# AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise

## 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
⇒ n2 – n = 20
⇒ n2 – n – 20 = 0
⇒ n2 – 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 x2 – 2x = 8
x2 – 2x – 8 = 0
⇒ x2 – 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 m2? 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 m2 ⇒ x2 + 64 + x2 – 16x = 16 × 2 = 32
⇒ 2x2 – 16x + 64 = 32
⇒ 2x2 – 16x + 32 = 0
⇒ 2(x2 – 8x + 16)= 0
⇒ x2 – 8x + 16 = 0
⇒ x2 – 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) = x2 + 5x
⇒ x2 – 7x – 30 = 0
⇒ x2 – 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 ax2 + bx + c = 0 is $$\frac{-b}{a}$$.
Answer:
Let the Q.E. = ax2 + bx + c = 0 (a ≠ 0)
⇒ ax2 + 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 ax2 + bx + c = 0 is $$\frac{c}{a}$$.
Answer:
Let the Q.E. = ax2 + bx + c = 0 (a ≠ 0)
⇒ ax2 + 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 = $$\frac{2x+1}{x}$$ 105x2 + 84x + 21 = 116x2 + 58x
11x2 – 26x – 21 = 0
11x2 – 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 – 4.9t2, 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 t2
The ball has to reach the ground in ‘t’ seconds, which means Height from ground H = 0
So 29.4 + 24.5t – 4.9t2 = 0 = H
⇒ 4.9 t2 – 24.5t – 29.4 = 0
⇒ 4.9 [t2 – 5t – 6] = 0
∴ t2 – 5t – 6 = 0
⇒ t2 – 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.