Chapter 2 Quadratic Equations Practice Set 2.6 Solutions

Let the present age of Pragati be x years. ∴ 2 years ago, Age of Pragati = (x – 2) years After 3 years, Age of Pragati = (x + 3) years According to the given condition, (x – 2) (x + 3) = 84 ∴ x(x + 3) – 2(x + 3) = 84 ∴ x2 + 3x – 2x – 6 = 84 ∴ x2 + x – 6 – 84 = 0 ∴ x2 + x – 90 = 0 x2 + 10x – 9x – 90 = 0 ∴ x(x + 10) – 9(x + 10) = 0 ∴ (x + 10)(x – 9) = 0 By using the property, if the product of two numbers is zero, then at least one of them is zero, we get ∴ x + 10 = 0 or x – 9 = 0 ∴ x = -10 or x = 9 But, age cannot be negative. ∴ x = 9 ∴ Present age of Pragati is 9 years.

Let the first even natural number be x. ∴ the next consecutive even natural number will be (x + 2). According to the given condition, x2 + (x + 2)2 = 244 ∴ x2 + x2 + 4x + 4 = 244 ∴ 2x2 + 4x + 4 – 244 = 0 ∴ 2x2 + 4x – 240 = 0 ∴ x2 + 2x – 120 = 0 …[Dividing both sides by 2] ∴ x2 + 12x – 10x – 120 = 0 ∴ x(x + 12) – 10 (x + 12) = 0 ∴ (x + 12) (x – 10) = 0 By using the property, if the product of two numbers is zero, then at least one of them is zero, we get ∴ x + 12 = 0 or x – 10 = 0 ∴ x = -12 or x = 10 But, natural number cannot be negative. ∴ x = 10 and x + 2 = 10 + 2 = 12 ∴ The two consecutive even natural numbers are 10 and 12.

i. Number of trees in a column is x. ii. Number of trees in a row = x + 5 iii. Total number of trees = x x (x + 5) iv. According to the given condition, x(x + 5) = 150 ∴ x2 + 5x = 150 ∴ x2 + 5x – 150 = 0 v. x2 + 15x – 10x – 150 = 0 ∴ x(x+ 15) – 10(x + 15) = 0 ∴ (x + 15)(x – 10) = 0 By using the property, if the product of two numbers is zero, then at least one of them is zero, we get ∴ x + 15 = 0 or x – 10 = 0 ∴ x = -15 or x = 10 But, number of trees cannot be negative. ∴ x = 10 vi. Number of trees in a column is 10. vii. Number of trees in a row = x + 5 = 10 + 5 = 15 ∴ Number of trees in a row is 15.

Let the present age of Kishor be x. ∴ Present age of Vivek = (x + 5) years According to the given condition, ∴ 6(2x + 5) = x(x + 5) ∴ 12x + 30 = x2 + 5x ∴ x2 + 5x – 12x – 30 = 0 ∴ x2 – 7x – 30 = 0 ∴ x2 – 10x + 3x – 30 = 0 ∴ x(x – 10) + 3(x – 10) = 0 ∴ (x – 10)(x + 3) = 0 By using the property, if the product of two numbers is zero, then at least one of them is zero, we get ∴ x – 10 = 0 or x + 3 = 0 ∴ x = 10 or x = – 3 But, age cannot be negative. ∴ x = 10 andx + 5 = 10 + 5 = 15 ∴ Present ages of Kishor and Vivek are 10 years and 15 years respectively.

Let the score of Suyash in the first test be x. ∴ Score in the second test = x + 10 According to the given condition, 5(x + 10) = x2 ∴ 5x + 50 = x2 ∴ x2 – 5x – 50 = 0 ∴ x2 – 10x + 5x – 50 = 0 ∴ x(x – 10) + 5(x – 10) = 0 ∴ (x – 10) (x + 5) = 0 By using the property, if the product of two numbers is zero, then at least one of them is zero, we get ∴ x – 10 = 0 or x + 5 = 0 ∴ x = 10 or x = – 5 But, score cannot be negative. ∴ x = 10 ∴ The score of Suyash in the first test is 10.

Let Mr. Kasam make x number of pots on daily basis. Production cost of each pot = ₹ (10x + 40) According to the given condition, x(10x + 40) = 600 ∴ 10x2 + 40x = 600 ∴ 10x2 + 40x- 600 = 0 ∴ x2 + 4x – 60 = 0 …[Dividing both sides by 10] ∴ x2 + 10x – 6x – 60 = 0 ∴ x(x + 10) – 6(x + 10) = 0 ∴ (x + 10) (x – 6) = 0 By using the property, if the product of two numbers is zero, then at least one of them is zero, we get ∴ x + 10 = 0 or x – 6 = 0 ∴ x = – 10 or x = 6 But, number of pots cannot be negative. ∴ x = 6 ∴ Production cost of each pot = 7(10 x + 40) = ₹ [(10×6)+ 40] = ₹(60 + 40) = ₹ 100 Production cost of one pot is ₹ 100 and the number of pots Mr. Kasam makes per day is 6.

Let the speed of water current be x km/hr. Speed of boat is 12 km/hr. (x < 12) In upstream, speed of the water current decreases the speed of the boat and it is the opposite in downstream. ∴ speed of the boat in upstream = (12 – x) km/hr and speed of the boat in downstream = (12 + x) km/hr. ∴ The speed of water current is 6 km/hr.

Let Nishu take x days to complete the work alone. ∴ Total work done by Nishu in 1 day = Also, Pintu takes (x + 6) days to complete the work alone. ∴ Total work done by Pintu in 1 day = ∴ Total work done by both in 1 day = ( + ) But, both take 4 days to complete the work together. ∴ Total work done by both in 1 day = According to the given condition, ∴ 4(2x + 6) = x(x + 6) ∴ 8x + 24 = x2 + 6x ∴ x2 + 6x – 8x – 24 = 0 ∴ x2 – 2x – 24 = 0 ∴ x2 – 6x + 4x – 24 = 0 ∴ x(x – 6)+ 4(x – 6) = 0 ∴ (x – 6) (x + 4) = 0 By using the property, if the product of two numbers is zero, then at least one of them is zero, we get ∴ x – 6 = 0 or x + 4 = 0 ∴ x = 6 or x = -4 But, number of days cannot be negative, ∴ x = 6 and x + 6 = 6 + 6 = 12 ∴ Number of days taken by Nishu and Pintu to complete the work alone is 6 days and 12 days respectively.

Let the natural number be x. ∴ Divisor = x, Quotient = 5x + 6 Also, Dividend = 460 and Remainder = 1 Dividend = Divisor × Quotient + Remainder By using the property, if the product of two numbers is zero, then at least one of them is zero, we get ∴ x – 9 = 0 or 5x + 51 = 0 ∴ x = 9 or x = But, natural number cannot be negative, ∴ x = 9 ∴ Quotient = 5x + 6 = 5(9) + 6 = 45 + 6 = 51 ∴ Quotient is 51 and Divisor is 9.

Solution: By using the property, if the product of two numbers is zero, then at least one of them is zero, we get