 4……..(2)

From the equation (1) we have,

x = ……..(3)

Substituting the equation (3)in equation (2)

+ 2y = 4

27-15y+16y = 32

y = 32-27

y = 5

Substituting the value of y = 5 in the equation (3)

x = -2

Hence the solution is x = -2 and y = 5

**Cross multiplication: **

The given equation can be written as

8x+5y-9 = 0……..(1)

3x+2y-4 = 0……..(2)

Comparing the above equations with the pair of equations

a_{1}x+b_{1}y +c_{1 }= 0

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

We have

a_{1 }= 8, b_{1} = 5 c_{1 }= -9

a_{2}= 3, b_{2 }= 2, c_{2 }= -4

Hence the solution is given by

and

x = -2 and y = 5

Hence the required solution is x = -2 and y = 5.

**4. Form the pair of linear equation in the following problems and find their solutions (if they exists) by any algebraic method.**

**(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay ₹1000 as hostel charges whereas a student B, who takes food for 26 days, pays ₹1180 as hostel charges. Find the fixed charges and the cost of food per day.**

**Solution:** Let x be the fixed charges and y be the cost of food per day.

Then using the given conditions we have,

x+20y = 1000………(1) and

x+26y = 1180………(2)

Subtracting equation (2) from equation (1) we have

6y = 180

y = 30

Substituting y = 30 in the equation (1)

x+20×30 = 1000

x+600 = 1000

x = 400

Hence the fixed charges of the hostel is ₹400 and the cost of food per day is ₹30.

**(ii) A fraction becomes when 1 is subtracted from the numerator and it becomes when 8 is added to its denominator. Find the fraction.**

**Solution:** Let x be the numerator and y be the denominator of a fraction. That is the fraction is .

Then using the given conditions, we get the equations as

=

y = 3x-3………(1)

And

=

y+8 = 4x……..(2)

Substituting the value of y of the equation (1) in the equation (2) we have,

3x-3+8 = 4x

x = 5

Substituting the value x = 5 in the equation (1) we get

y = 3×5-3

= 15-3

= 12

Therefore the fraction is =

**(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and loosing 1 marks for each wrong answer. Had 4 marks for being awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?**

**Solution:** Let Yash attained x questions right and y questions wrong. Then using the given conditions we have,

3x-1y = 40

y = 3x-40…….(1) and

4x-2y = 50…….(2)

Substituting the value of y from equation (1) in equation (2) we get

4x-2(3x-40) = 50

4x-6x+80 = 50

2x = 30

x = 15

Substituting the above value of x in the equation (1)

y = 3×15-40

= 45-40

= 5

Therefore Yash attained 15 questions right and 5 questions wrong.

Hence total number of questions in the test is 15+5 = 20.

**(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in same direction and different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the car?**

**Solution:** Let x km/hr be the speed of the car A and y km/hr be the speed of the car B.

Since total distance is 100 km.

Then relative speed when the cars travel in the same direction is = 20 km/hour.

Hence by the given condition,

x-y = 20……(1)

And when the cars travel towards each other the relative speed is = 100 km/hour.

Hence by the given condition,

x+y = 100…..(2)

Adding the equation (1) and the equation (2) we get

2x = 120

x = 60

Substituting the value of x in the equation (2)

60+y = 100

y = 40

Therefore, the speed of car A is 60 km/hr and speed of car B is 40 km/hr

**(v) The area of a rectangle reduced by 9 square units, if its length reduced by 5 units and breadth is increased by 3 units. If we increased the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.**

**Solution:** Let x be the length and y be the breadth of the rectangle.

Hence area is xy

When length reduced by 5 units and breadth is increased by 3 units, the length of new rectangle is x-5 and breadth is y+3

Area = (x-5)(y+3) = xy+3x-5y-15

Then using the given condition,

xy+3x-5y-15 = xy-9

3x-5y = 6……..(1)

When length increases by 3 units and the breadth increases by 2 units, the length and breadth of the new rectangle becomes (x+3) and (y+2) respectively.

Area = (x+3)(y+2) = xy+2x+3y+6

Hence by the given condition,

xy+2x+3y+6 = xy+67

2x+3y = 51……..(2)

Multiplying the equation (1) by 2 and the equation (2) by 3 we get

6x-10y = 12…….(3) and

6x+9y = 183……(4)

Subtracting the equation (4) from the equation (3) we get

19y = 171

y = 9

Substituting the value of y in the equation (1)

3x-5×9 = 6

3x-45 = 6

3x = 51

x = 17

Therefore, the length of the rectangle is 17 units and the breadth is 9 units.

**Download NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.5**