Warning: Undefined array key "https://nctbsolution.com/brilliants-composite-mathematics-class-6-solutions/" in /home/862143.cloudwaysapps.com/hpawmczmfj/public_html/wp-content/plugins/wpa-seo-auto-linker/wpa-seo-auto-linker.php on line 192
Brilliant’s Composite Mathematics Class 6 Solutions Chapter 7 Introduction to Linear Equations
Welcome to NCTB Solutions. Here with this post we are going to help 6th class students for the Solutions of Brilliant’s Composite Mathematics Class 6 Math Book, Chapter 7, Introduction to Linear Equations. Here students can easily find step by step solutions of all the problems for Introduction to Linear Equations, Exercise 7.1, 7.2, 7.3 and 7.4 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.
Introduction to Linear Equations Exercise 7.1 Solution
Question no – (1)
Solution :
(i) x – 8 = 4
L.H.S = (x – 8),
R.H.S = 4
(ii) 3y = y + 1
L.H.S = 3y,
R.H.S = y + 1
(iii) 3x + 5 = 11
L.H.S = 3X + 5,
R.H.S = 11
(iv) 2x – 7 = 2y + 3
L.H.S = 2x – 7,
R.H.S = 2y + 3
(v) y2 = 64
L.H.S = y2,
R.H.S = 64
Question no – (2)
Solution :
According to the given question :
(i) Five added to a number x gives 8
Equation = x + 5 = 8
Hence, the required equation is x + 5 = 8
(ii) A number x divided by 6 gives 2
Equation = x – 7 = 6
Thus, required equation is x – 7 = 6
(iii) 9 times a number x is 12
Equation, 9x = 12
Therefore, the required equation is 9x = 12
(iv) The square of x is 16
now in equation x² = 16
Therefore, the required equation is x² = 16
(v) Twice a number subtracted from 15 is 7
Now in equation 15 – 2x = 7
Hence, the required equation is 15 – 2x = 7
Introduction to Linear Equations Exercise 7.2 Solution
Question no – (1)
Solution :
In the given question,
x + 5 = 9
Trial and error method :
| X | LHS | RHS |
| 0 | 5 | 9 |
| 1 | 6 | 9 |
| 2 | 7 | 9 |
| 3 | 8 | 9 |
| 4 | 9 | 9 |
∴ Here x = 4
Question no – (2)
Solution :
In the given question,
5x = 20
Trial and error method :
| X | LHS | RHS |
| 1 | 5 | 20 |
| 3 | 15 | 20 |
| 4 | 20 | 20 |
∴ Here x = 4
Question no – (3)
Solution :
In the given question,
x – 8 = -4
Trial and error method :
| X | LHS | RHS |
| 0 | -8 | -4 |
| 2 | -6 | -4 |
| 4 | -4 | -4 |
∴ Here x = 4
Question no – (4)
Solution :
As per the given question,
5 – x = 9
Trial and error method :
| X | LHS | RHS |
| 0 | 5 | 9 |
| -1 | 6 | 9 |
| -2 | 7 | 9 |
| -3 | 8 | 9 |
| -4 | 9 | 9 |
∴ Here x = -4
Question no – (5)
Solution :
Given in the question,
19 = 7 + x
Trial and error method :
| X | LHS | RHS |
| 0 | 19 | 7 |
| 3 | 19 | 10 |
| 6 | 19 | 13 |
| 9 | 19 | 16 |
| 12 | 19 | 19 |
∴ Here x = 12
Question no – (6)
Solution :
In the given question,
x/8 = 9
Trial and error method :
| X | LHS | RHS |
| 0 | 0 | 9 |
| 8 | 1 | 9 |
| 24 | 3 | 9 |
| 48 | 6 | 9 |
| 72 | 9 | 9 |
∴ Here, x = 72
Question no – (7)
Solution :
In the given question,
2x – 1 = – 4 + x
Trial and error method :
| X | LHS | RHS |
| 0 | -1 | -4 |
| -1 | -3 | -5 |
| -2 | -5 | -6 |
| -3 | -7 | -7 |
∴ Here, x = -3
Question no – (8)
Solution :
From the question we get,
x – 1 = – 3 + 2x
Trial and error method :
| X | LHS | RHS |
| 0 | -1 | -3 |
| 1 | 0 | -1 |
| 2 | 1 | 1 |
∴ Here, x = 2
Introduction to Linear Equations Exercise 7.3 Solution
Question no – (1)
Solution :
Given, x + 7 = 5 …. (i)
Subtracting both side by 7
∴ x + 7 – 7 = 5 – 7
= x = -2
Check, put x = -2 in (i)
∴ LHS = -2 + 7 = 5 = RHS
Therefore, LHS = RHS
Question no – (2)
Solution :
Let, x – 5 = 7 ….. (i)
Adding both side by 5
∴ x – 5 + 5 = 7 + 5
= x = 12
Check, put x = 12 in (i)
∴ LHS 12 – 5 = 7 = RHS
∴ LHS= RHS…(Verified)
Question no – (3)
Solution :
Let, y + 3/2 = 6 ….. (i)
Subtracting both side by 3/2
∴ y = 6 – 3/2 = 12 – 3/2
= 9/2
Check, put y = 9/2 in (i)
∴ LHS, 9/2 + 3/2
= 9 + 3/2
= 12/2
= 6
Therefore, LHS = RHS…(Verified)
Question no – (4)
Solution :
Let, y – 5/2 = 8 …… (i)
adding both side by 5/2
∴ y – 5/2 + 5/2 = 8 + 5/2
y = 16 + 5/2 = 21/2
Check, put – y = 21/2 in …. (i)
∴ LHS, 21/2 – 5/2
= 21 – 5/2
= 16/2
= 8
∴ LHS = RHS …(Verified)
Question no – (5)
Solution :
Let, 3y = 6 …..(i)
Dividing both side by 3
3y/3 = 6/3
= y = 2
Check, put y = 2 in (i)
∴ LHS = 3 × 2
= 6
Hence, LHS = RHS…(Verified)
Question no – (6)
Solution :
Let, 8y = 88 …. (i)
Dividing both side by 8
8y/8 = 88/8
y = 11
Check, put y = 11 in …..(i)
∴ LHS = 8 × 11
= 88
∴ LHS = RHS …(Verified)
Question no – (7)
Solution :
Let, x/2 = 7 …..(i)
Multiplying both side by 2
∴ x/2 × 2 = 7 × 2
= x = 14
Check, put x = 14 in (i)
∴ LHS = 17/2
= 7
∴ LHS = RHS…(Verified)
Question no – (8)
Solution :
Let, -3x = 18 …… (i)
Dividing both side by – 3
∴ -3x/-3 = 18/-3
x = – 6
Putting x = -6 in (i)
∴ LHS = (-3) × (-6)
= 18
∴ L.H.S = R.H.S …(Verified)
Question no – (9)
Solution :
Let, 7 + 3Y = -14…..(i)
Subtracting both side by 7
∴ – 7 + 7 + 3y = -7 – 14
3y = -21
Dividing both side by 3
∴ 3y/3 = -21/3
y = – 7
Check, put y = -7 in (i)
∴ LHS = 7 + (3 × -7)
= 7 – 21
= -14
∴ L.H.S = R.H.S …(Verified)
Question no – (10)
Solution :
Let, 15x = 225 ….(i)
Dividing both side by 15
= 15x/15 = 225/15
= x = 15
Check Put x = 15 in (i)
∴ L.H.S = 15 × 15
= 225
∴ L.H.S = R.H.S…(Verified)
Question no – (11)
Solution :
Let, 2x + 3 = 19
= 2x + 3 = 19 – (i)
Subtracting = 19 – 3
= 2x = 16
Dividing both side by 2
= 2x/2 = 16/2
Check, Put x = 8 in …(i)
∴ L.H.S = 2 × 8 + 3
= 16 + 3
= 19
∴ L.H.S = R.H.S…(Verified)
Question no – (12)
Solution :
Let, 6x – 3 = 33
= 6x – 3 + 3 = 33 + 3
= 6x = 36
Dividing both side by 6
Check, put x = 6 in (i)
L.H.S = 6 × 6 – 3
= 36 – 3
= 33
∴ L.H.S = R.H.S…(Verified)
Question no – (13)
Solution :
Let, y/7 = 7
Multiplying both side by 7
= y/7 × 7 = 7 × 7
= y = 49
Check, put y = 49 in (i)
∴ L.H.S = 49/7
= 7
∴ L.H.S = R.H.S…(Verified)
Question no – (14)
Solution :
Let, 3x = 63 ….(i)
Dividing both side by 3
= 3x/3 = 63/3
= x = 21
Check, put x = 21 in ….(i)
∴ L.H.S = 3 × 21
= 63
∴ L.H.S = R.H.S …(Verified)
Question no – (15)
Solution :
Let, x – 3/2 = 7 …..(i)
Adding both side by 3/2
= x – 3/2 + 3/2 = 7 + 3/2
= x = 14 + 3/2
= x = 17/2
Check, put x = 17/2 in (i)
∴ L.H.S = 17/2 – 3/12
= 7
∴ L.H.S = R.H.S…(Verified)
Question no – (16)
Solution :
Let, y + 3/4 = 6 ….(i)
Subtracted both side by 3/4
= y + 3/4 – 3/4 = 6 – 3/4
= y = 24 – 3/4
= y 21/1
Check put y = 21/4 in (i)
∴ L.H.S = 21/4 + 3/4
= 24/4
= 6
∴ L.H.S = R.H.S …(Verified)
Question no – (17)
Solution :
Let, 2(x + 3) = 28 …..(i)
Dividing both side by 2
= 2(x + 3)/2 = 28/2
= x + 3 = 14
Subtracting both side by 3
= x + 3 – 3 = 14 – 3
= x = 11
Check, put x = 11 in (i)
∴ L.H.S = 2 (11 + 3)
= 2 × 14
= 28
∴ L.H.S = R.H.S …(Verified)
Question no – (18)
Solution :
Let, 3(x + 4) = 27 ….(i)
Dividing both side by 3
= 3(x + 4)/3 = 27/3
= x + 4 = 9
Subtracting both side by 4
x + 4 – 4 = 9 – 4
= x = 5
Check, put x = 5 in (i)
∴ L.H.S = 2 (5 × 4)
= 3 × 9
= 27
∴ L.H.S = R.H.S …(Verified)
Introduction to Linear Equations Exercise 7.4 Solution
Question no – (1)
Solution :
Let, the number be x
Given that,
= x + 8 = 15
= x = 15 – 8
= 7
Therefore, the number will be 7
Question no – (2)
Solution :
Let, the number be x
Given that, x × 15 = 90
= x = 90/15
= 6
Thus, the required number will be 6
Question no – (3)
Solution :
Let, the number be x
Given that,
= x/8 = 6
= x = 48
Therefore, the required number will be 48
Question no – (4)
Solution :
As per the question,
The sum of the ages of father and his son is = 75 years.
The age of the son is = 25 years
Let, the age of the father x
Given that,
= 25 + x = 75
= x = 50
Therefore, the age of his father will be 50 years.
Question no – (5)
Solution :
Let, one number be x
Another number is 3x
Given that,
= 3x – x = 24
= 24 = 24
= x = 12
Therefore, the required numbers are 12, 36
Question no – (6)
Solution :
Let, Ram’s age = x
Ram’s father age = 3x
Given that,
= 3x + x = 40
= 4x = 40
= x = 10
∴ Ram’s present age = 10 year.
∴ Ram’s father age = 30 Years.
Question no – (7)
Solution :
Let, number of boy’s in the school = x
Number of girls = x + 40
Given that,
= x + 40 + x = 2060
= 2x = 2060 – 40
= x = 2060/2
= 1010
∴ Boys in the school = 1010,
∴ Girls in the school,
= 1010 + 40
= 1050
Therefore, the number of girls will be 1050
Question no – (8)
Solution :
Let, the number of coins – x
Given that,
= x × 25 = 15 × 100
= x = 15 × 100/25
= 60
Therefore, there are total 60 coins.
Question no – (9)
Solution :
Let, there are x number of men employee
∴ Number of women employees 3x
Given that,
= 2x + x = 48
= 4x = 48
= x = 48/4 = 12
Hence, there are 12 numbers of men employees and 365 number of women employees.
Question no – (10)
Solution :
Let, the numbers are x and 2x
Given that,
= 2x – x = 15
= x = 15
Therefore, the required numbers are 15 and 30
Question no – (11)
Solution :
Let, the width is x and length is 3x
Perimeter of a rectangle,
= 2 (x + 3x)
= 2 × 4x
= 8x
Given that,
= 8x = 96
= x = 96/8
= x = 12
∴ Length of the rectangle,
= 3 × 12
= 36 m
∴ Width of the rectangle,
= 12 m
Question no – (12)
Solution :
Let, Jyoti age = x years
∴ Sanjoy’s age = x + 4 year
Given that,
= x + 4 + x = 18
= 2x = 18 – 4
= x = 14/2
= x = 7
∴ Jyoti’s age = 7 years.
∴ Sanjoy’s age = 7 + 4 = 11 years.
Hence, their ages are Jyoti 7 years and Sanjoy 11 years.
Question no – (13)
Solution :
Let, the consecutive numbers are = x and x + 1
Given that,
= x + x + 1 = 33
= 2x = 33 – 3
= x = 32/2
= x = 16
∴ The consecutive numbers are = 16 and (16 + 1) = 17
Therefore, the required numbers are 16 and 17
Question no – (14)
Solution :
Let, the even number are x, are x + 2
Given that,
= x + x + 2 = 86
= 2x = 86 – 2
= x = 48/2 = 42
On number = 42
Another number = 42 + 2 = 44
Therefore, the consecutive numbers are 42 and 44
Question no – (15)
Solution :
Let, the years of her birth is = x
Given that,
= x = x – 40 = 35
= x = 335 + 40
Thus, the year of her birth will be 375 years.
Question no – (16)
Solution :
Let, then present age = x
Given that,
= x + 15 = 4x
= 4x – x = 15
= 3x = 15
= x = 5
Hence, her present age 5 years
Question no – (17)
Solution :
Aruna is 3 times as old as her son Amit.
After 15 years Aruna will be twice as old as her son
Let, the age of Amit = x
∴ Aruna’s age = 3x
After 15 years = 3x + 15 , x + 15
Given that,
= 3x + 15 = 2 (x + 15)
= 3x + 15 = 2x + 30
= 3x – 2x = 30 – 15
= x = 15
Therefore, Amit’s age is 15 years and Aruna’s age 45 years.
Question no – (18)
Solution :
Let, the breadth = x and the length = 7 + x
∴ Given that
= 2 (X + 7 + X) = 86
= 2X + 7 = 86/2
= 2X = 43 – 7
= X = 36/2
= X = 18 m
∴ Length = 18 + 7 = 25 m
∴ Breadth = 18 m
Hence, the length will be 25 m and breadth will be 18 m .
Question no – (19)
Solution :
Let, the numbers are = x and = 3x
Given that’s,
= x + 3x = 124
= 4x = 124
= x = 124/4
= x= 31
Thus, the required numbers are 31 and 93
Question no – (20)
Solution :
Let, the number of girls = x
∴ Number of boys = 2x
Given that,
= 2x + x = 1242
= 3x = 1242
= x = 1242/3
= x = 414
∴ Girls students in the school = 414
∴ Boys students in the school,
= 414 × 2
= 828
Therefore, the number of boys will be 828 and girls will be 414.
Question no – (21)
Solution :
As per the question,
Let, his present age = x years
After 16 years = x + 16 years
Given that,
= 3x = x + 16
= 3x – x = 16
= 2x = 16
= x = 8
Hence, the present age of Rohit will be 8 years.
Next Chapter Solution :