Category: Class 10

Students can download Maths Chapter 3 Algebra Ex 3.12 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.12

Question 1.
If the difference between a number and its reciprocal is \(\frac { 24 }{ 5 } \), find the number.
Answer:
Let the number be “x” and its reciprocal is \(\frac { 1 }{ x } \)
By the given condition
x – \(\frac { 1 }{ x } \) = \(\frac { 24 }{ 5 } \)
\(\frac{x^{2}-1}{x}\) = \(\frac { 24 }{ 5 } \)
5x² – 5 = 24x
5x² – 24x – 5 = 0

5x² – 25x + x – 5 = 0 ⇒ 5x (x – 5) + 1(x – 5) = 0
(x – 5) (5x – 1) = 0 ⇒ x – 5 = 0 or 5x + 1 = 0
x = 5 or 5x = -1 ⇒ x = \(\frac { -1 }{ 5 } \)
The number is 5 or \(\frac { -1 }{ 5 } \)

Question 2.
A garden measuring 12m by 16m is to have a pedestrian pathway that is meters wide installed all the way around so that it increases the total area to 285 m². What is the width of the pathway?
Answer:
Let the width of the rectangle be “ω”
Length of the outer rectangle = 16 + (ω + ω)
16 + 2ω
Breadth of the outer rectangle = 12 + 2ω

By the given condition
(16 + 2ω) (12 + 2ω) = 285
192 + 32 ω + 24 ω + 4 ω² = 285
4 ω² + 56ω = 285 – 192
4 ω² + 56 ω = 93
4 ω²+ 56 ω – 93 = 0
Here a = 4, b = 56, c = -93

= 1.5 or -15.5 (Width is not negative)
∴ Width of the path way = 1.5 m

Question 3.
A bus covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more it would have taken 30 minutes less for the journey. Find the original speed of the journey.
Answer:
Let the original speed of the bus be “x” km/hr
Time taken to cover 90 km = \(\frac { 90 }{ x } \)
After increasing the speed by 15 km/hr
Time taken to cover 90 km = \(\frac { 90 }{ x+15 } \)
By the given condition
\(\frac { 90 }{ x } \) – \(\frac { 90 }{ x+15 } \) = \(\frac { 1 }{ 2 } \)
\(\frac{90(x+15)-90 x}{x(x+15)}\) = \(\frac { 1 }{ 2 } \)
90x + 1350 – 90x = \(\frac{x^{2}+15 x}{2}\)
1350 = \(\frac{x^{2}+15 x}{2}\)
2700 = x² + 15x

x² + 15x – 2700 = 0
(x + 60) (x – 45) = 0
x + 60 = 0 or x – 45 = 0
x = – 60 or x = 45
The speed will not be negative
∴ Original speed of the bus = 45 km/hr

Question 4.
A girl is twice as old as her sister. Five years hence, the product of their ages – (in years) will be 375. Find their present ages.
Answer:
Let the age of the sister be “x”
The age of the girl = 2x
Five years hence
Age of the sister = x + 5
Age of the girl = 2x + 5
By the given condition
(x + 5) (2x + 5) = 375
2x² + 5x + 10x + 25 = 375
2x² + 15x – 350 = 0
a = 2, b = 15, c = -350

Age will not be negative
Age of the girl = 10 years
Age of the sister = 20 years (2 × 10)

Question 5.
A pole has to be erected at a point on the boundary of a circular ground of diameter 20 m in such a way that the difference of its distances from two diametricallyopposite fixed gates P and Q on the boundary is 4 m. Is it possible to do so? If answer is yes at what distance from the two gates should the pole be erected?
Answer:
Let “R” be the required location of the pole
Let the distance from the gate P is “x” m : PR = “x” m
The distance from the gate Q is (x + 4)m
∴ QR = (x + 4)m
In the right ∆ PQR,

PR² + QR² = PQ² (By Pythagoras theorem)
x² + (x + 4)² = 20²
x² + x² + 16 + 8x = 400
2x² + 8x – 384 = 0

x² + 4x – 192 = 0(divided by 2)
(x + 16) (x – 12) = 0
x + 16 = 0 or x – 12 = 0 [negative value is not considered]
x = -16 or x = 12
Yes it is possible to erect
The distance from the two gates are 12 m and 16 m

Question 6.
From a group of 2x² black bees , square root of half of the group went to a tree. Again eight-ninth of the bees went to the same tree. The remaining two got caught up in a fragrant lotus. How many bees were there in total?
Answer:
Total numbers of black bees = 2x²
Half of the group = \(\frac { 1 }{ 2 } \) × 2x² = x²
Square root of half of the group = \(\sqrt{x^{2}}\) = x
Eight – ninth of the bees = \(\frac { 8 }{ 9 } \) × 2x² = \(\frac{16 x^{2}}{9}\)
Number ofbees in the lotus = 2
By the given condition
x + \(\frac{16 x^{2}}{9}\) + 2 = 2x²
(Multiply by 9) 9x + 16×2 + 18 = 18x²

18x² – 16x² – 9x – 18 = 0 ⇒ 2x² – 9x – 18 = 0
2x² – 12x + 3x – 18 = 0
2x(x – 6) + 3 (x – 6) = 0
(x – 6) (2x + 3) = 0
x – 6 = 0 or 2x + 3 = 0
x = 6 or 2x = -3 ⇒ x = \(\frac { -3 }{ 2 } \) (number of bees will not be negative)
Total number of black bees = 2x² = 2(6)²
= 72

Question 7.
Music is been played in two opposite galleries with certain group of people. In the first gallery a group of 4 singers were singing and in the second gallery 9 singers were singing. The two galleries are separated by the distance of 70 m. Where should a person stand for hearing the same intensity of the singers voice?
(Hint: The ratio of the sound intensity is equal to the square of the ratio of their corresponding distances).
Answer:
Number of singers in the first group = 4
Number of singers in the second group = 9
Distance between the two galleries = 70 m
Let the distance of the person from the first group be x
and the distance of the person from the second group be 70 – x
By the given condition
4 : 9 = x² : (70 – x)² (by the given hint)
\(\frac { 4 }{ 9 } \) = \(\frac{x^{2}}{(70-x)^{2}}\)
\(\frac { 2 }{ 3 } \) = \(\frac { x }{ 70-x } \) [taking square root on both sides]
3x = 140 – 2x
5x = 140
x = \(\frac { 140 }{ 5 } \) = 28
The required distance to hear same intensity of the singers voice from the first galleries is 28m
The required distance to hear same intensity of the singers voice from the second galleries is (70 – 28) = 42 m

Question 8.
There is a square field whose side is 10 m. A square flower bed is prepared in its centre leaving a gravel path all round the flower bed. The total cost of laying the flower bed and gravelling the path at ₹3 and ₹4 per square metre respectively is ₹364. Find the width of the gravel path.
Answer:
Let the width of the gravel path be ‘x’
Side of the flower bed = 10 – (x + x)
= 10 – 2x

Area of the path way = Area of the field – Area of the flower bed
= 10 × 10 – (10 – 2x) (10 – 2x) sq.m
= 100 – (100 + 4x² – 40x)
= 100 – 100 – 4x² + 40x
= 40x – 4x² sq.m
Area of the flower bed = (10 – 2x) (10 – 2x) sq.m.
= 100 + 4x² – 40x
By the given condition

3(100 + 4x² – 40x) + 4(40x – 4x²) = 364
300 + 12x² – 120x + 160x – 16x² = 364
-4x² + 40x + 300 – 364 = 0
-4x² + 40x – 64 = 0
(÷ by 4) ⇒ x² – 10x + 16 = 0
[The width must not be equal to 8 m since the side of the field is 10m]
(x – 8) (x – 2) = 0
x – 8 = 0 or x – 2 = 0 x = 8 or x = 2
Width of the gravel path = 2 m

Question 9.
Two women together took 100 eggs to a market, one had more than the other. Both sold them for the same sum of money. The first then said to the second: “If I had your eggs, I would have earned ₹ 15”, to which the second replied: “If I had your eggs, I would have earned ₹ 6 \(\frac{2}{3}\) How many eggs did each had in the beginning?
Solution:
Let the no. of eggs with woman 1 be x and woman 2 be y.
∴ x + y = 100
Let w₁ sell the eggs at ₹ ‘a’ per egg.
Let w₂ sell the eggs at ₹ ‘b’ per egg.
Case 1:
They sold them for same money.
∴ ax = by
Case 2:
ay = 15 and bx = \(\frac{20}{3}\)
∴ One woman had 40 eggs and the other had 60 eggs.

Question 10.
The hypotenuse of a right angled triangle is 25 cm and its perimeter 56 cm. Find the length of the smallest side.
Answer:
Perimeter of a right angle triangle = 56 cm
Sum of the two sides + hypotenuse = 56
Sum of the two sides = 56 – 25
= 31 cm
Let one side of the triangle be “x”

The other side of the triangle = (31 – x) cm
By Pythagoras theorem
AB² + BC² = AC²
x² + (31 – x)² = 25²
x² + 961 + x² – 62x = 625

2x² – 62x + 961 – 625 = 0
2x² – 62x + 336 = 0 ⇒ x² – 31x + 168 = 0
(x – 24) (x – 7) = 0
x – 24 = 0 (or) x – 7 = 0
x = 24 (or) x = 7
Length of the smallest side is 7 cm

Students can download Maths Chapter 3 Algebra Ex 3.2 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.2

Question 1.
Find the GCD of the given polynomials by Division Algorithm
(i) x⁴ + 3x³ – x – 3, x³ + x² – 5x + 3
Answer:
p(x) = x⁴ + 3x³ – x – 3
g(x) = x³ + x² – 5x + 3

3x² + 6x – 9 = 3(x² + 2x – 3)
Now dividing g(x) = x³ + x² – 5x + 3
by the new remainder
(leaving the constant 3)
we get x² + 2x – 3
G.C.F. = x² + 2x – 3

(ii) x⁴ – 1, x³ – 11x² + x – 11
p(x) = x⁴ – 1
g(x) = x³ – 11x² + x – 11

120x² + 120 = 120 (x² + 1)
Now dividing g(x) = x³ – 11x² + x – 11 by the new remainder (leaving the constant) we get x² + 1

G.C.D. = x² + 1

(iii) 3x⁴ + 6x³ – 12x² – 24x, 4x⁴ + 14x³ + 8x² – 8x
Answer:
p(x) = 3x⁴ + 6x³ – 12x² – 24x
= 3x (x³ + 2x² – 4x – 8)
g(x) = 4x⁴ + 14x³ + 8x² – 8x
= 2x (2x³ + 7x² + 4x – 4)
G.C.D. of 3x and 2x = x
Now g(x) is divide by p(x) we get

3x² + 12x + 12 = 3 (x² + 4x + 4)
Now dividing p(x) = x³ + 2x² – 4x – 8
by the new remainder
(leaving the constant)
x² + 4x + 4

G.C.D. = x(x² + 4x + 4) [Note x is common for p(x) and g(x)]

(iv) 3x³ + 3x² + 3x + 3, 6x³ + 12x² + 6x+12
p(x) = 3x³ + 3x² + 3x + 3
= 3(x³ + x² + x + 1)
g(x) = 6x³ + 12x² + 6x + 12
= 6(x³ + 2x² + x + 2)
G.C.D. of 3 and 6 = 3
Now g(x) is divided by p(x)

Now dividing p(x) by the remainder x² + 1
we get x + 1

∴ G.C.D. = 3(x² + 1) [3 is the G.C.D. of 3 and 6]

Question 2.
Find the LCM of the given polynomials
(i) 4x²y, 8x³y²
Answer:
4x² y = 2 × 2 × x² × y
8 x³ y² = 2 × 2 × 2 × x³ × y²
L.C.M. = 2³ × x³ × y²
= 8x³y²

Aliter: L.C.M of 4 and 8 = 8
L.C.M. of x²y and x³y² = x³y²
∴ L.C.M. = 8x³y²

(ii) -9a³b², 12a²b²c
Answer:
-9a³b² = -(3² × a³ × b²)
12a²b²c = 2² × 3 × a² × b² × c
L.C.M. = -(2² × 3² × a³ × b² × c)
= -36 a³b²c

(iii) 16m, -12m²n², 8n²
Answer:
16m = 24 × m
-12 m²n² = -(2² × 3 × m² × n²)
8n² = 2³ × n²
L.C.M. = -(2⁴ × 3 × m² × n²)
= -48 m²n²

(iv) p² – 3p + 2, p² – 4
Answer:
P² – 3p + 2 = p² – 2p – p + 2
= p(p – 2) – 1 (p – 2)
= (p – 2) (p – 1)

p² – 4 = p² – 2² (using a² – b² = (a + b) (a – b)]
= (p + 2) (p – 2)
L.C.M. = (p – 2) (p + 2) (p – 1)

(v) 2x² – 5x – 3,4.x² – 36
Answer:
2x² – 5x – 3 = 2x² – 6x + x – 3
= 2x (x – 3) + 1 (x – 3)
= (x – 3) (2x + 1)

= 4x² – 36 = 4 [x² – 9]
= 4 [x² – 3²]
= 4(x + 3) (x – 3)
L.C.M. = 4(x – 3) (x + 3) (2x + 1)

(vi) (2x² – 3xy)²,(4x – 6y)³,(8x³ – 27y³)
Answer:
(2x² – 3xy)² = x² (2x – 3y)²
(4x – 6y)³ = 2³ (2x – 3y)³
= 8 (2x – 3y)³
8x³ – 27y³ = (2x)³ – (3y)³
= (2x – 3y) [(2x)² + 2x × 3y + (3y²)]
[using a³ – b³ = (a – b) (a² + ab + b²)
(2x – 3y) (4x² + 6xy + 9y²)
L.C.M. = 8x² (2x – 3y)³ (4x² + 6xy + 9y)²

Students can download Maths Chapter 3 Algebra Ex 3.8 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.8

Question 1.
Find the square root of the following polynomials by division method
(i) x⁴ – 12x³ + 42x² – 36x + 9
Answer:

(ii) 31 x² – 28x³ + 4x⁴ + 42x + 9
Answer:
Rearrange the order we get
4x⁴ – 28x³ + 37x² + 42x + 9

(iii) 16x⁴ + 8x² + 1
Answer:

(iv) 121 x⁴ – 198x³ – 183x² + 216x + 144
Answer:

Question 2.
Find the square root of the expression
\(\frac{x^{2}}{y^{2}}\) – \(\frac { 10x }{ y } \) + 27 – \(\frac { 10y }{ x } \) + \(\frac{y^{2}}{x^{2}}\)
Answer:

Question 3.
Find the values of a and b if the following polynomials are perfect squares.
(i) 4x⁴ – 12x³ + 37x² + bx + a
Answer:

Since it is a perfect square
b + 42 = 0
b = – 42
a – 49 = 0
a = 49
∴ The value of a = 49 and b = – 42

(ii) ax⁴ + bx³ + 361x² + 220x + 100
Answer:
Re-arrange the order we get
100 + 220x + 361x² + bx³ + ax⁴

Since it is a perfect square
b – 264 = 0
b = 264
a – 144 = 0
a = 144
∴ The value of a = 144 and b = 264

Question 4.
Find the values of m and n if the following expressions are perfect sqaures.
(i) \(\frac{1}{x^{4}}\) – \(\frac{6}{x^{3}}\) + \(\frac{13}{x^{2}}\) + \(\frac { m }{ x } \) + n
Answer:

Since it is a perfect square
\(\frac { 1 }{ x } \) (m + 12) = 0
m + 12 = 0
m = -12
n – 4 = 0
n = 4
∴ The value of m = -12 and n = 4

(ii) x⁴ – 8x³ + mx² + nx + 16
Answer:

Since it is a perfect square
m – 16 – 8 = 0
m – 24 = 0
m = 24
n + 32 = 0
n = -32
∴ The value of m = 24 and n = -32

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.10 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.10

Multiple choice questions:

Question 1.
Euclid’s division lemma states that for positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy ………………….
(1) 1 < r < b
(2) 0 < r < b
(3) 0 < r < 6
(4) 0 < r < b
Ans.
(3) 0 < r < b

Question 2.
Using Euclid’s division lemma, if the cube of any positive integer is divided by 9 then the possible remainders are ………………….
(1) 0, 1, 8
(2) 1, 4, 8
(3) 0, 1, 3
(4) 1, 3, 5
Answer:
(1) 0, 1, 8
Hint: Let the +ve integer be 1, 2, 3, 4 …………
1³ = 1 when it is divided by 9 the remainder is 1.
2³ = 8 when it is divided by 9 the remainder is 8.
3³ = 27 when it is divided by 9 the remainder is 0.
4³ = 64 when it is divided by 9 the remainder is 1.
5³ = 125 when it is divided by 9 the remainder is 8.
The remainder 0, 1, 8 is repeated.

Question 3.
If the H.C.F of 65 and 117 is expressible in the form of 65m – 117 , then the value of m is
(1) 4
(2) 2
(3) 1
(4) 3
Answer:
(2) 2
Hint:
H.C.F. of 65 and 117
117 = 65 × 1 + 52
65 = 52 × 1 + 13
52 = 13 × 4 + 0
∴ 13 is the H.C.F. of 65 and 117.
65m – 117 = 65 × 2 – 117
130 – 117 = 13
∴ m = 2

Question 4.
The sum of the exponents of the prime factors in the prime factorization of 1729 is …………………….
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(3) 3
Hint: 1729 = 7 × 13 × 19
Sum of the exponents = 1 + 1 + 1
= 3

Question 5.
The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
(1) 2025
(2) 5220
(3) 5025
(4) 2520
Answer:
(4) 2520
Hint:

L.C.M. = 2³ × 3² × 5 × 7
= 8 × 9 × 5 × 7
= 2520

Question 6.
7^4k ≡ ______ (mod 100)
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(1) 1
Hint:
7^4k ≡______ (mod 100)
y^4k ≡ y^{4 × 1} = 1 (mod 100)

Question 7.
Given F₁ = 1 , F₂ = 3 and F_n = F_n-1 + F_n-2 then F₅ is ………….
(1) 3
(2) 5
(3) 8
(4) 11
Answer:
(4) 11
Hint:
F_n = F_n-1 + F_n-2
F₃ = F₂ + F₁ = 3 + 1 = 4
F₄ = F₃ + F₂ = 4 + 3 = 7
F₅ = F₄ + F₃ = 7 + 4 = 11

Question 8.
The first term of an arithmetic progression is unity and the common difference is 4. Which of the following will be a term of this A.P
(1) 4551
(2) 10091
(3) 7881
(4) 13531
Answer:
(3) 7881
Hint:
t₁ = 1
d = 4
t_n = a + (n – 1)d
= 1 + 4n – 4
4n – 3 = 4551
4n = 4554
n = will be a fraction
It is not possible.
4n – 3 = 10091
4n = 10091 + 3 = 10094
n = a fraction
4n – 3 = 7881
4n = 7881 + 3 = 7884
n = \(\frac{7884}{4}\), n is a whole number.
4n – 3 = 13531
4n = 13531 – 3 = 13534
n is a fraction.
∴ 7881 will be 1971st term of A.P.

Question 9.
If 6 times of 6^th term of an A.P is equal to 7 times the 7^th term, then the 13^th term of the A.P. is ………..
(1) 0
(2) 6
(3) 7
(4) 13
Answer:
(1) 0
Hint:
6 t₆ = 7 t₇
6(a + 5d) = 7 (a + 6d) ⇒ 6a + 30d = 7a + 42d
30 d – 42 d = 7a – 6a ⇒ -12d = a
t₁₃ = a + 12d (12d = -a)
= a – a = 0

Question 10.
An A.P consists of 31 terms. If its 16th term is m, then the sum of all the terms of this A.P. is
(1) 16 m
(2) 62 m
(3) 31 m
(4) \(\frac { 31 }{ 2 } \) m
Answer:
(3) 31 m
Hint:
t₁₆ = m
S₃₁ = \(\frac { 31 }{ 2 } \) (2a + 30d)
= \(\frac { 31 }{ 2 } \) (2(a + 15d))
(∵ t₁₆ = a + 15d)
= 31(t₁₆) = 31m

Question 11.
In an A.P., the first term is 1 and the common difference is 4. How many terms of the A.P must be taken for their sum to be equal to 120?
(1) 6
(2) 7
(3) 8
(4) 9
Answer:
(3) 8
Here a = 1, d = 4, S_n = 120
S_n = \(\frac { n }{ 2 } \)[2a + (n – 1)d]
120 = \(\frac { n }{ 2 } \) [2 + (n – 1)4] = \(\frac { n }{ 2 } \) [2 + 4n – 4)]
= \(\frac { n }{ 2 } \) [4n – 2)] = \(\frac { n }{ 2 } \) × 2 (2n – 1)
120 = 2n² – n
∴ 2n² – n – 120 = 0 ⇒ 2n² – 16n + 15n – 120 = 0
2n(n – 8) + 15 (n – 8) = 0 ⇒ (n – 8) (2n + 15) = 0
n = 8 or n = \(\frac { -15 }{ 2 } \) (omitted)
∴ n = 8

Question 12.
A = 2⁶⁵ and B = 2⁶⁴ + 2⁶³ + 2⁶² …. + 2⁰ which of the following is true?
(1) B is 2⁶⁴ more than A
(2) A and B are equal
(3) B is larger than A by 1
(4) A is larger than B by 1
Answer:
(4) A is larger than B by
A = 2⁶⁵
B = 2⁶⁴⁺⁶³ + 2⁶² + …….. + 2⁰
= 2
= 1 + 2² + 2² + ……. + 2⁶⁴
a = 1, r = 2, n = 65 it is in G.P.
S₆₅ = 1 (2⁶⁵ – 1) = 2⁶⁵ – 1
A = 2⁶⁵ is larger than B

Question 13.
The next term of the sequence \(\frac { 3 }{ 16 } \),\(\frac { 1 }{ 8 } \),\(\frac { 1 }{ 12 } \),\(\frac { 1 }{ 18 } \) is ………..
(1) \(\frac { 1 }{ 24 } \)
(2) \(\frac { 1 }{ 27 } \)
(3) \(\frac { 2 }{ 3 } \)
(4) \(\frac { 1 }{ 81 } \)
Answer:
(2) \(\frac { 1 }{ 27 } \)
Hint:
\(\frac { 3 }{ 16 } \),\(\frac { 1 }{ 8 } \),\(\frac { 1 }{ 12 } \),\(\frac { 1 }{ 18 } \)
a = \(\frac { 3 }{ 16 } \), r = \(\frac { 1 }{ 8 } \) ÷ \(\frac { 3 }{ 16 } \) = \(\frac { 1 }{ 8 } \) × \(\frac { 16 }{ 3 } \) = \(\frac { 2 }{ 3 } \)
The next term is = \(\frac { 1 }{ 18 } \) × \(\frac { 2 }{ 3 } \) = \(\frac { 1 }{ 27 } \)

Question 14.
If the sequence t₁,t₂,t₃ … are in A.P. then the sequence t₆,t₁₂,t₁₈ … is
(1) a Geometric Progression
(2) an Arithmetic Progression
(3) neither an Arithmetic Progression nor a Geometric Progression
(4) a constant sequence
Answer:
(2) an Arithmetic Progression
Hint:
If t₁, t₂, t₃, … is 1, 2, 3, …
If t₆ = 6, t₁₂ = 12, t₁₈ = 18 then 6, 12, 18 … is an arithmetic progression

Question 15.
The value of (1³ + 2³ + 3³ + ……. + 15³) – (1 + 2 + 3 + …….. + 15) is …………….
(1) 14400
(2) 14200
(3) 14280
(4) 14520
Answer:
(3) 14280
Hint:

1202 – 120 = 120(120 – 1)
120 × 119 = 14280

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.9 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.9

Question 1.
Find the sum of the following series
(i) 1 + 2 + 3 + …….. + 60
(ii) 3 + 6 + 9 + …….. +96
(iii) 51 + 52 + 53 + …….. + 92
(iv) 1 + 4 + 9 + 16 + …….. + 225
(v) 6² + 7² + 8² + …….. + 21²
(vi) 10³ + 11³ + 12³ + …….. + 20³
(vii) 1 + 3 + 5 + …… + 71
Solution:
(i) 1 + 2 + 3 + …….. + 60 = \(\frac{60 \times 61}{2}\)
[Using \(\frac{n(n+1)}{2}\) formula]
= 1830

(ii) 3 + 6 + 9 + …….. + 96 = 3(1 + 2 + 3 + ……… + 32)
= \(\frac{3 \times 32 \times 33}{2}\)
= 1584

(iii) 51 + 52 + 53 + …….. + 92 = (1 + 2 + 3 + ……. + 92) – (1 + 2 + 3 + …… + 50)

= 4278 – 1275
= 3003

(iv) 1 + 4 + 9 + 16 + …….. + 225 = 1² + 2² + 3² + 4² + ………… + 15²
\(\frac{15 \times 16 \times 31}{6}\)
[using \(\frac{n(n+1)(2 n+1)}{6}\)] formula
= 1240

(v) 6² + 7² + 8² + …….. + 21² = 1 + 2² + 3² + 4² + ………… + 21² – (1 + 2² + ………… + 5²)
= \(\frac{21 \times 22 \times 43}{6}\) – \(\frac{5 \times 6 \times 11}{6}\)
= 3311 – 55
= 3256

(vi) 10³ = 11³ + 12³ + …….. + 20³ = 1³ + 2³+ 3³ + ………… + 20³ – (1³ + 2³ + 3³ + …………. + 9³)

[Using (\(\frac{n(n+1)}{2}\))² formula]
= 210² – 45² = 44100 – 2025
= 42075

(vii) 1 + 3 + 5+ … + 71
Here a = 1; d = 3 – 1 = 2; l = 71

Question 2.
If 1 + 2 + 3 + …. + k = 325 , then find 1³ + 2³ + 3³ + …………. + k³
Answer:
1 + 2 + 3 + …. + k = 325
\(\frac{k(k+1)}{2}\) = 325 ……(1)

= 325² (From 1)
= 105625

Question 3.
If 1³ + 2³ + 3³ + ………… + K³ = 44100 then find 1 + 2 + 3 + ……. + k
Answer:
1³ + 2³ + 3³ + ………….. + k³ = 44100

\(\frac{k(k+1)}{2}\) = \(\sqrt { 44100 }\) = 210
1 + 2 + 3 + …… + k = \(\frac{k(k+1)}{2}\)
= 210

Question 4.
How many terms of the series 1³ + 2³ + 3³ + …………… should be taken to get the sum 14400?
Answer:
1³ + 2³ + 3³ + ……. + n³ = 14400

\(\frac{n(n+1)}{2}\) = \(\sqrt { 14400 }\)
\(\frac{n(n+1)}{2}\) = 120 ⇒ n² + n = 240

n² + n – 240 = 0
(n + 16) (n – 15) = 0
(n + 16) = 0 or (n – 15) = 0
n = -16 or n = 15 (Negative will be omitted)
∴ The number of terms taken is 15

Question 5.
The sum of the squares of the first n natural numbers is 285, while the sum of their cubes is 2025. Find the value of n.
Answer:
1² + 2² + 3² + …. + n² = 285

Question 6.
Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm, …, 24 cm. How much area can be decorated with these colour papers?
Answer:
Area of 15 square colour papers
= 10² + 11² + 12² + …. + 24²
= (1² + 2² + 3² + …. + 24²) – (1² + 2² + 9²)
= \(\frac{24 \times 25 \times 49}{6}-\frac{9 \times 10 \times 19}{6}\)
= 4 × 25 × 49 – 3 × 5 × 19
= 4900 – 285
= 4615
Area can be decorated is 4615 cm²

Question 7.
Find the sum of the series (2³ – 1)+(4³ – 3³) + (6³ – 15³) + …….. to
(i) n terms
(ii) 8 terms
Answer:
Sum of the series = (2³ – 1) + (4³ – 3³) + (6³ – 15³) + …. n terms
= 2³ + 4³ + 6³ + …. n terms – (1³ + 3³ + 5³ + …. n terms) …….(1)
2³ + 4³ + 6³ + …. n = ∑(2³ + 4³ + 6³ + ….(2n)³]
∑ 2³ (1³ + 2³ + 3³ + …. n³)
= 8 (\(\frac{n(n+1)}{2}\))²
= 2[n (n + 1)]²
1³ + 3³ + 5³ + ……….(2n – 1)³ [sum of first 2n cubes – sum of first n even cubes]

Substituting (2) and (3) in (1)
Sum of the series = 2n² (n + 1)² – n² (2n + 1)² + 2n²(n + 1)²
= 4n² (n + 1)² – n² (2n + 1)²
= n² [(4(n + 1)² – (2n + 1)²]
= n² [4n² + 4 + 8n – 4n² – 1 – 4n]
= n² [4n + 3]
= 4n³ + 3n²

(ii) when n = 8 = 4(8)³ + 3(8)²
= 4(512) + 3(64)
= 2240

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.8 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.8

Question 1.
Find the sum of first n terms of the G.P.
(i) 5, -3, \(\frac { 9 }{ 5 } \),-\(\frac { 27 }{ 25 } \), ……
(ii) 256,64,16,…….
Answer:
(i) 5,-3,\(\frac { 9 }{ 5 } \),\(\frac { 27 }{ 55 } \), ….. n terms

(ii) 256,64,16,…….
Answer:
Here a = 256, r = \(\frac { 64 }{ 256 } \) = \(\frac { 1 }{ 4 } \)

Question 2.
Find the sum of first six terms of the G.P. 5,15,45,…
Answer:
Here a = 5, r = \(\frac { 15 }{ 3 } \) = 3, n = 6

Sum of first 6 terms = 1820

Question 3.
Find the first term of the G.P. whose common ratio 5 and whose sum to first 6 terms is 46872.
Answer:
Common ratio (r) = 5
S₆ = 46872

The first term of the G.P. is 12.

Question 4.
Find the sum to infinity of (i) 9 + 3 + 1 + ….(ii) 21 + 14 + \(\frac { 28 }{ 3 } \) ……
Answer:
(i) 9 + 3 + 1 + ….
a = 9, r = \(\frac { 3 }{ 9 } \) = \(\frac { 1 }{ 3 } \)
Sum of infinity term = \(\frac { a }{ 1 – r } \) = \(\frac{9}{1-\frac{1}{3}}\)

Question 5.
If the first term of an infinite G.P. is 8 and its sum to infinity is \(\frac { 32 }{ 3 } \) then find the common ratio.
Answer:
Here a = 8, S∞ = \(\frac { 32 }{ 3 } \)
\(\frac { a }{ 1-r } \) = \(\frac { 32 }{ 3 } \)
\(\frac { 8 }{ 1-r } \) = \(\frac { 32 }{ 3 } \)
32 – 32 r = 24 ⇒ 32 r = 8
r = \(\frac { 8 }{ 32 } \) = \(\frac { 1 }{ 4 } \)
Common ration = \(\frac { 1 }{ 4 } \)

Question 6.
Find the sum to n terms of the series
(i) 0.4 + 0.44 + 0.444 + …… to n terms
Answer:


(ii) 3 + 33 + 333 + ………… to n terms
Answer:
S_n = 3 + 33 + 333 + …. to n terms
= 3[1 + 11 + 111 + …. to n terms]
= \(\frac { 3 }{ 9 } \) [9 + 99 + 999 + …. n terms]
= \(\frac { 1 }{ 3 } \) [(10 – 1) + (100 – 1) + (1000 – 1) + …… n terms]
= \(\frac { 1 }{ 3 } \) [10 + 100 + 1000 + ….. n terms – (1 + 1 + 1 ….. n terms)]

Question 7.
Find the sum of the Geometric series 3 + 6 + 12 + …….. + 1536
Answer:
3 + 6 + 12 …. +1536
a = 3, r = \(\frac { 6 }{ 3 } \) = 2
t_n = 1536


∴ Sum of the series is 3069

Question 8.
Kumar writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with the instruction that they continue the process similarly. Assuming that the process is unaltered and it costs ₹2 to mail ong letter, find the amount spent on postage when 8th set of letters is mailed.
Answer:
When kumar writes a letter to his friend.Friend writes a letter to another person.
It form a G.P
The G.P is 4, 16, 64,………
Here a = 4, r = 4
The last term is 4 (4)^8-1 = 4(4)⁷

Question 9.
Find the rational form of the number 0.123 .
Answer:
Let x = \(\overline { 0.123 } \)
= 0.123123123….
= 0.123 + 0.000123 + 000000123 + ….
This is an infinite G.P

Question 10.
If S_n = (x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + ………… n terms then prove that
Answer:
S_n = (x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + …….. n
Multiply by x
x S_n = x(x + y) + x(x² + xy + y²) + x(x³ + x²y + xy² + y³) + ……….. n
= x² + xy + x³ + x²y + xy² + x⁴ + x³y + x²y² + xy³ + …… n terms ……(1)
Multiply by y
yS_n = y(x + y) + y(x² + xy + y²) + y(x³ + x²y + xy² + y³) + ….. n
= xy + y² + x²y + xy² + y³ + x³y + x²y² + xy³ + y⁴ + ….. n terms
Subtract (1) and (2)
x S_n – y S_n = x² + xy + x³ + x²y + xy² + x⁴ + x³y + x²y² + xy³ + …….
– xy + y² + x²y + xy² + y³ + x³y + x²y² + xy³ + y⁴ + ……
(x – y) S_n = (x² + x³ + x⁴ + ……) – (y² + y³ + y⁴ + ……)
[ a = x²; r = x and a = y²; r = y, S_n = \(\frac{a\left(r^{n}-1\right)}{r-1}\)]

Hence it is proved.

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.9

Question 1.
Determine the quadratic equations, whose sum and product of roots are
(i) -9, 20
Answer:
Sum of the roots = -9 and Product of the roots = 20
The Quadratic equation is
x² – (sum of the roots) x + product of the roots = 0
x² – (-9) x + 20 = 0 ⇒ x² + 9x + 20 = 0

(ii) \(\frac { 5 }{ 3 } \), 4
Answer:
Sum of the roots = \(\frac { 5 }{ 3 } \); Product of the roots = 4
The Quadratic equation is
x² – (sum of the roots) x + product of the roots = 0
x² – (\(\frac { 5 }{ 3 } \)) x + 4 = 0 ⇒ x² – \(\frac { 5 }{ 3 } \) x + 4 = 0
3x² – 5x + 12 = 0

(iii) \(\frac { -3 }{ 2 } \), -1
Answer:
Sum of the roots = \(\frac { -3 }{ 2 } \); Product of the roots = -1
The Quadratic equation is
x² – (sum of the roots) x + product of the roots = 0
x² – (-\(\frac { 3 }{ 2 } \)) x + (-1) = 0 ⇒ x² + \(\frac { 3 }{ 2 } \) x – 1 = 0
2x² + 3x – 2 = 0

(iv) – (2 – a)², (a + 5)²
Answer:
Sum of the roots = – (2 – a)²; Product of the roots = (a + 5)²
x² – (sum of the roots) x + product of the roots = 0
x² – [-(2 – a)²] x + (a + 5)² = 0
x² + (2 – a)² x + (a + 5)² = 0

Question 2.
Find the sum and product of the roots for each of the following quadratic equations
(i) x² + 3x – 28 = 0
(ii) x² + 3x = 0
(iii) 3 + \(\frac{1}{a}=\frac{10}{a^{2}}\)
(iv) 3y² – y – 4 = 0
Solution:
(i) x² – (-3)x + (-28) = 0.
Comparing this with x² – (α + β)x + αβ = 0.
(α + β) = Sum of the roots = -3
αβ = product of the roots = -28

(ii) x² + 3x = 0 = x² – (-3)x + 0 = 0
x² – (α + β)x + αβ = 0
Sum of the roots α + β = -3
Products of the roots αβ =0

(iii) 3 + \(\frac { 1 }{ a } \) = \(\frac{10}{a^{2}}\)
Answer:
Multiply by a²
3a² + a = 10
3a² + a – 10 = 0
Sum of the roots (α + β) = \(\frac { -1 }{ 3 } \)
Product of the roots (α β) = \(\frac { -10 }{ 3 } \)

(iv) 3y² – y – 4 = 0
Answer:
Sum of the roots (α + β) = \(\frac { -(-1) }{ 3 } \) = \(\frac { 1 }{ 3 } \)
Product of the roots (α β) = \(\frac { -4 }{ 3 } \)

Students can download Maths Chapter 3 Algebra Ex 3.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.3

Question 1.
Find the LCM and GCD for the following and verify that f(x) × g(x) = LCM × GCD

(i) 21x²y, 35xy²
Answer:
p(x) = 21 x²y = 3 × 7 × x² × y
g(x) = 35xy² = 5 × 7 × x × y²
G.C.D = 7 xy
L.C.M = 3 × 5 × 7 x² × y²
= 105 x²y²
L.C.M × G.C.D = 105x²y² × 7xy
= 735 x³y³ ….(1)
p(x) × g(x) = 21x²y × 35xy²
= 735x³y³ ….(2)
From (1) and (2) we get
L.C.M × G.C.D. = p(x) × g(x)

(ii) (x³ – 1)(x + 1),(x³ + 1)
Answer:
p(x) = (x³ – 1) (x + 1) = (x – 1) (x² + x + 1) (x + 1)
g(x) = x³ + 1 = (x + 1) (x² – x + 1)
G.C.D = (x + 1)
L.C.M = (x + 1) (x – 1) (x² + x + 1) (x² – x + 1)
L.C.M × G.C.D = (x + 1) (x – 1)(x² + x + 1)(x² – x + 1)x(x + 1)
= (x + 1)² (x – 1) (x² + x + 1) (x² – x + 1) ……….(1)
p(x) × g(x) = (x – 1) (x² + x + 1) (x + 1) (x + 1) (x² – x + 1)
= (x + 1)² (x – 1) (x² + x + 1) (x² – x + 1) ……….(2)
From (1) and (2) we get
L.C.M × G.C.D. = p(x) × g(x)

(iii) (x²y + xy²), (x² + xy)
Answer:
p(x) = x²y + xy² = xy(x + y)
g(x) = x² + xy = x(x + y)
G.C.D = x(x+y)
L.C.M = xy (x +y).
L.C.M × G.C.D = xy(x + y) × x(x + y)
= x²y(x + y)² …..(1)
p(x) × g(x) = xy(x + y) × x(x + y)
= x²y(x + y)²
From (1) and (2) we get
L.C.M × G.C.D. = p(x) × g(x)

Question 2.
Find the LCM of each pair of the following polynomials
(i) a² + 4a – 12, a² – 5a + 6 whose GCD is a – 2
Answer:
p(x) = a² + 4a – 12
= a² + 6a – 2a – 12
= a (a + 6) – 2(a + 6)
= (a + 6) (a – 2)

g(x) = a² – 5a + 6
= a² – 3a – 2a + 6

= a(a – 3) – 2 (a – 3)
= (a – 3) (a – 2)

(ii) x⁴ – 27a³x, (x – 3a)² whose GCD is (x – 3a)
Answer:
p(x) = x⁴ – 27a³x = x[x³ – 27a³]
= x[x³ – (3a)³]
= x(x – 3a) (x² + 3ax + 9a²)
g(x) = (x – 3a)²
G.C.D. = x – 3a

L.C.M. = x (x – 3a)² (x² + 3ax + 9a²)

Question 3.
Find the GCD of each pair of the following polynomials
(i) 12(x⁴ – x³), 8(x⁴ – 3x³ + 2x²) whose LCM is 24x³ (x – 1)(x – 2)
Answer:
p(x) = 12(x⁴ – x³)
= 12x³(x- 1)
g(x) = 8(x⁴ – 3x³ + 2x²)

= 8x²(x² – 3x + 2)
= 8x²(x – 2)(x – 1)
L.C.M. = 24x³ (x – 1) (x – 2)

(ii) (x³ + y³), (x⁴ + x²y² + y⁴) whose LCM is (x³ + y³) (x² + xy + y²)
Answer:
p(x) = x³ + y³
= (x + y)(x² – xy + y²)
g(x) = x⁴ + x²y² + y⁴ = [x² + y²]² – (xy)²
= (x² + y² + xy) (x² + y² – xy)
L.C.M. = (x³ + y³) (x² + xy + y²)
(x + y) (x² – xy + y²) (x² + xy + y²)

G.C.D. = x² – xy + y²

Question 4.
Given the L.C.M and G.C.D of the two polynomials p(x) and q(x) find the unknown polynomial in the following table

Answer:
L.C.M. = a³ – 10a² + 11a + 70
= (a – 7) (a² – 3a – 10)
= (a – 7) (a – 5) (a + 2)

G.C.D. = (a – 7)
p(x) = a² -12a + 35
= (a – 5)(a – 7)

q(x) = \(\frac{\mathrm{LCM} \times \mathrm{GCD}}{p(x)}\)

(ii) L.C.M (x² + y²)(x⁴ + x²y² + y⁴)
(x² + y²)[(x² + y²)²-(xy)²]
(x² + y²) (x² + y² + xy) (x² + y² – xy)
G.C.D. = x² – y²
(x + y)(x – y)
q(x) = (x⁴ – y⁴) (x² + y² – xy)
= [(x²)² – (y²)²](x² + y² – xy)
= (x² + y²) (x² – y²) (x² + y² – xy)
(x² + y²) (x + y) (x – y) (x² + y² – xy)
P(x) = x² + y² + xy

Students can download Maths Chapter 2 Relations and Functions Unit Exercise 2 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Unit Exercise 2

Question 1.
Prove that n² – n divisible by 2 for every positive integer n.
Answer:
We know that any positive integer is of the form 2q or 2q + 1 for some integer q.
Case 1: When n = 2 q
n² – n = (2q)² – 2q = 4q² – 2q
= 2q (2q – 1)
In n² – n = 2r
2r = 2q(2q – 1)
r = q(2q + 1)
n² – n is divisible by 2

Case 2: When n = 2q + 1
n² – n = (2q + 1)² – (2q + 1)
= 4q² + 1 + 4q – 2q – 1 = 4q² + 2q
= 2q (2q + 1)
If n² – n = 2r
r = q (2q + 1)
∴ n² – n is divisible by 2 for every positive integer “n”

Question 2.
A milk man has 175 litres of cow’s milk and 105 litres of buffalow’s milk. He wishes to sell the milk by filling the two types of milk in cans of equal capacity. Calculate the following
(i) Capacity of a can
(ii) Number of cans of cow’s milk
(iii) Number of cans of buffalow’s milk.
Answer:
175 litres of cow’s milk.
105 litres of goat’s milk.
H.C.F of 175 & 105 by using Euclid’s division algorithm.
175 = 105 × 1 + 70, the remainder 70 ≠ 0

Again using division algorithm,

105 = 70 × 1 + 35, the remainder 35 ≠ 0
Again using division algorithm.
70 = 35 × 2 + 0, the remainder is 0.
∴ 35 is the H.C.F of 175 & 105.
(i) ∴ The milk man’s milk can’s capacity is 35 litres.
(ii) No. of cow’s milk obtained = \(\frac { 175 }{ 35 } \) = 5 cans
(iii) No. of buffalow’s milk obtained = \(\frac { 105 }{ 35 } \) = 3 cans

Question 3.
When the positive integers a, b and c are divided by 13 the respective remainders are 9,7 and 10. Find the remainder when a + 2b + 3c is divided by 13.
Answer:
Given the positive integer are a, b and c
a = 13q + 9 (divided by 13 leaves remainder 9)
b = 13q + 7
c = 13q + 10
a + 2b + 3c = 13q + 9 + 2(13q + 7) + 3 (13q + 10)
= 13q + 9 + 26q + 14 + 39q + 30
= 78q + 53
When compare with a = 3q + r
= (13 × 6) q + 53
The remainder is 53

Question 4.
Show that 107 is of the form 4q +3 for any integer q.
Solution:
107 = 4 × 26 + 3. This is of the form a = bq + r.
Hence it is proved.

Question 5.
If (m + 1)^th term of an A.P. is twice the (n + 1)^th term, then prove that (3m + 1)^th term is twice the (m + n + 1)^th term.
Answer:
t_n = a + (n – 1)d
Given t_m+1 = 2 t_n+1
a + (m + 1 – 1)d = 2[a + (n + 1 – 1)d]
a + md = 2(a + nd) ⇒ a + md =2a + 2nd
md – 2nd = a
d(m – 2n) = a ….(1)
To Prove t_{(3m + 1)} = 2(t_m+n+1)
L.H.S. = t_3m+1
= a + (3m + 1 – 1)d
= a + 3md
= d(m – 2n) + 3md (from 1)
= md – 2nd + 3md
= 4md – 2nd
= 2d (2m – n)
R.H.S. = 2(t_m+n+1)
= 2 [a + (m + n + 1 – 1) d]
= 2 [a + (m + n)d]
= 2 [d (m – 2n) + md + nd)] (from 1)
= 2 [dm – 2nd + md + nd]
= 2 [2 md – nd]
= 2d (2m – n)
R.H.S = L.H.S
∴ t_(3m+1) = 2 t_(m+n+1)
Hence it is proved.

Question 6.
Find the 12^th term from the last term of the A.P -2, -4, -6,… -100.
Answer:
The given A.P is -2, -4, -6, …. 100
d = -4 – (-2) = -4 + 2 = – 2
Finding the 12 term from the last term
a = -100, d = 2 (taking from the last term)
n = 12
t_n = a + (n – 1)d
t₁₂ = – 100 + 11 (2)
= -100 + 22
= -78
∴ The 12^th term of the A.P from the last term is – 78

Question 7.
Two A.P’s have the same common difference. The first term of one A.P is 2 and that of the other is 7. Show that the difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms.
Solution:
Let the two A.Ps be
AP₁ = a₁, a₁ + d, a₁ + 2d,…
AP₂ = a₂, a₂ + d, a₂ + 2d,…
In AP₁ we have a₁ = 2
In AP₂ we have a₂ = 7
t₁₀ in AP₁ = a₁ + 9d = 2 + 9d ………….. (1)
t₁₀ in AP₂ = a₂ + 9d = 7 + 9d …………… (2)
The difference between their 10th terms
= (1) – (2) = 2 + 9d – 7 – 9d
= -5 ………….. (I)
t₂₁ m AP₁ = a₁ + 20d = 2 + 20d …………. (3)
t₂₁ in AP₂ = a₂ + 20d = 7 + 20d ………… (4)
The difference between their 21 st terms is
(3) – (4)
= 2 + 20d – 7 – 20d
= -5 ……………. (II)
I = II
Hence it is Proved.

Question 8.
A man saved ₹16500 in ten years. In each year after the first he saved ₹100 more than he did in the preceding year. How much did he save in the first year?
Answer:
Amount of saving in ten years = ₹ 16500
S₁₀ = 16500, d= 100
S_n = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S₁₀ = \(\frac { 10 }{ 2 } \) [2a + 9d]
16500 = \(\frac { 10 }{ 2 } \) [2a + 900] = 5(2a + 900)
16500 = 10a + 4500 ⇒ 16500 – 4500 = 10a
12000 = 10a
a = \(\frac { 12000 }{ 10 } \) = 1200
Amount saved in the first year = ₹ 1200

Question 9.
Find the G.P. in which the 2nd term is \(\sqrt { 6 }\) and the 6th term is 9 \(\sqrt { 6 }\).
Answer:
2^nd term of the G.P = \(\sqrt { 6 }\)
t₂ = \(\sqrt { 6 }\)
[t_n = a r^n-1]
a.r = \(\sqrt { 6 }\) ….(1)
6^th term of the G.P. = 9 \(\sqrt { 6 }\)
a. r⁵ = 9\(\sqrt { 6 }\) ……..(2)

Question 10.
The value of a motorcycle depreciates at a rate of 15% per year. What will be the value of the motorcycle 3 year hence, which is now purchased for ₹45,000?
Solution:
a = ₹45000
Depreciation = 15% for ₹45000
= 45000 × \(\frac { 15 }{ 100 } \)
d = ₹6750 since it is depreciation
d = -6750
At the end of 1^st year its value = ₹45000 – ₹6750
= ₹38250,
Again depreciation = 38250 × \(\frac { 15 }{ 100 } \) = 5737.50
At the end of 2^nd year its value
= ₹38250 – ₹5737.50 = 32512.50
Again depreciation = 32512.50 × \(\frac { 15 }{ 100 } \) = 4876.88
At the end of the 3^rd year its value
= 32512.50 – 4876.88 = 27635.63
∴ The value of the automobile at the 3^rd year
= ₹ 27636

Subject Matter Experts at SamacheerKalvi.Guide have created Tamil Nadu State Board New Syllabus Samacheer Kalvi 10th Maths Model Question Papers 2020-2021 Pdf Free Download in English Medium and Tamil Medium of TN SSLC Class 10th Maths Model Question Papers, Previous Year Question Papers, Sample Papers are part of Samacheer Kalvi 10th Model Question Papers 2021 Tamilnadu.

Let us look at these Government of TN State Board Samacheer Kalvi 10th Std Maths Model Question Papers with Answers 2020-21 Tamil Medium Pdf. Students can view or download the Samacheer Kalvi Class 10th Maths New Model Question Papers 2021 Tamil Nadu English Medium Pdf for their upcoming TN SSLC board examinations. Students can also read Tamilnadu Samacheer Kalvi 10th Maths Guide.

Tamil Nadu Samacheer Kalvi 10th Maths Model Question Papers 2020 2021 Tamil English Medium

Tamil Nadu Samacheer Kalvi 10th Maths Model Question Papers English Medium

• Samacheer Kalvi 10th Maths Model Question Paper 1
• Samacheer Kalvi 10th Maths Model Question Paper 2
• Samacheer Kalvi 10th Maths Model Question Paper 3
• Samacheer Kalvi 10th Maths Model Question Paper 4
• Samacheer Kalvi 10th Maths Model Question Paper 5

Tamil Nadu Samacheer Kalvi 10th Maths Model Question Papers Tamil Medium

• Samacheer Kalvi 10th Maths Model Question Paper 1
• Samacheer Kalvi 10th Maths Model Question Paper 2
• Samacheer Kalvi 10th Maths Model Question Paper 3
• Samacheer Kalvi 10th Maths Model Question Paper 4
• Samacheer Kalvi 10th Maths Model Question Paper 5

Samacheer Kalvi 10th Maths Model Question Paper Design 2020-2021 Tamil Nadu

Types of Questions	Marks	No. of Questions to be Answered	Total Marks
Part-I Multiple Choice Questions	1	14	14
Part-II (Totally 14 questions will be given. Answer any Ten. Any one question should be answered compulsorily)	2	10	20
Part-Ill (Totally 14 questions will be given. Answer any Ten. Any one question should be answered compulsorily)	5	10	50
Part-IV	8	2	16
Total Marks			100

Samacheer Kalvi 10th Maths Model Question Paper Weightage of Marks

Purpose	Weightage
1. Knowledge	30%
2. Understanding	40%
3. Application	20%
4. Skill/Creativity	10%

It is necessary that students will understand the new pattern and style of Tamil Nadu Samacheer Kalvi 10th Maths Model Question Papers 2021 according to the latest exam pattern. These State Board 10th Standard Maths Public Exam Model Question Papers 2020-21 Tamil Nadu in Tamil Medium and English Medium are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for TN SSLC Board Exams and Score More marks.

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