## AP State Syllabus SSC 10th Class Maths Solutions 5th Lesson Quadratic Equations Optional Exercise

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 5 Quadratic Equations Optional Exercise Textbook Questions and Answers.

### 10th Class Maths 5th Lesson Quadratic Equations Optional Exercise Textbook Questions and Answers

Question 1.

Some points are plotted on a plane. Each point is joined with all remaining points by line segments. Find the number of points if the number of line segments are 10.

Answer:

Number of distinct line segments that can be formed out of n-points = \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\)

Given: No. of line segments

\(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\) = 10

⇒ n^{2} – n = 20

⇒ n^{2} – n – 20 = 0

⇒ n^{2} – 5n + 4n – 20 = 0

⇒ n(n – 5) + 4(n – 5) = 0

⇒ (n – 5) (n + 4) = 0

⇒ n – 5 = 0 (or) n + 4 = 0

⇒ n = 5 (or) -4

∴ n = 5 [n – can’t be negative]

Question 2.

A two digit number is such that the product of its digits, is 8. When 18 is added to the number, they interchange their places. Determine the number.

Answer:

Let the digit in the units place = x

Let the digit in the tens place = y

∴ The number = 10y + x

By interchanging the digits the number becomes 10x + y

By problem (10x + y) – (10y + x) = 18

⇒ 9x – 9y = 18

⇒ 9(x – y) =18

⇒ x – y = \(\frac{18}{9}\) = 2

⇒ y = x – 2

(i.e.) digit in the tens place = x – 2

digit in the units place = x

Product of the digits = (x – 2) x

By problem x^{2} – 2x = 8

x^{2} – 2x – 8 = 0

⇒ x^{2} – 4x + 2x – 8 = 0

⇒ x(x – 4) + 2(x – 4) = 0

⇒ (x – 4) (x + 2) = 0

⇒ x – 4 = 0 (or) x + 2 = 0

⇒ x = 4 (or) x = -2

∴ x = 4 [∵ x can’t be negative]

∴ The number is 24.

Question 3.

A piece of wire 8m in length is cut into twp pieces and each piece is bent into a square. Where should the cut in the wire be made if the sum of the areas of these squares is to be 2 m^{2}?

Answer:

Let the length of the first peice = x m

Then length of the second piece = 8 – x m

∴ Side of the 1st square = \(\frac{x}{4}\) m and

Side of the second square = \(\frac{8-x}{4}\) m

sum of the areas = 2 m^{2}

⇒ x^{2} + 64 + x^{2} – 16x = 16 × 2 = 32

⇒ 2x^{2} – 16x + 64 = 32

⇒ 2x^{2} – 16x + 32 = 0

⇒ 2(x^{2} – 8x + 16)= 0

⇒ x^{2} – 8x + 16 = 0

⇒ x^{2} – 4x – 4x + 16 = 0

⇒ x(x – 4) – 4(x – 4) = 0

⇒ (x – 4) (x – 4) = 0

∴ x = 4

∴ The cut should be made at the centre making two equal pieces of length 4 m, 4 m.

Question 4.

Vinay and Praveen working together can paint the exterior of a house in 6 days. Vinay by himself can complete the job in 5 days less than Praveen. How long will it take Vinay to complete the job by himself?

Answer:

Let the time taken by Vinay to complete the job = x days

Then the time taken by Praveen to complete the job = x + 5 days

Both worked for 6 days to complete a job.

∴ Total Work done by them is

⇒ 6(2x + 5) = x^{2} + 5x

⇒ x^{2} – 7x – 30 = 0

⇒ x^{2} – 10x + 3x – 30 = 0

⇒ x(x – 10) + 3(x – 10) = 0

⇒ (x – 10) (x + 3) = 0

⇒ x – 10 = 0 (or) x + 3 = 0

⇒ x = 10 (or) x = -3

∴ x = 10 (∵ x can’t be negative)

∴ Time taken by Vinay = x = 10 days

Time taken by Praveen = x + 5 = 15 days.

Question 5.

Show that the sum of the roots of a quadratic equation ax^{2} + bx + c = 0 is \(\frac{-b}{a}\).

Answer:

Let the Q.E. = ax^{2} + bx + c = 0 (a ≠ 0)

⇒ ax^{2} + bx = -c

∴ Sum of roots of a Q.E. is \(\frac{-b}{a}\)

Question 6.

Show that the product of the roots of a quadratic equation ax^{2} + bx + c = 0 is \(\frac{c}{a}\).

Answer:

Let the Q.E. = ax^{2} + bx + c = 0 (a ≠ 0)

⇒ ax^{2} + bx = -c

Question 7.

The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2\(\frac{16}{21}\) find the fraction.

Answer:

Let the numerator = x

then denominator = 2x + 1

Then the fraction = \(\frac{x}{2x+1}\)

Its reciprocal = \(\frac{2x+1}{x}\)

105x^{2} + 84x + 21 = 116x^{2} + 58x

11x^{2} – 26x – 21 = 0

11x^{2} – 33x + 7x – 21 = 0

11x (x – 3) + 7 (x – 3) = 0

(x – 3) (11x + 7) = 0

⇒ x – 3 = 0 (or) 11x + 7 = 0

⇒ x = 3 (or) \(\frac{-7}{11}\)

∴ x = 3

Numerator = 3;

Denominator = 2 × 3 + 1 = 7

Fraction = \(\frac{3}{7}\).

Question 8.

A ball is thrown vertically upwards from the top of a building of height 29.4m and with an initial velocity 24.5m/sec. If the height H of the ball from the ground level is given by H = 29.4 + 24.5t – 4.9t^{2}, then find the time taken by the ball to reach the ground.

Answer:

Initial velocity ‘U’ = 24.5

height of the ball from the ground can be expressed as

H = 29.4 + 24.5 t – 4.9 t^{2}

The ball has to reach the ground in ‘t’ seconds, which means Height from ground H = 0

So 29.4 + 24.5t – 4.9t^{2} = 0 = H

⇒ 4.9 t^{2} – 24.5t – 29.4 = 0

⇒ 4.9 [t^{2} – 5t – 6] = 0

∴ t^{2} – 5t – 6 = 0

⇒ t^{2} – 6t + t – 6 = 0

⇒ t(t – 6) + 1 (t – 6) = 0

(t – 6) (t + 1) = 0

⇒ t – 6 = 0

∴ t = 6 or t + 1 = 0

⇒ t = -1 but ‘t’ cannot be negative

So t = 6

it means in 6 seconds of time the ball reaches ground.