NCERT Solutions Maths Class 10 Chapter 3

# NCERT Solutions Maths Class 10 Chapter 3

## NCERT Solutions Maths Class 10 Exercise 3.5

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

(i) x – 3y – 3 = 0

3x – 9y – 2 = 0

(ii) 2= 5

3+ 2= 8

(iii) 3− 5= 20

6− 10= 40

(iv) − 3– 7 = 0

3− 3– 15 = 0

Solution: (i) − 3– 3 = 0 ….. (1)

3− 9– 2 = 0 ….. (2)

Comparing equation − 3– 3 = 0 with a1x + b1c1 = 0 and 3− 9– 2 = 0 with a2x + b2y + c2 = 0, we get

a1 = 1, b1 = –3, c1 = –3, a2 = 3, b2 = –9, c2 = –2

Now, a1/a2 = 1/3, b1/b2 = 1/3 and c1/c2 = 3/2

Here, a1/a2 = b1/b2 ≠ c1/c2

It means that the two lines are parallel.

Therefore, there is no solution for the given equations.

(ii) 2= 5 ….. (1)

3+ 2= 8 ….. (2)

Comparing equation 2= 5 with a1x + b1y + c1 = 0 and 3+ 2= 8 with a2x + b2y + c2 = 0, we get

a1 = 2, b1 = 1, c1 = –5, a2 = 3, b2 = 2, c2 = –8

Now, a1/a2 = 2/3 and b1/b2 = 1/2

Here, a1/a2 ≠ b1/b2

This means that there is a unique solution for the given equations.

Using cross-multiplication method,

= 2 and = 1

(iii) 3− 5= 20 ….. (1)

6− 10= 40 ….. (2)

Comparing equation 3− 5= 20 with a1x + b1y + c1 = 0 and 6− 10= 40 with a2x + b2y + c2 = 0, we get

a1 = 3, b1 = –5, c1 = –20, a2 = 6, b2 = – 10, c2 = –40

Now, a1/a2 = 1/2, b1/b2 = 1/2 and c1/c2 = 1/2

Here, a1/a2 = b1/b2 = c1/c2

It means the lines coincide with each other.

Hence, there are infinitely many solutions to the given equations.

(iv) − 3– 7 = 0 ….. (1)

3− 3– 15 = 0 ….. (2)

Comparing equation − 3– 7 = 0 with a1x + b1y + c1 = 0 and 3− 3– 15 = 0 with a2x + b2y + c2 = 0, we get

a1 = 1, b1 = –3, c1 = –7, a2 = 3, b2 = –3, c2 = –15

Now, a1/a2 = 1/3 and b1/b2 = 1

Here, a1/a2 ≠ b1/b2

It means that we have a unique solution for the given equations.

= 4 and = –1

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

2+ 3= 7

(− b+ (b= 3– 2

(ii) For which value of k will the following pair of linear equations have no solution?

3= 1

(2− 1)+ (− 1)= 2+ 1

Solution: (i) 2+ 3= 7 ….. (1)

(− b)+ (b)= 3– 2 ….. (2)

Comparing equation 2+ 3– 7 = 0 with a1x + b1y + c1 = 0 and (− b)+ (b)− 3– + 2 = 0 with a2x + b2y + c2 = 0, we get

a1 = 2, b1 = 3, c1 = −7, a2 = (a – b), b2 = (a + b) and c2 = 2 – b – 3a

Linear equations have infinitely many solutions if a1/a2 = b1/b2 = c1/c2

2+ 2= 3− 3and 6 − 3− 9= −7− 7b

= 5b ….. (1) and 2= 4+ 6 ….. (2)

Putting = 5b from equation (1) in equation (2), we get

2(5b) = 4+ 6

10− 4= 6

6= 6

= 1

Putting the value of b in equation (1), we get

= 5= 5(1) = 5

Therefore, = 5 and = 1.

(ii) 3= 1 …… (1)

(2− 1)+ (− 1)= 2+ 1 ….. (2)

Comparing equation 3– 1 = 0 with a1x + b1y + c1 = 0 and (2− 1)+ (− 1)−2– 1 = 0, with a2x + b2y + c2 = 0, we get

a1 = 3, b1 = 1 and c1 = –1, a2 = (2k –1), b2 = (k –1) and c2 = −2− 1

Linear equations have no solution if a1/a2 = b1/b2 ≠ c1/c2

3(− 1) = 2– 1

3– 3 = 2− 1

= 2

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

8+ 5= 9

3+ 2= 4

Solution: Substitution Method:

8+ 5= 9 … (1)

3+ 2= 4 … (2)

From equation (1), we have

5= 9 − 8x

= (9 − 8x)/5

Putting this value of y in equation (2), we get

3+ 2(9 − 8x)/5 = 4

3+ (18 − 16x)/5 = 4

3– 16x/5 = 4 – 18/5

15− 16= 20 – 18

= −2

Putting the value of x in equation (1), we get

8(−2) + 5= 9

5= 9 + 16 = 25

= 5

Therefore, = −2 and = 5.

Cross-multiplication Method:

8+ 5= 9 … (1)

3+ 2= 4 … (2)

= −2 and = 5

4. Form the pair of linear equations in the following problems and find their solutions (if they exist) 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 Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been 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?

(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 the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase 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: (i) Let the fixed monthly charge is Rs x and the charge of food for one day be Rs y.

According to the given conditions, we have

+ 20= 1000 … (1)

and + 26= 1180 … (2)

Subtracting equation (1) from equation (2), we get

6= 180

= 30

Putting the value of y in equation (1), we get

+ 20(30) = 1000

= 1000 – 600

x = 400

Therefore, the fixed monthly charge is Rs 400 and the charge of food for one day is Rs 30.

(ii) Let the numerator of the required fraction be and the denominator be y.

According to the given conditions, we have

3– 3 = y … (1) 4+ 8 … (2)

3– = 3 … (1) 4– = 8 … (2)

Subtracting equation (1) from equation (2), we get

4– − (3− y) = 8 – 3

= 5

Putting the value of x in equation (1), we get

3(5) – = 3

15 – = 3

= 12

Therefore, the numerator is 5 and the denominator is 12.

Thus, fraction = 5/12

(iii) Let the number of correct answers be x and the number of wrong answers be y.

According to the given conditions, we have

3– = 40 … (1)

And, 4− 2= 50 … (2)

From equation (1), = 3− 40

Putting this value of y in equation (2), we get

4– 2(3− 40) = 50

4− 6+ 80 = 50

2= −30

= 15

Putting the value of x in equation (1), we get

3(15) – = 40

45 – = 40

= 45 – 40

= 5

Therefore, the number of correct answers is 15 and the number of wrong answers is 5.

Total number of questions = = 15 + 5 = 20

(iv) Let the speed of the car which starts from point A be x km/h and the speed of the car which starts from point B be y km/h.

When the cars travel in the same direction, relative speed = (xy) km/h

When the cars travel in the opposite directions, relative speed = (x + y) km/h

According to the first condition, we have

T = D/S

5 = 100/(xy)          (Assuming x > y)

5− 5= 100

– = 20 … (1)

According to the second condition, we have

T = D/S

1 = 100/(x + y)

= 100 … (2)

Adding equations (1) and (2), we get

2= 120

= 60 km/h

Putting the value of x in equation (1), we get

60 – = 20

= 60 – 20

y = 40 km/h

Therefore, the speed of the car starting from point A is 60 km/h.

And, the speed of the car starting from point B is 40 km/h.

(v) Let the length of the rectangle be x units and the breadth of the rectangle be y units.

Area = xy square units.

According to the first condition, we have

xy – 9 = (− 5) (+ 3)

xy – 9 = xy + 3− 5– 15

3− 5= 6 … (1)

According to the second condition, we have

xy + 67 = (+ 3) (+ 2)

xy + 67 = xy + 2+ 3+ 6

2+ 3= 61 … (2)

From equation (1), we get

3= 6 + 5y

= (6 + 5y)/3

Putting this value of x in equation (2), we get

2(6 + 5y)/3 + 3= 61

12 + 10y + 9y = 183

19= 171

= 9 units

Putting the value of y in equation (2), we get

2+ 3(9) = 61

2= 61 – 27 = 34

= 17 units

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