Chapter 1 Linear Equations in Two Variables Practice Set 1.5 Solutions

Let the greater number be x and the smaller number be y. According to the first condition, x – y = 3 …(i) According to the second condition, 3x + 2y = 19 …(ii) Multiplying equation (i) by 2, we get 2x – 2y = 6 …(iii) Adding equations (ii) and (iii), we get Substituting x = 5 in equation (i), we get 5 – y = 3 ∴ 5 – 3 = y ∴ y = 2 ∴ The required numbers are 5 and 2.

Opposite sides of a rectangle are equal. ∴ 2x + y + 8 = 4x – y ∴ 8 = 4x – 2x – y – y ∴ 2x – 2y = 8 ∴ x – y = 4 …(i)[Dividingboth sides by 2] Also, x + 4= 2y ∴ x – 2y = -4 …(ii) Subtracting equation (ii) from (i), we get Substituting y = 8 in equation (i), we get x – 8 = 4 ∴ x = 4 + 8 ∴ x = 12 Now, length of rectangle = 4x – y = 4(12) – 8 = 48 – 8 ∴ Length of rectangle = 40 Breadth of rectangle = 2y = 2(8) = 16 Perimeter of rectangle = 2(length + breadth) = 2(40 + 16) = 2(56) ∴ Perimeter of rectangle =112 units Area of rectangle = length × breadth = 40 × 16 ∴ Area of rectangle = 640 sq. units ∴ x = 12 and y = 8, Perimeter of rectangle is 112 units and area of rectangle is 640 sq. units.

Let the present ages of father and son be x years and y years respectively. According to the first condition, x + 2y = 70 …(i) According to the second condition, 2x + y = 95 …(ii) Multiplying equation (i) by 2, we get 2x + 4y = 140 …(iii) Subtracting equation (ii) from (iii), we get Substituting y = 15 in equation (i), we get x + 2(15) = 7O ⇒ x + 30 = 70 ⇒ x = 70 – 30 ∴ x = 40 ∴ The present ages of father and son are 40 years and 15 years respectively.

Let the numerator of the fraction be x and the denominator be y. ∴ Fraction = According to the first condition, y = 2x + 4 ∴ 2x – y = -4 …(i) According to the second condition, (y – 6)= 12(x – 6) ∴ y – 6 = 12x – 72 ∴ 12x – y = 72 – 6 ∴ 12x – y = 66 …(ii) Subtracting equation (i) from (ii), we get

Let the weights of box of type A be x kg and that of box of type B be y kg. 1 ton = 1000 kg ∴ 10 tons = 10000 kg According to the first condition, 150x + 100y = 10000 ∴ 3x + 2y = 200 …(i) [Dividing both sides by 50] According to the second condition, 260x + 40y = 10000 ∴ 13x + 2y = 500 …(ii) [Dividing both sides by 20] Subtracting equation (i) from (ii), we get ∴ The weights of box of type A is 30 kg and that of box of type B is 55 kg.

Let the distance Vishal travelled by bus be x km and by aeroplane be y km. According to the first condition, x + y = 1900 …(i) ∴ Time required to cover x km by bus = hr Time required to cover y km by aeroplane = hr According to the second condition, Multiplying equation (i) by 6, we get 6x + 6y= 11400 …(iii) Subtracting equation (iii) from (ii), we get ∴ The distance Vishal travelled by bus is 150 km.