Quadratic Equations Exercise 9.11 – Class X

Quadratic Equations Exercise 9.11 – Questions

  1. Find two consecutive positive odd numbers such that the sum of their squares is equal to 130.
  2. Find the whole number such that four times the number subtracted from three times the square of the number makes 15.
  3. The sum of two natural numbers is 8. Determine the numbers, if the sum of their reciprocals is 8/15
  4. A two digit number is such that the product of the digits is 12. When 36 is added to this number the digits interchange their places. Determine the number
  5. Find three consecutive positive integers such that the sum of the square of the first and the product of other two is 154.
  6. The ages of Kavya and Karthik are 11 years and 14 years. In how many years times will the product of their ages be 304?
  7. The age of the mother is twice the square of the age of her son. Eight years hence, the age of the mother will be 4 years more than three times the age of her son. Find their present age.
  8. The area of a rectangle is 56 cm2. If the measure of its base is represented by (x + 5) and the measure of its height by (x – 5), find the dimensions of the rectangle.
  9. The altitude of a triangle is 6 cm greater than its base. If its area is 108cm2 . Find its base and height.
  10. In rhombus ABCD, the diagonals AC and BD intersect at E. If AE = x, BE = (x + 7) and AB = (x + 8), find the lengths of the diagonals AC and BD.
  11. In an isosceles triangle ABC, AB = BC and BD is the altitude to base AC.IF DC = x , BD = 2x – 1 and BC = 2x + 1, fibd the lengths of all three sides of the triangle.
  12. A motor boat whose speed is 15km/hr in still water 30 km downstream and comes back in a total of 4 hours 30 minutes. Determine the speed of the stream.
  13. A dealer sells an article for Rs. 24 and gains as much percent as the cost price of the article. Find the cost price of the article.
  14. Shwetha takes 6 ways less than the number of days taken by Ankitha to complete a piece of work. If both Shwetha and Ankitha together can complete the same work in 4 days, in how many days will Ankitha alone complete the work?

Quadratic Equations Exercise 9.11 – Solutions:

  1. Find two consecutive positive odd numbers such that the sum of their squares is equal to 130.

Solution:

Let one of the odd positive number be x.

The other odd positive number be (x + 2).

The sum of the numbers = x2 + (x + 2)2 = 130

x2 + x2 + 2x + 4 = 130

2x2 + 2x + 4 = 130

2x2 + 2x – 126 = 0

x2 + x – 63 = 0

x2 + 9x – 7x – 63 = 0

x(x + 9) -7(x + 9) = 0

(x – 7)(x + 9) = 0

x = 7 or x = -9

Therefore, one of the odd positive number, x = 7 and the other odd positive number is x + 2 = 7 + 2 = 9

Therefore, two consecutive positive odd numbers 7 and 9.

  1. Find the whole number such that four times the number subtracted from three times the square of the number makes 15.

Solution:

Let the whole number be x.

Four times the number be 4x and three times the number be 3x2

Therefore,

4x – 3x2 + 15 = 0

-3x2 + 4x + 15 = 0

-3x2 + 9x – 5x + 15 = 0

3x(x – 3)-5(x – 3) = 0

(x – 3)(3x – 5) = 0

x = 3 or 3x = 5

x = 3 or x = 5/3

Therefore the whole number is 3.

  1. The sum of two natural numbers is 8. Determine the numbers, if the sum of their reciprocals is 8/15

Solution:

The sum of two natural numbers is 8. i.e., x + y = 8

x = 8 – y

The sum of their reciprocals is 8/15 i.e., 1x + 1/y = 8/15

1x + 1/y = 8/15

y + x/xy = 8/15

(x + y)15 = 8xy

15 x 8 = 8 xy

xy = 15

(8 – y)y = 15

8y – y2 = 15

-y2 + 8y – 15 = 0

-y2 + 3y + 5y – 15 = 0

-y(y – 3)+5(y – 3) = 0

(y – 3)(-y + 5) = 0

y = 3 or –y + 5 = 0

y = 3 or y = 5

Therefore the numbers be 3, 5 .

  1. A two digit number is such that the product of the digits is 12. When 36 is added to this number the digits interchange their places. Determine the number

Solution:

Let ten’s digit of the number = x and its unit digit = y. Then two digit number is 10x + y

xy = 12 and 10x + y + 36 = 10y + x

y = 12/x and 9x + 36 = 9y i.e., x + 4 = y

Substituting y = 12/x in x + 4 = 4; we get,

x + 4 = 12/x

x2 +  4x – 12 = 0

x2 +6x – 2x – 12 = 0

x(x + 6)-2(x + 6) = 0

(x + 6)(x – 2) = 0

x = -6 and x = 2

Since x is a digit, therefore, x = 2 and y = 12/x = 12/2 = 6

Therefore the required number is 10x + y = 10×2 + 6 = 26

  1. Find three consecutive positive integers such that the sum of the square of the first and the product of other two is 154.

Solution:

Let three consecutive numbers be x, x + 1, x + 2

x2 +(x+1)(x+2) = 154

x2 + (x2 + 2x + x + 2) = 154

2x2 + 3x + 2 = 154

2x2 + 3x – 152 = 0

2x2 + 19x – 16x – 152 = 0

x(2x + 19) – 8(2x +19) = 0

(2x + 19)(x – 8) = 0

2x + 19 = 0 or x – 8 = 0

2x = -19 or x = 8

x = –19/2 or x = 8

Therefore, three consecutive numbers be 8, 9 and 10.

  1. The ages of Kavya and Karthik are 11 years and 14 years. In how many years time will the product of their ages be 304?

Solution:

The age of Kavya be 11 years and the age of the Karthik be 14 years.

In x years the product of their ages will be 304. i.e., (x + 11)(x + 14) = 304

x2 + 14x + 11x + 154 = 304

x2 + 25x + 154 – 304 = 0

x2 + 25x + 150 = 0

x2 + 30x – 5x + 150 = 0

x(x + 30) – 5(x + 30) = 0

(x + 30)(x – 5) = 0

x + 30 = 0 or x – 5 = 0

x = -30 or x = 5

Therefore, in 5 years the product of their ages will be 304.

  1. The age of the mother is twice the square of the age of her son. Eight years hence, the age of the mother will be 4 years more than three times the age of her son. Find their present age.

Solution:

Let the age of the mother be x.

Hence the present age of son = 2x2

Given the age of the mother is twice the square of the age of her son i.e., x = 2x2

Age of son after 8 years = (x + 8)years

Age of mother after 8 years = (2x2 + 8) years

Given that, (2x2 + 8) = 3(x + 8) + 4

2x2 + 8 = 3x + 24 + 4

2x2 – 3x + 8 – 28 = 0

2x2 – 3x – 20 =0

2x2 – 8x + 5x – 20 = 0

2x(x – 4) +5(x – 4) = 0

(2x + 5)(x – 4) = 0

x – 4 = 0 or 2x + 5 = 0

x = 4 or 2x = -5

x = 4 or x = –5/2

Thus, x = 4

Hence present age of son is 4 years and present age of  mother is 2x2 = 2(4)2 = 32.

  1. The area of a rectangle is 56 cm2. If the measure of its base is represented by (x + 5) and the measure of its height by (x – 5), find the dimensions of the rectangle.

Solution:

A = 56 m2

56 = (x + 5)(x – 5)

56 = x2 – 5x + 5x – 25

56 = x2 – 25

56 + 25 = x2

81 = x2

x = √81 = 9

Therefore, base of the rectangle = x + 5 = 9 + 5 = 14

height of the rectangle = x – 5 = 9 – 5 = 4

  1. The altitude of a triangle is 6 cm greater than its base. If its area is 108cm2. Find its base and height.

Solution:

Let the base of the triangle be x.

Then altitude of the triangle = x + 6

Area of the triangle = 108 cm2

Formula to find the area of triangle = 1/2 x base x height

108 = 1/2 x(x + 6)

216 = x(x + 6)

216 = x2 + 6x

x2 + 6x – 216 = 0

x2 + 18x – 12x – 216 = 0

x(x + 18) -12(x + 18) = 0

(x – 12)(x + 18) = 0

x – 12 = 0 or x + 18 = 0

x = 12

Let the base of the triangle be 12cm

Then height of the triangle 18cm

  1. In rhombus ABCD, the diagonals AC and BD intersect at E. If AE = x, BE = (x + 7) and AB = (x + 8), find the lengths of the diagonals AC and BD.

Solution:

. In rhombus ABCD, the diagonals AC and BD intersect at E. Therefore, angles at E are right angles. Hence, all the triangles formed by the diagonals AC and BD are right angled triangle.

In right angled triangle ABE, by Pythagoras theorem. We have

AB2 = AE2 + EB2

(x + 8)2 = x2 + (x + 7)2

x2 + 16x + 64 = x2 + x2 + 14x + 49

x2 + 16x + 64 – x2 – x2 – 14x – 49 = 0

-x2 + 2x + 15 = 0

-x2 + 5x – 3x + 15 = 0

x(-x + 5) +3(-x + 5) = 0

(-x + 5)(x + 3) = 0

x + 3 = 0 or –x +5 = 0

x = -3 or x = 5

The diagonal AC = AE + EC = 2(AE) = 2x = 10 cm

The diagonal BD = BE + ED = 2(BE)= 2(x+7) = 2(5 +7) = 2(12) = 24 cm

 

  1. In an isosceles triangle ABC, AB = BC and BD is the altitude to base AC.IF DC = x , BD = 2x – 1 and BC = 2x + 1, find the lengths of all three sides of the triangle.

Solution:

Given, AB = BC = 2x + 1

In an isosceles triangle ABC, AB = BC and BD is the altitude to base AC.

Therefore triangle BCD is a right angled triangle.

By Pythagoras theorem, BC2 = BD2 + CD2

(2x + 1)2 = (2x – 1)2 + x2

4x2 + 4x  + 1 = 4x2 – 4x + 1 + x2

4x2 + 4x  + 1 – 4x2 + 4x – 1 – x2 = 0

-x2 + 8x = 0

x(-x + 8) = 0

-x + 8 = 0

x = 8

We know, AB = BC = 2x + 1 = 2(8) + 1 = 17cm

AC = AD + DC = 2(DC) = 2x = 2(8) = 16 cm

  1. A motor boat whose speed is 15km/hr in still water 30 km downstream and comes back in a total of 4 hours 30 minutes. Determine the speed of the stream.

Solution:

Let the speed of the stream be x km/hr. Then speed downstream = (15 + x) kn/hr

Speed upstream = (15 –x )km/hr

given, it took 4 hours 30 minutes to travel back to same place.

So, we have,

30/15 + x30/15 – x = 41/2

30*30/(15+x)(15-x) = 41/2

900/225 – x^2 = 9/2

900 *2 = (225 – x2)*9

1800 = 2025 – 9x2

9x2 = – 1800 + 2025

9x2 = 225

x2 = 225/9

x2 = 25

x = 5km/hr

  1. A dealer sells an article for Rs. 24 and gains as much percent as the cost price of the article. Find the cost price of the article.

Solution:

Let x be the cost price of the article.

The selling price of the article Rs. 24

x = 24 – x/x * 100

x2 = (24 – x)100

x2 = 2400 – 100 x

x2 + 100x – 2400 = 0

x2 + 120x – 20x – 2400 = 0

x(x + 120) -20(x + 120) = 0

(x + 120)(x – 20) = 0

x = -120 or x = 20

Therefore, cost price of the article is 20 Rs.

  1. Shwetha takes 6 ways less than the number of days taken by Ankitha to complete a piece of work. If both Shwetha and Ankitha together can complete the same work in 4 days, in how many days will Ankitha alone complete the work?

Solution:

Let Shwetha take x – 6 days and Ankitha takes x days.

Shwetha one day work = 1/(x – 6) and Ankitha one day work = 1/x

Given, (A + B) can finish work in 4 day.

1/x + 1/(x – 6)  = 1/4

(x-6)+x/x(x-6) = 1/4

2x -6/x(x – 6) = 1/4

4(2x – 6) = x(x – 6)

8x – 24 = x2 – 6x

x2 – 6x – 8x + 24 = 0

x2 – 14x + 24 = 0

x2 – 12x – 2x + 24 = 0

x(x – 12)-2(x – 12) = 0

(x – 12)(x – 2) = 0

x = 12 days taken by Ankitha alone.