NCERT Solutions for Class 10 Maths Exercise 3.3

NCERT Solutions for Class 10 Maths Exercise 3.3 Class 10 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 10 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

NCERT solutions for Maths Pair of Linear Equations in Two Variables Download as PDF

 NCERT Solutions for Class 10 Maths Pair of Linear Equations in Two Variables

1. Solve the following pair of linear equations by the substitution method.

(i) x + y = 14

x – y = 4

(ii) s – t = 3

(iii) 3x – y = 3

9x − 3y = 9

(iv)0.2x + 0.3y = 1.3

0.4x + 0.5y = 2.3

(v)

(vi)

Ans. (i) x + y = 14 …(1)

x – y = 4 … (2)

x = 4 + y from equation (2)

Putting this in equation (1), we get

4 + y + y = 14

⇒ 2y = 10⇒ y = 5

NCERT Solutions for Class 10 Maths Exercise 3.3

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

x + 5 = 14

⇒ x = 14 – 5 = 9

Therefore, x = 9 and y = 5

(ii) s – t = 3 … (1)

…(2)

Using equation (1), we can say that s = 3 + t

Putting this in equation (2), we get

⇒

⇒ 5t + 6 = 36

⇒ 5t = 30⇒ t = 6

Putting value of t in equation (1), we get

s – 6 = 3⇒ s = 3 + 6 = 9

Therefore, t = 6 and s = 9

(iii) 3x – y = 3 … (1)

9x − 3y = 9 … (2)

Comparing equation 3x – y = 3 with and equation 9x − 3y = 9 with ,

We get

Here

Therefore, we have infinite many solutions for x and y

(iv) 0.2x + 0.3y = 1.3 … (1)

0.4x + 0.5y = 2.3 … (2)

Using equation (1), we can say that

0.2x = 1.3 − 0.3y

⇒ x =

Putting this in equation (2), we get

0.4 + 0.5y = 2.3

⇒ 2.6 − 0.6y + 0.5y = 2.3

⇒ −0.1y = −0.3 ⇒ y = 3

Putting value of y in (1), we get

0.2x + 0.3 (3) = 1.3

⇒ 0.2x + 0.9 = 1.3

⇒ 0.2x = 0.4 ⇒ x = 2

Therefore, x = 2 and y = 3

(v) ……….(1)

……….(2)

Using equation (1), we can say that

x =

NCERT Solutions for Class 10 Maths Exercise 3.3

Putting this in equation (2), we get

⇒

⇒ ⇒ y = 0

Putting value of y in (1), we get x = 0

Therefore, x = 0 and y = 0

(vi) … (1)

… (2)

Using equation (2), we can say that

x =

⇒ x =

Putting this in equation (1), we get

⇒

⇒

⇒

⇒ ⇒ y = 3

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

⇒

⇒

⇒ x = 2

Therefore, x = 2 and y = 3

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NCERT Solutions for Class 10 Maths Exercise 3.3

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

y = mx + 3.

Ans. 2x + 3y = 11 … (1)

2x − 4y = −24 … (2)

Using equation (2), we can say that

2x = −24 + 4y

⇒ x = −12 + 2y

Putting this in equation (1), we get

2 (−12 + 2y) + 3y = 11

⇒ −24 + 4y + 3y = 11

⇒ 7y = 35 ⇒ y = 5

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

2x + 3 (5) = 11

⇒ 2x + 15 = 11

⇒ 2x = 11 – 15 = −4⇒ x = −2

Therefore, x = −2 and y = 5

Putting values of x and y in y = mx + 3, we get

5 = m (−2) + 3

⇒ 5 = −2m + 3

⇒ −2m = 2 ⇒ m = −1

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NCERT Solutions for Class 10 Maths Exercise 3.3

3. Form a pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(iii)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.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

(v) A fraction becomes , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and denominator it becomes . Find the fraction.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Ans. (i) Let first number be x and second number be y.

According to given conditions, we have

x – y = 26 (assuming x > y) … (1)

x = 3y(x > y)… (2)

Putting equation (2) in (1), we get

3y – y = 26

⇒ 2y = 26

⇒ y = 13

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

x = 3y =

Therefore, two numbers are 13 and 39.

(ii) Let smaller angle =x and let larger angle =y

According to given conditions, we have

y = x + 18 … (1)

Also, (Sum of supplementary angles) … (2)

Putting (1) in equation (2), we get

x + x + 18 = 180

⇒ 2x = 180 – 18 = 162

⇒

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

y = x + 18 = 81 + 18 =

Therefore, two angles are .

NCERT Solutions for Class 10 Maths Exercise 3.3

(iii) Let cost of each bat = Rs x and let cost of each ball = Rs y

According to given conditions, we have

7x + 6y = 3800 … (1)

And,3x + 5y = 1750 … (2)

Using equation (1), we can say that

7x = 3800 − 6y ⇒ x =

Putting this in equation (2), we get

3 + 5y = 1750

⇒ + 5y = 1750

⇒

⇒

⇒ 17y = 850 ⇒ y = 50

Putting value of y in (2), we get

3x + 250 = 1750

⇒ 3x = 1500 ⇒ x = 500

Therefore, cost of each bat = Rs 500 and cost of each ball = Rs 50

(iv) Let fixed charge = Rs x and let charge for every km = Rs y

According to given conditions, we have

x + 10y = 105… (1)

x + 15y = 155… (2)

Using equation (1), we can say that

x = 105 − 10y

Putting this in equation (2), we get

105 − 10y + 15y = 155

⇒ 5y = 50 ⇒ y = 10

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

x + 10 (10) = 105

⇒ x = 105 – 100 = 5

Therefore, fixed charge = Rs 5 and charge per km = Rs 10

To travel distance of 25 Km, person will have to pay = Rs (x + 25y)

= Rs (5 + 25 × 10)

= Rs (5 + 250) = Rs 255

NCERT Solutions for Class 10 Maths Exercise 3.3

(v) Let numerator = x and let denominator = y

According to given conditions, we have

… (1)

… (2)

Using equation (1), we can say that

11 (x + 2) = 9y + 18

⇒ 11x + 22 = 9y + 18

⇒ 11x = 9y – 4

⇒ x =

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

6 = 5 (y + 3)

⇒

⇒

⇒

⇒ y = 9

Putting value of y in (1), we get

⇒ x + 2 = 9 ⇒ x = 7

Therefore, fraction =

(vi) Let present age of Jacob = x years

Let present age of Jacob’s son = y years

According to given conditions, we have

(x + 5) = 3 (y + 5) … (1)

And, (x − 5) = 7 (y − 5) … (2)

From equation (1), we can say that

x + 5 = 3y + 15

⇒ x = 10 + 3y

NCERT Solutions for Class 10 Maths Exercise 3.3

Putting value of x in equation (2) we get

10 + 3y – 5 = 7y − 35

⇒ −4y = −40

⇒ y = 10 years

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

x + 5 = 3 (10 + 5) =

⇒ x = 45 – 5 = 40 years

Therefore, present age of Jacob = 40 years and, present age of Jacob’s son = 10 years

NCERT Solutions for Class 10 Maths Exercise 3.3

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