CBSE Class 10 Mathematics Chapter 3 - Algebraic methods of solving a pair of linear equations - Exercise 3.2

CBSE Class 10 Mathematics  Chapter 3 - Algebraic methods of solving a pair of linear equations  - Exercise 3.2

Q1.	Solve the following pair of linear equations by the substitution method.

Ans. (i)

Given equation are: x + y = 14 ………………… (i) x – y = 4   ………………… (ii) From equation (ii) x = 4 + y Substitute the value from equation (i) (4 + y) + y = 14 Þ 4 + 2y = 14 Þ 2y = 14 – 4 Þ 2y = 10 Þ y = 5 Put y value in equation (ii) Þ x – 5 = 4 Þ x  = 4 + 5 Þ x  = 9 So, the solution is x = 9 and y = 5

Ans. (ii)

Given equation are: s – t = 3 ………………… (i) 2s + 3t = 36  ………………… (ii) From equation (i) s = 3 + t Substitute the value from equation (ii) 2(3+t) + 3t = 36 Þ 6 + 2t + 3t = 36 Þ 2t + 3t = 36 — 6 Þ 5t = 30 Þ t = 6 Put t value in equation (i) Þ s – 6 = 3 Þ s  = 3 + 6 Þ s  = 9 So, the solution is s = 9 and t = 6

Ans. (iii)

Given equation are: 3x – y = 3 ………………… (i) 9x – 3y = 9 ………………… (ii) From equation (ii) 9x – 3y = 9  (simplified this by dividing 3) 3x – y = 3 Equation (i) and (ii) are same, which means there are infinitely many solutions. This is because both equations represent the same line on the coordinate plane.

Ans. (iv)

Given equation are: 0.2x + 0.3y = 1.3  (by multiply both side by 10) 2x + 3y = 13 ………………… (i) 0.4x + 0.5y = 2.3   (by multiply both side by 10) 4x + 5y = 23 ………………… (ii) From equation (i) 2x + 3y = 13 2x = 13 — 3y Þ 2 (13—3y) + 5y = 23 Þ 26—6y + 5y = 23 Þ  26—y = 23 Þ  y = 26—23 Þ  y = 3 Put y value in equation (i) Þ 2x + 3(3) = 13 Þ 2x + 9 = 13 Þ 2x = 13 — 9 = 4 Þ x  = 2 So, the solution is x = 2 and y = 3

Ans. (v)

Given equation are: √2 x + √3 y = 0 ………………… (i) √3 x — √8 y = 0 ………………… (ii) From equation (i) √2 x + √3 y = 0 √2 x = — √3 y So, the solution is x = 0 and y = 0, which means there are infinitely many solutions. This is because both equations represent the same line on the coordinate plane.

Ans. (vi)

Q2.

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

Ans.

Given equation are: 2x + 3y = 11 ………………… (i) 2x – 4y = – 24   ………………… (ii) From equation (i) 2x + 3y = 11 2x = 11 — 3y Substitute the value from equation (ii) Þ 11 — 3y — 4y = —24 Þ  — 3y — 4y = —24—11 Þ  — 7y = —35 Þ y = 5 Put y value in equation (i) Þ 2x + 3(5) = 11 Þ 2x + 15 = 11 Þ 2x  = 11—15 Þ 2x  = —4 Þ x  = —2 So, the solution is x = —2 and y = 5 As per question we have to find value of m y = mx + 3 by putting the value of x and y 5 = m(—2) + 3 5 = —2m + 3 5 — 3 = —2m 2 = —2m M = —1 So, the value of m = —1

Q3.	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.
Ans.	Let bigger number is x and smaller number is y. Then according to questions condition first: x — y = 26 …………. (i) Condition second: x = 3y ……………… (ii) Put x value from equation (ii) to equation (i) 3y — y = 26 2y = 26 y = 13 put y value in equation (ii) x = 3 (13) x = 39 So, the value of x (bigger number) is 39 and y (smaller number) is 13.

(ii)

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

Ans.

Let bigger angle is x and smaller angle is y. According to question First condition: x — y = 18 …………. (i) because larger angle exceed smaller angle by 180 Condition second : as given angles are supplementary, so sum of angle must be 1800. x + y = 180 …………….. (ii) From equation (i) x = 18 + y substitute from equation (ii) (18 + y) + y = 180 18 + 2y = 180 2y = 180 —18 2y = 162 y = 81 Put y value in equation (i) x—81 = 18 x = 18 + 81 x = 99 so, the value of x (bigger angle) is 99 and y (smaller angle) is 81.

(iii)

The coach of a cricket team buys 7 bats and 6 balls for rupees 3800. Later, she buys 3 bats and 5 balls for rupees 1750. Find the cost of each bat and each ball.

Ans.

let cost of per bat is x and cost of per ball is y. According to questions, condition first: 7x + 6y = 3800 ………….. (i) Second condition: 3x + 5y = 1750 ………….. (ii) From equation (i) 7x = 3800 — 6y substitute from equation (ii) 3(3800—6y) + 35y = 12250 11400—18y + 35y = 12250 —18y + 35y = 12250 — 11400 17y = 850 y = 50 Put y value in equation (i) 7x + 6(50) = 3800 7x + 300 = 3800 7x  = 3800 — 300 7x  = 3500 x = 500 So, the value of x (each bat) is 500 and y (each ball) is 50.

(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 Rupees 105 and for a journey of 15 km, the charge paid is Rupees 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?

Ans.

Let Fix charge (in rupees) is x and charge per km. is y (in rupees). According to questions, condition first: x + 10y = 105 ………….. (i) Second condition: x + 15y = 155 ………….. (ii) First, subtract the (i) equation from the (ii) equation to eliminate x (x + 15y) — (x + 10y) = 155 — 105 5y = 50 y = 10 Put y value in equation (i) x + 10(10) = 105 x + 100 = 105 x  = 105 — 100 x = 5 So, the value of x (Fixed charge) is 5 rupees and y (charge per km.) is 10 rupees. Now, to find out how much a person has to pay for traveling a distance of 25 km, we can use the formula: Total charges = Fixed charges + (Charges per km.) × Distance Put the value in formula Total charges = 5 + 10 × 25 Total Charges = 5 + 250 Total Charges = 255 rupees So, A person has to pay 255 Rupees for traveling 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 5/6. Find the fraction.
Ans.

(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.

Let Jacob’s present age is J and his son present age is S. Then according to questions condition first: Five years hence, the age of Jacob will be three times that of his son J + 5 = 3(S + 5) J + 5 = 3S + 15 J  = 3S + 15 – 5 J  = 3S + 10  …………. (i) Condition second: Five years ago, Jacob's age was seven times that of his son J – 5 = 7(S − 5) J – 5 = 7S − 35 J =  7S – 35 + 5 J =  7S – 30 ……………… (ii) we have two equation (i) and (ii)  for J, we can set them equal to each other: 3S + 10 = 7S – 30 10 + 30 = 7S – 3S 40 = 4S S = 10 (Jacob Son present age) put S value in equation (i) J  = 3(10) + 10 J  = 30 + 10 J  = 40 Therefore, Jacob's present age is 40, and his son's present age is 10.