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Class ten maths chapter 3 Ncert solutions

Chapter 3: PAIR OF LINEAR EQUATIONS IN TWO VARIABLES NCERT SOLUTIONS

TEXTBOOK’S EXERCISE 3.3

3. Form the 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 ` 3800. Later, she buys 3 bats and 5 balls for ` 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 ` 105 and for a journey of 15 km, the charge paid is ` 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 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 56. 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?

Sol. (i) Let numbers be x and y.

As per condition :

x – y = 26 …(i)

x = 3y …(ii)

Substituting the value of x in equation (i), we have

3y – y = 26

⇒ 2y = 26

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

x = 3 × 13 = 39

Hence, the numbers are 39, 13.

(ii) Let x and y be the required angles.

As per condition :

x + y = 180° …(i)

x = y + 18° …(ii)

Substituting the value of x in equation (i), we have

y + 18° + y = 180°

⇒ 2y = 180° – 18°

⇒ 2y = 162°

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

x = 81° + 18° = 99°

Hence, the required angles are 99°, 81°.

(iii) Let cost of one bat = ` x and cost of one ball = Rs y

As per condition :

7x + 6y = 3800 …(i)

3x + 5y = 1750 …(ii)

From equation (i), we have,

7x = 3800 – 6y

Substituting the value of x in equation (ii), we have,

11400 + 17y = 1750 × 7

⇒ 17y = 12250 – 11400

⇒ 17y = 850, y = 850/17=50

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

Hence, cost of one bat = Rs 500 and cost of one ball = Rs 50

(iv) Let fixed charges = Rs x, and charge per km = Rs y

As per condition

x + 10y = 105 …(i)

x + 15y = 155 …(ii)

From equation (i)

x = 105 – 10y …(iii)

Substituting the value of x in equation (ii), we have,

105 – 10y + 15y = 155

⇒ 5y = 155 – 105 ⇒ 5y = 50

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

x = 105 – 10×10 = 105 – 100 = 5

Hence, fixed charges = Rs 5 and charge per km = Rs 10

Hence for travelling 25 km, a person has to pay

= Rs (10 × 25) + Rs 5 = ` (250 + 5) = Rs 255

(v) Let the fraction be x/y .

As per condition

⇒ 11x + 22 = 9y + 18

⇒ 11x = 9y + 18 – 22

⇒ 11x = 9y – 4

⇒ 6 (x + 3) = 5 (y + 3)

⇒ 6x + 18 = 5y+15 ⇒ 6x–5y = 15 – 18

⇒ 6x – 5y = – 3 …(ii)

Substituting the value of x in equation (ii), we have

⇒ – y – 24 = – 3 × 11

⇒ – y = – 9

y = 9

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

Hence, required fraction is 7/9

(vi) Let present age of Jacob = x years, and present age of his son = y years.

5 years hence age of jacob = (x + 5) years

5 years hence age of his son = (y + 5) years

5 years ago age of Jacob = (x – 5) years

5 years ago age of his son = (y – 5) years

Now, (x + 5) =3 (y + 5)

x + 5 = 3y + 15

x = 3y + 10 …(i)

Also, (x – 5) = 7 (y – 5)

x – 5 = 7y – 35

x – 7y = –30 …(ii)

Substituting the value of x in equation (ii), we have

3y +10 –7 y = – 30

⇒ – 4 y = – 40

y = 10

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

x = 3 × 10 + 10

x = 30 + 10

x = 40

Hence, present age of Jacob = 40 years, and present age of his son = 10 years.

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