Class 8 ICSE Maths Solutions Chapter 14 Linear Equations in One Variable (Selina Concise)
Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Selina Class 8 ICSE Math Book, Chapter 14, Linear Equations in One Variable. Here students can easily find step by step solutions of all the problems for Linear Equations in One Variable, Exercise 14A, 14B and 14C 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. Here in this post students will get chapter 14 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.
Linear Equations in One Variable Exercise 14(A) Solution :
Question no – (1)
Solution :
Given, 20 = 6 + 2x
∴ 2x = 20 – 6 = 14
= x = 14/2
= x = 7
Question no – (2)
Solution :
Given equation, 15 + x = 5x + 3
∴ x – 5x = 3 – 15
= – 4x = – 12
= x = 12/4
= x = 3
Question no – (3)
Solution :
Given equation, 3x + 2 / x – 6 = -7
∴ 3x + 2 = – 7x + 42
= 3x + 7x = 42 – 2
= 10x = 40
= x = 40/10
= x = 4
Question no – (4)
Solution :
Given equation, 3a – 4 = 2 (4 – a)
∴ 3a – 4 = 8 – 2a
= 3a + 2a = 8 + 4
= 5a = 12
= a = 12/5
= a = 2.4
Question no – (5)
Solution :
In the question, 3(b – 4) = 2(4 – b)
∴ 3b – 12 = 8 – 2b
= 3b + 2b = 8 + 12
= 5b = 20
= b = 20/5
= b = 4
Question no – (6)
Solution :
Given equation, x + 2 / 9 = x + 4 / 11
∴ 11x + 22 = 9x + 36
= 11x – 9x = 36 – 22
= 2x = 14
= x = 14/2
= x = 7
Question no – (7)
Solution :
Given equation, x – 8/5 = x – 12/9
∴ 9x – 72 = 5x – 60
= 9x – 5x = 72 – 60
= 4x = 12
= x = 12/4
= x = 3
Question no – (8)
Solution :
Given equation, 5 (8x + 3) = 9 (4x + 7)
∴ 40x + 15 = 36x + 63
= 40x – 36x = 63 – 15 = 48
= 4x = 48
= x = 48/4
= x = 12
Question no – (9)
Solution :
Given equation, 3(x + 1) = 12 + 4 (x – 1)
∴ 3x + 3 = 12 + 4x – 4
= 3x – 4x = 12 – 4 – 3
= – x = 5
= x = -5
Question no – (10)
Solution :
In the question, 3x/4 – 1/4 (x – 20) = x/4 + 32
∴ 3x/4 – x/4 – x/4 = 32 – 5
= 3x – x – x/4 = 27
= x/4 = 27
= x = 27 × 4
= x = 108
Question no – (11)
Solution :
Given equation, 3a – 1/5 = a/5 + 5 2/5
∴ 3a – a/5 = 27/5 + 1/5
= 15a – a/5 = 27 + 1 / 5
= 14a = 18
= a = 28/14
= a = 2
Question no – (12)
Solution :
Given equation, x/3 – 2 1/2 = 4x/9 – 2x/3
∴ x/3 – 5/2 = 4x/9 – 2x/3
= x/3 – 4x/9 + 2x/3 = 5/2
= 3x – 4x + 6x /9 = 5/2
= 3x/9 = 15/2
= x = 9/2
= x = 4.5
Question no – (13)
Solution :
Given equation, 4(y + 2)/5 = 7 + 5y/13
∴ 4y + 8/5 = 91 + 5y /13
= 52y + 104 = 455 + 25y
= 52y – 252y = 455 – 104
= y = 351/27
= x = 13
Question no – (14)
Solution :
Given equation, a + 5/6 – a + 1/9 = a + 3 /4
∴ 3a + 15 – 2a – 2/18 = a + 3 / 4
= a + 13 / 18 = a + 3 / 4
= 4a + 52 = 18a + 54
= 4a – 18a = 52 – 12
= a = 42/-14
Question no – (16)
Solution :
Given equation, 6(6x – 5) -5 (7x – 8) = 12(4 – x) + 1
∴ 36x – 30 – 35x + 40 = 48 – 12x +1
= x + 12x = 49 – 10
= 13x = 39
= x = 39/13
= x = 3
Question no – (17)
Solution :
Given equation, (x – 5) (x + 3) = (x – 7) (x + 4)
x² + 3x – 5x – 15 = x² + 4x – 7x – 28
= – 2x – 15 = – 3x – 28
= 3x – 2x = 15 – 28
= x = – 13
Question no – (18)
Solution :
Given, (x – 5)2 – (x + 2)2 = -2
= (x2 – 10x + 25) – (x2 + 4x + 4) = -2
= x2 – 10x + 25 – x2 – 4x – 4 = -2
= -10x – 4x + 25 – 4 = -2
= – 14x = 4 – 2 – 25 = -23
= x = -23/-14
= x = 23/14
Question no – (19)
Solution :
Given equation, (x – 1) (x + 6) – (x – 2) (x – 3) = 3
∴ x² – x + 6x – 6 – x2 + 3x + 2x – 6 = 3
= – x + 6x + 3x + 2x – 6 – 6 = 3
= 10x = 15
= x = 15/10
= x = 3/2
Question no – (20)
Solution :
Given equation, 3x/x + 6 – x/x + 5 = 2
∴ 3×2 + 15x – x2 – 6x / (x + 6) (x + 5) = 2
= 2×2 + 9x = 2 (x + 6) (x +5)
= 2 (x2 + 11x + 30)
= 2×2 + 9x = 2×2 + 22x + 60
= + 9x – 22x = 60
= – 13x = 60
∴ x = 60/13
Question no – (21)
Solution :
Given equation, 1/x – 1 + 2 /x -2 = 3/x – 3
∴ x – 2 + 2x – 2/(x – 1) (x – 2) = 3/x – 3
= 3x – 4 / x2 – 3x + 2 = 3/x – 3
= 3x2 – 4x – 9x + 12 = 3x2 – 9x + 6
= 3x2 – 13x – 3x2 + 9x = 6 – 12
= – 4x = -6
= x = 6/4 = 3/2
= x = 1 1/2
Question no – (22)
Solution :
Given equation, x – 1 /7x – 14 = x – 3/7x – 26
∴ 7x2 – 7x – 26x + 26 = 7x2 – 14x – 21x + 42
= – 33x + 26 = 35x + 42
= 2x = 42 – 26 = 16
= x = 16/2
= x = 8
Question no – (23)
Solution :
Given equation, 1 /x -1 – 1/x = 1/x +3 – 1/x + 4
∴ x – (x -1) / x (x -1) = (x + 4) – (x + 3) / (x + 3) (x + 4)
= 1 / x (x- 1) = 1 / (x + 3) (x + 4)
= (x + 3) (x + 4) = x (x – 1)
= x2 + 4x + 3x + 12 = x2 – x
= x2 + 7x – x2 + x = -12
= 8x = -12
= x = – 12/8
= x = – 3/2
= x = -1 1/2
Question no – (26)
Solution :
Given, 0.25x + 1.95 = 0.9x
= 0,9x – 0.25x = 1.95
= 0.65x = 1.95
= x = 1.95/0.65
= x = 3…(Solved)
Linear Equations in One Variable Exercise 14(B) Solution :
Question no – (1)
Solution :
Let, the number = x
4 times of that no = 4x
15 les than 4 times the no = 4x – 15
∴ Required number,
= 4x – 15 = 9
= 4x = 9 + 15
= x = 24/4
= x = 6
Therefore, the required number will be 6.
Question no – (2)
Solution :
Let, Megha’s age = x year
∴ x + 3x = 60
= 4x = 60
= x = 60/4
= x = 15
Therefore, Megha’s age will be 15 years.
Question no – (3)
Solution :
Let, the number ‘x’
∴ 4x – 12 = 28
= 4x = 28 + 12
= x = 40/4
= x = 10
Therefore, required number will be 10.
Question no – (4)
Solution :
Let, number is = x
According to the question,
= 3x – 5 = – 20
= 3x = 5 – 20
= x = – 15/3
= x = -5
Therefore, the required number will be 5.
Question no – (5)
Solution :
Let, Neetu’s age = x years
According to the question,
4x = 3x + 15
= 4x – 3x = 15
= x = 15
Therefore, Neetu’s age will be 15 years.
Question no – (6)
Solution :
Let, required number = x
∴ x – 30 = 14 – 3x
= x + 3x = 30 + 14
= 4x = 44
= x = 44/4
= x = 11
Therefore, the required number will be 11.
Question no – (7)
Solution :
Let, B’s salary = x
A’s salary = 4x
∴ x + 4x = 3750 Rs
= 5x = 3750
= x = 3750/5
= x = 750 Rs
∴ B’s salary = 750 Rs
∴ A’s salary = (4 × 750) = 3000 Rs
Hence, the salary of A will be Rs 3000 and salary of B will be 750 Rs.
Question no – (8)
Solution :
Let, 1st part = x
2nd part = 178 – x
∴ x = 2 (178 – x) – 8
= x = 356 – 2x – 8
= x + 2x = 356 – 8
= 3x = 348
= x = 348/3
= x = 116
∴ 1st part = 116
∴ 2nd part,
= (178 – 116)
= 62
Question no – (9)
Solution :
Let, number = x
1/4th of number = 1/4 x
∴ 2x/5 = 6 + x/4
= 2x/5 – x/4 = 6
= 8x – 5x / 20 = 6
= 3x = 6 × 20
= x = 6 × 20/3
= x = 40
Therefore, the required number will be 40.
Question no – (10)
Solution :
Let, width = x cm
Length = 2x cm…(according to question)
∴ Perimeter = 2 (x + 2x)
= 2 × 3x = 6x
∴ 6x = 54
= x = 54/6
= x = 9 cm
∴ Width = 9 cm
∴ Length = (2 × 9) = 18 cm
Therefore, it ‘s length will be 18 cm.
Question no – (11)
Solution :
Let, width of original rectangle = x cm
Length = (2x – 5) cm
Length of new rectangle,
= (2x – 5 – 5)
= (2x – 10) cm
∴ New width = x + 2
∴ New perimeter = 2 [2x – 10 + x + 2]
= 2 [3x – 8]
= (6x – 16)cm
∴ 6x – 16 = 74
=> 6x = 74 + 16
=> x = 74 + 16 / 6
= 15 cm
∴ Length of original rectangle,
= 2x – 5
= (2 × 15 – 5)
= 25 cm
Hence, in the original rectangle the length will be 25 cm and width will be 15 cm.
Question no – (12)
Solution :
Let, three consecutive odd prime numbers,
= x , x + 2 , x + 4
∴ x + x + 2 + x + 4 = 57
= 3x = 57 – 6
= x = 51/3
= x = 17
Therefore, three consecutive odd numbers are 17, 19 and 21
Question no – (13)
Solution :
Let, son’s present age = x years
man’s present age = 3x years
After 12 yrs, son’s age = (x + 12)
Man’s age = (3x + 12) yrs
∴ 3x + 12 = 2 (x + 12)
= 3x + 12 = 2x + 24
= 3x – 2x = 24 – 12
= x = 12 (son’s age)
Therefore, 3x = (3 × 12) = 36 years (man’s age)
Question no – (14)
Solution :
Let, the year is x years,
∴ 12 + x = 42 + x/2
= 24 + 2x = 42 +x
= 2x – x = 42 – 24
= x = 18 years
Therefore, after 18 years, the son’s age will half the age of the man.
Question no – (15)
Solution :
Let, at 18 km/hr distance cover x km
at 15 km/hr distance cover (136 – x) km
∴ x/18 + 136 – x/15 = 8
= 5x + 816 – 6x/90 = 8
= 5x – 6x = 720 – 816
= -x = 96
= x = 96 km
Therefore, 96 km of the trip covered at 18 km/hr.
Question no – (16)
Solution :
Let, 1st number = x
2nd number = x + 3
∴ (x + 3)² – x² = 69
= x² + 2 . x .3+ 9 – x² = 69
= 6x = 69 – 9
= x = 60/6 = 10
∴ One number = 10
2nd number = (10 + 3) = 13
Hence, the numbers are 10 and 13.
Question no – (17)
Solution :
Let, two consecutive natural numbers = x , x + 1
∴ x/4 = x +1/5 + 1
= x/4 – x +1/ 5 = 1
= 5x – 4x – 4 / 20 = 1
= x = 20 + 4 = 24
∴ x + 1 = 24 + 1 = 25
Therefore, the required numbers will be 24 and 25.
Question no – (18)
Solution :
Let, three consecutive numbers are, x , x + 1 , x + 2
∴ x/5 + x+1/3 + x+2/4 = 40
= 12x + 20x + 20 + 15x + 30 /60 = 40
= 47x = 2400 – 20 – 30
= x = 2350/47
= x = 50
∴ x + 1 = 51
x + 2 = 50 + 2 = 52
Therefore, the three required numbers are 50, 51, and 52
Question no – (20)
Solution :
Let, son’s present age = x year.
Father age = 2x yr.
∴ 2x + 8 /x + 8 = 7/4
= 8x + 32 = 7x + 56
= 8x – 7x = 56 – 32
= x = 24
∴ Son’s present age = 24 yr.
∴ Now, father’s age will be,
= 2x = (2 × 24)
= 48 yr.
Therefore, present age of son will be 24 years, and present age of father will be 48 years.
Linear Equations in One Variable Exercise 14(C) Solution :
Question no – (1)
Solution :
(i) 1/3x – 6 = 5/2
= x/3 = 5/2 + 6
= 5 + 12/2
= x/3 = 17/2
= 2x = 3 × 17
= x = 3 × 17/2
= 25 1/2
(ii) 2x/3 – 3x/8 = 7/12
∴ 16x – 9x/24 = 7/12
= 7x/24 = 7/12
= x = 24 × 7/7 × 12
= 2
(iii) (x + 2) (x + 3) + (x – 3) (x – 2) – 2x (x + 1) = 0
= x² + 5x + 6 + x² – 5x + 6 – 2x² – 2x = 0
= 12 – 2x = 0
= x = 12/2
= 6
(iv) 1/10 – 7/x = 35
= – 7/x = 35 – 1/10
= 350 – 1/10
= – 7/x = 349/10
= 1/x = – 7 × 10/349
= x = – 349/70
(v) 13(x – 4) – 3 (x – 9) – 5(x + 4) = 0
∴ 13x – 52 – 3x + 27 – 5x – 20 = 0
= 13x – 3x – 5x – 52 + 27 – 20 = 0
= 13x – 8x – 72 + 27 = 0
= 5x – 45 = 0
= x = 45/5
= 9
(vi) x + 7 – 8x/3 = 17x/6 – 5x/8
∴ 3(x + 7) – 8x/3 = 17x × 4 – 5x × 3/24
= 3x + 21 – 8x/3 = 68x – 15x/24
= – 5x + 21/3 = 53x/24
= 159x = -120x + 504
= 159x + 120x = 504
= 279x = 504
= x = 504/279
= 1 25/31
(vii) 3x – 2/4 – 2x + 3/3 = 2/3 – x
∴ 3x – 2/4 – 2x + 3/3 = 2 – 3x/3
= 9x – 6 – 8x – 12/12 = 2- 3x/3
= x – 18/12 = 2 – 3x/3
= 3x – 54 = 24 – 36x
= 3x + 36x = 54 + 24
= 39x = 78
= x = 78/39
= 2
(viii) x + 2/6 – (11 – x/3 – 1/4) = 3x – 4/12
∴ x + 2/6 – 44 + 4x + 3/12 = 3x – 4/12
= 2x + 4 – 41 + 4x/12 = 3x – 4/12
= 6x – 37/12 = 3x – 4/12
= 6x – 3x = – 4 + 37
= 3x = 33
= x = 33/3
= 11
(ix) 2/5x – 5/3x = 1/15
∴ 6 – 25/15x = 1/15
= -19/15x = 1/15
= 15x = – 19 × 15
= x = – 19 × 15/15
= – 19
(x) x + 2/3 – x + 1/5 = x – 3/4 – 1
∴ 5x + 10 – 3x – 3/15 = x – 3 – 4/4
= 5x + 10 – 3x – 3/15 = x – 7/4
= 2x + 7/15 = x – 7/4
= 8x + 28 = 15x – 105
= 8x – 15x = -105 -28
= – 7x = – 133
= x = 133/7
= x = 19
(xi) 3x – 2/3 + 2x + 3/2 = x + 7/6
∴ 6x – 4 + 6x + 9/6 = 6x + 7/6
= 12x + 5/6 = 6x + 7/6
= 12x – 6x = 7 – 5
= x = 2/6
= 1/3
(xii) x – x – 1/2 = 1 – x – 2/3
∴ 2x – x + 1/2 = 3 – x + 2/3
= x + 1/2 = 5 – x/3
= 3x + 3 = 10 – 2x
= 3x + 2x = 10 – 3
= x = 7/5
(xiii) 9x + 7/2 – (x – x – 2/7) = 36
∴ 9x + 7/2 – (7x – x – 2/7) = 36
= 9x + 7/2 – (6x – 2/7)
= 63 + 49 – 12x + 4/14
= 51x + 53 = 14 × 36
= 51x = 504 – 53
= x = 459/51
= 9
(xiv) 6x + 1/2 + 1 = 7x – 3/3
∴ 6x + 1 + 2/2 = 7x – 3/3
= 6x + 3/2 = 7x – 3/3
= 18x + 9 = 14x – 6
= 18x – 14x = – 9 – 6
= 4x = -15
= x = – 15/4
Question no – (2)
Solution :
Let, present age 2 years,
∴ (x + 2) = 3 (x – 4)
= x + 12 = 3x – 12
= x – 3x = – 12 – 12
= – 2x = – 24
= x = 24/2
= 12
Therefore, Present age will be 12 years.
Question no – (3)
Solution :
Let, the Cost Price (C.P) is = x
∴ 110x/100 = 396
= x = 396 × 100/110
= 360
Hence, the cost price of article will be 360 Rs.
Question no – (4)
Solution :
Let, 1st number = x
2nd number = y
∴ x + y = 4500
and 10% of x = 12.5% of y
= 10x/100 = 12.5 y/100
= 10x = 12.5y
= x = 12.5y/10
Now, substitute the value,
12.5y/10 + y = 4500
= 12.5y + 10y = 4500
= 22.5y = 4500
= y = 4500/22.5
= 2000
Now, x = (12.5/10 × 2000)
= 2500
Therefore, the required numbers will be 2500 and 2000.
Question no – (5)
Solution :
Let, 1st no x,
2nd number y
x + y = 405
and 8x/7y = 1
= 8x = 7y
= x = 7/8y
Now, substitute the value,
7/8 y + y = 405
= 7x + 8y/8 = 405
= 15y/8 = 405
= y = 8 × 405/15
= 216
Now, put the value is equal
x = 7/8 × 216
= 189
Therefore, the required numbers will be 189 and 216.
Question no – (6)
Solution :
Let, Age of A = 7x years
Age B = 5x years
After 10 years Age A = 7x + 10 years
Age B = 5x + 10 years
∴ 7(7x + 10) = 9(5x + 10)
= 49x + 70 = 45x + 90
= 49x – 45 = 90 – 70
= 4x = 20
= x = 20/4
= 5
∴ A’s present age,
= 7x = 7 × 5
= 35 years
∴ B’s present age,
= 5x = 5 × 5
= 25 years.
Therefore, the present age of A will be 35 years, and present age of B will be 25 years.
Question no – (7)
Solution :
Let, the number will be ‘x’
2x = x/2 + 45
= 2x – x/2 = 45
= 4x – x/2 = 45
= 4x – x/2 = 45
= 3x = 90
= x = 90/3
= x = 30
Hence, the required number will be 30
Question no – (8)
Solution :
Let, numbers are ‘x’ x + 1
∴ (x + 1)² – x² = 31
= x² + 2x + 1 – x² = 31
= 2x + 1 = 31
= 2x = 31 – 1
= 30
= x = 30/2
= 15
∴ Numbers are 15 and (15 + 1) = 16
Therefore, the required numbers are 15 and 16.
Question no – (9)
Solution :
Let, a number is x.
According to question
(5x – 5) = 2x + 4
or, 5x – 5 = 2x + 4
or, 5x – 2x = 5v + 4 = 9
or, 3x = 9
or, x =9/3
or, x = 3
Therefore, the required number will be 3
Question no – (10)
Solution :
Fraction is = x – 5/x
∴ x – 5 + 3/x + 3 = 4/5
= x – 2/x + 3 = 4/5
= 5x – 10 = 12 + 10
= x = 22
∴ Original Fraction,
= 22 – 5/22
= 17/22
Therefore, the original fraction will be 17/22.
Next Chapter Solution :
👉 Chapter 15 👈
