Samacheer Kalvi 10th Maths Chapter 3 Algebra Notes PDF Download: Tamil Nadu STD 10th Maths Chapter 3 Algebra Notes 
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Samacheer Kalvi 10th Maths 
Subject 
10th Maths 
Chapter 
Chapter 3 Algebra 
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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Unit Exercise 3
10th Maths Unit Exercise 3 Question 1.
Solve 13 (x + y – 5) = y – z = 2x – 11 = 9 – (x + 2 z).
Solution:
Given
3z = 3 ⇒ z = 1
(3) becomes, 3x + 2 = 20 ⇒ 3x = 20 – 2 = 18
x = 183 = 6
(1) becomes, 6 – 2y + 3(1) = 5 ⇒ 9 – 2y = 5
⇒ 9 – 5 = 2y ⇒ 2y = 4
∴ y = 42 = 2
∴ Solution set is {6, 2, 1}
Unit Exercise 3 Question 2.
One hundred and fifty students are admitted to a school. They are distributed over three sections A, B and C. If 6 students are shifted from section A to section C, the sections will have equal number of students. If 4 times of students of section C exceeds the number of students of section A by the number of students in section B, find the number of students in the three sections.
Solution:
Let the students in section A, B, C be a, b, c, respectively.
Samacheer Kalvi 10th Maths Exercise 3.1 Question 3.
In a threedigit number, when the tens and the hundreds digit are interchanged the new number is 54 more than three times the original number. If 198 is added to the number, the digits are reversed. The tens digit exceeds the hundreds digit by twice as that of the tens digit exceeds the unit digit. Find the original number.
Solution:
Let the three digits numbers be 100a +10b + c.
100b + 10a + c = 3(100a + 10b + c) + 54 ………. (1)
100a + 106 + c + 198 = 100c + 106 + a ……… (2)
(b – a) = 2(b – c) ……… (3)
(1) ⇒ 100b + 10a + c = 300a + 30b + 3c + 54
⇒ 290a – 70b + 2c = 54
(2) ⇒ 99a – 99c = 198 ⇒ a – c = 2
⇒ a = c – 2
(3) ⇒ a + b – 2c = 0 ⇒ a + b = 2c
⇒ b = 2c – c + 2
⇒ b = c + 2
Substituting a, b in (1)
290(c – 2) – 70 (c + 2) + 2c = 54
290c – 580 – 70c – 140 + 2c = 54
222c = 666 ⇒ c = 3
a = 1, 6 = 5
∴ The number is 153.
10th Maths Unit Exercise 2 Question 4.
Find the least common multiple of
xy (k^{2} + 1) + k(x^{2} + y^{2}) and
xy(k^{2} – 1) + k (x^{2} – y^{2})
Answer:
xy (k^{2} + 1) + k(x^{2} + y^{2}) = k^{2}xy + xy + kx^{2} + ky^{2}
= (k^{2}xy + kx^{2}) + (ky^{2} + xy)
= kx(ky + x) + y (ky + x)
= (ky + x) (kx + y)
xy (k^{2} – 1) + k(x^{2} – y^{2}) = k^{2}xy – xy + kx^{2} – ky^{2}
= (k^{2}xy + kx^{2}) – xy – ky^{2}
= kx(ky + x) y (ky + x)
= (ky + x) (kx – y)
L.C.M. = (ky + x) (kx + y) (kx – y)
= (ky + x)(k^{2}x^{2} – y^{2})
The least common multiple is
(ky + x) (k^{2}x^{2} – y^{2})
Unit Exercise Question 5.
Find the GCD of the following by division algorithm 2x^{4} + 13.x^{3} + 21 x^{2} + 23x + 7, x^{3} + 3x^{2} + 3x + 1, x^{2} + 2x + 1.
Solution:
2x^{4} + 13x^{3} + 27x^{2} + 23x + 7,
x^{3} + 3x^{2} + 3x + 1, x^{2} + 2x + 1.
By division algorithm, first divide
∴ (x + 1)^{2} is G.C.D of x^{3} + 3x^{2} + 3x + 1 and x^{2} + 2x + 1.
Next let us divide
2x^{4} + 13x^{3} + 27x^{2} + 23x + 7 by x^{2} + 2x + 1
∴ G.C.D of 2x^{4} + 13x^{3} + 21 x^{2} + 23x + 7, x^{3} + 3x^{2} + 3x + 1, x^{2} + 2x + 1 is (x + 1)^{2}.
10th Maths Example Sums Question 6.
Reduce the given Rational expressions to its lowest form
Solution:
Samacheer Kalvi 10th Maths Solution Question 7.
Solution:
Samacheer Kalvi 10th Maths Answers Question 8.
Arul, Ravi and Ram working together can clean a store in 6 hours. Working alone, Ravi takes twice as long to clean the store as Arul does. Ram needs three times as long as Arul does. How long would it take each if they are working alone?
Solution:
Let Aral’s speed of working be x
Let Ravi’s speed of working be y
Let Ram’s speed of working be z
given that they are working together. ,
Let V be the quantum of work, x + y + z = w6 …………. (1)
Also given that Ravi takes twice the time as Aral for finishing the work.
Also Ram takes 3 times the time as Aral for finishing the work.
∴ wz = 3 × wx
∴ x = 3z ∴ z = x3
Substitute (2) and (3) in (1),
Exercise 1.6 Class 10 Maths Samacheer Question 9.
Find the square root of 289x^{4} – 612x^{3} + 970x^{2} – 684x + 361
Solution:
10th Maths Exercise 1.6 Answers Question 10.
Solve y+1−−−−√+2y−5−−−−−√ = 3.
Solution:
Squaring both sides
9y^{2} – 78y + 169 = 4 (y + 1)(2y – 5)
9y^{2} – 78y + 169 = 4 (2y^{2} + 2y – 5y – 5)
9y^{2} – 78y + 169 = 8y^{2} + 8y – 20y – 20
9y^{2} – 78y + 169 – 8y^{2} + 12y + 20 = 0
y^{2} – 66y + 189 = 0
y^{2} – 63y – 3y + 189 = 0
y(y – 63) – 3(y – 63) = 0
(y – 63)(y – 3) = 0
y = 63, 3
10th Maths Book Example Sums Question 11.
A boat takes 1.6 hours longer to go 36 kms up a river than down the river. If the speed of the water current is 4 km per hr, what is the speed of the boat in still water?
Solution:
Let the speed of boat in still water be ‘v’
⇒ 36(v + 4) – 36(v – 4) = 85 (v – 4) (v + 4)
⇒ 36v + 144 – 36v + 144 = 85 (v^{2} – 4v + 4v – 16)
⇒ 288 = 85 v^{2} – 1285 ⇒ 8v^{2} – 128 = 1440
⇒ 8v^{2} = 1568 ⇒ v^{2} = 196 v = ±14
∴ Speed of the boat = 14 km/hr. (∵ speed cannot be ve)
10th Algebra Solutions Question 12.
Is it possible to design a rectangular park of perimeter 320 m and area 4800 m^{2}? If so find its length and breadth.
Solution:
Let the length and breadth of the rectangle be lm and bm
Given 2(1 + b)
⇒ l + b = 160 ………. (1)
Also l b = 4800
∴ Length and breadth of the rectangular park is 120m and 40 m
10th Maths Samacheer Question 13.
At t minutes past 2 pm, the time needed to 3 pm is 3 minutes less than t24 Find t.
Solution:
60 – t = t24 – 3
⇒ t^{2} – 12 = 240 – 4t
⇒ t^{2} + 4t – 252 = 0
⇒ t^{2} + 18t – 14t – 252 = 0
⇒ t(t + 18) – 14(t + 18) = 0
⇒ (t + 18) (t – 14) = 0
∴ t = 14 or t = 18 is not possible.
Samacheer Kalvi 10th Solutions Question 14.
The number of seats in a row is equal to the total number of rows in a hall. The total number of seats in the hall will increase by 375 if the number of rows is doubled and the number of seats in each row is reduced by 5. Find the number of rows in the hall at the beginning.
Solution:
Let the no of seats in each row be x
⇒ 2x^{2} – 10x = x^{2} + 375
⇒ x^{2} – 10x – 375 = 0
⇒ x^{2} – 25x + 15x – 375 = 0
⇒ x (x – 25) + 15 (x – 25) = 0
⇒ (x – 25) (x + 15) = 0
⇒ x = 25, x = 15, x > 0
∴ 25 rows are in the hall.
3rd Standard Samacheer Maths Book Solutions Question 15.
If a and b are the roots of the polynomial f(x) = x^{2} – 2x + 3, find the polynomial whose roots are
(i) Î± + 2, Î² + 2
(ii) Î±−1Î±+1,Î²−1Î²+1
Solution:
(i) Î± + 2, Î² + 2 are the roots (given)
Sum of the roots = Î± + 2 + Î² + 2
= Î± + Î² + 4
= 2 + 4 = 6
Product of the roots = (Î± + 2) (Î² + 2)
= Î±Î² + 2Î± + 2Î² + 4
= Î±Î² + 2(Î± + Î²) + 4
= 3 + 2 × 2 + 4
= 3 + 4 + 4 = 11
∴ The required equation = x^{2} – 6x + 11 = 0.
⇒ 3x^{2} – 2x + 1 = 0
Class 3 Samacheer Kalvi Solutions Question 16.
If 4 is a root of the equation
x^{2} + px – 4 = 0 and if the equation
x^{2} + px + q = 0 has equal roots, find the values of p and q.
Answer:
Let p(x) = x^{2} + px – 4
– 4 is the root of the equation
P(4) = 0
16 – 4p – 4 = 0
4p + 12 = 0
4p = 12
p = 124 = 3
The equation x^{2} + px + q = 0 has equal roots
x^{2} + 3 x + q = 0
Here a = 1, b = 3, c = q
since the roots are real and equal
b^{2} – 4 ac = 0
3^{2} – 4(1)(q) = 0
9 – 4q = 0
9 = 4q
q = 94
The value of p = 3 and q = 94
10thmaths Question 17.
Two farmers Senthil and Ravi cultivates three varieties of grains namely rice, wheat and ragi. If the sale (in ₹) of three varieties of grains by both the farmers in the month of April is given by the matrix.
May month sale (in ₹) is exactly twice as that of the April month sale for each variety.
(i) What is the average sales of the months April and May.
(ii) If the sales continues to increase in the same way in the successive months, what will be sales in the month of August?
Solution:
Samacheer Kalvi 10th Maths Book Answers Question 18.
Solution:
10th Maths Question 19.
Solution:
10th Algebra Chapter 3 Question 20.
Solution:
Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Additional Questions
Question 1.
Solve the following system of linear equations in three variables. x + y + z = 6; 2x + 3y + 4z = 20;
3x + 2y + 5z = 22
Solution:
x + y + z = 6 ………….. (1)
2x + 3y + 4z = 20 ………… (2)
3x + 2y + 5z = 22 …………(3)
Sub. z = 3 in (5) ⇒ y – 2(3) = 4
y = 2
Sub. y = 2, z = 3 in (1), we get
x + 2 + 3 = 6
x = 1
x = 1, y = 2, z = 3
Question 2.
Using quadratic formula solve the following equations.
(i) p^{2}x^{2} + (p^{2} – q^{2}) x – q^{2} = 0
(ii) 9x^{2} – 9 (a + b)x + (2a^{2} + 5ab + 2b^{2}) = 0
Solution:
(i) p^{2}x^{2} + (p^{2} – q^{2})x – q^{2} = 0
Comparing this with ax^{2} + bx + c = 0, we have
a = p^{2}
b = p^{2} – q^{2}
c = q^{2}
Î” = b^{2} – 4ac
= (p^{2} – q^{2})4 × p^{2} × q^{2}
= (p^{2} – q^{2})^{2} + 4p^{2} q^{2}
= (p^{2} + q^{2})^{2} > 0
So, the given equation has real roots given by
Î±=−b−Î”√2a=−(p2−q2)+(p2+q2)2p2=q2p2
Î²=−b−Î”√2a=−(p2−q2)−(p2+q2)2p2
= 1
(ii) 9x^{2} – 9(a + b)x + (2a^{2} + 5ab + 2b^{2}) = 0
Comparing this with ax^{2} + bx + c = 0.
a =9
b = 9 (a + b)
c = (2a^{2} + 5 ab + 2b^{2})
Î” = B^{2} – 4AC
⇒ 81 (a + b)^{2} – 36(2a^{2} + 5ab + 2b^{2})
⇒ 9a^{2} + 9b^{2} – 18ab
⇒ 9(a – b)^{2} > 0
∴ the roots are real and given by
Question 3.
Find the HCF of x^{3} + x^{2} + x + 1 and x^{4} – 1.
Answer:
x^{3} + x^{2} + x + 1 = x^{2} (x + 1) + 1 (x + 1)
= (x + 1) (x^{2} + 1)
x^{4}1 = (x^{2})^{2 }– 1
= (x^{2} + 1) (x^{2}– 1)
= (x^{2} + 1) (x + 1) (x – 1)
H.C.F. = (x^{2} + 1)(x + 1)
Question 4.
Prove that the equation x^{2}(a^{2} + b^{2}) + 2x(ac + bd) + (c^{2} + d^{2}) = 0 has no real root if ad ≠ bc
Solution:
Î” = b^{2} – 4ac
⇒ 4(ac + bd)^{2} – 4(a^{2} + b^{2})(c^{2} + d^{2})
⇒ 4[(ac + bd)^{2} – (a^{2} + b^{2})(c^{2} + d^{2})]
⇒ 4(a^{2}c^{2} + b^{2}d^{2} + 2acbd – a^{2}c^{2}b^{2}c^{2} – a^{2}d^{2} – b^{2}d^{2}]
⇒ 4[2acbd – a^{2}d^{2} – b^{2}c^{2}]
⇒ 4 [a^{2}c^{2} + b^{2}c^{2} – 2adbc]
⇒ 4[ad – bc]^{2}
We have ad ≠ bc
∴ ad – bc > 0
⇒ (ad – bc)^{2} > 0
⇒ 4(ad – bc)^{2} < 0 ⇒ Î” < 0
Hence the given equation has no real roots.
Question 5.
Find the L.C.M of 2(x^{3} + x^{2} – x – 1) and 3(x^{3} + 3x^{2} – x – 3)
Answer:
2[x^{3} + x^{2} – x – 1] = 2[x^{2}(x+ 1) 1 (x + 1)]
= 2(x + 1) (x^{2} – 1)
= 2(x + 1) (x + 1) (x – 1)
= 2(x + 1)^{2} (x – 1)
3[x^{3} + 3x^{2} – x – 3] = 3[x^{2}(x + 3) 1 (x + 3)]
= 3[(x + 3)(x^{2} – 1)]
= 3(x + 3)(x + 1) (x – 1)
L.C.M. = 6(x + 1)^{2}(x – 1) (x + 3)
Question 6.
A two digit number is such that the product of its digits is 12. When 36 is added to the number the digits interchange their places. Find the number.
Solution:
Let the ten’s digit of the number be x. It is given that the product of the digits is 12.
Unit’s digit = 12x
Number = 10x + 12x
If 36 is added to the number the digits interchange their places.
x = 6, 2.
But a number can never be (ve). So, x = 2.
The number is 10 × 2 + 122 = 26
Question 7.
Seven years ago, Vanin’s age was five times the square of swati’s age. Three years hence Swati’s age will be two fifth of Varun’s age. Find their present ages.
Solution:
Seven years ago, let Swathi’s age be x years.
Seven years ago, let Varun’s age was 5x^{2} years.
Swathi’s present age = x + 7 years
Varun’s present age = (5x^{2} + 7) years
3 years hence, we have Swathi’s age = x + 7 + 3 years
= x + 10 years
Varun’s age = 5x^{2} + 7 + 3 years
= 5x^{2} + 10 years
It is given that 3 years hence Swathi’s age will be 25 of Varun’s age.
∴ x + 10 = 25 (5x^{2} + 10)
⇒ x + 10 = 2x^{2} + 4
⇒ 2x^{2} – x – 6 = 0
⇒ 2x(x – 2) + 3(x – 2) = 0
⇒ (2x + 3)(x – 2) = 0
⇒ x – 2 = 0
⇒ x = 2 (∵ 2x + 3 ≠ 0 as x > 0)
Hence Swathi’s present age = (2 + 7) years
= 9 years
Varun’s present age = (5 × 2^{2} + 7) years = 27 years
Question 8.
A chess board contains 64 equal squares and the area of each square is 6.25 cm^{2}. A border round the board is 2 cm wide find its side.
Solution:
Let the length of the side of the chess board be x cm. Then,
Area of 64 squares = (x – 4)^{2}
(x – 4)^{2} = 64 × 6.25
⇒ x^{2} – 8x+ 16 = 400
⇒ x^{2} – 8x384 = 0
⇒ x^{2} – 24x + 16x – 384 = 0
⇒ (x – 24)(x + 16) = 0
⇒ x = 24 cm.
Question 9.
Find two consecutive natural numbers whose product is 20.
Solution:
Let a natural number be x.
The next number = x + 1
x (x + 1) = 20
x^{2} + x – 20 = 0
(x + 5)(x – 4) = 0
x = 5, 4
∴ x = 4 (∵ x ≠ 5, x is natural number)
The next number = 4 + 1 = 5
Two consecutive numbers are 4, 5
Question 10.
A two digit number is such that the product of its digits is 18, when 63 is subtracted from the number, the digits interchange their places. Find the number.
Solution:
Let the tens digit be x. Then the units digits = 18x
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