**Chapter 2 Quadratic Equations Set 2.6**

## Chapter 2 Quadratic Equations Set 2.6

**Question 1.Product of Pragati’s age 2 years ago and years hence is 84. Find her present age.Solution:**

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

∴ x

^{2}+ 3x – 2x – 6 = 84

∴ x

^{2}+ x – 6 – 84 = 0

∴ x

^{2}+ x – 90 = 0

x

^{2}+ 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.

**Question 2.The sum of squares of two consecutive even natural numbers is 244; find the numbers.Solution:**

Let the first even natural number be x.

∴ the next consecutive even natural number will be (x + 2).

According to the given condition,

x

^{2}+ (x + 2)2 = 244

∴ x

^{2}+ x

^{2}+ 4x + 4 = 244

∴ 2x

^{2}+ 4x + 4 – 244 = 0

∴ 2x

^{2}+ 4x – 240 = 0

∴ x

^{2}+ 2x – 120 = 0 …[Dividing both sides by 2]

∴ x

^{2}+ 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.

**Question 3.In the orange garden of Mr. Madhusudan there are 150 orange trees. The number of trees in each row is 5 more than that in each column. Find the number of trees in each row and each column with the help of following flow chart.**

Solution:

Solution:

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

∴ x

^{2}+ 5x = 150

∴ x

^{2}+ 5x – 150 = 0

v. x

^{2}+ 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.

**Question 4.Vivek is older than Kishor by 5 years. The Find their present ages is Find their Present agesSolution:**

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

∴ x

^{2}+ 5x – 12x – 30 = 0

∴ x

^{2}– 7x – 30 = 0

∴ x

^{2}– 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.

**Question 5.Suyash scored 10 marks more in second test than that in the first. 5 times the score of the second test is the same as square of the score in the first test. Find his score in the first test.Solution:**

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) = x

^{2}

∴ 5x + 50 = x

^{2}

∴ x

^{2}– 5x – 50 = 0

∴ x

^{2}– 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.

**Question 6.‘Mr. Kasam runs a small business of making earthen pots. He makes certain number of pots on daily basis. Production cost of each pot is ₹ 40 more than 10 times total number of pots, he makes in one day. If production cost of all pots per day is ₹ 600, find production cost of one pot and number of pots he makes per day.Solution:**

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

∴ 10x

^{2}+ 40x = 600

∴ 10x

^{2}+ 40x- 600 = 0

∴ x

^{2}+ 4x – 60 = 0 …[Dividing both sides by 10]

∴ x

^{2}+ 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.

**Question 7.Pratik takes 8 hours to travel 36 km downstream and return to the same spot. The speed of boat in still water is 12 km. per hour. Find the speed of water current.Solution:**

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.

**Question 8.Pintu takes 6 days more than those of Nishu to complete certain work. If they work together they finish it in 4 days. How many days would it take to complete the work if they work alone.Solution:**

According to the given condition,

∴ 4(2x + 6) = x(x + 6)

∴ 8x + 24 = x

^{2}+ 6x

∴ x

^{2}+ 6x – 8x – 24 = 0

∴ x

^{2}– 2x – 24 = 0

∴ x

^{2}– 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.

**Question 9.If 460 is divided by a natural number, quotient is 6 more than five times the divisor and remainder is 1. Find quotient and divisor.Solution:**

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.

**Question 10.In the given fig. []ABCD is a trapezium, AB || CD and its area is 33 cm2. From the information given in the figure find the lengths of all sides of the []ABCD. Fill in the empty boxes to get the solution.Solution:**

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