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AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 5 Quadratic Equations Ex 5.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 5th Lesson Quadratic Equations Exercise 5.3

10th Class Maths 5th Lesson Quadratic Equations Ex 5.3 Textbook Questions and Answers

Question 1.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
i) 2x² + x – 4 = 0
Answer:
Given: 2x² + x – 4 = 0
⇒ 2x² + x = 4
⇒ (√2x)² + x = 4
⇒ (√2x)² + 2.√2.x.\(\frac{1}{2 \sqrt{2}}\) = 4
Now LHS is in the form a² + 2ab
where b = \(\frac{1}{2 \sqrt{2}}\)
Adding b² = \(\left(\frac{1}{2 \sqrt{2}}\right)^{2}\) on both sides we get

ii) 4x² + 4√3x + 3 = 0
Answer:
Given: 4x² + 4√3x + 3 = 0
⇒ 4x² + 4√3x = -3
⇒ (2x)² + 2(2x)√3 = -3
LHS is of the form a² + 2ab where
where b = √3.
∴ Adding b² = (√3)² = 3 on both sides, we get
(2x)² + 2(2x)(√3) + (√3)² = -3 + (√3)²
⇒ (2x + √3)² = -3 + 3 = 0
∴ (2x + √3)² = 0
⇒ 2x + √3 = 0
⇒ 2x = -√3
⇒ x = \(\frac{-\sqrt{3}}{2}\)
∴ The roots are \(\frac{-\sqrt{3}}{2}\), \(\frac{-\sqrt{3}}{2}\).

iii) 5x² – 7x – 6 = 0
Given quardratic equation = 5x² – 7x – 6 = 0
∴ 5x² – 7x – 6
⇒ x² – \(\frac{7}{5}\)x = \(\frac{6}{5}\), it can be re-written as
x² – 2.\(\frac{7}{10}\)x = \(\frac{6}{5}\) now it is in the form
of a² – 2ab where a = x, and b = \(\frac{7}{10}\)
Now adding b² = \(\left(\frac{7}{10}\right)^{2}\) on both sides, we get

Note: If we take the Q.E. as 5x² – 7x + 6 = 0, then we get the T.B. answer.

iv) x² + 5 = -6x
Answer:
The given Q.E. is x² + 5 = -6x
⇒ x² + 6x = -5
⇒ (x)² + 2.(x).3 = -5
Now L.H.S. is of the form a² + 2ab where b = 3.
Adding b² = 3² on both sides we get
x² + 2(x)(3) + 3² = -5 + 3²
(x + 3)² = -5 + 9 = 4
∴ x + 3 = 74 = ± 2
⇒ x = +2 – 3 or – 2 – 3
= -1 or -5 are the roots of the given Q.E.

Question 2.
Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula,
i) 2x² + x – 4 = 0
Answer:
Comparing this Q.E. with ax² + bx + c = 0
a = 2; b = 1; c = -4
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

ii) 4x² + 4√3x + 3 = 0
Answer:
Given: 4x² + 4√3x + 3 = 0
Here a = 4; b = 4√3 ; c = 3

iii) 5x² – 7x – 6 = 0
Answer:
Given: 5x² – 7x – 6 = 0
Here a = 5; b = -7 and c = -6

iv) x² + 5 = -6x
Answer:
Given: x² + 5 = -6x
⇒ x² + 6x + 5 = 0
Here a = 1; b = 6; c = 5

Question 3.
Find the roots of the following equations:
i) x – \(\frac{1}{x}\) = 3, x ≠ 0
Answer:
Given: x – \(\frac{1}{x}\) = 3
⇒ x² + 6x + 5 = 0
⇒ \(\frac{x^{2}-1}{x}\) = 3
⇒ x² – 1 = 3x
⇒ x² – 3x – 1 = 0
Here a = 1; b = -3; c = -1

ii) \(\frac{1}{x+4}\) – \(\frac{1}{x-7}\) = \(\frac{11}{30}\), x ≠ -4, 7
Answer:
Given: \(\frac{1}{x+4}\) – \(\frac{1}{x-7}\) = \(\frac{11}{30}\)

⇒ x² – 3x – 28 = -30
⇒ x² – 3x – 28 + 30 = 0
⇒ x² – 3x + 2 = 0
⇒ x² – 2x – x + 2 = 0
⇒ x(x – 2) – 1(x – 2) = 0
⇒ (x – 2) (x – 1) = 0
⇒ x – 2 = 0 (or) x – 1 = 0
⇒ x = 2 or x = 1
⇒ x = 2 or 1.

Question 4.
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\). Find his present age.
Answer:
Let the present age of Rehman be x years.
3 years ago Rehman’s age = x – 3 and its reciprocal is \(\frac{1}{x-3}\)
Rehman’s age 5 years from now = x + 5 and its reciprocal is \(\frac{1}{x+5}\)
The sum of the reciprocals


⇒ x² + 2x – 15 = 3(2x + 2)
⇒ x² + 2x – 15 = 6x + 6
⇒ x² + 2x – 15 – 6x – 6 = 0
⇒ x² – 4x – 21 =0
⇒ x² – 7x + 3x – 21 =0
⇒ x(x – 7) + 3(x – 7) 0
⇒ (x – 7) (x + 3) = 0
⇒ x – 7 = 0 or x + 3 = 0
⇒ x = 7 or x = -3
But x can’t be negative, x = 7
i.e., Present age of Rehman = 7 years.

Question 5.
In a class test, the sum of Moulika’s marks in Mathematics and English is 30. If she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in the two subjects.
Answer:
Sum of the marks in Mathematics and English = 30
Let Moulika’s marks in Mathematics be x Then her marks in English = 30 – x
If she got 2 more marks in Mathematics, then her marks would be x + 2.
If she got 3 marks less in English then her marks would be 30 – x – 3 = 27 – x
By problem (x + 2) (27 – x) = 210
⇒ x(27 – x) + 2(27 – x) = 210
⇒ 27x – x² + 54 – 2x = 210
⇒ -x² + 25x + 54 = 210
⇒ x² – 25x – 54 + 210 = 0
⇒ x² – 25x + 156 = 0
⇒ x² – 12x – 13x + 156 = 0
⇒ x(x – 12) – 13(x 12) = 0
⇒ (x – 12) (x – 13) = 0
⇒ x – 12 = 0 or x – 13 = 0
⇒ x = 12 or x = 13
If x = 12, then marks in Mathematics = 12 English = 30 – 12 = 18
If x = 13, then marks in Mathematics = 13 English = 30 – 13 = 17

Question 6.
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Answer:
Let the shorter side of the rectangular field = x m.
Then its longer side = x + 30 m.

The diagonal of a rectangle is also the hypotenuse of the lower triangle Here the diagonal = x + 60
∴ By Pythagoras Theorem
(side)² + (side)² = (hypotenuse)²
⇒ (x + 30)² + x² = (x + 60)²
⇒ x² + 60x + 900 + x² = x² + 120x + 3600
⇒ x² – 60x – 2700 = 0
⇒ x² – 90x + 30x – 2700 = 0
⇒ x(x – 90) + 30 (x – 90) = 0
⇒ (x – 90) (x + 30) = 0
⇒ x – 90 = 0 (or) x + 30 = 0
⇒ x = +90 (or) x = -30 But ‘x’ can’t be negative.
∴ x = 90 m
i.e., the shorter side x = 90 m Longer side x + 30 = 90 + 30 = 120 m.

Question 7.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Answer:
Let the large number be x.
8 times larger number = Square of the srnall number = 8x
Square of the larger number = x²
By problem, x² – 8x = 180
⇒ x² – 8x – 180 = 0
⇒ x²– 18x + 10x – 180 = 0
⇒ x(x – 18) + 10(x – 18) = 0
⇒ (x + 10)(x – 18) = 0
⇒ x + 10 = 0 (or) x – 18 = 0
⇒ x = -10 (or) x = 18
If x = 18, then larger number =18;
(small number)2 = 8 × (+18) = 144
∴ Small number = √144 = 12
The numbers are 18, 12
Note: Discard x = -10.

Question 8.
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Answer:
The distance travelled = 360 km.
Let the speed of the train = x kmph.
Time taken to complete a journey = \(\frac{\text { distance }}{\text { speed }}\)

⇒ x² + 5x = 1800
⇒ x² + 5x – 1800 = 0
⇒ x² + 45x – 40x – 1800 = 0
⇒ x(x + 45) – 40(x + 45) = 0
⇒ (x + 45) (x – 40) = 0
x + 45 = 0 or x -40 = 0
x = -45 or x = +40
But x can’t be negative.
∴ The speed of the train = 40 kmph.

Question 9.
Two water taps together can fill a tank in 9\(\frac{3}{8}\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Answer:
Let the time taken to fill the tank by smaller tap = x (hours)
So the part filled by smaller tap in
1 hour = \(\frac{1}{x}\) × \(\frac{75}{8}\) = \(\frac{75}{8x}\) ……. (1)
Again then the time taken to fill the tank by larger tap = (x – 10) hours
∴ the part of tank that can be filled by larger tap alone in one hour of time = \(\frac{1}{x-10}\)
∴ In \(\frac{75}{8}\) hours the part filled by larger tap = \(\frac{75}{8}\left(\frac{1}{x-10}\right)\)
∴ By both taps together

⇒ 150x – 750 = 8x² – 80x
⇒ 8x² – 80x – 150x + 750 = 0
⇒ 8x² – 230x + 750 = 0
⇒ 4x² – 115x + 375 = 0
⇒ 4x² – 100x – 15x + 375 = 0
⇒ 4x(x – 25) – 15(x – 25) = 0
∴ (4x – 15) (x – 25) = 0 15
⇒ 4x = 15, x = \(\frac{15}{4}\) or x = 25
x = 25 hours.
then time taken to fill by larger tap = x – 10 = 25 – 10 = 15 hours
(x cannot be \(\frac{15}{4}\) since we have considered ‘x’ as time taken by smaller tap, which is to be higher one)

Question 10.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/hr more than that of the passenger train, find the average speed of the two trains.
Answer:
Let the speed of the passenger train = x kmph.
Then speed of the express train = x + 11 kmph.
Distance travelled = 132 km
We know that time = \(\frac{\text { distance }}{\text { speed }}\)

⇒ x² + 11x = 13 × 11
⇒ x² + 11x – 1452 = 0
⇒ x² + 44x – 33x – 1452 = 0
⇒ x(x + 44) – 33 (x + 44) = 0
⇒ (x + 44) (x – 33) = 0
⇒ x + 44 = 0 (or) x – 33 = 0
⇒ x = -44 (or) x = 33
But x can’t be negative.
∴ Speed of the passenger train = x = 33 kmph.
Speed of the express train = x + 11 = 44 kmph.

Question 11.
Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24m, find the sides of the two squares.
(OR)
If the sum of the areas of two squares is 468 m² and the difference of their perimeters is 24m, then find the measurements of their sides.
Answer:
Let the side of first square = x m say Then perimeter of the first square = 4x [∵ P = 4 . side]
By problem, perimeter of the second square = 4x + 24 (or) 4x – 24
∴ Side of the second square =

Now sum of the areas of the two squares is given as 468 m²
x² + (x + 6)² = 468
⇒ x² + x² + 12x + 36 = 468
⇒ 2x² + 12x + 36 – 468 = 0
⇒ 2x² + 12x – 432 = 0
⇒ x² + 6x – 216 = 0
⇒ x² + 18x – 12x – 216 = 0
⇒ x(x + 18)- 12(x + 18) = 0
⇒ (x + 18) (x – 12) = 0
⇒ x + 18 = 0 (or) x – 12 = 0
⇒ x = -18 (or) 12
But x can’t be negative.
∴ x = 12
i.e., side of the first square = 12
∴ Perimeter = 4 × 12 = 48
∴ Perimeter of the second square = 48 + 24 = 72
∴ Side of the second square = \(\frac{72}{4}\) = 18 m.
(or)
x² + (x – 6)² = 468
⇒ x² + x² – 12x + 36 = 468
⇒ 2x² – 12x – 432 – 0
⇒ x² – 6x – 216 = 0
⇒ x² – 18x + 12x – 216 = 0
⇒ x(x-18) + 12(x-18) = 0
⇒ (x – 18) (x + 12) = 0
⇒ x – 18 = 0 (or) x + 12 = 0
⇒ x = 18 (or) – 12
But x can’t be negative.
∴ x = 18
i.e., side of the first square = 18 m
∴ Perimeter = 4 × 18 = 72
Perimeter of the second square = 72 – 24 = 48
∴ Side of the second square = \(\frac{48}{4}\) = 12 m.
i.e., In any way, the sides of the squares are 12m, 18m.

Question 12.
If a polygon of ‘n’ sides has \(\frac{1}{2}\)n(n – 3) diagonals. How many sides will a polygon having 65 diagonals? Is there a polygon with 50 diagonals?
Answer:
Given: Number of diagonals of a polygon with n-sides = \(\frac{n(n-3)}{2}\)
No. of diagonals of a given polygon = 65
i.e., \(\frac{n(n-3)}{2}\) = 65
where n is number of sides of the polygon
⇒ n² – 3n = 2 × 65
⇒ n² – 3n – 130 = 0
⇒ n² – 13n + 10n – 130 = 0
⇒ n(n – 13) + 10(n – 13) = 0
⇒ (n – 13) (n + 10) = O
⇒ n – 13 = 0 (or) n + 10 = 0
⇒ n = 13 (or) n = -10
But n can’t be negative.
∴ n = 13 (i.e.) number of sides = 13.
Also to check 50 as the number of diagonals of a polygon
∴ \(\frac{n(n-3)}{2}\) = 50
⇒ n² – 3n = 100
⇒ n² – 3n – 100 = 0
There is no real value of n for which the above equation is satisfied.
∴ There can’t be a polygon with 50 diagonals.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 5 Quadratic Equations Ex 5.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 5th Lesson Quadratic Equations Exercise 5.2

10th Class Maths 5th Lesson Quadratic Equations Ex 5.2 Textbook Questions and Answers

Question 1.
Find the roots of the following quadratic equations by factorisation,
i) x² – 3x – 10 = 0
Answer:
Given: x² – 3x – 10 = 0
x² – 5x + 2x- 10 = 0
⇒ x(x – 5) + 2 (x – 5) = 0
⇒ (x – 5) (x + 2) = 0
⇒ x – 5 = 0 or x + 2 = 0
⇒ x = 5 or x = -2
⇒ x = 5 or -2
are the roots of the given Q.E.

ii) 2x² + x – 6 = 0
Answer:
Given: 2x² + x – 6 = 0
⇒ 2x² + 4x – 3x – 6 = 0
⇒ 2x(x + 2) – 3(x + 2) = 0
⇒ (x + 2) (2x – 3) = 0
⇒ (x + 2) or 2x – 3 = 0
⇒ x = -2 or 2x = 3
⇒ x = -2 or \(\frac{3}{2}\)
are the roots of the given Q.E.

iii) √2x² + 7x + 5√2 =0
Answer:
Given: √2x² + 7x + 5√2 =0
⇒ √2x² + 5x + 2x + 5√2 = 0
⇒ x(√2x + 5)+ √2(√2x + 5) = 0
⇒ (√2x + 5) (x + √2) = 0
⇒ √2x + 5 = 0 or x + √2 = 0
⇒ √2x = -5 or x = -√2
⇒ x = \(\frac{-5}{\sqrt{2}}\) = or -√2
are the roots of √2 the given Q.E.

iv) 2x² – x + \(\frac{1}{8}\) = 0
Answer:
Given: 2x² – x + \(\frac{1}{8}\) = 0
⇒ \(\frac{16 x^{2}-8 x+1}{8}\) = 0
⇒ 16x² – 8x + 1 =0
⇒ 16x² – 4x – 4x + 1 = 0
⇒ 4x(4x – 1) – l(4x – 1) = 0
⇒ (4x – 1) (4x – 1) – 0
⇒ 4x – 1 = 0
⇒ 4x = l
⇒ x = \(\frac{1}{4}\), \(\frac{1}{4}\)
are the roots of given Q.E.

v) 100x² – 20x + 1 = 0
Answer:
Given : 100x² – 20x + 1 =0
⇒ 100x² – 10x – 10x + 1 = 0
⇒ 10x(10x – 1) – l(10x – 1) = 0
⇒ (10x – 1) (10x – l) = 0
⇒ 10x – 1 = 0
⇒ 10x = 1
⇒ x = \(\frac{1}{10}\), \(\frac{1}{10}\)
are the roots of the given Q.E.

vi) x(x + 4) = 12
Answer:
Given: x(x + 4) = 12
⇒ x² + 4x = 12
⇒ x² + 4x – 12 = 0
⇒ x² + 6x – 2x – 12 = 0
⇒ x(x + 6) – 2(x + 6) = 0
⇒ (x + 6) (x – 2) = 0
⇒ x + 6 = 0 or x – 2 = 0
⇒ x = -6 or x = 2
⇒ x = -6 or 2
are the roots of the given Q.E.

vii) 3x² – 5x + 2 = 0
Answer:
Given: 3x² – 5x + 2 = 0
⇒ 3x² – 3x – 2x + 2 = 0
⇒ 3x(x – 1) – 2(x – 1) = 0
⇒ (x – 1) (3x – 2) = 0
⇒ x – 1 = 0 or 3x – 2 = 0
⇒ x = 1 or \(\frac{2}{3}\),
⇒ x = 1 or \(\frac{2}{3}\) are the roots of the given Q.E.

viii) x – \(\frac{3}{x}\) = 2
Answer:
Given: x – \(\frac{3}{x}\) = 2
⇒ \(\frac{x^{2}-3}{x}\) = 2
⇒ x² – 3 = 2x
⇒ x² – 2x – 3 = 0
⇒ x² – 3x + x – 3 = 0
⇒ x(x – 3) + l(x – 3) = 0
⇒ (x – 3) (x + 1) = 0
⇒ (x – 3) = 0 or (x + 1) = 0
⇒ x = 3 or x = -1
⇒ x = 3 or -1 are the roots of the given Q.E.

ix) 3(x – 4)² – 5(x – 4) = 12
Answer:
Take (x – 4) = a, then the given Q.E. reduces to 3a² – 5a = 12
⇒ 3a² – 5a – 12 = 0
⇒ 3a² – 9a + 4a – 12 = 0
⇒ 3a(a – 3) + 4(a – 3) = 0
⇒ (a – 3) (3a + 4) = 0
⇒ a – 3 = 0 or 3a + 4 = 0
⇒ a = 3 or a = \(\frac{-4}{3}\)
but a = x – 4
x – 4 = 3 (or) x – 4 = \(\frac{-4}{3}\)
⇒ x = 7 or x = 4 – \(\frac{-4}{3}\) = \(\frac{8}{3}\)
∴ x = 7 or \(\frac{8}{3}\)
are the roots of the given Q.E.

Question 2.
Find two numbers whose sum is 27 and product is 182.
Answer:
Let a number be x.
Then the other number = 27 – x
Product of the numbers = x(27 – x) = 27x – x²
By problem 27x – x² = 182
⇒ x² – 27x + 182 = 0
⇒ x² – 14x – 13x + 182 = 0
⇒ x(x- 14) – 13(x – 14) = 0
⇒ (x – 13) (x – 14) = 0
⇒ x – 13 = 0 or x – 14 = 0
⇒ x = 13 or 14.
∴ The numbers are 13; 27 – 13 = 14 or 14 and 27 – 14 = 13.

Question 3.
Find two consecutive positive integers, sum of whose squares is 613.
Answer:
Let a positive integer be x.
Then the second integer = x + 1
Sum of the squares of the above integers = x² + (x + 1)²
= x² + x² + 2x + 1
= 2x² + 2x + 1
By problem 2x² + 2x + 1 = 613
⇒ 2x² + 2x – 612 = 0
⇒ x² + x – 306 = 0
⇒ x² + 18x – 17x – 306 = 0
⇒ x(x + 18) – 17(x + 18) = 0
⇒ (x – 17) (x + 18) = 0
⇒ x – 17 = 0 (or) x + 18 = 0
⇒ x = 17 (or) -18,
we do not consider -18
Then the numbers are (17, 17 + 1)
i.e., 17, 18 are the required two consecutive positive integers.

Question 4.
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Answer:
Let the base of the right triangle = x cm
Then its altitude = x – 7 cm
By Pythagoras Theorem

(base)² + (height)² = (hypotenuse)²
⇒ x² + (x – 7)² = 13²
⇒ x² + x² – 14x + 49 = 169 .
⇒ 2x² – 14x + 49 – 169 = 0
⇒ 2x² – 14x – 120 = 0
⇒ x² – 7x – 60 = 0
⇒ x² – 12x + 5x – 60 = 0
⇒ x(x – 12) + 5(x – 12) = 0
⇒ (x – 12) (x + 5) = 0
⇒ x – 12 = 0 (or) x + 5 = 0
⇒ x = 12 (or) x = -5 But x can’t be negative.
∴ x = 12
x – 7 = 12 – 7 = 5
The two sides are 12 cm and 5 cm.

Question 5.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs. 90, find the number of articles produced and the cost of each article.
Answer:
Let the number of articles produced be x.
Then the cost of each article = 2x + 3
Total cost of the articles produced = x [2x + 3] = 2x² + 3x
By problem 2x² + 3x = 90
⇒ 2x² + 3x – 90 = 0
⇒ 2x² + 15x – 12x – 90 = 0
⇒ x (2x + 15) – 6 (2x + 15) = 0
⇒ (2x + 15) (x – 6) = 0
⇒ 2x + 15 = 0 (or) x – 6 = 0
⇒ x = \(\frac{-15}{2}\) or x = 6
But x can’t be negative.
∴ x = 6
2x + 3 = 2 × 6 + 3 = 15
∴ Number of articles produced = 6 Cost of each article = Rs. 15.

Question 6.
Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40 square meters.
Answer:
Let the length of the rectangle = x
Given perimeter = 2(1 + b) = 28
⇒ (1 + b) = \(\frac{28}{2}\) = 14
Breadth of the rectangle = 14 – x
Area = length . breadth = x (14 – x)
= 14x – x²
By problem, 14x – x² = 40.
⇒ x² – 14x + 40 = 0
⇒ x² – 10x – 4x + 40 = 0
⇒ x(x – 10) – 4(x – 10) = 0
⇒ (x – 10) (x – 4) = 0
⇒ x – 10 = 0 (or) x – 4 = 0
⇒ x = 10 (or) 4
∴ Length = 10 m or 4 m
Then breadth = 14 – 10 = 4 m (or) 14 – 4 = 10 m

Question 7.
The base of a triangle is 4 cm longer than its altitude. If the area of the triangle is 48 sq.cm, then find its base and altitude.
Answer:
Let the altitude of the triangle h = x cm
Then its base ‘b’ = x + 4.
Area = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\)(x + 4)(x)
= \(\frac{x^{2}+4 x}{2}\)
By problem \(\frac{x^{2}+4 x}{2}\) = 48
⇒ x² + 4x = 2 × 48
⇒ x² + 4x – 96 = 0
⇒ x² + 12x – 8x – 96 = 0
⇒ x(x + 12) – 8(x + 12) = 0
⇒ (x + 12)(x – 8) = 0
⇒ x + 12 = 0 (or) x – 8 = 0
⇒ x = -12 (or) x = 8
But x can’t be negative.
∴ x = 8 and x + 4 = 8 + 4 = 12
Hence altitude = 8 cm and base = 12 cm.

Question 8.
Two trains leave a railway station at the same time. The first train travels towards west and the second train towards north. The first train travels 5 km/hr faster than the second train. If after two hours they are 50 km. apart, find the average speed of each train.
Answer:
Let the speed of the slower train = x kmph
Then speed of the faster train = x + 5 kmph.

Distance = Speed × Time
Distance travelled by the first train = 2(x + 5) = 2x + 10
Distance travelled by the second train = 2.x = 2x
By Pythagoras Theorem
(hypotenuse)² = (side)² + (side)²
⇒ (2x)² + (2x + 10)2² = 50²
⇒ 4x² + (4x² + 40x + 100) = 2500
⇒ 4x² + 4x² + 40x + 100 = 2500
⇒ 8x² + 40x – 2400 = 0
⇒ x² + 5x – 300 = 0
⇒ x² + 20x – 15x – 300 = 0
⇒ x (x + 20) – 15 (x + 20) = 0
⇒ (x + 20) (x – 15) = 0
∴ x – 15 = 0 (or) x + 20 = 0
⇒ x = 15 (or) – 20
But x can’t be negative.
∴ Speed of the slower train x = 15 kmph.
Speed of the faster train x + 5 = 15 + 5 = 20 kmph.

Question 9.
In a class of 60 students, each boy contributed rupees equal to the number of girls and each girl contributed rupees equal to the number of boys. If the total money then collected was Rs. 1600, how many boys are there in the class?
Answer:
Let the number of boys in the class = x
Then number of girls in the class = 60 – x [∵ total students = 60]
Money contributed by the boys = x(60 – x) = 60x – x² [∵ given]
Money contributed by the girls = (60 – x)x = 60x – x²
∴ Money contributed by the class = 120x – 2x²
By problem 120x -2x² = 1600
⇒ 2x²– 120x + 1600 = 0
⇒ x² – 60x + 800 = 0
⇒ x² – 40x – 20x + 800 = 0
⇒ x(x – 40) – 20 (x – 40) = 0
⇒ (x – 40) (x – 20) = 0
⇒ x = 40 (or) 20
∴ Boys = 40 or 20 Girls = 20 or 40.

Question 10.
A motor boat heads upstream a distance of 24 km on a river whose current is running at 3 km per hour. The trip up and back takes 6 hours. Assuming that the motor boat maintained a constant speed, what was its speed ?
Answer:
Let the speed of the boat in still water be x kmph.
Speed of the current = 3 kmph
Then speed of the boat in upstream = (x – 3) kmph
Speed of the boat in downstream = (x + 3) kmph
By problem total time taken = 6h.

⇒ 24(2x) = 6(x² – 9)
⇒ 8x = x² – 9
⇒ x² – 8x – 9 = 0
⇒ x² – 9x + x-9 = 0
⇒ x (x – 9) + 1 (x – 9) = 6
⇒ (x – 9) (x + 1) = 0
⇒ x – 9 = 0 or x + 1 = 0
x can’t be negative,
∴ x = 9
i.e., speed of the boat in still water = 9 kmph.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 5 Quadratic Equations Ex 5.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 5th Lesson Quadratic Equations Exercise 5.1

10th Class Maths 5th Lesson Quadratic Equations Ex 5.1 Textbook Questions and Answers

Question 1.
Check whether the following are quadratic equations.
i) (x + l)² = 2(x-3)
Answer:
Given: (x + l)² = 2(x – 3)
⇒ x² + 2x + 1 = 2(x – 3) = 2x – 6
⇒ x² + 2x + l – 2x + 6 = 0
⇒ x² + 7 = 0 is a Q.E.

ii) x² – 2x = (-2) (3 – x)
Answer:
Given: x² – 2x = -2(3 – x)
⇒ x² – 2x = -6 + 2x
⇒ x² – 4x + 6 = 0 is a Q.E.

iii) (x-2) (x + 1) = (x- 1) (x + 3)
Answer:
Given: (x – 2) (x + 1) = (x – 1) (x + 3)
⇒ x (x + 1) – 2 (x +1)
= x (x + 3) – 1 (x + 3)
Note : Compare the coefficients of x2 on both sides. If they are equal it is not a Q.E.
⇒ x² + x – 2x – 2 = x² + 3x – x -3
⇒ x² – x – 2 = x² + 2x – 3
⇒ 3x – 1 = 0 is not a Q.E.

iv) (x – 3) (2x + 1) = x(x + 5)
Answer:
Given: (x – 3) (2x + 1) = x(x + 5)
⇒ x (2x + 1) – 3 (2x + 1) = x . x + 5 . x
⇒ 2x² + x – 6x – 3 = x² + 5x
⇒ 2x² – 5x – 3 – x² – 5x = 0
⇒ x² – 10x – 3 = 0 is a Q.E.
(or)
Comparing the coefficients of x2 on both sides.
x . 2x and x . x
⇒ 2x² and x²
2x² ≠ x²
Hence it’s a Q.E.

v) (2x – 1) (x – 3) = (x + 5) (x – 1)
Answer:
Given: (2x – 1) (x – 3) = (x + 5) (x – 1)
⇒ 2x (x – 3) -1 (x – 3) = x (x – 1) + 5(x – 1)
⇒ 2x² – 6x – x + 3 = x² – x + 5x – 5
⇒ 2x² -7x + 3 – x² – 4x + 5 = 0
⇒ x² – 11x + 8 = 0
Hence it’s a Q.E.
(or)
Co.eff. of x² on L.H.S. = 2 × 1 = 2
Co.eff. of x² on R.H.S = 1 × 1 = 1
LHS ≠ RHS Hence it is a Q.E.

vi) x² + 3x + 1 = (x – 2)²
Answer:
Given: x² + 3x + 1 = (x – 2)²
⇒ x² + 3x + 1 = x² – 4x + 4
⇒ 7x – 3 = 0 is not a Q.E.

vii) (x + 2)³ = 2x (x² – 1)
Answer:
Given: (x + 2)³ = 2x(x² – 1)
⇒ x³ + 6x² + 12x + 8 = 2x³ – 2x [∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
⇒ -x³ + 6x² + 14x + 8 = 0
is not a Q.E. [∵ degree = 3]

viii) x³ – 4x² – x + 1 = (x – 2)³
Answer:
Given : x³ – 4x² – x + 1 = (x – 2)³
⇒ x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8
⇒ 6x² – 12x + 8 – 4x² – x + 1 = 0
⇒ 2x² – 13x + 9 = 0 is a Q.E.

Question 2.
Represent the following situations in the form of quadratic equations:
i) The area of a rectangular plot is 528 m². The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.
Answer:
Let the breadth of the rectangular plot be x m.
Then its length (by problem) = 2x + 1.
Area = l . b = (2x + 1) . x = 2x² + x
But area = 528 m² (∵ given)
∴ 2x² + x = 528
⇒ 2x² + x – 528 = 0 where x is the breadth of the rectangle.

ii) The product of two consecutive positive integers is 306. We need to find the integers.
Answer:
Let the consecutive integers be x and x + 1.
Their product = x(x + 1) = x² + x
By problem x² + x = 306
⇒ x² + x – 306 = 0
where x is the smaller integer.

iii) Rohan’s mother is 26 years older than him. The product of their ages after 3 years will be 360 years. We need to find Rohan’s present age.
Answer:
Let the present age of Rohan be x years.
Then age of Rohan’s mother = x + 26
After 3 years:
Age of Rohan would be = x + 3
Rohan’s mother’s age would be = (x + 26) + 3 = x + 29
By problem (x + 3) (x + 29) = 360
⇒ x(x + 29) + 3(x + 29) = 360
⇒ x² + 29x + 3x + 87 = 360
⇒ x² + 32x + 87 – 360 = 0
⇒ x² + 32x – 273 = 0
⇒ x² + 39x – 7x – 273 = 0
⇒ x (x + 39) – 7 (x + 39) = 0
⇒ (x – 7) (x + 39) = 0
⇒ x = 7 or x = -39 ‘x’ being age cannot be negative.
∴ x = Present age of Rohan = 7 years.

iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Answer:
Let the speed of the train be x km/h.
Then time taken to travel a distance of distance of 480 km = \(\frac{\text { distance }}{\text { speed }}\) = \(\frac{480}{x}\)
If the speed is 8km/h less, then time needed to cover the same distance would be \(\frac{480}{x-8}\)

⇒ x² – 8x = 1280
⇒ x² – 8x – 1280 = 0
where x is the speed of the train.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.5 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.5

10th Class Maths 6th Lesson Progressions Ex 6.5 Textbook Questions and Answers

Question 1.
For each geometric progression find the common ratio ‘r’, and then find a_n.
i) 3, \(\frac{3}{2}\), \(\frac{3}{4}\), \(\frac{3}{8}\), …….
Answer:
Given G.P.: 3, \(\frac{3}{2}\), \(\frac{3}{4}\), \(\frac{3}{8}\), …….

ii) 2, -6, 18, -54, …….
Answer:
Given G.P. = 2, -6, 18, -54, …….
a = 2, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{-6}{2}\) = -3
a_n = a . r^n-1 = 2 × (-3)^n-1
∴ r = -3; a_n = 2(-3)^n-1

iii) -1, -3, -9, -27, ……
Given G.P. = -1, -3, -9, -27, ……
a = -1, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{-3}{-1}\) = 3
a_n = a . r^n-1 = (-1) × 3^n-1
∴ r = 3; a_n = (-1) × 3^n-1

iv) 5, 2, \(\frac{4}{5}\), \(\frac{8}{25}\), …….
Given G.P. = 5, 2, \(\frac{4}{5}\), \(\frac{8}{25}\), …….
a = 5, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{2}{5}\)
a_n = a . r^n-1 = 5 × \(\left(\frac{2}{5}\right)^{n-1}\)
∴ r = \(\frac{2}{5}\); a_n = 5\(\left(\frac{2}{5}\right)^{n-1}\)

Question 2.
Find the 10^th and nth term of G.P.: 5, 25, 125,…..
Answer:
Given G.P.: 5, 25, 125,…..
a = 5, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{25}{5}\) = 5
a_n = a . r^n-1 = 5 × (5)^n-1 = 5^1+n-1 = 5ⁿ
a₁₀ = a . r⁹ = 5 × 5⁹ = 5¹⁰
∴ a₁₀ = 5¹⁰; a_n = 5ⁿ

Question 3.
Find the indicated term of each geometric progression.
i) a₁ = 9; r = \(\frac{1}{3}\); find a₇.
Answer:
a_n = a . r^n-1

ii) a₁ = -12; r = \(\frac{1}{3}\); find a₆.
Answer:
a_n = a . r^n-1

Question 4.
Which term of the G.P.
i) 2, 8, 32,….. is 512?
Answer:
Given G.P.: 2, 8, 32,….. is 512
a = 2, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{8}{2}\) = 4
Let the n^th term of G.P. be 512

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2⁹
∴ 2n – 1 = 9
[∵ bases are equal, exponents are also equal]
∴ 2n = 9 + 1 = 10
n = \(\frac{10}{2}\) = 5
∴ 512 is the 5^th term of the given G.P.

ii) √3, 3, 3√3, …….. is 729?
Answer:
Given G.P.: √3, 3, 3√3, …….. is 729
a = √3, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{3}{\sqrt{3}}\) = √3
now a_n = a . r^n-1 = 729
⇒ (√3)(√3)^n-1 = 729
⇒ (√3)ⁿ = 3⁶ = (√3)¹²
⇒ n = 12
So 12th term of GP √3, 3, 3√3, …….. is 729.

iii) \(\frac{1}{3}\), \(\frac{1}{9}\), \(\frac{1}{27}\), ……. is \(\frac{1}{2187}\)?
Answer:
Given G.P.: \(\frac{1}{3}\), \(\frac{1}{9}\), \(\frac{1}{27}\), ……. is \(\frac{1}{2187}\)

Let \(\frac{1}{2187}\) be the n^th term of the G.P., then
a_n = a . r^n-1 = \(\frac{1}{2187}\)

[∵ bases are equal, exponents are also equal]
7^th term of G.P is \(\frac{1}{2187}\).

Question 5.
Find the 12^th term of a G.P. whose 8 term is 192 and the common ratio is 2.
Answer:
Given a G.P. such that a₈ = 192 and r = 2
a_n = a . r^n-1
a₈ = a . (2)^8-1 = 192

= 3 × 2¹⁰ = 3 × 1024 = 3072.

Question 6.
The 4^th term of a geometric progression is \(\frac{2}{3}\) and the seventh term is \(\frac{16}{81}\). Find the geometric series.
Answer:
Given: In a G.P.

Now substituting r = \(\frac{2}{3}\) in equation (1)
we get,

Question 7.
If the geometric progressions 162, 54, 18, ….. and \(\frac{2}{81}\), \(\frac{2}{27}\), \(\frac{2}{9}\),….. have their n^th term equal, find the value of n.
Answer:
Given G.P.: 162, 54, 18, ….. and \(\frac{2}{81}\), \(\frac{2}{27}\), \(\frac{2}{9}\),……

Given that n^th terms are equal
a_n = a . r^n-1

⇒ 3^n-1+n-1 = 81 × 81
⇒ 3^2n-2 = 3⁴ × 3⁴
⇒ 3^2n-2 = 3⁸ [∵ a^m . aⁿ = a^m+n]
⇒ 2n – 2 = 8
[∵ bases are equal, exponents are also equal]
2n = 8 + 2
⇒ n = \(\frac{10}{2}\) = 5
The 5^th terms of the two G.P.s are equal.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.4

10th Class Maths 6th Lesson Progressions Ex 6.4 Textbook Questions and Answers

Question 1.
In which of the following situations, does the list of numbers involved in the form a G.P.?
i) Salary of Sharmila, when her salary is Rs. 5,00,000 for the first year and expected to receive yearly increase of 10% .
Answer:
Given: Sharmila’s yearly salary = Rs. 5,00,000.
Rate of annual increment = 10 %.

Here, a = a₁ = 5,00,000
a₂ = 5,00,000 × \(\frac{11}{10}\) = 5,50,000
a₃ = 5,00,000 × \(\frac{11}{10}\) × \(\frac{11}{10}\) = 6,05,000
a₄ = 5,00,000 × \(\frac{11}{10}\) × \(\frac{11}{10}\) × \(\frac{11}{10}\) = 6,65,000

Every term starting from the second can be obtained by multiplying its pre¬ceding term by a fixed number \(\frac{11}{10}\).
∴ r = common ratio = \(\frac{11}{10}\)
Hence the situation forms a G.P.

ii) Number of bricks needed to make each step, if the stair case has total 30 steps. Bottom step needs 100 bricks and each successive step needs 2 bricks less than the previous step.
Answer:
Given: Bricks needed for the bottom step = 100.
Each successive step needs 2 bricks less than the previous step.
∴ Second step from the bottom needs = 100 – 2 = 98 bricks.
Third step from the bottom needs = 98 – 2 = 96 bricks.
Fourth step from the bottom needs = 96 – 2 = 94 bricks.
Here the numbers are 100, 98, 96, 94, ….
Clearly this is an A.P. but not G.P.

iii) Perimeter of the each triangle, when the mid-points of sides of an equilateral triangle whose side is 24 cm are joined to form another triangle, whose mid-points in turn are joined to form still another triangle and the process continues indefinitely.

Answer:
Given: An equilateral triangle whose perimeter = 24 cm.
Side of the equilateral triangle = \(\frac{24}{3}\) = 8 cm.
[∵ All sides of equilateral are equal] ……. (1)
Now each side of the triangle formed by joining the mid-points of the above triangle in step (1) = \(\frac{8}{2}\) = 4 cm
[∵ A line joining the mid-points of any two sides of a triangle is equal to half the third side.]
Similarly, the side of third triangle = \(\frac{4}{2}\) = 2 cm
∴ The sides of the triangles so formed are 8 cm, 4 cm, 2 cm,
a = 8

Thus each term starting from the second; can be obtained by multiplying its preceding term by a fixed number \(\frac{1}{2}\).
∴ The situation forms a G.P.

Question 2.
Write three terms of the G.P. when the first term ‘a’ and the common ratio ‘r’ are given.
i) a = 4 ; r = 3.
Answer:
The terms are a, ar, ar², ar³, ……..
∴ 4, 4 × 3, 4 × 3² , 4 × 3² , ……
⇒ 4, 12, 36, 108, ……

ii) a = √5 ; r = \(\frac{1}{5}\)
Answer:
The terms are a, ar, ar², ar³, ……..

iii) a = 81 ; r = –\(\frac{1}{3}\)
Answer:
The terms of a G.P are:
a, ar, ar², ar³, ……..

⇒ 81, -27, 9,

iv) a = \(\frac{1}{64}\); r = 2.
Answer:
Given: a = \(\frac{1}{64}\); r = 2.

∴ The G.P is \(\frac{1}{64}\), \(\frac{1}{32}\), \(\frac{1}{16}\), …….

Question 3.
Which of the following are G.P. ? If they are G.P, write three more terms,
i) 4, 8, 16, ……
Answer:
Given: 4, 8, 16, ……
where, a₁ = 4; a₂ = 8; a₃ = 16, ……
\(\frac{a_{2}}{a_{1}}=\frac{8}{4}=2\)
\(\frac{a_{3}}{a_{2}}=\frac{16}{8}=2\)
∴ r = \(\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}\) = 2
Hence 4, 8, 16, … is a G.P.
where a = 4 and r = 2
a₄ = a . r³ = 4 × 2³ = 4 × 8 = 32
a₅ = a . r⁴ = 4 × 2⁴ = 4 × 16 = 64
a₆ = a . r⁵ = 4 × 2⁵ = 4 × 32 = 128

ii) \(\frac{1}{3}\), –\(\frac{1}{6}\), \(\frac{1}{12}\), …….
Answer:
Given: t₁ = \(\frac{1}{3}\), t₂ = –\(\frac{1}{6}\), t₃ = \(\frac{1}{12}\), ….
\(\frac{1}{3}\), –\(\frac{1}{6}\), \(\frac{1}{12}\), …….

Hence the ratio is common between any two successive terms.
∴ \(\frac{1}{3}\), –\(\frac{1}{6}\), \(\frac{1}{12}\), ……. is G.P.
where a = \(\frac{1}{3}\) and r = –\(\frac{1}{2}\)

iii) 5, 55, 555, ……..
Answer:
Given: t₁ = 5, t₂ = 55, t₃ = 555, ….

∴ 5, 55, 555, …….. is not a G.P.

iv) -2, -6, -18, ……
Given: t₁ = -2, t₂ = -6, t₃ = -18

∴ -2, -6, -18, is a G.P.
where a = -2 and r = 3
a_n = a . r^n-1 =
a₄ = a . r³ = (-2) × 3³ = -2 × 27 = -54
a₅ = a . r⁴ = (-2) × 3⁴ = -2 × 81 = -162
a₆ = a . r⁵ = (-2) × 3⁵ = -2 × 243 = -486

v) \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{6}\), …….
Answer:

i.e., \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{6}\), ….. is not a G.P.

vi) 3, -3², 3³, ……
Given: t₁ = 3, t₂ = -3², t₃ = 3³, ……


i.e., every term is obtained by multiplying its preceding term by a fixed number -3.
3, -3², 3³, …… forms a G.P,
where a = 3; r = -3
a_n = a . r^n-1
a₄ = a . r³ = 3 × (-3)³ = 3 × (-27) = -81
a₅ = a . r⁴ = 3 × (-3)⁴ = 3 × 81 = 243
a₆ = a . r⁵ = 3 × (-3)⁵ = 3 × (-243) = -729

vii) x, 1, \(\frac{1}{x}\), …….
Answer:
Given: t₁ = x, t₂ = 1, t₃ = \(\frac{1}{x}\), ……

Hence x, 1, \(\frac{1}{x}\), …. forms a G.P.
where a = x; r = \(\frac{1}{x}\)

viii) \(\frac{1}{\sqrt{2}}\), -2, \(\frac{8}{\sqrt{2}}\), …….
Answer:
Given: t₁ = \(\frac{1}{\sqrt{2}}\), t₂ = -2, t₃ = \(\frac{8}{\sqrt{2}}\), ……

ix) 0.4, 0.04, 0.004, ……..
Answer:
Given: t₁ = 0.4, t₂ = 0.04, t₃ = 0.004, ……

∴ 0.4, 0.04, 0.004, …….. forms a G.P.
where a = 0.4; r = \(\frac{1}{10}\)

Question 4.
Find x so that x, x + 2, x + 6 are consecutive terms of a geometric progression.
Answer:
Given x, x + 2 and x + 6 are in G.P. but read it as x, x + 2 and x + 6.
∴ r = \(\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}\) = \(\frac{\mathrm{t}_{3}}{\mathrm{t}_{2}}\)
⇒ \(\frac{x+2}{x}\) = \(\frac{x+6}{x+2}\)
⇒(x + 2)² = x(x + 6)
⇒ x² + 4x + 4 = x² + 6x
⇒ 4x – 6x = – 4 = -2x = -4
∴ x = 2

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.3

10th Class Maths 6th Lesson Progressions Ex 6.3 Textbook Questions and Answers

Question 1.
Find the sum of the following APs:
i) 2, 7, 12,…, to 10 terms.
Answer:
Given A.P: 2, 7, 12, …… to 10 terms
a = 2; d = a₂ – a₁ = 7 – 2 = 5; n = 10
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S₁₀ = \(\frac{10}{2}\)[2 × 2 + (10 – 1)5]
= 5 [4 + 9 × 5]
= 5 [4 + 45]
= 5 × 49
= 245

ii) -37, -33, -29,…, to 12 terms.
Answer:
Given A.P: -37, -33, -29,…, to 12 terms.
a = -37; d = a₂ – a₁ = (-33) – (-37) = -33 + 37 = 4; n = 12
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S₁₂ = \(\frac{12}{2}\)[2 × (-37) + (12 – 1)4]
= 6 [-74 + 11 × 4]
= 6 [-74 + 44]
= 6 × (-30)
= -180

iii) 0.6, 1.7, 2.8,…, to 100 terms.
Answer:
Given A.P : 0.6, 1.7, 2.8,…. S₁₀₀
a = 0.6; d = a₂ – a₁ = 1.7 – 0.6 = 1.1; n = 100
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S₁₀₀ = \(\frac{100}{2}\)[2 × 0.6 + (100 – 1)1.1]
= 50 [1.2 + 99 × 1.1]
= 50 [1.2 + 108.9]
= 50 × 110.1
= 5505

iv) \(\frac{1}{15}\), \(\frac{1}{12}\), \(\frac{1}{10}\),…, to 11 terms.
Answer:
Given A.P: \(\frac{1}{15}\), \(\frac{1}{12}\), \(\frac{1}{10}\),…, S₁₁
a = \(\frac{1}{15}\); d = a₂ – a₁ = \(\frac{1}{12}\) – \(\frac{1}{15}\) = \(\frac{5-4}{60}\) = \(\frac{1}{60}\); n = 11
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]

2. Find the sums given below =:
i) 7 + 10\(\frac{1}{2}\) + 14 + …. + 84
Answer:
Given A.P : 7 + 10\(\frac{1}{2}\) + 14 + …. + 84
a = 7; d = a₂ – a₁ = 10\(\frac{1}{2}\) – 7 = 3\(\frac{1}{2}\) and the last term l = a_n = 84
But, a_n = a + (n – 1) d
∴ 84 = 7 + (n – 1) 3\(\frac{1}{2}\)
⇒ 84 – 7 = (n – 1) × \(\frac{7}{2}\)
⇒ n – 1 = 77 × \(\frac{2}{7}\) = 22
⇒ n = 22 + 1 = 23
Now, S_n = \(\frac{n}{2}\)(a + l) where a = 7; l = 84
S₂₃ = \(\frac{23}{2}\)(7 + 84)
= \(\frac{23}{2}\) × 91
= \(\frac{2093}{2}\)
= 1046\(\frac{1}{2}\)

ii) 34 + 32 + 30 + … + 10
Answer:
Given A.P: 34 + 32 + 30 + … + 10
a = 34; d = a₂ – a₁ = 32 – 34 = -2 and the last term l = a_n = 10
But, a_n = a + (n – 1) d
∴ 10 = 34 + (n – 1) (-2)
⇒ 10 – 34 = -2n + 2
⇒ -2n = -24 – 2
⇒ n = \(\frac{-26}{-2}\) = 13
∴ n = 13
Also, S_n = \(\frac{n}{2}\)(a + l)
where a = 34; l = 10
S₁₃ = \(\frac{13}{2}\)(34 + 10)
= \(\frac{13}{2}\) × 44
= 13 × 22
= 286

iii) -5 + (-8) + (-11) + … + (-230)
Answer:
Given A.P: -5 + (-8) + (-11) + … + (-230)
Here first term, a = -5;
d = a₂ – a₁ = (-8) – (-5) = -8 + 5 = -3 and the last term l = a_n = 10
But, a_n = a + (n – 1) d
∴ (-230) = -5 + (n – 1) (-3)
⇒ -230 + 5 = -3n + 3
⇒ -3n + 3 = -225
⇒ -3n = -225 – 3
⇒ 3n = 228
⇒ n = \(\frac{228}{3}\) = 76
∴ n = 76
Now, S_n = \(\frac{n}{2}\)(a + l)
where a = -5; l = -230
S₇₆ = \(\frac{76}{2}\)((-5) + (-230))
= 38 × (-235)
= -8930

Question 3.
In an AP:
i) Given a = 5, d = 3, a_n = 50. find n and S_n.
Answer:
Given :
a = 5; d = 3;
a_n = a + (n – 1)d = 50
⇒ 50 = 5 + (n – 1) 3
⇒ 50 – 5 = 3n – 3
⇒ 3n = 45 + 3
⇒ n = \(\frac{48}{3}\) = 16
Now, S_n = \(\frac{n}{2}\)(a + l)
S₁₆ = \(\frac{16}{2}\)(5 + 50)
= 38 × 55
= 440

ii) Given a = 7, a₁₃ = 35, find d and S₁₃.
Answer:
Given: a = 7;
a₁₃ = a + 12d = 35
⇒ 7 + 12d = 35
⇒ 12d = 35 – 7
⇒ n = \(\frac{28}{12}\) = \(\frac{7}{3}\)
Now, S_n = \(\frac{n}{2}\)(a + l)
S₁₃ = \(\frac{13}{2}\)(7 + 35)
= \(\frac{13}{2}\) × 42
= 13 × 21
= 273

iii) Given a₁₂ = 37, d = 3 find a and S₁₂.
Answer:
Given:
a₁₂ = a + 11d = 37
d = 3
So, a₁₂ = a + 11 × 3 = 37
⇒ a + 33 = 37
⇒ a = 37 – 33 = 4
Now, S_n = \(\frac{n}{2}\)(a + l)
S₁₂ = \(\frac{12}{2}\)(4 + 37)
= 6 × 41
= 246

iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀.
Answer:
Given:
a₃ = a + 2d = 15
⇒ a = 15 – 2d ……… (1)
S₁₀ = 125 but take S₁₀ as 175
i.e., S₁₀ = 175
We know that,

⇒ 35 = 2 (15 – 2d) + 9d [∵ a = 15 – 2d]
⇒ 35 = 30 – 4d + 9d
⇒ 35 – 30 = 5d
⇒ d = \(\frac{5}{5}\) = 1
Substituting d = 1 in equation (1) we get
a = 15 – 2 × 1 = 15 – 2 = 13
Now, a_n = a + (n – 1) d
a₁₀ = a + 9d = 13 + 9 × 1 = 13 + 9 = 22
∴ a₁₀ = 22; d = 1

v) Given a = 2, d = 8, S_n = 90, find n and a_n.
Answer:
Given a = 2, d = 8, S_n = 90
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]

⇒ 90 = 2n [2n – 1]
⇒ 4n² – 2n = 90
⇒ 4n² – 2n – 90 = 0
⇒ 2(2n² – n – 45) = 0
⇒ 2n² – n – 45 = 0
⇒ 2n² -10n + 9n – 45 = 0
⇒ 2n(n – 5) + 9(n – 5) = 0
⇒ (n – 5)(2n + 9) = 0
⇒ n – 5 = 0 (or) 2n + 9 = 0
⇒ n = 5 (or) n = \(\frac{-9}{2}\) (discarded)
∴ n = 5
Now a_n = a₅ = a + 4d = 2 + 4 x 8
= 2 + 32 = 34

vi) Given a_n = 4, d = 2, S_n = -14, find n and a.
Answer:
Given a_n = a + (n – 1) d = 4 ……. (1)
d = 2; S_n = – 14
From (1); a + (n – 1) 2 = 4
a = 4 – 2n + 2
a = 6 – 2n
Given a = 2, d = 8, S_n = 90
S_n = \(\frac{n}{2}\)[a + a_n]
-14 = \(\frac{n}{2}\)[(6-2n) + 4] [∵ a = 6 – 2n]
-14 × 2 = n (10 – 2n)
⇒ 10n – 2n² = – 28
⇒ 2n² – 10n – 28 = 0
⇒ n² – 5n – 14 = 0
⇒ n² – 7n + 2n – 14 = 0
⇒ n (n – 7) + 2 (n – 7) = 0
⇒ (n – 7) (n + 2) = 0
⇒ n = 7 (or) n = – 2
∴ n = 7
Now a = 6 – 2n = 6 – 2 × 7
= 6 – 14 = -8
∴ a = – 8; n = 7

vii) Given l = 28, S = 144, and there are total 9 terms. Find a.
Answer:
Given:
l = a₉ = a + 8d = 28 and S₉ = 144 But,
Now, S_n = \(\frac{n}{2}\)(a + l)
144 = \(\frac{9}{2}\)(a + 28)
⇒ 144 × \(\frac{2}{9}\) = a + 28
⇒ a + 28 = 32
⇒ a = 4

Question 4.
The first and the last terms of an A.P are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Answer:
Given A.P in which a = 17
Last term = l = 350
Common difference, d = 9
We know that, a_n = a + (n – 1) d
350 = 17 + (n- 1) 9
⇒ 350 = 17 + 9n – 9
⇒ 9n = 350 – 8
⇒ n = \(\frac{342}{9}\) = 38
Now, S_n = \(\frac{n}{2}\)(a + l)
S₃₈ = \(\frac{38}{2}\)(17 + 350)
= 19 × 367 = 6973
∴ n = 38; S_n = 6973

Question 5.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Answer:
Given A.P in which

Substituting d = 4 in equation (1),
we get a + 4 = 14
⇒ a = 14 – 4 = 10

Question 6.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Answer:
Given :
A.P such that S₇ = 49; S₁₇ = 289
We know that,

Substituting d = 2 in equation (1), we get,
a + 3 × 2 = 7
⇒ a = 7 – 6 = 1
∴ a = 1; d = 2

∴ Sum of first n terms S_n = n².
Shortcut: S₇ = 49 = 7²
S₁₇ = 289 = 17²
∴ S_n = n²

Question 7.
Show that a₁, a₂ …,a_n, …. form an AP where a_n is defined as below:
i) a = 3 + 4n
ii) a_n = 9 – 5n. Also find the sum of the first 15 terms in each case.
Answer:
Given a_n = 3 + 4n
Then a₁ = 3 + 4 × l = 3 + 4 = 7
a₂ = 3 + 4 × 2 = 3 + 8 = 11
a₃ = 3 + 4 × 3 = 3 + 12 = 15
a₄ = 3 + 4 × 4 = 3 + 16 = 19
Now the pattern is 7, 11, 15, ……
where a = a₁ = 7; a₂ = 11; a₃ = 15, ….. and
a₂ – a₁ = 11 – 7 = 4;
a₃ – a₂ = 15 – 11 = 4;
Here d = 4
Hence a₁, a₂, ….., a_n ….. forms an A.P.

ii) a_n = 9 – 5n
Given: a_n = 9 – 5n.
a₁ = 9 – 5 × l = 9 – 5 = 4
a₂ = 9 – 5 × 2 = 9 – 10 = -1
a₃ = 9 – 5 × 3 = 9 – 15 = -6
a₄ = 9 – 5 × 4 = 9 – 20 = -11
Also
a₂ – a₁ = -1 – 4 = -5;
a₃ – a₂ = -6 – (-1) = – 6 + 1 = -5
a₄ – a₃ = -11 – (-6) = -11 + 6 = -5
∴ d = a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = …. = -5
Thus the difference between any two successive terms is constant (or) starting from the second term, each term is obtained by adding a fixed number ‘-5’ to its preceding term.
Hence {a_n} forms an A.P.

Question 8.
If the sum of the first n terms of an AP is 4n – n², what is the first term (remember the first term is S₁)? What is the sum of first two terms? What is the second term? Similarly, find the 3^rd, the 10^th and the n^th terms.
Answer:
Given an A.P in which S_n = 4n – n²
Taking n = 1 we get
S₁ = 4 × 1- 12 = 4 – 1 = 3
n = 2; S₂ = a₁ + a₂ = 4 × 2 – 2² = 8 – 4 = 4
n = 3; S₃ = a₁ + a₂ + a₃ = 4 × 3 -3² = 12 – 9 = 3
n = 4; S₄ = a₁ + a₂ + a₃ + a₄ = 4 × 4 – 4² = 16 – 16 = 0
Hence, S₁ = a₁ = 3
a₂ = S₂ – S₁ = 4 – 3 = 1
a₃ = S₃ – S₂ = 3 – 4 = -1
a₄ = S₄ – S₃ = 0 – 3 = -3
So, d = a₂ – a₁ = l – 3 = -2
Now, a₁₀ = a + 9d  [∵ a_n = a + (n – 1) d]
= 3 + 9 × (- 2)
= 3 – 18 = -15
a_n = 3 + (n – 1) × (-2)
= 3 – 2n + 2
= 5 – 2n

Question 9.
Find the sum of the first 40 positive integers divisible by 6.
Answer:
The given numbers are the first 40 positive multiples of 6
⇒ 6 × 1, 6 × 2, 6 × 3, ….., 6 × 40
⇒ 6, 12, 18, ….. 240
S_n = \(\frac{n}{2}\)(a + l)
S₄₀ = \(\frac{40}{2}\)(6 + 240)
= 20 × 246
= 4920
∴ S₄₀ = 4920

Question 10.
A sum of Rs. 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs. 20 less than its preceding prize, find the value of each of the prizes.
Answer:
Given:
Total/Sum of all cash prizes = Rs. 700
Each prize differs by Rs. 20
Let the prizes (in ascending order) be x, x + 20, x + 40, x + 60, x + 80, x + 100, x + 120
∴ Sum of the prizes = S₇ = \(\frac{n}{2}\)(a + l)
⇒ 700 = \(\frac{7}{2}\)[x + x + 120]
⇒ 700 × \(\frac{2}{7}\) = 2x + 120
⇒ 100 = x + 60
⇒ x = 100 – 60 = 40
∴ The prizes are 160, 140, 120, 100, 80, 60, 40.

Question 11.
In a school, students thought of plant¬ing trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Answer:
Given: Classes: From I to XII
Section: 3 in each class.
∴ Trees planted by each class = 3 × class number

∴ Total trees planted = 3 + 6 + 9 + 12 + …… + 36 is an A.P.
Here, a = 3 and l = 36; n = 12
∴ S_n = \(\frac{n}{2}\)(a + l)
S₁₂ = \(\frac{12}{2}\)[3 + 36]
= 6 × 39
= 234
∴ Total plants = 234

Question 12.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, … as shown in figure. What is the total length of such a spiral made up of thirteen
consecutive semicircles? (Take π = \(\frac{22}{7}\))

[Hint: Length of successive semicircles is l₁, l₂, l₃, l₄,….. with centres at A, B, A, B,…, respectively.]
Answer:
Given: l₁, l₂, l₃, l₄,….., l₁₃ are the semicircles with centres alternately at A and B; with radii
r₁ = 0.5 cm [1 × 0.5]
r₂ = 1.0 cm [2 × 0.5]
r₃ = 1.5 cm [3 × 0.5]
r₄ = 2.0 cm [4 × 0.5] [∵ Radii are in A.P. as aj = 0.5 and d = 0.5]
……………………………
r₁₃ = 13 × 0.5 = 6.5
Now, the total length of the spiral = l₁ + l₂ + l₃ + l₄ + ….. + l₁₃ [∵ 13 given]
But circumference of a semi-cirle is πr.
∴ Total length of the spiral = π × 0.5 + π × 1.0 + ………. + π × 6.5
= π × \(\frac{1}{2}\)[l + 2 + 3 + ….. + 13]
[∵ Sum of the first n – natural numbers is \(\frac{n(n+1)}{2}\)
= \(\frac{22}{7} \times \frac{1}{2} \times \frac{13 \times 14}{2}\)
= 11 × 13
= 143 cm.

Question 13.
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

Answer:
Given: Total logs = 200
Number of logs stacked in the first row = 20
Number of logs stacked in the second row = 19
Number of logs stacked in the third row = 18
The number series is 20, 19, 18,….. is an A.P where a = 20 and
d = a₂ – a₁ = 19 – 20 = -1
Also, S_n = 200

400 = 41n – n²
⇒ n² – 41n + 400 = 0
⇒ n² – 25n – 16n + 400 = 0
⇒ n(n – 25) – 16(n – 25) = 0
⇒ (n – 25) (n – 16) = 0
⇒ n – 25 (or) 16
There can’t be 25 rows as we are starting with 20 logs in the first row.
∴ Number of rows must be 16.
∴ n = 16

Question 14.
In a bucket and ball race, a bucket is placed at the starting point, which is 5 m from the first ball, and the other balls are placed 3 m apart in a straight line. There are ten balls in the line.

A competitor starts from the bucket, picks up the nearest ball, runs back with it, drops it in the bucket, runs back to pick up the next ball, runs to the bucket to drop it in, and she continues in the same way until all the balls are in the bucket. What is the total distance the competitor has to run?
[Hint: To pick up the first ball and the second ball, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]
Answer:
Given: Balls are placed at an equal distance of 3 m from one another.
Distance of first ball from the bucket = 5 m
Distance of second ball from the bucket = 5 + 3 = 8 m (5 + 1 × 3)
Distance of third ball from the bucket = 8 + 3 = 11 m (5 + 2 × 3)
Distance of fourth ball from the bucket = 11 + 3 = 14 m (5 + 3 × 3)
………………………………
∴ Distance of the tenth ball from the bucket = 5 + 9 × 3 = 5 + 27 = 32 m.
1st ball: Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 5 = 10 m.
2nd ball: Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 8 = 16 m.
3rd ball: Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 11 = 22 m.
………………………………
10th ball: Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 32 = 64 m.
Total distance = 10 m + 16 m + 22 m + …… + 64 m.
Clearly, this is an A.P in which a = 10; d = a₂ – a₁ = 16 – 10 = 6 and n = 10.
∴ S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S₁₀ = \(\frac{10}{2}\)[2 × 10 + (10 – 1)6]
= 5 [20 + 54]
= 5 × 74
= 370 m
∴ Total distance = 370 m.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.2

10th Class Maths 6th Lesson Progressions Ex 6.2 Textbook Questions and Answers

Question 1.
Fill in the blanks in the following table, given that ‘a’ is the first term, d the common difference and an the nth term of the A.P:

Answer:

Question 2.
Find the i) 30th term of the A.P.: 10, 7, 4,……
ii) 11th term of the A.P.: -3, –\(\frac{1}{2}\), 2,…
Answer:
i) Given A.P. = 10, 7, 4, …….
a₁ = 10; d = a₂ – a₁ = 7 – 10 = – 3
a_n = a + (n – 1) d
a₃₀ = 10 + (30 – 1) (- 3) = 10 + 29 × (- 3) = 10 – 87 = – 77

ii) Given A.P. = – 3, –\(\frac{1}{2}\), 2,…
a₁ = -3; d = a₂ – a₁ = –\(\frac{1}{2}\) – (-3) = – 3
= –\(\frac{1}{2}\) + 3
= \(\frac{-1+6}{2}\)
= \(\frac{5}{2}\)
a_n = a + (n – 1) d
= -3 + (11-1) × \(\frac{5}{2}\)
= -3 + 10 × \(\frac{5}{2}\)
= -3 + 5 × 5
= -3 + 25
= 22

Question 3.
Find the respective terms for the following APs.
i) a₁ = 2; a₃ = 26, find a₂.
Answer:
Given: a₁ = a = 2 …….. (1)
a₃ = a + 2d = 26 …….. (2
Equation (2) – equation (1)
⇒ (a + 2d) – a = 26 – 2
⇒ 2d = 24
d = \(\frac{24}{2}\) = 12
Now a₂ = a + d = 2 + 12 = 14

ii) a₂ = 13; a₄ = 3, find a₁, a₃.
Answer:
Given: a₂ = a + d = 13 ….. (1)
a₄ = a + 3d = 3 ….. (2)
Solving equations (1) and (2);

∴ Substituting d = – 5 in equation (1) we get
a + (-5) = 13
∴ a = 13 + 5 = 18 i.e., a₁ = 18
a₃ = a + 2d = 18 + 2(- 5)
= 18 – 10 = 8

iii) a₁ = 5; a₄ = 9\(\frac{1}{2}\), find a₂, a₃.
Answer:
Given: a₁ = a = 5 ….. (1)
a₄ = a + 3d = 9\(\frac{1}{2}\) ….. (2)
Solving equations (1) and (2);

⇒ 3d = 4\(\frac{1}{2}\)
⇒ 3d = \(\frac{9}{2}\)
⇒ d = \(\frac{9}{2 \times 3}\) = \(\frac{3}{2}\)
∴ a₂ = a + d = 5 + \(\frac{3}{2}\) = \(\frac{13}{2}\)
a₃ = a + 2d = 5 + 2 × \(\frac{3}{2}\) = 5 + 3 = 8

iv) a₁ = -4; a₆ = 6, find a₂, a₃, a₄, a₅.
Answer:
Given: a₁ = a = -4 ….. (1)
a₆ = a + 5d = 6 ….. (2)
Solving equations (1) and (2);
(-4) + 5d = 6
⇒ 5d = 6 + 4
⇒ 5d = 10
⇒ d = \(\frac{10}{5}\)
Now
∴ a₂ = a + d = -4 + 2 = -2
a₃ = a + 2d = -4 + 2 × 2 = -4 + 4 = 0
a₄ = a + 3d = -4 + 3 × 2 = -4 + 6 = 2
a₅ = a + 4d = -4 + 4 × 2 = -4 + 8 = 4

v) a₂ = 38; a₆ = -22, find a₁, a₃, a₄, a₅.
Answer:
Given: a₂ = a + d = 38 ….. (1)
a₆ = a + 5d = -22 ….. (2)
Subtracting (2) from (1) we get

Now substituting, d = – 15 in equation (1), we get
a + (- 15) = 38 ⇒ a = 38 + 15 = 53
Thus,
a₁ = a = 53;
a₃ = a + 2d = 53 + 2 × (- 15) = 53 – 30 = 23;
a₄ = a + 3d = 53 + 3 × (- 15) = 53 – 45 = 8;
a₅ = a + 4d = 53 + 4 × (- 15) = 53 – 60 = – 7

Question 4.
Which term of the AP:
3, 8, 13, 18,…, is 78?
Answer:
Given: 3, 8, 13, 18, ……
Here a = 3; d = a₂ – a₁ = 8 – 3 = 5
Let ‘78’ be the nth term of the given A.P.
∴ a_n = a + (n – 1) d
⇒ 78 = 3 + (n – 1) 5
⇒ 78 = 3 + 5n – 5
⇒ 5n = 78 + 2
⇒ n = \(\frac{80}{2}\) = 16
∴ 78 is the 16th term of the given A.P.

Question 5.
Find the number of terms in each of the following APs:
i) 7, 13, 19, ….., 205
Answer:
Given: A.P: 7, 13, 19, ……….
Here a₁ = a = 7; d = a₂ – a₁ = 13 – 7 = 6
Let 205 be the n^th term of the given A.P.
Then, a_n = a + (n – 1) d
205 = 7 + (n- 1)6
⇒ 205 = 7 + 6n – 6
⇒ 205 = 6n + 1
⇒ 6n = 205 – 1 = 204
∴ n = \(\frac{204}{6}\) = 34
∴ 34 terms are there.

ii) 18, 15\(\frac{1}{2}\), 13, …, -47
Answer:
Given: A.P: 18, 15\(\frac{1}{2}\), 13, …….
Here a₁ = a = 18;
d = a₂ – a₁ = 15\(\frac{1}{2}\) – 18 = -2\(\frac{1}{2}\) = –\(\frac{5}{2}\)
Let ‘-47’ be the nth term of the given A.P.
a_n = a + (n – 1) d

⇒ -94 = 36 – 5n + 5
⇒ 5n = 94 + 41
⇒ n = \(\frac{135}{5}\) = 27
∴ 27 terms are there.

Question 6.
Check whether, -150 is a term of the AP: 11, 8, 5, 2…
Answer:
Given: A.P. = 11, 8, 5, 2…
Here a₁ = a = 11;
d = a₂ – a₁ = 8 – 11 = -3
If possible, take – 150 as the n^th term of the given A.P.
a_n = a + (n – 1) d
⇒ -150 = 11 + (n – 1) × (-3)
⇒ -150 = 11 – 3n + 3
⇒ 14 – 3n = – 150
⇒ 3n= 14 + 150 = 164
∴ n = \(\frac{164}{3}\) = 54\(\frac{2}{3}\)
Here n is not an integer.
∴ -150 is not a term of the given A.P.

Question 7.
Find the 31^st term of an A.P. whose 11^th term is 38 and the 16^th term is 73.
Answer:
Given: An A.P. whose

⇒ -5d = -35
⇒ d = \(\frac{-35}{-5}\) = 7
Substituting d = 7 in the equation (1)
we get,
a + 10 x 7 = 38
⇒ a + 70 = 38
⇒ a = 38 – 70 = -32
Now, the 31^st term = a + 30d
= (-32) + 30 × 7
= -32 + 210 = 178

Question 8.
If the 3rd and the 9^th terms of an A.P are 4 and -8 respectively, which term of this A.P is zero?
Answer:
Given: An A.P. whose

Substituting d = -2 in equation (1) we get
a + 2 × (-2) = 4
⇒ a – 4 = 4
⇒ a = 4 + 4 = 8
Let n^th term of the given A.P be equal to zero.
an = a + (n – 1)d
⇒ 0 = 8 + (n – 1) × (-2)
⇒ 0 = 8 – 2n + 2
⇒ 10 – 2n = 0
⇒ 2n = 10 and n = \(\frac{10}{2}\) = 5
∴ The 5^th term of the given A.P is zero.

Question 9.
The 17^th term of an A.P exceeds its 10 term by 7. Find the common difference.
Answer:
Given an A.P in which a₁₇ = a₁₀ + 7
⇒ a₁₇ – a₁₀ = 7
We know that a_n = a + (n – 1)d

⇒ d = \(\frac{7}{7}\) = 1

Question 10.
Two APs have the same common difference. The difference between their 100^th terms is 100, what is the difference between their 1000^th terms?
Answer:
Let the first A.P be:
a, a + d, a + 2d, ……..
Second A.P be:
b, b + d, b + 2d, b + 3d, ………
Also, general term, a_n = a + (n – 1)d
Given that, a₁₀₀ – b₁₀₀ = 100
⇒ a + 99d – (b + 99d) = 100
⇒ a – b = 100
Now the difference between their 1000^th terms,

∴ The difference between their 1000^th terms is (a – b) = 100.
Note: If the common difference for any two A.Ps are equal then difference between n^th terms of two A.Ps is same for all natural values of n.

Question 11.
How many three-digit numbers are divisible by 7?
Answer:
The least three digit number is 100.

∴ The least 3 digit number divisible by 7 is 100 + (7 – 2) = 105
The greatest 3 digit number is 999

∴ The greatest 3 digit number divisible by 7 is 999 – 5 = 994.
∴ 3 digit numbers divisible by 7 are
105, 112, 119,….., 994.
a₁ = a = 105; d = 7; a_n = 994
a_n = a + (n – 1)d
⇒ 994 = 105 + (n – 1)7
⇒ (n – 1)7 = 994 – 105
⇒ (n – 1)7 = 889
⇒ n – 1 = \(\frac{889}{7}\) = 127
∴ n = 127 + 1 = 128
∴ There are 128, 3 digit numbers which are divisible by 7.
(or)
\(\frac{\text { last number – first number }}{7}\)
\(\frac{999-100}{7}\)
≃ 128.4 = 128 numbers divisible by 7.

Question 12.
How many multiples of 4 lie between 10 and 250?
Answer:
Given numbers: 10 to 250

∴ Multiples of 4 between 10 and 250 are
First term: 10 + (4 – 2) = 12
Last term: 250 – 2 = 248
∴ 12, 16, 20, 24, ….., 248
a = a₁ = 12; d = 4; a_n = 248
a_n = a + (n – 1)d
248 = 12 + (n – 1) × 4
⇒ (n – 1)4 = 248 – 12
⇒ n – 1 = \(\frac{236}{4}\) = 59
∴ n = 59 + 1 = 60
There are 60 numbers between 10 and 250 which are divisible by 4.

Question 13.
For what value of n, are the n^th terms of two APs: 63, 65, 67, ….. and 3, 10, 17,… equal?
Answer:
Given : The first A.P. is 63, 65, 67, ……
where a = 63, d = a₂ – a₁,
⇒ d = 65 – 63 = 2
and the second A.P. is 3, 10, 17, …….
where a = 3; d = a₂ – a₁ = 10 – 3 = 7
Suppose the nth terms of the two A.Ps are equal, where a_n = a + (n – 1)d
⇒ 63 + (n – 1)2 = 3 + (n – 1)7
⇒ 63 + 2n – 2 = 3 + 7n – 7
⇒ 61 + 2n = 7n – 4
⇒ 7n – 2n = 61 + 4
⇒ 5n = 65
⇒ n = \(\frac{65}{5}\) = 13
∴ 13^th terms of the two A.Ps are equal.

Question 14.
Determine the AP whose third term is 16 and the 7^th term exceeds the 5^th term by 12.
Answer:
Given : An A.P in which
a₃ = a + 2d = 16 …… (1)
and a₇ = a₅ + 12
i.e., a + 6d = a + 4d + 12
⇒ 6d – 4d = 12
⇒ 2d = 12
⇒ d = \(\frac{12}{2}\) = 6
Substituting d = 6 in equation (1) we get
a + 2 × 6 = 16
⇒ a = 16 – 12 = 4
∴ The series/A.P is
a, a + d, a + 2d, a + 3d, …….
⇒ 4, 4 + 6, 4 + 12, 4 + 18, ……
⇒ A.P.: 4, 10, 16, 22, …….

Question 15.
Find the 20^th term from the end of the AP: 3, 8, 13,…, 253.
Answer:
Given: An A.P: 3, 8, 13, …… , 253
Here a = a₁ = 3
d = a₂ – a₁ = 8 – 3 = 5
a_n = 253, where 253 is the last term
a_n = a + (n – l)d
∴ 253 = 3 + (n – 1)5
⇒ 253 = 3 + 5n – 5
⇒ 5n = 253 + 2
⇒ n = \(\frac{255}{5}\) = 51
∴ The 20^th term from the other end would be
1 + (51 – 20) = 31 + 1 = 32
∴ a₃₂ = 3 + (32 – 1) × 5
= 3 + 31 × 5
= 3 + 155 = 158

Question 16.
The sum of the 4^th and 8^th terms of an AP is 24 and the sum of the 6^th and 10^th terms is 44. Find the first three terms of the AP.
Answer:
Given an A.P in which a₄ + a₈ = 24
⇒ a + 3d + a + 7d = 24
⇒ 2a + 10d = 24
⇒ a + 5d = 12 ……. (1)
and a₆ + a₁₀ = 44
⇒ a + 5d + a + 9d = 44
⇒ 2a + 14d = 44
⇒ a + 7d = 22 ……. (2)
Also a + 5d = 12
⇒ a + 5(5) = 12
⇒ a + 25 = 12
⇒ a = 12 – 25 = -13
∴ The A.P is a, a + d, a + 2d, ……
i.e., – 13, (- 13 + 5), (-13 + 2 × 5)…
⇒ -13, -8, -3, …….

Question 17.
Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs. 200 each year. In which year did his income reach Rs. 7000?
Answer:
Given: Salary of Subba Rao in 1995 = Rs. 5000
Annual increment = Rs. 200
i.e., His salary increases by Rs. 200 every year.

Clearly 5000, 5200, 5400, forms an A.P in which a = 5000 and d = 200.
Now suppose that his salary reached Rs. 7000 after x – years.
i.e., a_n = 7000
But, a_n = a + (n – 1)d
7000 = 5000 + (n – 1)200
⇒ 7000 – 5000 = (n – 1)200
⇒ n – 1 = \(\frac{2000}{200}\) = 10
⇒ n = 10 + 1
∴ In 11^th year his salary reached Rs. 7000.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.1

10th Class Maths 6th Lesson Progressions Ex 6.1 Textbook Questions and Answers

Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
i) The taxi fare after each km when the fare is Rs. 20 for the first km and rises by Rs. 8 for each additional km.
Answer:
Fare for the first km = Rs. 20 = a
Fare for each km after the first = Rs. 8 = d
∴ The fares would be 20, 28, 36, 44, …….
The above list forms an A.P.
Since each term in the list, starting from the second can be obtained by adding ‘8’ to its preceding term.

ii) The amount of air present in a cylinder when a vacuum pump removes \(\frac{1}{4}\)th of the air remaining in the cylinder at a time.
Answer:
Let the amount of air initially present in the cylinder be 1024 lit.
First it removes \(\frac{1}{4}\)th of the volume
i.e., \(\frac{1}{4}\) × 1024 = 256
∴ Remaining air present in the cylinder = 768
At second time it removes \(\frac{1}{4}\)th of 768
i.e., \(\frac{1}{4}\) × 768 = 192
∴ Remaining air in the cylinder = 768 – 192 = 576
Again at third time it removes \(\frac{1}{4}\)th of 576
i.e., \(\frac{1}{4}\) × 576 = 144
Remaining air in the cylinder = 576 – 144 = 432
i.e., the volume of the air present in the cylinder after 1st, 2nd, 3rd,… times is 1024, 768, 576, 432, …..
Here, a₂ – a₁ = 768 – 1024 = – 256
a₃ – a₂ = 576 – 768 = – 192
a₄ – a₃ = 432 – 576 = – 144 .
Thus the difference between any two successive terms is not equal to a fixed number.
∴ The given situation doesn’t show an A.P.

iii) The cost of digging a well, after, every metre of digging, when it costs ? 150 for the first metre and rises by ? 50 for each subsequent metre.
Answer:
Cost for digging the first metre = Rs. 150
Cost for digging subsequent metres = Rs. 50 each.
i.e.,

The list is 150, 200, 250, 300, 350, ……..
Here d = a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = ……. = 50
∴ The given situation represents an A.P.

iv) The amount of money in the account every year, when Rs. 10000 is deposited at compound interest at 8 % per annum.
Answer:
Amount deposited initially = P = Rs. 10,000
Rate of interest = R = 8% p.a [at C.I.]
∴ \(A=P\left(1+\frac{R}{100}\right)^{n}\)

The terms 10800, 11664, 12597.12, ……. a₂ – a₁ = 800
Here, a = 10,000                                     a₃ – a₂ = 864
But, a₂ – a₁ ≠ a₃ – a₂ ≠ a₄ – a₃                a₄ – a₃ = 953.12
∴ The given situation doesn’t represent an A.P.

Question 2.
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
i) a = 10, d = 10
ii) a = -2, d = 0
iii) a = 4, d = – 3
iv) a = – 1, d = 1/2
v) a = – 1.25, d = – 0.25
Answer:

Question 3.
For the following A.Ps, write the first term and the common difference:
i) 3, 1, – 1, – 3,….
ii) – 5, – 1, 3, 7,….
iii) \(\frac{1}{3}\), \(\frac{5}{3}\), \(\frac{9}{3}\), \(\frac{13}{3}\), ……..
iv) 0.6, 1.7, 2.8, 3.9,…
Answer:

Question 4.
Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
i) 2, 4, 8, 16, …….
ii) 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), …….
iii) – 1.2, – 3.2, – 5.2, – 7.2,……
iv) -10,-6, -2, 2, …….
v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, …….
vi) 0.2, 0.22, 0.222, 0.2222, ……
vii) 0, -4, -8, -12, …….
viii) –\(\frac{1}{2}\), –\(\frac{1}{2}\), –\(\frac{1}{2}\), –\(\frac{1}{2}\)
ix) 1, 3, 9, 27,…..
x) a, 2a, 3a, 4a,….
xi) a, a², a³, a⁴, …..
xii) √2, √8, √18, √32, …….
xiii) √3, √6, √9, √12, …….
Answer:

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Ex 7.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.4

10th Class Maths 7th Lesson Coordinate Geometry Ex 7.4 Textbook Questions and Answers

Question 1.
Find the slope of the line joining the two given points.
i) (4,-8) and (5,-2).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{-2+8}{5-4}\)
= 6

ii) (0, 0) and (√3, 3)
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{3-0}{\sqrt{3}-0}\)
= \(\frac{3}{\sqrt{3}}\)
= \(\frac{\sqrt{3} \times \sqrt{3}}{\sqrt{3}}\)
= √3

iii) (2a, 3b) and (a, -b).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{-b-3b}{a-2a}\)
= \(\frac{-4b}{-a}\)
= \(\frac{4b}{a}\)

iv) (a, 0) and (0, b).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{b-0}{0-a}\)
= \(\frac{-b}{a}\)

v) A (-1.4, -3.7), B (-2.4, 1.3).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{1.3+3.7}{-2.4+1.4}\)
= \(\frac{5.0}{-1}\)
= -5

vi) A (3, -2), B (-6, -2).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{-2+2}{-6-3}\)
= 0

vii) A (-3\(\frac{1}{2}\), 3), B (-7, 2\(\frac{1}{2}\)).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{2 \frac{1}{2}-3}{-7+3 \frac{1}{2}}\)
= \(\frac{-\frac{1}{2}}{-3 \frac{1}{2}}\)
= \(\frac{1}{2}\) × \(\frac{2}{7}\)
= \(\frac{1}{7}\)

viii) A(0, 4), B(4, 0)
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{0-4}{4-0}\)
= \(\frac{-4}{4}\)
= -1

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Ex 7.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.3

10th Class Maths 7th Lesson Coordinate Geometry Ex 7.3 Textbook Questions and Answers

Question 1.
Find the area of the triangle whose vertices are
i) (2, 3), (-1, 0), (2,-4)
Answer:
Given: A (2, 3), B (- 1, 0) and C (2, – 4) are the vertices of a △ABC.
Area of the triangle ABC = \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|2(0+4)-1(-4-3)+2(3-0)|\)
= \(\frac{1}{2}|8+7+6|\)
= \(\frac{21}{2}\)
= 10\(\frac{1}{2}\) sq.units

ii) (-5, -1), (3, -5), (5, 2)
Answer:
Given: A (- 5, – 1), B (3, – 5) and C (5, 2) are the vertices of △ABC.
Area of the △ABC
= \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|-5(-5-2)+3(2+1)+5(-1+5)|\)
= \(\frac{1}{2}|35+9+20|\)
= \(\frac{64}{2}\)
= 32 sq.units

iii) (0, 0), (3, 0), (0, 2)
Answer:
Given: O (0, 0), A (3, 0) and B (0, 2) are the vertices of a triangle, △AOB.
Area of the △AOB
= \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|0(0-2)+3(2-0)+0(0-0)|\)
= \(\frac{1}{2}|6|\)
= 3 sq.units

Or

△AOB = \(\frac{1}{2}\) × OA × OB
= \(\frac{1}{2}\) × 3 × 2
= 3 sq.units

Question 2.
Find the value of ‘K’ for which the points are collinear.
i) (7, -2), (5, 1), (3, K)
Answer:
Given: A (7, – 2), B (5, 1) and C (3, K) are collinear.
∴ Area of △ABC = 0
But area of triangle
\(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
⇒ \(\frac{1}{2} \mid 7(1-\mathrm{K})+5(\mathrm{~K}+2)+3(-2-1)\) = 0
⇒ \(|7-7 K+5 K+10-9|\) = 0
⇒ \(|-2 \mathrm{~K}+8|\) = 0
⇒ -2K + 8 = 0
⇒ -2K = -8
⇒ K = \(\frac{8}{2}\)
i.e., K = 4

ii) (8, 1), (K,-4), (2,-5)
Answer:
Given: A (8, 1), B(K, – 4) and C (2, – 5) are collinear.
∴ Area of △ABC = 0
⇒ \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\) = 0
⇒ \(\frac{1}{2}|8(-4+5)+\mathrm{K}(-5-1)+2(1+4)|\) = 0
⇒ \(|8-6 \mathrm{~K}+10|\) = 0
⇒ \(|18-6 \mathrm{~K}|\) = 0
⇒ 18 – 6K = 0
⇒ 6K = 18
⇒ K = \(\frac{18}{6}\)
i.e., K = 3

iii) (K,K), (2, 3), and (4,-1)
Answer:
A (K, K), B (2, 3) and C (4, – 1) are collinear.
∴ Area of △ABC = 0
⇒ \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\) = 0
⇒ \(\frac{1}{2}|\mathrm{~K}(3+1)+2(-1-\mathrm{K})+4(\mathrm{~K}-3)|\) = 0
⇒ \(|4K-2-2K+4K-12|\) = 0
⇒ \(|6 \mathrm{~K}-14|\) = 0
⇒ 6K – 14 = 0
⇒ 6K = 14
⇒ K = \(\frac{14}{6}\) = \(\frac{7}{3}\)
∴ K = \(\frac{7}{3}\)

Question 3.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Answer:

Given: A (0, – 1), B (2, 1) and C (0, 3) are the vertices of △ABC.
Let D, E and F be the midpoints of the sides \(\overline{\mathrm{AB}}\), \(\overline{\mathrm{BC}}\) and \(\overline{\mathrm{AC}}\).

Area of a triangle ABC =
\(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|0(1-3)+2(3+1)+0(-1-1)|\)
= \(\frac{1}{2}|8|\)
= 4 sq.units
Area of △DEF = \(\frac{1}{2}|1(2-1)+1(1-0)+0(0-2)|\)
= \(\frac{1}{2}|1+1|\)
= \(\frac{2}{2}\)
= 1 sq.units
Ratio of areas = △ABC : △DEF = 4 : 1.
△ADF ≅ △BED ≅ △DEF ≅ △CEF
∴ △ABC : △DEF = 4 : 1

Question 4.
Find the area of the quadrilateral whose vertices taken inorder are (-4, -2), (-3, -5),(3, -2) and (2, 3).
Answer:

Given: A (- 4, – 2), B (- 3, – 5), C (3, – 2) and D (2, 3) are the vertices of the quadrilateral ▱ ABCD.
Area of ▱ ABCD = △ABC + △ACD.
Area of a triangle =
\(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)

Question 5.
Find the area of the triangle formed by the points by using Heron’s formula.
i) (1, 1), (1, 4) and (5, 1)
ii) (2, 3), (-1,3) and (2, -1)
Answer:
i) (1, 1) (1, 4) and (5, 1)
let A (1, 1) B(l, 4) and C(5, 1) are the vertices then length of sides can be calculated using the formula
\(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
now

now formula for area of triangle using Heron’s formula = △ = \(\sqrt{s(s-a)(s-b)(s-c)}\)
where s = \(\frac{a+b+c}{2}\)
∴ s = \(\frac{3+4+5}{2}\) = \(\frac{12}{2}\) = 6
∴ △ = \(\sqrt{6(6-5)(6-4)(6-3)}\)
= \(\sqrt{6 \times 1 \times 2 \times 3}\)
= \(\sqrt{6 \times 6}\)
= 6 sq. units
∴ area of given triangle = 6 sq units

ii) let the vertices of given triangle A (2, 3), B (-l, 3) and C (2, -1)

∴ a = 5, b = 4, c = 3 units
now from using Heron’s formula area of triangle
= △ = \(\sqrt{s(s-a)(s-b)(s-c)}\)
where s = \(\frac{a+b+c}{2}\)
= \(\frac{5+4+3}{2}\)
= \(\frac{12}{2}\) = 6
∴ △ = \(\sqrt{6(6-5)(6-4)(6-3)}\)
= \(\sqrt{6 \times 1 \times 2 \times 3}\)
= \(\sqrt{6 \times 6}\)
= 6 sq. units
∴ area of given triangle = 6 sq units

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Ex 7.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.2

10th Class Maths 7th Lesson Coordinate Geometry Ex 7.2 Textbook Questions and Answers

Question 1.
Find the coordinates of the point which divides the line segment joining the points (-1, 7) and (4, -3) in the ratio 2 :3.
Answer:
Given points P (-1, 7) and Q (4, – 3). Let ‘R’ be the required point which divides \(\overline{\mathrm{PQ}}\) in the ratio 2:3. Then

Question 2.
Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Answer:
Given points A (4, – 1) and B (- 2, – 3) Let P and Q be the points of trisection
of \(\overline{\mathrm{AB}}\), then AP = PQ = QB.

∴ P divides \(\overline{\mathrm{AB}}\) internally in the ratio 1 : 2.

Also, Q divides \(\overline{\mathrm{AB}}\) in the ratio 2 : 1 internally.

Question 3.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Answer:
Let the point (-1, 6) divides the line segment joining the points (-3, 10) and (6, -8) in a ratio of m : n

⇒ 6m – 3n = -(m + n) = -m – n
⇒ 6m + m = – n + 3n
⇒ 7m = 2n
⇒ \(\frac{m}{n}=\frac{2}{7}\)
⇒ m : n = 2 : 7
∴ The point (-1, 6) divides the given line segment in a ratio of 2 : 7.

Question 4.
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Answer:

Given: ▱ ABCD is a parallelogram where A (1, 2), B (4, y), C (x, 6) and D (3, 5).
In a parallelogram, diagonals bisect each other.
i.e., the midpoints of the diagonals coincide with each other.
i.e.,midpoint of AC = midpoint of BD

⇒ 1 + x = 7 and 8 = y + 5
⇒ x = 7 – 1 and y = 8 – 5
∴ x = 6 and y = 3.

Question 5.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).
Answer:

Given:
A circle with centre ‘C’ (2, -3). \(\overline{\mathrm{AB}}\) is a diameter where
B = (1, 4); A = (x, y).
C is the midpoint of AB.
[∵ Centre of a circle is the midpoint of the diameter]

4 = x + 1 and – 6 = y + 4
⇒ x = 4 – 1 = 3 and y = -6 – 4 = -10
A (x, y) = (3, -10)

Question 6.
If A and B are (-2, -2) and (2, -4) respectively. Find the coordinates of P such that AP = \(\frac{3}{7}\) AB and P lies on the segment AB.
Answer:
Given: A (- 2, – 2) and B (2, – 4)
P lies on AB such that AP = latex]\frac{3}{7}[/latex] AB

i.e., P divides \(\overline{\mathrm{AB}}\) in the ratio 3 : 4 By section formula,

Question 7.
Find the coordinates of points which divide the line segment joining A (-4, 0) and B (0, 6) into four equal parts.
Answer:
Given, A (- 4, 0) and B (0, 6).
Let P, Q and R be the points which divide AB into four equal parts.

P divides \(\overline{\mathrm{AB}}\) in the ratio 1 : 3, Q → 1 : 1 and R → 3 : 1 Use section formula to find P, Q and R.
Then, Q is the midpoint of \(\overline{\mathrm{AB}}\)
P is the midpoint of \(\overline{\mathrm{AQ}}\)
R is the midpoint of \(\overline{\mathrm{QB}}\)

Question 8.
Find the coordinates of the points which divides the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
Answer:
Given, A (- 2, 2) and B (2, 8).
Let P, Q and R be the points which divide AB into four equal parts.

Then, Q is the midpoint of \(\overline{\mathrm{AB}}\)
P is the midpoint of \(\overline{\mathrm{AQ}}\)
R is the midpoint of \(\overline{\mathrm{QB}}\)

Question 9.
Find the coordinates of the point which divide the line segment joining the points (a + b, a-b) and (a-b, a + b) in the ratio 3 : 2 internally.
Answer:
Given : A (a + b, a – b) and B (a – b, a + b).
Let P (x, y) divides \(\overline{\mathrm{AB}}\) in the ratio 3 : 2 internally.
Section formula

Question 10.
Find the coordinates of centroid of the triangle with following vertices:
i) (-1, 3), (6, -3) and (-3, 6)
Answer:
Given: △ABC in which- A (- 1, 3), B (6, -3) and C (-3, 6)

ii) (6, 2), (0, 0) and (4, -7)
Answer:
Given: The three vertices of a triangle are A (6, 2), B (0, 0) and C (4, – 7).
Centroid (x, y)

iii) (1,-1), (0, 6) and (-3, 0)
Answer:
Given: (1, -1), (0, 6) and (-3, 0) are the vertices of a triangle.
Centroid (x, y)

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Ex 7.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.1

10th Class Maths 7th Lesson Coordinate Geometry Ex 7.1 Textbook Questions and Answers

Question 1.
Find the distance between the following pair of points,
(i) (2, 3) and (4, 1)
Answer:
Distance = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
= \(\sqrt{(4-2)^{2}+(1-3)^{2}}\)
= \(\sqrt{4+4}\)
= √8 = 2√2 units

ii) (- 5, 7) and (-1, 3)
Answer:
Distance = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
= \(\sqrt{(-1+5)^{2}+(3-7)^{2}}\)
= \(\sqrt{4^{2}+(-4)^{2}}\)
= \(\sqrt{16+16}\)
= √32 = 4√2 units

iii) (- 2, -3) and (3, 2)
Answer:

iv) (a, b) and (- a, – b)
Answer:

Question 2.
Find the distance between the points (0, 0) and (36, 15).
Answer:
Given: Origin O (0, 0) and a point P (36, 15).
Distance between any point and origin = \(\sqrt{x^{2}+y^{2}}\)
∴ Distance = \(\sqrt{36^{2}+15^{2}}\)
= \(\sqrt{1296+225}\)
= \(\sqrt{1521}\)
= 39 units

∴ 1521 = 3² × 13²
\(\sqrt{1521}\) = 3 × 13 = 39

Question 3.
Verify that the points (1, 5), (2, 3) and (-2, -1) are collinear or not.
Answer:
Given: A (1, 5), B (2, 3) and C (- 2, – 1)

Here the sum of no two segments is equal to third segment.
Hence the points are not collinear.
!! Slope of AB, m₁ = \(\frac{3-5}{2-1}\) = -2
Slope of BC, m₂ = \(\frac{-1-3}{-2-2}\) = 1
m₁ ≠ m₂
Hence A, B, C are not collinear.

Question 4.
Check whether (5, -2), (6, 4) and (7,-2) are the vertices of an isosceles triangle.
Answer:
Let A = (5, – 2); B = (6, 4) and C = (7, – 2).

Now we have, AB = BC.
∴ △ABC is an isosceles triangle,
i.e., given points are the vertices of an isosceles triangle.

Question 5.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in figure. Jarina and Phani walk into the class and after observing for a few minutes Jarina asks Phani “Don’t you think ABCD is a square?” Phani disagrees. Using distance formula, find which of them is correct. Why?

Answer:
Given: Four friends are seated at A, B, C and D where A (3, 4), B (6, 7), C (9, 4) and D (6, 1).
Now distance

BD = \(\sqrt{(6-6)^{2}+(1-7)^{2}}\) = √36 = 6
Hence in □ ABCD four sides are equal
i.e., AB = BC = CD = DA
= 3√2 units
and two diagonals are equal.
i.e., AC = BD = 6 units.
∴ □ ABCD forms a square.
i.e., Jarina is correct.

Question 6.
Show that the following points form an equilateral triangle A(a, 0), B(- a, 0), C(0, a√3).
Answer:
Given: A (a, 0), B (- a, 0), C (0, a√3).

Now, AB = BC = CA.
∴ △ABC is an equilateral triangle.

Question 7.
Prove that the points (-7, -3), (5, 10), (15, 8) and (3, -5) taken in order are the corners of a parallelogram.
Answer:
To show that the given points form a parallelogram.
We have to show that the mid points of each diagonal are same. Since diagonals of a parallelogram bisect each other.
Now let A(-7, -3), B(5, 10), C(15, 8) and D(3, -5)
Then midpoint of diagonal

∴ (1) = (2)
Hence the given are vertices of a parallelogram.

Question 8.
Show that the points (-4, -7), (-1, 2), (8, 5) and (5, -4) taken in order are the vertices of a rhombus. And find its area.
(Hint: Area of rhombus = \(\frac{1}{2}\) × product of its diagonals)
Answer:
Given in ▱ ABCD , A(-4, – 7), B (- 1, 2), C (8, 5) and D (5,-4)


∴  In ▱ ABCD, AB = BC = CD = AD [from sides are equal]
Hence ▱ ABCD is a rhombus.
Area of a rhombus = \(\frac{1}{2}\) d₁d₂
= \(\frac{1}{2}\) × 12√2 × 6√2
= 72 sq. units.

Question 9.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.
i) (-1,-2), (1,0), (-1,2), (-3,0)
Answer:
Let A (- 1, -2), B (1, 0), C (- 1, 2), D (- 3, 0) be the given points. Distance formula

In ▱ ABCD, AB = BC = CD = AD – four sides are equal.
AC = BD – diagonals are equal.
Hence, the given points form a square,

ii) (-3, 5), (1, 10), (3, 1), (-1,-4).
Answer:
Let A(-3, 5), B(l,10), C(3, 1), D(-l, -4) then

In ▱ ABCD, \(\overline{\mathrm{AB}}\) = \(\overline{\mathrm{CD}}\) and \(\overline{\mathrm{BC}}\) = \(\overline{\mathrm{AD}}\) (i.e., both pairs of opposite sides are equal) and \(\overline{\mathrm{AC}}\) ≠ \(\overline{\mathrm{BD}}\).
Hence ▱ ABCD is a parallelogram,
i.e., The given points form a parallelogram.

iii) (4, 5), (7, 6), (4, 3), (1, 2).
Answer:
Let A (4, 5), B (7, 6), C (4, 3) and D (1, 2) be the given points.
Distance formula

In ▱ ABCD, AB = CD and BC = AD (i.e., both pairs of opposite sides are equal) and AC ≠ BD.
Hence ▱ ABCD is a parallelogram, i.e., The given points form a parallelogram.

Question 10.
Find the point on the X-axis which is equidistant from (2, -5) and (-2,9).
Answer:
Given points, A (2, – 5), B (- 2, 9).
Let P (x, 0) be the point on X – axis which is equidistant from A and B. i.e., PA = PB.
Distance formula = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

But PA = PB.
⇒ \(\sqrt{x^{2}-4 x+29}=\sqrt{x^{2}+4 x+85}\)
Squaring on both sides, we get
x² – 4x + 29 = x² + 4x + 85
⇒ – 4x – 4x = 85 – 29
⇒ – 8x = 56
⇒ x = \(\frac{56}{-8}\) = -7
∴ (x, 0) = (- 7, 0) is the point which is equidistant from the given points.

Question 11.
If the distance between two points (x, 7) and (1, 15) is 10, find the value of x.
Answer:
Formula for distance between two points = \(\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}\)
Now distance between (x, 7) and (1,15) is 10.
∴ \(\sqrt{(x-1)^{2}+(7-15)^{2}}\) = 10
∴ (x – l)² + (-8)² = 10²
⇒ (x – l)² = 100 – 64 = 36
∴ x – 1 = √36 = ± 6
∴ x – 1 = 6 or x – 1 = -6
⇒ x = 6 + 1 = 7 or x = -6 + 1 = -5
∴ x = 7 or x = – 5

Question 12.
Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Answer:
Given: P (2, – 3), Q (10, y) and
\(\overline{\mathrm{PQ}}\) = 10.
Distance formula = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

⇒ y² + 6y + 73 = 100
⇒ y² + 6y – 27 = 0
⇒ y² + 9y – 3y – 27 = 0
⇒ y (y + 9) – 3 (y + 9) = 0
⇒ (y + 9) (y – 3) = 0
⇒ y + 9 = 0 or y – 3 = 0
⇒ y = -9 or y = 3
⇒ y = – 9 or 3.

Question 13.
Find the radius of the circle whose centre is (3, 2) and passes through (-5,6).
Answer:

Given: A circle with centre A (3, 2) passing through B (- 5, 6).
Radius = AB
[∵ Distance of a point from the centre of the circle]
Distance formula = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

Question 14.
Can you draw a triangle with vertices (1, 5), (5, 8) and (13, 14)? Give reason.
Answer:
Let A (1, 5), B (5, 8) and C (13, 14) be the given points.
Distance formula

Here, AC = AB + BC.
∴ △ABC can’t be formed with the given vertices.
[∵ Sum of the any two sides of a triangle must be greater than the third side].

Question 15.
Find a relation between x and y such that the point (x, y) is equidistant from the points (-2, 8) and (-3, -5).
Answer:
Let A (- 2, 8), B (- 3, – 5) and P (x, y). If P is equidistant from A, B, then PA = PB.
Distance formula =


Squaring on both sides we get, x² + y² + 4x – 16y + 68
= x² + y² + 6x +10y + 34
⇒ 4x – 16y – 6x – 10y = 34-68
⇒ – 2x – 26y = -34
⇒ x + 13y = 17 is the required condition.