10th Maths - Book Back Answers - Chapter 3 Unit Exercise 3 - English Medium Guides

  

 

SSLC / 10th - Maths - Book Back Answers - Chapter 3 Unit Excercise 3 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 3 Unit Exercise 3: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 3 Unit Exercise 3 – Maths from the Tamil Nadu State Board 10th Standard textbook. These detailed answers have been carefully prepared by our expert teachers at KalviTips.com.

    We have explained each answer in a simple, easy-to-understand format, highlighting important points step by step under the relevant subtopics. Students are advised to read and memorize these subtopics thoroughly. Once you understand the main concepts, you’ll be able to connect other related points with real-life examples and confidently present them in your tests and exams.

    By going through this material, you’ll gain a strong understanding of Chapter 3 Unit Exercise 3 along with the corresponding book back questions and answers (PDF format).

Question Types Covered:

• 1 Mark Questions: Choose the correct answer, Fill in the blanks, Identify the correct statement, Match the following 
• 2 Mark Questions: Answer briefly 
• 3, 4, and 5 Mark Questions: Answer in detail

All answers are presented in a clear and student-friendly manner, focusing on key points to help you score full marks.

All the best, Class 10 students! Prepare well and aim for top scores. Thank you!

Chapter 3 Algebra Unit Ex 3

1. Solve
$\frac{1}{3}$(x + y – 5) = y – z = 2x – 11 = 9 – (x + 2z)
Answer Key:
$\frac{1}{3}$(x + y – 5) = y – z
x + y – 5 = 3y – 3z
x + y – 3y + 3z = 5
x – 2y + 3z = 5 ….(1)
y – z – 2x – 11
-2 x + y – z = -11
2x – y + z = 11 …..(2)
2x – 11 = 9-(x + 2 z)
2x – 11 = 9 – x – 2z
2x + x + 2z = 9 + 11
3x + 2z = 20 ….(3)

3x – z = 17 …. (5)

Substitute the value of z = 1 in (3)
3x + 2(1) = 20
3x = 20 – 2
3x = 18
x = $\frac{18}{3}$ = 6
substitute the value of x = 6, z = 1 in (2)
2(6) – y + 1 = 11
12 – y + 1 = 11
13 – y = 11
-y = 11 – 13
-y = -2
y = 2
∴ The value of x = 6, y = 2 and z = 1

 

2. One hundred and fifty students are admitted to a school. They are distrbuted 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.
Answer Key:
Let the number of students in section A be “x”
Let the number of students in section B be “y”
Let the number of students in section C be “z”
By the given first condition
x + y + z = 150 ……(1)
again by the second condition
x – 6 = z + 6
x – z = 6 + 6
x – z = 12 ….(2)
again by the third condition
x + y = 4z
x + y – 4z = 0
x + y – 4z = 0 ….(3)
Subtracting (1) and (3)

Substitute the value of z = 30 in (2)
x – 30 = 12
x = 12 + 30
= 42
Substitute the value of x = 42 and z = 30 in (1)
42 + y + 30 = 150
y + 72 = 150
y = 150 – 72
= 78
Number of students in section A, B and C are = 42, 78 and 30.

 

3. In a three-digit 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.
Answer Key:
Let the hundred digit be x
the tens digit be y and the unit digit be z
∴ The number is 100x + 10y + z
By the given first condition
100y + 10x + z = 54 + 3 (100x + 10y + z)
100y + 10x + z = 54 + 300x + 30y + 3z
-290x + 70y – 2z = 54 (÷ -2)
145x-35y + z = -27 ….(1)
Again by the second condition
198 + 100x + 10y + z = 100z + 10y + x
99x – 99z = -198 (÷ 99)
x – z = -2 ….(2)
Again by the third condition
y – x = 2(y – z)
y – x = 2y – 2z
– x – y + 2z = 0
x + y – 2z = 0 ….(3)

substitute the value of x = 1 in …….(2)
1 – z = -2
3 = z
∴ z = 3
substitute the value of x = 1 and z = 3 in …….(3)
1 – y – 6 = 0
y – 5 = 0
y = 5
∴ The number is 153

 

4. Find the least common multiple of xy(k² +1) + k(x² + y²) and xy(k² – 1) + k(x² – y²).
Answer Key:
 xy (k² + 1) + k (x² + y²) …………… (1)
xy(k² – 1) + k(x² – y²) …………… (2)
(1) ⇒ xyk² + xy + kx² + ky²
(2) ⇒ xyk² – xy + kx² – ky²
(1) ⇒ yk (xk + y) + x (xk + y)
= (xk + y) (x + yk)
(2) ⇒ yk (xk – y) + x (xk – y)
= (x + yk) (xk – y)
∴ L.C.M. : (x + yk) (xk + y) (xk – y)
= (x + yk) (x²k² – y²)

 

5. Find the GCD of the following by division algorithm
2x⁴ + 13x³ + 27x² + 23x + 7,
x³ + 3x² + 3x + 1, x² + 2x + 1
Answer Key:
p(x) = 2x⁴ + 13x³ + 27x² + 23x + 7
g(x) = x³ + 3x² + 3x + 1
r(x) = x² + 2x + 1
(i) Find the G.C.D. of p(x) and g(x)

(ii) Find the G.C.D. of r(x) and the G.C.D. of p(x) and g(x)

∴ G.C.D.= x² + 2x + 1
∴ G.C.D. of the three
polynomials = x² + 2x + 1

 

6. Reduce the given Rational expressions to its lowest form
(i) $\frac{x^{3a}-8}{x^{2a}+2x^a+4}$
Answer Key:
x^3a – 8 = (x^a)³ – 2³
(using the formula a³ – b³ = (a – b)(a² + ab + b²)
= (x^a – 2)[(x^a)² + x^a × 2 + 2²]
= (x^a – 2) (x^2a + 2x^a + 4)

  

(ii) $\frac{10x^3-25x^2+4x+10}{-4-10x^2}$
Answer Key:
10x³ – 25x² + 4x – 10 = 5x²(2x – 5) + 2 (2x – 5)
= (2x – 5) (5x² + 2)
– 4 – 10x² = -2 (2 + 5x²)
= -2(5x² + 2)

 

7. Simplify

Answer Key:

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?
Answer Key:
Let the time taken by Arul be “x” hours
Let the time taken by Ravi be “y” hours
Let the time taken by Ram be “z” hours
By the given first condition
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{5}$
Again by the given second condition
$\frac{1}{x}=2\times \frac{1}{y}$
$\frac{1}{x}-\frac{2}{y}=0$
By the given third condition
$3\times \frac{1}{\mathrm{z}}=\frac{1}{\mathrm{x}}$
$-\frac{1}{\mathrm{x}}+\frac{3}{\mathrm{z}}=0$
Let $\frac{1}{x}$ = a, $\frac{1}{y}$ = b, $\frac{1}{z}$ = c
a + b + c = $\frac{1}{6}$
6a + 6b + 6c = 1 …….(1)
a – 2b = 0 ……….(2)
-a + 3c = 0 …………(3)

Arul take 11 hours, Ravi take 22 hours and Ram takes 33 hours.

 

9. Find the square root of 289x⁴ – 612x³ + 970x² – 684x + 361.
Answer Key:

10. Solve $\sqrt{\mathrm{y}+1}+\sqrt{2\mathrm{y}-5}$= 3
Answer Key:
$\sqrt{\mathrm{y}+1}+\sqrt{2\mathrm{y}-5}$= 3
(squaring on bothsides)

8y² – 9y² – 12y + 78y – 20 – 169 = 0
-y² – 66y – 189 = 0
y² – 66y + 189 = 0
(y – 3) (y – 63) = 0

y – 3 or y = 63
The value of y is 3 and 63

 

11. A boat takes 1.6 hours longer to go 36 kins 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?
Answer Key:
Let the speed of the boat in still water be “x”
Time taken to go for up of a river = $\frac{36}{\mathrm{x}+4}$
By the given condition
$\frac{36}{\mathrm{x}-4}-\frac{36}{\mathrm{x}+4}$= 1.6

The speed of the boat in still water = 14 km/hr

 

12. Is it possible to design a rectangular park of perimeter 320 m and area 4800 m2? If so find its length and breadth.
Answer Key:
Let the length of the rectangular park be “l”
and the breadth of the rectangular park be “b”
Perimeter of the park = 320 m
2 (l + b) = 320
l + b = 160
l = 160 – b ……….(1)
Area of the park = 4800 m²
l × b = 4800 ….(2)
substitute the value of l = 160 – b in (2)
(160 – b)b = 4800
160b – b² = 4800
b² – 160b + 4800 = 0
(b – 120) (b – 40) = 0

b = -120 = 0 or b – 40 = 0
b = 120 or b = 40
If breadth is 120 length is 40
If breadth is 40 length is 120
Length of the park = 120 m
Breadth of the park = 40 m

 

13. At t minutes past 2 pm, the time needed to 3 pm is 3 minutes less than$\frac{t^2}{4}$Find t.
Answer Key:
Time needed by the minutes hand to show
3 pm = (60 – 1) minutes
By the given condition

∴ The value of t = 14 minutes 

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.
Answer Key:
Let the number of rows in the hall be “x”
∴ Total number of rows = x
Total number of seats in the hall is “x2”
By the given condition
x² + 375 = 2x (x – 5)
x² + 375 = 2x²– 10x
x² – 2x² + 10x + 375 = 0
– x² + 10x + 375 = 0
– x² – 10x – 375 = 0
(x – 25) (x + 15)
x – 25 = 0 or x + 15 = 0
x = 25 or x = – 15

Number of rows in the hall = 25

 

15. If α and β are the roots of the polynomial f(x) – x² – 2x + 3, find the polynomial whose roots are
(i) α + 2, β + 2
Answer Key:
α and β are the roots of the polynomial
x² – 2x + 3 = 0
α + β = 2; αβ = 3
(i) Sum of the roots = α + 2 + β + 2
= α + β + 4
= 2 + 4
= 6
Product of the roots = (α + 2) (β + 2)
= αβ + 2α + 2β + 4
= αβ + 2(α + β) + 4
= 3 + 4 + 4
= 11
The quadratic polynomial
x² – (sum of the roots) x + product of the roots = 0
x² – (6) x + 11 = 0
x² – 6x + 11 = 0

 

(ii) $\frac{\alpha -1}{\alpha +1},\frac{\beta -1}{\beta +1}$
Answer Key:
Sum of the roots

The quadratic polynomial is
x² – (sum of the roots) + products of the roots = 0
x² – ($\frac{2}{3}$) x + $\frac{1}{3}$ = 0
3x² – 2x + 1 = 0

 

16. If -4 is a root of the equation x² + px – 4 = 0 and if the equation x² + px + q = 0 has equal roots, find the values of p and q.
Answer Key:
f(x) = x² + px – 4 = 0
If -4 is a root, then
f(-4) = (-4)² + P(-4) – 4 = 16 – 4p – 4 = 0
12 – 4p = 0
-4p = -12
p = 3
x² + 3x + q =0 has equal roots,
∆ = b² – 4ac = 0
3² – 4 × 1 × q = 0
9 – 4q = 0
-4 q = -9
q = $\frac{9}{4}$
p = 3, q = $\frac{9}{4}$

 

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.

and the 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.
Answer Key:
(i) Let A represent the sale on April

 Let B represent the sale on May

Average sale of the month April and May

 

(ii) If the sales continue to increase in the same way in the successive months, what will be sales in the month of August?
Answer Key:
If it increasing in the successive months of
May sale is 2 (April sale)
June sale is 4 (April sale)
July sale is 8 (April sale)
August sale is 16 (April sale)
Sales in the month of August

 

18. 

Answer Key:

 

19. Given

and if BA = C², find p and q
Answer Key:

∴The value of p = 8 and q = 4

 

20. find the matrix D, such that CD – AB = 0
Answer Key:
Given