10th Maths - Book Back Answers - Chapter 3 Exercise 3.1 - English Medium Guides

  

 

SSLC / 10th - Maths - Book Back Answers - Chapter 3 Excercise 3.1 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 3 Excercise 3.1: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 3 Excercise 3.1 – Maths from the Tamil Nadu State Board 10th Standard Maths 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 Excercise 3.1 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 Ex 3.1

1. Solve the following system of linear equations in three variables

(i) x + y + z = 5

2x – y + z = 9
x – 2y + 3z = 16
Answer Key:
x + y + z = 5 ….(1)
2x – y + z = 9 ….(2)

x – 2y + 3z = 16 ….(3)

by adding (1) and (2)

Substituting z = 4 (4)
3x + 2(-4) = 14
3x – 8 = 14
3x = 14 – 8
3x = 6
x = $\frac{6}{3}$ = 2 

Substituting x = 2 and z = 4 in (1)
2 + y + 4 = 5
y + 6 = 5
y = 5 – 6
= -1
∴ The value of x = 2, y = -1 and z = 4

 

$\mathbf{\left(}\mathbf{i}\mathbf{i}\mathbf{\right)}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{-}\frac{\mathbf{2}}{\mathbf{y}}\mathbf{+}\mathbf{4}\mathbf{=}\mathbf{0}\mathbf{,}$

$\frac{2}{z}+\frac{3}{x}=14$

Answer Key:

Let $\frac{1}{x}$ = p, $\frac{1}{y}$ = q and $\frac{1}{z}$ = r
p – 2q + 4 = 0
p – 2q = -4 ……(1)
q – r + 1 = 0
q – r = -1 ……(2)
3p + 2r = 14 ……(3)

Substituting the value of p = 2 in (1)
2 – 2q = -4
-2q = – 4 – 2
-2q = -6
q = $\frac{6}{2}$ = 3
Substituting the value of q = 3 in (2)
3 – r = 1
– r = – 1 – 3
r = 4

The value of x = $\frac{1}{2}$, y = $\frac{1}{3}$ and z = $\frac{1}{4}$

(iii) x + 20 = $\frac{3y}{2}$+ 10 = 2z + 5 = 110 – (y + z)
Answer Key:

x + 20 = $\frac{3y}{2}$ + 10
Multiply by 2
2x + 40 = 3y + 20
2x – 3y = -40 + 20
2x – 3y = -20 ……(1)

$\frac{3y}{2}$ + 10 = 2z + 5
Multiply by 2
3y + 20 = 4z + 10
3y – 4z = 10 – 20
3y – 4z = -10 ……(2)

2z + 5 = 110 – (y + z)
2z + 5 = 110 – y – z
y + 3z = 110 – 5
y + 3z = 105 ….(3)

Substitute the value of z = 25 in (2)
3y – 4(25) = -10
3y – 100 = – 10
3y = – 10 + 100
3y = 90
y = $\frac{90}{3}$ = 30
∴ The value of x = 35, y = 30 and z = 25

Substitute the value of y = 30 in (1)
2x – 3(30) = -20
2x – 90 = -20
2x = -20 + 90
2x = 70
x = $\frac{70}{2}$ = 35

 

2. Discuss the nature of solutions of the following system of equations
(i) x + 2y – z = 6, – 3x – 2y + 5z = -12 , x – 2z = 3
Answer Key:
x + 2y – z = 6 …..(1)
-3x – 2y + 5z = -12 …..(2)
x – 2z = 3 ……(3)
Adding (1) and (2) we get

The above statement tells us that the system has an infinite number of solutions.

 

(ii) 2y + z = 3(- x + 1) ,-x + 3y – z = -4, 3x + 2y + z =$-\frac{1}{2}$
2y + z = 3 (- x + 1)
2y + z = -3x + 3 ……(1)
3x + 2y + z = $-\frac{1}{2}$

-x + 3y – z = – 4

x – 3y + z = 4 …..(2)

3x + 2y + z = – 12

Hence we arrive at a contradiction as 0 ≠ 7
This means that the system is inconsistent and has no solution.

 

(iii) $\frac{\mathrm{y}+\mathrm{z}}{4}=\frac{\mathrm{z}+\mathrm{x}}{3}=\frac{\mathrm{x}+\mathrm{y}}{2}$, x + y + z = 27

Answer Key:
$\frac{\mathrm{y}+\mathrm{z}}{4}=\frac{\mathrm{z}+\mathrm{x}}{3}$
3y + 3z = 4z + 4x
-4x + 3y + 3z – 4z = 0
-4x + 3y – z = 0
4x – 3y + z = 0 ………(1)

z+x3 = x+y2
2z + 2x = 3x + 3y
-3x + 2x – 3y + 2z = 0
-x – 3y + 2z = 0
x + 3y – 2z = 0 ………(2)

Substituting the value of x in (5)
6 + 5z = 81
5z = 81 – 6
5z = 75
z = $\frac{75}{5}$ = 15
Substituting the value of x = 3
and z = 15 in (3)
3 + y + 15 = 27
y + 18 = 27
y = 27 – 18
= 9
The value of x = 3, y = 9 and z = 15

This system of equations have unique solution.

 

3. Vani, her father and her grand father have an average age of 53. One-half of her grand father’s age plus one-third of her father’s age plus one fourth of Vani’s age is 65. If 4 years ago Vani’s grandfather was four times as old as Vani then how old are they all now?
Answer Key:

Let the age of Vani be”x” years
Vani father age = “y” years
Vani grand father = “z” years
By the given first condition.
$\frac{x+y+z}{3}$ = 53
x + y + z = 159 ….(1)
By the given 2^nd Condition.
$\frac{1}{2}$ z + $\frac{1}{3}$y + $\frac{1}{4}$x = 65
Multiply by 12
6z + 4y + 3x = 780
3x + 4y + 6z = 780 ….(2)
By the given 3^rd condition
z – 4 = 4 (x – 4) ⇒ z – 4 = 4x – 16
– 4x + z = – 16 + 4
4x – z = 12 ……(3)

Vani age = 24 years
Vani’s father age = 51 years
Vani grand father age = 84 years
Substitute the value of x = 24 in (3)
4 (24) – z = 12
96 – z = 12
-z = 12 – 96
z = 84
Substitute the value of
x = 24 and z = 84 in (1)
24 + y + 84 = 159
y + 108 = 159
y = 159 – 108
= 51

 

4. The sum of the digits of a three-digit number is 11. If the digits are reversed, the new number is 46 more than five times the old number. If the hundreds digit plus twice the tens digit is equal to the units digit, then find the original three digit number ?
Answer Key:
Let the hundreds digit be x and the tens digit be ”y” and the unit digit be “z”
∴ The number is 100x + 10y + z
If the digits are reversed the new number is 100z + 10y + x
By the given first condition
x + y + z = 11 ….(1)
By the given second condition
100z + 10y + x = 5 (100x + 10y + z) + 46
= 500x + 50y + 5z + 46
x – 500x + 10y – 50y + 100z – 5z = 46
– 499x – 40y + 95z = 46
499x + 40y – 95z = -46 ….(2)
By the given third condition
x + 2y = z
x + 2y – z = 0 ….(3)

∴ The number is 137
Subtituting the value of y = 3 in (5)
2x + 3(3) = 11
2x = 11 – 9
2x = 2
x = $\frac{2}{2}$ = 1
Subtituting the value of x = 1, y = 3 in (1)
1 + 3 + z = 11
z = 11 – 4
= 7

 

5. There are 12 pieces of five, ten and twenty rupee currencies whose total value is ₹105. But when first 2 sorts are interchanged in their numbers its value will be increased by ₹20. Find the number of currencies in each sort.
Answer Key:

Let the number of ₹5 currencies be “x”
Let the number of ₹10 currencies be “y”
and the number of ₹20 currencies be “z”
By the given first condition
x + y + z = 12 ………(1)
By the given second condition
5x + 10y + 20z = 105
x + 2y + 4z = 21 (÷5) ……….(2)
By the given third condition
10x + 5y + 20z = 105 + 20
10x + 5y + 20z = 125
2x + y + 4z = 25 ………..(3)

Substituting the value of x = 7 in (5)
7 – y = 4 ⇒ -y = 4 – 7
-y = -3 ⇒ y = 3
Substituting the value of x = 7, y = 3 in …. (1)
7 + 3 + z = 12
z = 12 – 10 = 2
x = 7, y = 3, z = 2
Number of currencies in ₹ 5 = 7
Number of currencies in ₹ 10 = 3
Number of currencies in ₹ 20 = 2