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

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

10th Class Maths 7th Lesson Coordinate Geometry Optional Exercise Textbook Questions and Answers

Question 1.
Centre of the circle Q is on the Y-axis. And the circle passes through the points (0, 7) and (0, -1). Circle intersects the positive X-axis at (p, 0). What is the value of ‘p’ ?
Answer:

Given points A, B are on y – axis.
(∵ their x – coordinate is zero) then they will be the end points (0, 7) and (0, -1).
Centre (Q) = \(\left(\frac{0+0}{2}, \frac{7-1}{2}\right)\) = (0, 3)
then radius = AQ = BQ
= \(\sqrt{(0-0)^{2}+(3-(-1))^{2}}\)
= \(\sqrt{0^{2}+4^{2}}\) = √16 = 4
now P(p, 0) is also another point on circle then \(\overline{\mathrm{QP}}\) is also radius
∴ \(\overline{\mathrm{QP}}\) = \(\sqrt{(0-p)^{2}+(3-0)^{2}}\) = 4
⇒ \(\sqrt{p^{2}+3^{2}}\) = 4
⇒ p² = 4² – 3² = 16 – 9 = 7
∴ P = ± √7 .

Question 2.
A triangle ABC is formed by the points A(2, 3), B(-2, -3), C(4, -3). What is the point of intersection of side BC and angular bisector of angle A?
Answer:
Given: △ABC, where A (2, 3), B (- 2, – 3), C (4, – 3).
Let AD be the bisector to ∠A meeting BC at D.

Then BD : DC = AB : AC
[∵ The bisector of vertical angle of triangle divides the base in the ratio of other two sides.]

Now D is a point which divides \(\overline{\mathrm{BC}}\) in the ratio √l3 : √10 internally section formula (x, y) =

(By rationalising the denominator of the x-coordinate).

Question 3.
The side BC of an equilateral △ABC is parallel to X – axis. Find the slopes of line along sides BC, CA and AB.
Answer:

Given: △ABC; BC // X – axis.
Slope of BC = tan θ
where θ is the angle made of BC with + ve X – axis.
= tan ‘0’ = ‘0’
Slope of AB = tan 60° = √3
[∵ Each angle of an equilateral triangle is 60°]
Slope of AC = tan 60° = √3

Question 4.
A right triangle has sides ‘a’ and ‘b’ where a > b. If the right angle is bisected then find the distance between ortho centres of the smaller triangles using coordinate geometry.
Answer:
Given : △ABC; ∠B = 90°

Question 5.
Find the centroid of the triangle formed by the line 2x + 3y – 6 = 0 with the coordinate axes.
Answer:
Given : △AOB formed by the line 2x + 3y – 6 = 0 with axes.
Let A lie on X – axis and B on Y – axis.
∴ Y – coordinate of A = 0
X – coordinate of B = 0
∴ A (3, 0), B (0, 2)

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

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

10th Class Maths 6th Lesson Progressions Optional Exercise Textbook Questions and Answers

Question 1.
Which term of the AP:
121, 117, 113,…, is the first negative term? [Hint: Find n for a_n < 0]
Answer:
Given A.P: 121, 117, 113, ……
a = 121 and d = a₂ – a₁
= 117 – 121 = -4
Let the n^th term be the first negative term of the given G.P.
Then, a_n < 0
⇒ a_n = a + (n – 1) d < 0
⇒ 121 + (n – 1) (-4) < 0
⇒ 121 – 4n + 4 < 0
⇒ 125 – 4n < 0
⇒ -4n < -125
⇒ 4n >125 125
⇒ n > \(\frac{125}{4}\)
n > 31.25
∴ When n = 32, the term becomes negative, (or)
32 nd term is the first negative term of the given A.P.

Question 2.
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
Answer:
Let the 3^rd term of AP = a + 2d
and the 7^th term of AP = a + 6d
∴ Sum of 3rd and 7^th terms = a + 2d + a + 6d = 6
⇒ 2a + 8d = 6
⇒ a + 4d = 3 …… (1)
Now product of above two terms = (a + 2d) (a + 6d) = 8
we can re-write above terms as following
(a + 4d – 2d) (a + 4d + 2d) = 8
⇒ (3 – 2d) (3 + 2d) = 8
⇒ 9 – 4d² = 8
⇒ 4d² = 9 – 8 = 1
∴ d² = \(\frac{1}{4}\)
⇒ d = ± \(\frac{1}{2}\) ……. (2)
Now putting d = \(\frac{1}{2}\) in eq(1) we get
a + 4d = a + 4(\(\frac{1}{2}\)) = 3 ⇒ a = 1
so a = 1, d = \(\frac{1}{2}\)
(or) now putting d = \(\frac{-1}{2}\), we get
a + 4d = a + 4(\(\frac{-1}{2}\)) = 3
⇒ a – 2 = 3
⇒ a = 5
∴ a = 5, d = \(\frac{-1}{2}\)
∴ Sum of sixteen terms =

So S₁₆ = 20 or 76

Question 3.
A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2\(\frac{1}{2}\) m apart, what is the length of the wood required for the rungs. [Hint: Number of rungs = \(\frac{250}{25}\) + 1]

Answer:
Given: A ladder with rungs separated by a distance of 25 cm.
Total distance between the rungs = 2\(\frac{1}{2}\) m = 250 m
∴ Number of rungs = \(\frac{250}{25}\) + 1 = 10 + 1 = 11
Length of the bottom and top rungs = 45 cm and 25 cm
where a = 45; last term l = a₁₁ = 25 and d = 2 cm
∴ S_n = \(\frac{n}{2}\)(a + l)
= \(\frac{11}{2}\)(45 + 25)
= \(\frac{11}{2}\) × 70
= 35 × 11
= 385 cm
∴ Length of wood required for total rungs = 385 cm.

Question 4.
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. And find this value of x. [Hint: S_x-1 = S₄₉ – S_x]
Answer:
Given: Houses with numbers from 1 to 49.
x is a number x such that,
S_x-1 = S₄₉ – S_x


⇒ x(x – 1) + x(x + 1) = 2 × 1225
⇒ x² – x + x² + x = 2450
⇒ 2x² = 2450
⇒ x² = \(\frac{2450}{2}\) = 1225
⇒ x = √1225 = ±35
∴ x = 35 [∵ x is a counting number]

Question 5.
A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete.
Each step has a rise of \(\frac{1}{4}\) m and a tread of \(\frac{1}{2}\) m. (see Fig.).
Calculate the total volume of concrete required to build the terrace.
[Hint: Volume of concrete required to build the first step = \(\frac{1}{4}\) × \(\frac{1}{2}\) × 50 m ]

Answer:
Length of each step = 50 m = l
Rise/height of each step = \(\frac{1}{4}\) m = h
Tread of each step = \(\frac{1}{2}\) m = b
Given volume of concrete required to build the first step = \(\frac{1}{4}\) × \(\frac{1}{2}\) × 50 m³ = lbh
We can compare the shape of each step with a cuboid.
Volume of the cuboid = l . b . h
Volumes of concrete required to build the 15 steps are

[∵ 1, 2, 3, 15 is in A.P. where a = 1; d = 1, n = 15]

∴ The total volume of concrete required to build the terrace is 750 m³.

Question 6.
150 workers were engaged to finish a piece of work in a certain number of days. Four workers dropped from the work in the second day. Four workers dropped in third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.
[Let the no.of days to finish the work is ‘x’ then 150x = \(\frac{x+8}{2}\)[2 × 150 + (x + 8 – 1) (-4)]
Answer:
Given: Number of workers engaged initially = 150.
4 workers were dropped each day.
Let the total work was to be completed initially was in x days.
∴ Work done by 150 workers in x days = 150.x.
But due to the dropping of 4 workers each day it took 8 more days.
Work done in this case is
150 × 1 + 146 × 1 + 142 × 1 + …. (x + 8) terms,

= (x + 8) (136 – 2x)
= -2x² + 136x + 1088 – 16x
= -2x² + 120x + 1088
∴ 150x = -2x² + 120x + 1088
⇒ 2x² + 30x – 1088 = 0
⇒ x² + 15x – 544 = 0
⇒ x² + 32x – 17x – 544 =0
⇒ x(x + 32) – 17 (x + 32) = 0
⇒ (x + 32) (x – 17) = 0
⇒ x + 32 = 0 (or) x – 17 = 0
⇒ x = – 32 (or) x = 17
x can’t be negative.
∴ x = 17.
i.e., The total work was completed in x + 8 days = 17 + 8 = 25 days.

Question 7.
A machine costs Rs. 5,00,000. If the value depreciates 15% in the first year, 13\(\frac{1}{2}\)% in the second year, 12% in the third year and so on. What will be its value at the end of 10 years, when all the percentages will be applied to the original cost?
Answer:
Given: Cost price of a machine = Rs. 5,00,000
Depreciation during the years

Sum of the depreciations =

∴ Cost after 10 years = (100 – 82.5)% of 5,00,000
= 17.5 % of 5,00,000
= \(\frac{17.5 × 500000}{2}\)
= Rs. 87,500
∴ The value at the end of 10 years will be Rs. 87,500.

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.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Optional Exercise

10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables Optional Exercise Textbook Questions and Answers

Question 1.
i) \(\frac{2x}{a}\) + \(\frac{y}{b}\) = 2
\(\frac{x}{a}\) – \(\frac{y}{b}\) = 4
Answer:

Substituting x = 2a in the equation (1) we get
\(\frac{2}{a}\)(2a) + \(\frac{y}{b}\) = 2
⇒ 4 + \(\frac{y}{b}\) = 2
⇒ \(\frac{y}{b}\) = -2
⇒ y = -2b
∴ The solution (x, y) = (2a, -2b)

ii) \(\frac{x+1}{2}\) + \(\frac{y-1}{3}\) = 8
\(\frac{x-1}{3}\) + \(\frac{y+1}{2}\) = 9
Answer:
Given: \(\frac{x+1}{2}\) + \(\frac{y-1}{3}\) = 8

⇒ 3x + 2y + 1 = 48
⇒ 3x + 2y = 47 …… (1)
and \(\frac{x-1}{3}\) + \(\frac{y+1}{2}\) = 9

⇒ 2x + 3y + 1 = 54
⇒ 2x + 3y = 53 …… (2)

⇒ y = \(\frac{-65}{-5}\) = 13
Substituting y = 13 in equation (1) we get
3x + 2(13) = 47
⇒ 3x = 47 – 26
⇒ 3x = 21
⇒ x = \(\frac{21}{3}\) = 7
∴ The solution (x, y) = (7, 13)

iii) \(\frac{x}{7}\) + \(\frac{y}{3}\) = 5
\(\frac{x}{2}\) – \(\frac{y}{9}\) = 6
Answer:
Given: \(\frac{x}{7}\) + \(\frac{y}{3}\) = 5 and \(\frac{x}{2}\) – \(\frac{y}{9}\) = 6
⇒ \(\frac{3x+7y}{21}\) = 5 and \(\frac{9x-2y}{18}\) = 6
⇒ 3x + 7y = 105 …….. (1) and
9x – 2y = 108 …….. (2)

⇒ y = \(\frac{207}{23}\) = 9
Substituting y = 9 in equation (1) we get
3x + 7(9) = 105
⇒ 3x = 105 – 63
⇒ 3x = 42
⇒ x = \(\frac{42}{3}\) = 14
∴ The solution (x, y) = (14, 9)

iv) √3x + √2y = √3
√5x + √3y= √3
Answer:
Given that √3x + √2y = √3 …… (1)
√5x + √3y = √3 …… (2)
By following elimination method

Now again following elimination method

∴ The solution x = \(\frac{3-\sqrt{6}}{3-\sqrt{10}}\) and y = \(\frac{3-\sqrt{15}}{3-\sqrt{10}}\)

v) \(\frac{ax}{b}\) + \(\frac{by}{a}\) = a + b
ax – by = 2ab
Answer:
Given: \(\frac{ax}{b}\) + \(\frac{by}{a}\) = a + b ……. (1)
ax – by = 2ab …….. (2)

Substituting y = – a in equation (2)
we get ax – b(-a) = 2ab
ax + ab = 2ab
ax = 2ab – ab = ab
⇒ x = \(\frac{ab}{a}\) = b
∴ (x, y) = (b, -a)

vi) 2^x + 3^y = 17
2^x+2 – 3^y+1 = 5
Answer:
Given: 2^x + 3^y = 17 and
2^x+2 – 3^y+1 = 5
Take 2^x = a and 3^y = b then the given equations reduce to
2^x + 3^y = 17 ⇒ a + b = 17 …… (1)
2^x . 2² – 3^y . 3 = 5 ⇒ 4a – 3b = 5 …… (2)

Substituting b = 9 in equation (1) we get
a + 9 = 17 ⇒ a = 17 – 9 = 8
But a = 2^x – 8 and b = 3^y = 9
⇒ 2^x = 2³ and 3^y = 3²
⇒ x = 3 and y = 2
∴ The solution (x, y) is (3, 2)

Question 2.
Animals in an experiment are to be kept on a strict diet. Each animal is to receive among other things 20g of protein and 6g of fat. The laboratory technicians purchased two food mixes, A and B. Mix A has 10% protein and 6% fat. Mix B has 20% protein and 2% fat. How many grams of each mix should be used?
Answer:
Let x gms of mix A and y gms of mix B are to be mixed, then


Substituting y = 60 in equation (1)
we get x + 2 × 60 = 200
⇒ x + 120 = 200
⇒ x = 200 – 120 = 80 gm
∴ Quantity of mix. A = 80 gms.
Quantity of mix. B = 60 gms.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials Optional Exercise Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Optional Exercise

10th Class Maths 3rd Lesson Polynomials Optional Exercise Textbook Questions and Answers

Question 1.
Verify that the numbers given along-side the cubic polynomials below are their zeroes. Also verify the relation-ship between the zeroes and the coefficients in each case:
i) 2x³ + x² – 5x + 2; (\(\frac{1}{2}\), 1,-2)
ii) x³ + 4x² + 5x – 2 ; (1, 1, 1)
Answer:
i) Given polynomial 2x³ + x² – 5x + 2
Comparing the given polynomial with ax³ + bx² + cx + d,
we get a = 2, b = 1, c = – 5 and d = 2

P(1) = 2(1)³ + (1)² – 5(1) – 2
= 2 + 1 – 5 + 2 = 0
p(-2) = 2(-2)³ + (-2)² – 5(-2) + 2
= 2(-8) + 4 + 10 + 2
= – 16 + 16 = 0
∴ \(\frac{1}{2}\), 1 and – 2 are the zeroes of 2x³ + x² – 5x + 2
So, α = \(\frac{1}{2}\), β = 1 and γ = – 2 Therefore,

ii) Given polynomial x³ + 4x² + 5x – 2
Comparing the given polynomial with ax³ + bx² + cx + d,
we get a = 1, b = 4, c = 5 and d = – 2.
Given zeroes are (1, 1, 1)
p(1) = (1)³ + 4(1)² + 5(1) – 2
= 1 + 4 + 5 – 2
= 10 – 2 = 8
∴ (1, 1, 1) are not zeroes of the given polynomial p(x).

Question 2.
Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.
Answer:
Let the cubic polynomial be
ax³ + bx² + cx + d, and its zeroes be α, β and γ.
Then,
α + β + γ = 2 = \(\frac{-(-2)}{1}\) = \(\frac{-b}{a}\)
αβ + βγ + γα = -7 = \(\frac{-7}{1}\) = \(\frac{c}{a}\)
αβγ = – 14 = \(\frac{-14}{1}\) = \(\frac{-d}{a}\)
a = 1, then b = -2, c = -7 and d = 14.
So, one cubic polynomial which satisfies the given conditions will be x³ – 2x² – 7x + 14.

Question 3.
If the zeroes of the polynomial x³ – 3x² + x + 1 are a – b, a, a + b, find a and b.
Answer:
Given polynomial x³ – 3x² + x + 1
Since, (a – b), a, (a + b) are the zeroes of the polynomial x³ – 3x² + x + 1.
Therefore, sum of the zeroes
= (a – b) + a + (a + b) = \(\frac{-(-3)}{1}\) = 3
So, 3a = 3 ⇒ a = 1
∴ Sum of the products of its zeroes taken two at a time.
= a(a – b) + a(a + b) + (a + b) (a – b) = \(\frac{1}{1}\) = 1
⇒ a² – ab + a² + ab + a² – b² = 1
⇒ 3a² – b² = 1
So, 3(1)² – b² = 1 ⇒ 3 – b² = 1
⇒ b² = 2
⇒ b = √2 = ± √2
Here, a = 1 and b = ± √2

Question 4.
If two zeroes of the polynomial x⁴ – 6x³ – 26x² + 138x – 35 are 2 ± √3 , find other zeroes.
Answer:
Let the other, two zeroes are α, β.
Then the sum of zeroes of given polynomial = 2 + √3 + 2 – √3 + α + β
= \(\frac{-b}{a}\) = \(\frac{-(-6)}{1}\) = 6
4 + α + β = 6
⇒ α + β = 2 ….. (1)
Now product of zeroes is
(2 + √3 ) (2 – √3) (α) (β)= \(\frac{e}{a}\) = \(\frac{-35}{1}\)
(4 – 3) (αβ) = – 35
⇒ αβ = – 35 …… (2)
Now (α – β)² = (α + β)² – 4αβ
= (2)² – 4(-35) = 4 + 140 = 144
⇒ (α – β) = ± 12 ….. (3)
Now solving (1) & (3) we get

⇒ α = 7; then α + β = 7 + β = 2
⇒ β = -5
The remaining zeroes are α, β
= 7, -5
So total zeroes of given polynomial are 2 + √3, 2 – √3, 7, – 5.

Question 5.
If the polynomial x⁴ – 6x³ – 16x² + 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.
Answer:
Given polynomial x⁴ – 6x³ – 16x² + 25x + 10 and another polynomial is x² – 2x + k.
Remainder is x + a
Let us divide
x⁴ – 6x³ – 16x² + 25x + 10 by x² – 2x + k

∴ Remainder
= x(4k + 25 – 2k – 48) + 10 + k(k + 24)
= x(2k – 23) + (k² + 24k + 10)
Given remainder is x + a
on comparing the coefficients of x and constant terms on both sides
2k – 23 = 1 ……. (1)
2k = 1 + 23 = 24
⇒ k = \(\frac{24}{2}\) = 12
k² + 24k + 10 = a …….. (2)
Substitute ‘k’ value in equation (2)
(12)² + 24(12) + 10 = a
144 + 288 + 10 = a
⇒ a = 442
∴ Required k = 12 and ‘a’ = 442

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers Optional Exercise Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers Optional Exercise

10th Class Maths 1st Lesson Real Numbers Optional Exercise Textbook Questions and Answers

Question 1.
Can the number 6ⁿ, n being a natural number, end with the digit 5? Give reason.
Answer:
Given number = 6ⁿ ; n ∈ N
6ⁿ to be end in 5; it should be divisible by 5
6ⁿ = (2 × 3)ⁿ
The prime factors of 6ⁿ are 2 and 3.
It can’t end with the digit 5.

Question 2.
Is 7 × 5 × 3 × 2 + 3 a composite number? Justify your answer.
Answer:
Given:
7 × 5 × 3 × 2 + 3
= 3 (7 × 5 × 2 + 1)
= 3 × (70 + 1)
= 3 × 71
∴ The given number has two factors namely 3 and 71.
Hence it is a composite number.

Question 3.
Prove that (2√3 + √5 ) is an irrational number. Also check whether (2√3 + √5) (2√3 – √5) is rational or irrational.
Answer:
To prove:
2√3 + √5 is an irrational number. On contrary, let us suppose that 2√3 + √5 be a rational number.
Then 2√3 + √5 = \(\frac{p}{q}\)
Squaring on both sides, we get


L.H.S = an irrational number.
R.H.S = p, q being integers, \(\frac{p^{2}-17 q^{2}}{4 q^{2}}\) is a rational number.
This is a contradiction to the fact that √l5 is an irrational. This is due to our assumption that 2√3 + √5 is a rational. Hence our assumption is wrong and 2√3 + √5 is an irrational number. Also,
(2√3 + √5) (2√3 – √5)
= (2√3)² – (√5)²
[∵ (a + b) (a – b) = a² – b²]
= 4 × 3 – 5
= 12 – 5 = 7, a rational number.

Question 4.
If x² + y² = 6xy, prove that 2 log (x + y) = log x + log y + 3 log 2.
Answer:
Given: x² + y² = 6xy
x² + y² + 2xy = 6xy + 2xy
(x + y)² = 8xy
Taking logarithms on both sides log (x + y)² = log8xy
⇒ 2log(x + y)= log8 + logx + logy [∵ logx^m = mlogx]
[∵ logxy = logx + logy]
= log2³ + logx + logy
⇒ 2log(x + y) = logx + logy + 3log2

Question 5.
Find the number of digits in 42013, if log₁₀2 = 0.3010.
Answer:
Given:
log₁₀2 = 0.3010
4²⁰¹³ = (2²)²⁰¹³ = 2⁴⁰²⁶
∴ log₁₀ 2⁴⁰²⁶ = 4026 log₁₀2
[∵ log x^m = m log x]
= 4026 × 0.3010 = 1211.826.
So 1211 + 1 = 1212
∴ 4²⁰¹³ has 1212 digits in its expansion.
(∵ characteristic 1211)

AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers InText Questions and Answers.

10th Class Maths 1st Lesson Real Numbers InText Questions and Answers

Do this

Question 1.
Find q and r for the following pairs of positive integers a and b, satisfying a = bq + r. (Page No. 3)
i) a = 13, b = 3
Answer:
13 = 3 × 4 + 1
here q = 4 ; r = 1
ii) a = 8, b = 80
Answer:
Take a = 80, b = 8
80 = 8 × 10 + 0 here q = 10 ; r = 0
iii) a = 125, b = 5
Answer:
125 = 5 × 25 + 0
here q = 25 ; r = 0
iv) a = 132, b = 11
Answer:
132 = 11 × 12 + 0
here q = 12 ; r = 0

Question 2.
Find the HCF of the following by using Euclid division lemma,
i) 50 and 70 (Page No. 4)
Answer:
For given two positive integers a > b;
there exists unique pair of integers q and r satisfying a = bq + r; 0≤r<b.
∴ 70 = 50 × 1 + 20
Here a = 70, b = 50, q = 1, r = 20.
Now consider 50, 20
50 = 20 × 2 + 10
Here a = 50, b = 20, q = 2, r = 10.
Now taking 20 and 10.
20 = 10 × 2 + 0
Here the remainder is zero.
∴ 10 is the HCF of 70 and 50.

ii) 96 and 72
Answer:
96 = 72 × 1 + 24
72 = 24 × 3 + 0
∴ HCF = 24

iii) 300 and 550
Answer:
550 = 300 × 1 + 250
300 = 250 × 1 + 50
250 = 50 × 5 + 0
∴ HCF = 50

iv) 1860 and 2015
Answer:
2015 = 1860 × 1 + 155
1860 = 155 × 12 + 0
∴ HCF = 155

Think & Discuss

Question 1.
From the above questions in ‘DO THIS’, what is the nature of q and r? (Page No. 3)
Answer:
Given: a = bq + r
q > 0 and r lies in between 0 and b
i.e. q > 0 and 0 ≤ r < b

Question 2.
Can you find the HCF of 1.2 and 0.12? Justify your answer. (Page No. 4)
Answer:
Given: 1.2 and 0.12
we have 1.2 = \(\frac{12}{10}\) = \(\frac{120}{100}\)
0.12 = \(\frac{12}{100}\)
Now considering the numerators 12 and 120, their HCF is 12.
∴ HCF of 1.2 and 0.12 is \(\frac{12}{100}\) = 0.12
i.e., if x is a factor of y then x is the HCF of x and y.

Question 3.
If r = 0, then what is the relationship between a, b and q in a = bq + r of Euclid divison lemma? (Page No. 6)
Answer:
Given: r = 0 in a = bq + r then a = bq
i.e., b divides a completely.
i.e., b is a factor of a.

Do this

Question 1.
Express 2310 as a product of prime factors. Also see how your friends have factorized the number. Have they done it as you ? Verify your final product with your friend’s result. Try this for 3 or 4 more numbers. What do you conclude? (Page No. 7)
Answer:
Given: 2310
2310 = 2 × 1155
= 2 × 3 × 385
= 2 × 3 × 5 × 77
2310 = 2 × 3 × 5 × 7 × 11

We notice that this prime factorization is unique.
And also notice that prime factorization of any number is unique i.e., every composite number can be expressed as a product of primes and this factorization is unique.
E.g: 144 = 2 × 72
= 2 × 2 × 36
= 2 × 2 × 2 × 18
= 2 × 2 × 2 × 2 × 9
= 2 × 2 × 2 × 2 × 3 × 3
= 24 × 32
320 = 2 × 160
= 2 × 2 × 80
= 2 × 2 × 2 × 40
= 2 × 2 × 2 × 2 × 20
= 2 × 2 × 2 × 2 × 2 × 10
= 2 × 2 × 2 × 2 × 2 × 2 × 5
= 26 × 5
125 = 5 × 25
= 5 × 5 × 5
= 5³

Question 2.
Find the HCF and LCM of the following given pairs of numbers by prime factorization, (Page No. 8)
i) 120, 90
Answer:
We have 120 = 2 × 2 × 2 × 3 × 5
= 2³ × 3 × 5
90 = 2 × 3 × 3 × 5
= 2 × 3² × 5

∴ HCF = 2 × 3 × 5 = 30
LCM = 2³ × 3² × 5 = 360

ii) 50, 60
Answer:
We have
50 = 2 × 5 × 5 = 2 × 5²
60 = 2 × 2 × 3 × 5 = 2² × 3 × 5

∴ HCF = 2 × 5 = 10
LCM = 2² × 3 × 5² = 300

iii) 37, 49
Answer:
We have
37 = 1 × 37
49 = 7 × 7 = 7²
∴ HCF = 1
LCM = 37 × 7²
Note: H.C.F. of two relatively prime numbers is 1 and LCM is equal to product of the numbers.

Try this

Question 1.
Show that 3ⁿ × 4^m cannot end with the digit 0 or 5 for any natural numbers ‘n’ and’m’. (Page No. 8)
Answer:
Given number is 3⁴ × 4^m.
So the prime factors to it are 3 and 2 only.
I: but if a number want to be end with zero it should have 2 and 5 as its prime factors, but the given hasn’t ‘5’ as its prime factor.
So it cannot be end with zero.
II : now if a number went to be end with 5 it should have ‘5’ as its one of prime factors. But given 3ⁿ × 4^m do not have 5 as a factor.
So it cannot be end with 5.
Hence proved.

Do this

Question 1.
Write the following terminating decimals in the form of p/q, q ≠ 0 and p, q are co-primes.
i) 15.265
ii) 0.1255
iii) 0.4
iv) 23.34
v) 1215.8
What can you conclude about the denominators through this process? (Page No. 10)
Answer:
i) 15.265

ii) 0.1255

iii) 0.4
0.4 = \(\frac{4}{10}\) = \(\frac{2}{5}\)
iv) 23.34

v) 1215.8

Two and five are the factors for the denominator.

Question 2.
Write the following rational numbers in the form of p/q, where q is of the form 2ⁿ.5^m where n, m are non-negative integers and then write the numbers in their decimal form. (Page No. 11)
i) \(\frac{3}{4}\)
ii) \(\frac{7}{25}\)
iii) \(\frac{51}{64}\)
iv) \(\frac{14}{25}\)
v) \(\frac{80}{100}\)
Answer:
i) \(\frac{3}{4}\)
\(\frac{3}{4}\) = \(\frac{3}{2 \times 2}\) = \(\frac{3}{2^{2}}\)

Decimal form of \(\frac{3}{4}\) = 0.75

ii) \(\frac{7}{25}\)
\(\frac{7}{25}\) = \(\frac{7}{5 \times 5}\) = \(\frac{7}{5^{2}}\)

Decimal form of \(\frac{7}{25}\) = 0.28

iii) \(\frac{51}{64}\)
\(\frac{51}{64}\) = \(\frac{51}{2^{6}}\)
[∵ 64 = 2 × 32
= 2² × 16
= 2³ × 8
= 2⁴ × 4 = 2⁵ × 2 = 2⁶]

Decimal form of \(\frac{51}{64}\) = 0.796875

iv) \(\frac{14}{25}\)

v) \(\frac{80}{100}\)
\(\frac{80}{100}\) = \(\frac{80}{2^{2} \times 5^{2}}\) = \(\frac{80}{10^{2}}\) = 0.80

Question 3.
Write the following rational numbers as decimal form and find out the block of repeating digits in the quotient. (Page No. 11)
i) \(\frac{1}{3}\)
ii) \(\frac{2}{7}\)
iii) \(\frac{5}{11}\)
iv) \(\frac{10}{13}\)
Answer:
i) \(\frac{1}{3}\)
\(\frac{1}{3}\) = 0.3333…. = \(0 . \overline{3}\)
Block of digits, repeating in the quotient = period = 3.

ii) \(\frac{2}{7}\)
Decimal form of \(\frac{2}{7}\) = 0.285714….
Repeating part/period = 285714
∴ \(\frac{2}{7}\) = \(0 . \overline{285714}\)

iii) \(\frac{5}{11}\)
Period = 45
Decimal form of \(\frac{5}{11}\) = 0.454545.
= \(0 . \overline{45}\)

iv) \(\frac{10}{13}\)
Decimal form of \(\frac{10}{13}\) = 0.769230.
= \(0 . \overline{769230}\)
Period = 769230

Do this

Question 1.
Verify the statement proved above for p = 2, p = 5 and for a² = 1, 4, 9, 25, 36, 49, 64 and 81. (Page No. 14)
Answer:


From the above we can conclude that if a prime number ‘p’ divides a², then it also divides a.

Think and Discuss

Question 1.
Write the nature of y, a and x in y = a^x. Can you determine the value of x for a given y? Justify your answer. (Page No. 17)
Answer:
y = a^x here a ≠ 0
We can determine the value of ‘x’ for a given y.
for example y = 5, a = 2
We cannot express y = a^x for y = 5, a = 2 and for y = 7, a = 3, we cannot express seven (7) as a power of 3.

Question 2.
You know that 2¹ = 2, 4¹ = 4, 8¹ = 8 and 10¹ = 10. What do you notice about the values of log₂ 2, log₄ 4, log₈ 8 and log₁₀ 10? What can you generalise from this?  (Page No. 18)
Answer:
From the graph log₂ 2 = log₄ 4 = log₈ 8 = log₁₀ 10 = 1
We conclude that log_a a = 1 where a is a natural number.

Question 3.
Does log₁₀ 0 exist? (Page No. 18)
Answer:
No, log₁₀ 0 doesn’t exist, i.e a^x ≠ 0 ∀ a, x ∈ N.

Question 4.
We know that, if 7 = 2^x then x = log₂ 7. Then what is the value of \(2^{\log _{2} 7}\)? Justify your answer. Generalise the above by taking some more examples for \(\mathbf{a}^{\log _{\mathrm{a}} \mathbf{N}}\). (Page No. 21)
Answer:
We know that if 7 = 2^x then x = log₂ 7
We want to find the value of \(2^{\log _{2} 7}\);
Now put log₂ 7 = x in the given
∴ \(2^{\log _{2} 7}\) = 2^x = 7 (given)
∴ \(2^{\log _{2} 7}\) = 7
Thus \(\mathbf{a}^{\log _{\mathrm{a}} \mathbf{N}}\) = N

a) \(3^{\log _{3} 8}\)
Answer:
If x = \(3^{\log _{3} 8}\) then
log₃ x = log₃ 8
⇒ x = 8

b) \(5^{\log _{5} 10}\)
Answer:
If y = \(5^{\log _{5} 10}\)
then log₅ y = log₅ 10
⇒ y = 10

Do this

Question 1.
Write the powers to which the bases to be raised in the following.  (Page No. 18)
i) 64 = 2^x
Answer:
64 = 2^x
We know that
64 = 2 × 32
= 2 × 2 × 16
= 2 × 2 × 2 × 8
= 2 × 2 × 2 × 2 × 4
= 2 × 2 × 2 × 2 × 2 × 2
64 = 2⁶
⇒ x = 6

ii) 100 = 5^b
Answer:
Here also 100 cannot be written as any power of 5.
i.e., there exists no integer for b such that 5^b = 100

iii) \(\frac{1}{81}\) = 3^c
Answer:
We know that 81 = 3 x 27
= 3 × 3 × 9
= 3 × 3 × 3 × 3
= 34
∴ \(\frac{1}{81}\) = 3^-4   [∵ a^-m = \(\frac{1}{\mathrm{a}^{\mathrm{m}}}\)]
∴ c = – 4

iv) 100 = 10^z
Answer:
100 = 10²
z = 2

v) \(\frac{1}{256}\) = 4^a
Answer:
We know that 256 = 4 × 64
= 4 × 4 × 16
= 4 × 4 × 4 × 4
∴ \(\frac{1}{256}\) = 4^-4
∴ a = – 4

Question 2.
Express the logarithms of the following into sum of the logarithms.   (Page No. 19)
i) 35 × 46
Answer:
log xy = log x + log y
log₁₀35 × 46 = log₁₀35 + log₁₀46

ii) 235 × 437
Answer:
log₁₀235 × 437 = log₁₀235 + log₁₀437   [∵ log xy = log x + log y]

iii) 2437 × 3568
Answer:
log₁₀ 2437 × 3568 = log₁₀2437 + log₁₀3568   [∵ log xy = log x + log y]

Question 3.
Express the logarithms of the follow¬ing into difference of the logarithms.   (Page No. 20)
i) \(\frac{23}{34}\)
Answer:
log₁₀ = \(\frac{23}{34}\) = log₁₀ 23 – log₁₀ 34
[∵ log \(\frac{x}{y}\) = log x – log y]

ii) \(\frac{373}{275}\)
Answer:
log₁₀ = \(\frac{373}{275}\) = log₁₀ 373 – log₁₀ 275
[∵ log \(\frac{x}{y}\) = log x – log y]

iii) \(\frac{4525}{3734}\)
Answer:
log₁₀ = \(\frac{4525}{3734}\) = log₁₀ 4525 – log₁₀ 3734
[∵ log \(\frac{x}{y}\) = log x – log y]

iv) \(\frac{5055}{3303}\)
Answer:
log₁₀ = \(\frac{5055}{3303}\) = log₁₀ 5055 – log₁₀ 3303
[∵ log \(\frac{x}{y}\) = log x – log y]

Question 4.
By using the formula log_axⁿ = n log_a x, convert the following.   (Page No. 21)
i) log₂ 7²⁵
Answer:
log₂ 7²⁵ = 25 log₂ 7

ii) log₅ 8⁵⁰
Answer:
log₅ 8⁵⁰ = 50 log₅ 8 = 50 log₅ 23
= 3 × 50 log₅2 = 150 log₅2

iii) log 5²³
Answer:
log 5²³ = 23 log 5

iv) log 1024
Answer:
log 1024 = log 2¹⁰ [∵ 1024 = 2¹⁰]
= 10 log 2

Try this

Question 1.
Write the following relation in exponential form and find the values of respective variables.   (Page No. 18)
i) log₂32 = x
Answer:
log₂32 = x
⇒ log₂2⁵ = x     [∵ 32 = 2⁵]
⇒ 5 log₂2 = x     [∵ log a^m = m log a]
⇒ 5 × 1 = x      [∵ log_a a = 1]
∴ x = 5

ii) log₅625 = y
Answer:
log₅625 = y
⇒ log₅4 = y    [∵ 625 = 5⁴]
⇒ 4 log₅ 5 = y     [∵ log a^m = m log a]
⇒ 4 × 1 = y     [∵ log_a a = 1]
∴ y = 4

iii) log₁₀10000 = z
Answer:
log₁₀10000 = z
=> log₁₀10⁴ = z     [∵ 10000 = 10 × 10 × 10 × 10 = 10⁴]
=> 4 log₁₀10 = z     [∵ log a^m = m log a]
=> 4 × 1 = z     [∵ log_a a = 1]
∴ z = 4

iv) \(\log _{7} \frac{1}{343}\) = -a
Answer:

Question 2.
i) Find the value of log₂32. (Page No. 21)
Answer:
log₂ 32 = log₂ 2⁵
[∵ 32 = 2 × 2 × 2 × 2 × 2 = 2⁵]
= 5 log₂ 2 [∵ log a^m = m log a]
= 5 × 1 [∵ log_a a = 1]
= 5

ii) Find the value of log_c √c.
Answer:

= \(\frac{1}{2}\) × 1 [∵ log_a a = 1]
= \(\frac{1}{2}\)

iii) Find the value of log₁₀0.001
Answer:

iv) Find the value of \(\log _{\frac{2}{3}} \frac{8}{27}\)
Answer:

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers Ex 1.5 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.5

10th Class Maths 1st Lesson Real Numbers Ex 1.5 Textbook Questions and Answers

Question 1.
Determine the values of the following,
i) log₂₅5
Answer:

ii) log₈₁3
Answer:

iii) log₂(\(\frac{1}{16}\))
Answer:

iv) log₇1
Answer:
log₇1 = log₇7⁰ = 0 log₇7 = 0

v) log_x√x
Answer:

vi) log₂512
Answer:
log₂512 = log₂2⁹    [∵ 512 = 2⁹]
= 9log₂2   [∵ log x^m = m log x]
= 9 × 1    [∵ log_aa = 1]
= 9

vii) log₁₀0.01
Answer:

viii) \(\log _{\frac{2}{3}}\left(\frac{8}{27}\right)\)
Answer:

ix) \(2^{2+\log _{2} 3}\)
Answer:
\(2^{2+\log _{2} 3}\) = 2² . \(2^{\log _{2} 3}\)   [∵ a^m . aⁿ = a^m+n]
= 4 × 3 [∵ \(\log _{\mathrm{a}} \mathrm{N}\) = N]
= 12

Question 2.
Write the following expressions as log N and find their values.
i) log 2 + log 5
Answer:
log 2 + log 5
= log 2 × 5   [∵ log m + log n = log mn]
= log 10
= 1

ii) log₂ 16 – log₂ 2
Answer:

iii) 3 log₆₄4
Answer:

iv) 2 log 3 – 3 log 2
Answer:
2 log 3 – 3 log 2
= log 3² – log 2³
= log 9 – log 8
= log \(\frac{9}{8}\)

v) log 10 + 2 log 3 – log 2
Answer:
log 10 + 2 log 3 – log 2
= log 10 + log 3² – log 2
= log 10 + log 9 – log 2    [∵ m log a = log a^m]
= log \(\frac{10 \times 9}{2}\)    [∵ log a + log b = log ab; log a – log b = log \(\frac{a}{b}\)]
= log 45

Question 3.
Evaluate each of the following in terms of x and y, if it is given x = log₂3 and y = log₂ 5.
i) log₂15
Answer:
log₂15 = log₂ 3 × 5
= log₂3 + log₂5   [∵ log mn = log m + log n]
= x + y

ii) log₂7.5
Answer:

iii) log₂60
Answer:
log₂60 = log₂2² × 3 × 5
= log₂2² + log₂3 + log₂5
= 2 log₂2 + x + y
= 2 + x + y

iv) log₂6750
Answer:

log₂6750
= log₂2 × 3³ × 5³
= log₂2 + log₂33 + log₂5³
= 1 + 3 log₂3 + 3 log₂5
= 1 + 3x + 3y

Question 4.
Expand the following,
i) log 1000
Answer:
log 1000 = log 10³
= 3 log 10
= 3 × 1
= 3

ii) \(\log \left[\frac{128}{625}\right]\)
Answer:

iii) log x²y³z⁴
Answer:
log x²y³z⁴ = logx² + logy³ + logz⁴ [∵ log ab = log a + log b]
= 2 log x + 3 log y + 4 log z
[∵ log a^m = m log a]

iv) \(\log \frac{\mathbf{p}^{2} \mathbf{q}^{3}}{\mathbf{r}}\)
Answer:

iv) \(\log \sqrt{\frac{x^{3}}{y^{2}}}\)
Answer:

Question 5.
If x² + y² = 25xy, then prove that 2 log (x + y) = 3log3 + logx + logy.
Answer:
Given: x² + y² = 25xy
We know that (x + y)² = x² + y² + 2xy
= 25xy + 2xy    [∵ x² + y² = 25xy given]
(x + y)² = 27xy
Taking ‘log’ on both sides
log (x + y)² = log 27xy
2 log (x + y) = log 27 + log x + log y
= log 3³ + log x + log y
⇒ 2 log (x + y) = 3log3 + log x + log y

Question 6.
If \(\log \left(\frac{\mathbf{x}+\mathbf{y}}{3}\right)\) = \(\frac{1}{2}\) (log x + log y), then find the value of \(\frac{x}{y}+\frac{y}{x}\).
Answer:

(squaring on both sides)
⇒ (x + y)² = (3√xy)²
⇒ x² + y² + 2xy = 9xy
⇒ x² + y² = 9xy – 2xy = 7xy

Question 7.
If (2.3)^x = (0.23)^y = 1000 then find the value of \(\frac{1}{x}-\frac{1}{y}\).
Answer:
Given (2.3)^x = (0.23)^y = 1000 = 10³

Question 8.
If 2^x+1 = 3^1-x then find the value of x.
Answer:
Given: 2^x+1 = 3^1-x
log 2^x+1 = log 3^1-x
(x + 1) log 2 = (1 – x) log 3
x log 2 + log 2 = log 3 – x log 3
x log 2 + x log 3 = log 3 – log 2
x (log 3 + log 2) = log 3 – log 2

Question 9.
Is
i) log 2 is rational or irrational? Justify your answer.
Answer:
Let log₁₀2 = x
Then 10^x = 2
But 2 can’t be written as 10^x for any value of x
∴ log 2 is irrational.

ii) log 100 is rational or irrational? Justify your answer.
Answer:
Let log₁₀100 = x
⇒ log₁₀10² = x
⇒ 2 log₁₀10 = x = 2
∴ log 100 is rational.
∴ log 100 = 2
Hence rational.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers Ex 1.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.4

10th Class Maths 1st Lesson Real Numbers Ex 1.4 Textbook Questions and Answers

Question 1.
Prove that the following are irrational,
i) \(\frac{1}{\sqrt{2}}\)
ii) √3 + √5
iii) 6 + √2
iv) √5
v) 3 + 2√5
Answer:
i) \(\frac{1}{\sqrt{2}}\)
On the contrary, suppose that is a \(\frac{1}{\sqrt{2}}\) rational number;
then \(\frac{1}{\sqrt{2}}\) is of the form where \(\frac{p}{q}\) and q are integers.
∴ \(\frac{1}{\sqrt{2}}\) = \(\frac{p}{q}\)
⇒ \(\frac{\sqrt{2}}{1}\) = \(\frac{q}{p}\)
(i.e.,) √2 is a rational number and it is a contradiction. This contradiction arised due to our supposition that \(\frac{1}{\sqrt{2}}\) is a rational number.
Hence \(\frac{1}{\sqrt{2}}\) is an irrational number.

ii) Suppose √3 + √5 is not an irrational number.
Then √3 + √5 must be a rational number.
√3 + √5 = \(\frac{p}{q}\), q ≠ 0 and p, q ∈ Z
Squaring on both sides

but √15 is an irrational number.
\(\frac{p^{2}-8 q^{2}}{2 q^{2}}\) is a rational number
(p² – 8q², 2q² ∈ Z, 2q² ≠ 0)
but an irrational number can’t be equal to a rational number, so our supposition that √3 + √5 is not an irrational number is false.
∴ √3 + √5 is an irrational number.

iii) 6 + √2
To prove: 6 + √2 is an irrational number.
Let us suppose that 6 + √2 is a rational number.
∴ 6 + √2 = \(\frac{p}{q}\), q ≠ 0
⇒ √2 = \(\frac{p}{q}\) – 6
⇒ √2 = Difference of two rational numbers
⇒ √2 = rational number But this contradicts the fact that √2 is an irrational number.
∴ Our supposition is wrong.
Hence the given statement is true.
6 + √2 is an irrational number.

iv) √5
To prove: √5 is an irrational number.
On the contrary, let us assume that √5 is a rational number.
∴ √5 = \(\frac{p}{q}\), q ≠ 0
If p, q have a common factor, on cancelling the common factor let it be
reduces to \(\frac{a}{b}\) where a, b are co-primes.
Now √5 = \(\frac{a}{b}\), where HCF (a, b) = 1
Squaring on both sides we get
⇒ (√5)² = \(\left(\frac{a}{b}\right)^{2}\)
⇒ 5 = \(\frac{\mathrm{a}^{2}}{\mathrm{~b}^{2}}\)
⇒ 5b² = a²
⇒ 5 divides a² and thereby 5 divides 5
Now, take a = 5c
then, a² = 25c²
i.e., 5b² = 25c²
⇒ b² = 5c²
⇒ 5 divides b² and thereby b.
⇒ 5 divides both b and a.
This contradicts that a and b are co-primes.
This contradiction arised due to our assumption that √5 is a rational number.
Hence our assumption is wrong and the given statement is true, i.e., √5 is an irrational number,

v) 3 + 2√5
To Prove: 3 + 2√5 is an irrational.
On the contrary, let us assume that 3 + 2√5 is a rational number.

Here p, q being integers we can say that \(\frac{p-3q}{2q}\) is a rational number.
This contradicts the fact that √5 is an irrational number. This is due to our assumption “3 + 2√5 is a rational number”.
Hence our assumption is wrong.
∴ 3 + 2√5 is an irrational number.

Question 2.
Prove that √p + √q is an irrational, where p, q are primes.
Answer:
Given that p, q are primes.
Hence fp and fq are irrationals.
[∵ p, q have no factors other than 1 ∵ they are primes.]
Now √p + √q = sum of two irrational numbers = an irrational number
Hence proved.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers Ex 1.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.3

10th Class Maths 1st Lesson Real Numbers Ex 1.3 Textbook Questions

Question 1.
Write the following rational numbers in their decimal form and also state which are terminating and which have non-terminating repeating decimals.
i) \(\frac{3}{8}\)
ii) \(\frac{229}{400}\)
iii) 4\(\frac{1}{5}\)
iv) \(\frac{2}{11}\)
v) \(\frac{8}{125}\)
Answer:
i) \(\frac{3}{8}\)

[!! Denominator 8 = 2³, consists of only 2’s. Hence a terminating decimal.]
∴ \(\frac{3}{8}\) = 0.375 is a terminating decimal.

ii) \(\frac{229}{400}\)

[!! Denominator 400 = 2⁴ × 5² = 2ⁿ × 5^m. Hence a terminating decimal.]
∴ \(\frac{229}{400}\) = 0.5725 is a terminating decimal.

iii) 4\(\frac{1}{5}\)
4\(\frac{1}{5}\) = 4 + \(\frac{1}{5}\)

[!! Denominator is 5. Hence a terminating decimal.]
∴ 4\(\frac{1}{5}\) = 4.2 is a terminating decimal.

iv) \(\frac{2}{11}\)

[!! Denominator is not of the form 2^m × 5ⁿ. Hence a non-terminating repeating decimal.]
∴ \(\frac{2}{11}\) = \(0 . \overline{18}\) is a non terminating, repeating decimal.

v) \(\frac{8}{125}\)

[!! Denominator 125 = 5³. Hence a terminating decimal.]
∴ \(\frac{8}{125}\) = 0.064 is a terminating decimal.

Question 2.
Without performing division, state whether the following rational numbers will have a terminating decimal form or a non-terminating, repeating decimal form.
i) \(\frac{13}{3125}\)
ii) \(\frac{11}{12}\)
iii) \(\frac{64}{455}\)
iv) \(\frac{15}{1600}\)
v) \(\frac{29}{343}\)
vi) \(\frac{23}{2^{3} \cdot 5^{2}}\)
vii) \(\frac{129}{2^{2} \cdot 5^{7} \cdot 7^{5}}\)
viii) \(\frac{9}{15}\)
iX) \(\frac{36}{100}\)
X) \(\frac{77}{210}\)
Answer:
i) \(\frac{13}{3125}\)
Note: We check whether the denominator is of the form 2ⁿ . 5^m or not? If yes, the rational number can be expressed as a terminating decimal. If not, it can’t be expressed as a terminating decimal.

[!! Denominator is of the form 2^m × 5ⁿ. Hence a terminating decimal.]
3125 = 5⁵
∴ \(\frac{13}{3125}\) is a terminating decimal.

ii) \(\frac{11}{12}\)
The denominator 12 is not a factor of 11. Moreover 12 = 2² × 3.
[!! Denominator is not of the form 2^m × 5ⁿ.]
∴ \(\frac{11}{12}\) is a non terminating, repeating decimal.

iii) \(\frac{64}{455}\)

[!! Denominator is not of the form 2^m × 5ⁿ. Hence a non terminating decimal.]
∴ 455 = 5 × 7 × 13
Hence\(\frac{64}{455}\) is a non terminating, repeating decimal.

iv) \(\frac{15}{1600}\)

∴ 1600 = 2⁶ × 5² [∵ The denominator is of the form 2ⁿ . 5^m]
Hence \(\frac{15}{1600}\) is a terminating decimal.

v) \(\frac{29}{343}\)

343 = 7³ [Not of the form 2ⁿ . 5^m]
∴ \(\frac{29}{343}\) is a non terminating, repeating decimal.

vi) \(\frac{23}{2^{3} \cdot 5^{2}}\)
\(\frac{23}{2^{3} \cdot 5^{2}}\) is a terminating decimal.
[∵ The denominator is of the form 2ⁿ . 5^m]

vii) \(\frac{129}{2^{2} \cdot 5^{7} \cdot 7^{5}}\)
\(\frac{129}{2^{2} \cdot 5^{7} \cdot 7^{5}}\) is a non terminating, repeating decimal.

viii) \(\frac{9}{15}\)
\(\frac{9}{15}\) = \(\frac{3}{5}\)
Denominator is of the form 2ⁿ . 5^m.
∴ \(\frac{9}{15}\) = \(\frac{3}{5}\) is a terminating decimal.

ix) \(\frac{36}{100}\)
100 = 2² × 5² is of the form 2ⁿ . 5^m
Hence \(\frac{36}{100}\) is a terminating decimal.

x) \(\frac{77}{210}\)

210 = 2 × 3 × 5 × 7 is not of the form 2ⁿ . 5^m
Given fraction has a non-terminating, repeating decimal expansion.

Question 3.
Write the following rationals in decimal form using Theorem 1.4.
i) \(\frac{13}{25}\)
ii) \(\frac{15}{16}\)
iii) \(\frac{23}{2^{3} \cdot 5^{2}}\)
iv) \(\frac{7218}{3^{2} \cdot 5^{2}}\)
v) \(\frac{143}{110}\)
Answer:
i) \(\frac{13}{25}\)

ii) \(\frac{15}{16}\)
\(\frac{15}{16}\) = \(\frac{15}{2 \times 2 \times 2 \times 2}\)

iii) \(\frac{23}{2^{3} \cdot 5^{2}}\)

iv) \(\frac{7218}{3^{2} \cdot 5^{2}}\)

v) \(\frac{143}{110}\)

Question 4.
The decimal form of some real numbers are given below. In each case, decide whether the number is rational or not. If it is rational, and expressed in form p/q, what can you say about the prime factors of q?
i) 43.12345678?
ii) 0.120120012000120000 ……….
iii) \(43 . \overline{123456789}\)
Answer:
i) 43.123456789
The given decimal expansion is terminating. Hence it is a rational number and the denominator q is of the form 2ⁿ . 5^m.

ii) 0.120120012000120000 …………
The given decimal expansion is neither terminating nor repeating.
Hence it is not a rational number. It represents an irrational number.

iii) \(43 . \overline{123456789}\)
The given real number is a repeating decimal with period 123456789. Hence it is a rational number.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers Ex 1.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.2

10th Class Maths 1st Lesson Real Numbers Ex 1.2 Textbook Questions and Answers

Question 1.
Express each of the following numbers as a product of its prime factors.
i) 140
ii) 156
iii) 3825
iv) 5005
v) 7429
Answer:
i) 140

∴ 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7

ii) 156

∴ 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13

iii) 3825

∴ 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17

iv) 5005

∴ 5005 = 5 × 7 × 11 × 13

v) 7429

∴ 7429 = 17 × 19 × 23

Question 2.
Find the L.C.M and H.C.F of the following integers by the prime factorization method.
i) 12, 15 and 21
ii) 17, 23 and 29
iii) 8, 9 and 25
iv) 72 and 108
v) 306 and 657
Answer:
i) 12, 15 and 21
12 = 2 × 2 × 3 = 22 × 3
15 = 3 × 5
21 = 3 × 7
L.C.M = 2² × 3 × 5 × 7 = 420
H.C.F = 3

ii) 17, 23 and 29
The given numbers 17, 23 and 29 are all primes.
L.C.M = their product
= 17 × 23 × 29 = 11339
∴ H.C.F = 1

iii) 8, 9 and 25
8 = 2 × 2 × 2 = 2³
9 = 3 × 3 = 3²
25 = 5 × 5 = 5²
L.C.M = 2³ × 3² × 5² = 1800
(or)
8, 9 and 25 are relatively prime, therefore L.C.M is equal to their product,
(i.e.,) L.C.M = 8 × 9 × 25 = 1800
H.C.F = 1

iv) 72 and 108

72 = 2³ × 3²
108 = 2² × 3³
L.C.M = 2³ × 3³ = 8 × 27 = 216
H.C.F = 2² × 3² = 4 × 9 = 36

v) 306 and 657

306 = 2 × 3² × 17
657 = 32 × 73
L.C.M = 2 × 3² × 17 × 73 = 22338
H.C.F = 3² = 9

Question 3.
Check whether 6ⁿ can end with the digit ‘0’ for any natural number n.
Answer:
Given number = 6ⁿ = (2 × 3)ⁿ
The prime factors here are 2 and 3 only.
To be end with 0; 6ⁿ should have a prime factor 5 and also 2.
So, 6ⁿ can’t end with zero.

Question 4.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Answer:
Given numbers are 7 × 11 × 13
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
⇒ 13(7 × 11 + 1) and
5(7 × 6 × 4 × 3 × 2 × 1 + 1)
⇒ 13 K and 5 L, where K = 78 and L = 7 × 6 × 4 × 3 × 2 × 1 + 1 = 1009
As the given numbers can be written as product of two numbers, they are composite.

Question 5.
How will you show that (17 × 11 × 2) + (17 × 11 × 5) is a composite number? Explain.
Answer:
(17 × 11 × 2) + (17 × 11 × 5)
= (17 × 11) (2 + 5)
= (17 × 11) (7)
= 187 × 7
Now the given expression is written as a product of two integers and hence it is a composite number.

Question 6.
What is the last digit of 6100?
Answer: We know that
6¹ = 6
6² = 36
6³ = 216
6⁴ = 1296
6⁵ = 7776
We see that 6ⁿ for any positive integer n ends is 6.
i.e., unit digit is always 6.
∴ Unit digit of 6¹⁰⁰ is 6.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers Ex 1.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.1

10th Class Maths 1st Lesson Real Numbers Ex 1.1 Textbook Questions and Answers

Question 1.
Use Euclid’s division algorithm to find the HCF of
i) 900 and 270
Answer:
900 = 270 × 3 + 90
270 = 90 × 3 + 0
∴ HCF = 90

ii) 196 and 38220
Answer:
38220 = 196 × 195 + 0
∴ 196 is the HCF of 196 and 38220.

iii) 1651 and 2032
Answer:
2032 = 1651 × 1 + 381
1651 = 381 × 4 + 127
381 = 127 × 3 + 0
∴ HCF = 127

Question 2.
Use Euclid division lemma to show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integers.
Answer:
Let ‘a’ be an odd positive integer.
Let us now apply division algorithm with a and b = 6.
∵ 0 ≤ r < 6, the possible remainders are 0, 1, 2, 3, 4 and 5.
i.e., ’a’ can be 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5, where q is the quotient.
But ‘a’ is taken as an odd number.
∴ a can’t be 6q or 6q + 2 or 6q + 4.
∴ Any odd integer is of the form 6q + 1, 6q + 3 or 6q + 5.

Question 3.
Use Euclid’s division lemma to show that the square of any positive integer is of the form 3p, 3p + 1.
Answer:
Let ‘a’ be the square of an integer.
Applying Euclid’s division lemma with a and b = 3
Since 0 ≤ r < 3, the possible remainders are 0, 1, and 2.
∴ a = 3q (or) 3q + 1 (or) 3q + 2
∴ Any square number is of the form 3q, 3q + 1 or 3q + 2, where q is the quotient.
(or)
Let ‘a’ be a positive integer
So it can be expressed as a = bq + r (from Euclideans lemma)
now consider b = 3 then possible values of ‘r’ are ‘0’ or ‘1’ or 2.
then a = 3q + 0 = 3q (or) 3q + 1 or 3q + 2 now square of given positive integer (a²) will be
Case – I: a² – (3q)² = 9q²=3(3q²) = 3p (p = 3q²)
Case-II: a² = (3q + l)² = 9q² + 6q+ 1
= 3[3q² + 2q] + 1 = 3p+l (Where p = 3q² + 2q) or
Case – III: a² = (3q + 2)² = 9q² + 12q + 4 = 9q² + 12q + 3 + 1
= 3[3q² + 4q + 1] + 1
= 3p + 1 (where ‘p’ = 3q² + 4q + 1)
So from above cases 1, 2, 3 it is clear that square of a positive integer (a) is of the form 3p or 3p + 1
Hence proved.

Question 4.
Use Euclid’s division lemma to show that the cube of a positive integer is of the form 9m, 9m + 1 or 9m + 8.
(OR)
Show that the cube of any positive integer is of form 9m or 9m + 1 or 9m + 8, where m is an integer.
Answer:
Let ‘a’ be positive integer. Then from Euclidean lemma a = bq + r;
now consider b = 9 then 0 ≤ r < 9, it means remainder will be 0, or 1, 2, 3, 4, 5, 6, 7, or 8
So a = bq + r
⇒ a = 9q + r (for b = 9)
now cube of a = a³ + (9q + r)³
= (9q)³ + 3.(9q)³r + 3. 9q.r + r³
= 9³q³ + 3.9²(q²r) + 3.9(q.r) + r³
= 9[9².q³ + 3.9.q²r + 3.q.r] + r³
a³ = 9m + r³ (where ‘m’ = 9²q³ + 3.9.q2r + 3.q.r)
if r = 0 ⇒ r³ = 0 then a³ = 9m + 0 = 9m
and for r = 1 ⇒ r³ = l³ then a³ = 9m + 1
and for r = 2 ⇒ r³ = 2³ then a³ = 9m + 8
for r = 3 ⇒ r³, = 3³ ⇒ a³ = 9m + 27 = 9(m) where m = (9m +3)
for r = 4 ⇒ r³ = 4³ ⇒ a³ = 9m + 64 = (9m + 63) + 1 = 9m + 1
for r = 5 ⇒ r³ = 125 ⇒ a³ = 9m + 125 = (9m + 117) + 8 = 9m + 8
for r = 6 ⇒ r³ — 216 ⇒ a³ = 9m + 216 = 9m + 9(24) = 9m
for r = 7 ⇒ r³ = 243
⇒ a³ = 9m + 9(27) = 9m
for r = 8 ⇒ r³ = 512
⇒ a³ = 9m + 9(56) + 8 = 9m + 8
So from the above it is clear that a³ is either in the form of 9m or 9m + 1 or 9m + 8.
Hence proved.

Question 5.
Show that one and only one out of n, n + 2 or n + 4 is divisible by 3, where n is any positive integer.
(Or)
Show that one and only one out of a, a + 2 and a + 4 is divisible by 3 where ‘a’ is any positive integer.
Answer:
Let ‘n’ be any positive integer.
Then from Euclidean’s lemma n = bq + r (now consider b = 3)
⇒ n = 3q + r (here 0 ≤ r < 3) which means the possible values of ‘r’ = 0 or 1 or 2
Now consider r = 0 then ‘n’ = 3q (divisible by 3)
and n + 2 = 3q + 2 (not divisible by 3)
n + 4 = 3q + 4 (not divisible by 3)
Case – II: For r = 1
n = 3q + 1 (not divisible by 3)
n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + l) divisible by 3
n + 4 = 3q + 1 + 4 = 3q + 5 not divisible by 3
Case – III: For r = 2,
n = 3q + 2 not divisible by 3
n + 2 = 3q + 2 + 2 = 3q + 4, not divisible by 3
n + 4 = 3q + 2 + 4 = 3q + 6 = 3(q + 2) divisible by 3
So in all above three cases we observe, only one of either (n) or (n + 1) or (n + 4) is divisible by 3.
Hence proved.