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Maths Wiz Class 8 Solutions Chapter 7 Linear Equations
Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Maths Wiz Class 8 Math Book, Chapter 7, Linear Equations. Here students can easily find step by step solutions of all the problems for Linear Equations, Exercise 7A and 7B 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 7 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.
Linear Equations Exercise 7A Solution :
Question no – (1)
Solution :
8x + 14 = 5x + 44
= 8x – 5x = 44 – 14
= 3x = 30
= x = 30/3
= x = 10 …(Solved)
Question no – (2)
Solution :
5m + 7 = 10m – 3
= 10m – 5m = 7+3
= 5m = 10
= m = 10/5
= m = 2 …(Solved)
Question no – (3)
Solution :
8y – 14 = 6 – 2y
= 8y + 2y = 6 + 14
= 10y = 20
= y = 20/10
= y = 2 …(Solved)
Question no – (4)
Solution :
6p – 21 = p + 4
= 6p – p = 4 + 21
= 5p = 25
= p = 25/5
= p = 5 …(Solved)
Question no – (5)
Solution :
2(x – 1) = 12
= 2x – 2 = 12
= 2x = 12 + 2
= x = 14/2
= x = 7 …(Solved)
Question no – (6)
Solution :
7 + 3 (t + 5) = 31
= 3t + 15 = 31 – 7
= 3t = 24 – 15
= 3t = 9
= t = 9/3
= t = 3 …(Solved)
Question no – (7)
Solution :
15 + 3 (4x – 30) = 33
= 12x – 90 = 33 – 15
= 12x = 18 + 90
= x = 108/12
= x = 9 …(Solved)
Question no – (8)
Solution :
5(x + 2) – 9 (x – 2) = 0
= 5x + 10 – 9x + 18 = 0
= – 4x + 28 = 0
= 4x = 28
= x = 28/4
= x = 7 …(Solved)
Question no – (9)
Solution :
3x + 5 = 0.5x – 7
= 3x – 0.5x = – 7 – 5
= 2.5x = – 12
= x = – 12/2.5 × 10
= x = – 24/5 = 4.8
= x = – 4.8 …(Solved)
Question no – (10)
Solution :
0.1(t – 3) = 0.15(t – 4)
= 0.t – 0.3 = 0.15t – 0.6
= 0.15t – 0.t = 0.6 – 0.3
= 0.05t = 0.3
= t = 0.3/0.05 × 100/10
= t = 6 …(Solved)
Question no – (11)
Solution :
2(x – 2) + 5 = 4(x – 6) + 3
= 2x – 4 + 5 = 4x – 24 + 3
= 2x + 1 = 4x – 21
= 4x – 2x = 21 + 1
= 2x = 22
= x = 11 …(Solved)
Question no – (12)
Solution :
2m (m – 4) – 8 = m (2m – 12)
= 2m2 – 8m – 8 = 2m2 – 12m
= 12m – 8m = 8
= 4m = 8
= m = 2 …(Solved)
Question no – (13)
Solution :
3a + 2(a – 9) = 6 – (2a – 3)
= 3a + 2a – 18 = 6 – 2a + 3
= 53a + 2a = 9 + 18
= 75a = 27
= a = 27/7
= a = 3 6/7 …(Solved)
Question no – (14)
Solution :
8x – 2 (2 + 3x) = 2 (1 – 2x) – 3 (x – 1)
= 8x – 4 6x = 2 – 4x – 3x + 3
= 2x – 4 = 5 – 7x
= 2x + 7x = 5 + 4
= 9x = 9
= x = 1 …(Solved)
Question no – (15)
Solution :
x/2 + x/3 + x/6 = 18
= 6x + 4x + 2x/12 = 18
= 12x = 18 × 12
= x = 18 × 12/12
= x = 18 …(Solved)
Question no – (16)
Solution :
2x/3 + x/4 + x/2 = 34
= 8x + 3x + 6x/12 = 34
= 17x = 34 × 12
= x = 34 × 12/17
= x = 24 …(Solved)
Question no – (17)
Solution :
P+1/2 – 2(P-1)/3 = 0
= P + 1/2 = 2(P – 1)/3
= 3P + 3 = 4P – 4
= 4P – 3P = 3 + 4
= P = 7 …(Solved)
Question no – (18)
Solution :
6 – 3(n – 7)/4 = 1/2 n
= 2y – 3n + 21 = 4/2 n
= 45 – 3n = 2n
= 2n + 3n = 45
= 5n = 45
= n = 9 …(Solved)
Question no – (19)
Solution :
10 – 5(1 + x)/3 + 3x + 1/5 = 0
= 150 – 25(1 + x) + 3 (3x + 1)/15 = 0
= 150 – 25 – 25x + 9x + 3 = 0
= 128 – 16x = 0
= 16x = 128
= x = 128/16
= x = 8 …(Solved)
Question no – (20)
Solution :
2b + 1/5 – b – 1/3 = 1
= 6b + 3 – 5b + 5/15 = 1
= b + 8 = 15
= b = 15 – 8
= b = 7 …(Solved)
Question no – (21)
Solution :
5(1 – z)/3 – 3z – 1/5 = 1/6
= 25(1-z) -3 (3z – 1)/15 = 1/6
= 25 – 25z – 9z + 3 = 15/6
= -34z + 28 = 5/2
= -34z = 5/2 – 28
= -34z = 5-56/2
= z = -51/2 × 34
= z = 3/4 …(Solved)
Question no – (22)
Solution :
x + 5/15 – x – 5/10 = 1 + 2x/15
= 2x + 10 – 3x + 15/30 = 15 + 2x/15
= – x + 25 = 30 + 4x
= 4x + x = – 30 + 25
= 5x = – 5
= x = – 1 …(Solved)
Question no – (23)
Solution :
3x – 4/6 – 2x + 3/8 = 2x – 7/24
= 4(3x – 4) – 3 (2x + 3)/24 =2x – 7/24
= 12x – 16 – 6x – 9 = 2x – 7
= 6x – 2x = 16 + 9 – 7
= 4x = 18
= x = 18/4
= x = 4.5 …(Solved)
Question no – (24)
Solution :
(x + 0.5) + 1/2 (3x – 1/3) = 1/3 (x + 1)
= (x + 0.5) + 1/2 (9x – 1/3) = x + 1/3
= 6(x + 0.5) + (9x – 1) / 6 = x+1/3
= 6x + 3 + 9x – 1 /6 = x + 1/3
= 15x + 2 = 2x + 2
= 15x – 2x = 2 – 2
= 13x = 0
= x = 0 …(Solved)
Question no – (25)
Solution :
9x – 5/7 = 6x + 2 / 5
= 45x – 25 = 42x + 14
= 45x – 42x = 14+25
= 3x = 39
= x = 39/3
= x = 13 …(Solved)
Question no – (26)
Solution :
7x – 10/5 = 8x – 5/7
= 49x – 70 = 40x – 25
= 49x – 40x = -25 + 70
= 9x = 45
= x = 45/9
= x = 5 …(Solved)
Question no – (27)
Solution :
2x – 7/x + 4 = 3/4
= 8x – 28 = 3x + 12
= 8x – 3x = 12 + 28
= 5x = 40
= x = 40/5
= x = 8 …(Solved)
Question no – (28)
Solution :
3m + 4/2 – 6m = -2/5
= 15m + 20 = -4 + 12m
= 15m – 12m = – 4 – 20
= 3m = – 24
= m = – 24/3
= m = – 8 …(Solved)
Question no – (29)
Solution :
2n – 3/2n – 1 = 3n – 1/3n + 1
= 6n2 – 9n + 2n – 3 = 6n2 – 3n – 2n + 1
= – 7n – 3 = – 5n + 1
= 7n – 5n = -3-1
= 2m = – 4
= m = – 2 …(Solved)
Linear Equations Exercise 7B Solution :
Question no – (1)
Solution :
(a) If x is an integer, then (x + 1) (x + 2) are the next two larger integers.
(b) If x is an even integer, then (x + 2) (x + 4) are the next two larger even integer.
(c) One integer is x, let the another one is y
Given, sum of two integer is = – 8
∴ x + y = – 8
= y = – x – 8
= – (x + 8)
(d) Let the another number be x
∴ Given, difference between x and y is 20
∴ y – x = 20
= x = y – 20
(e) Priya is n year old now
she will (n + 1) years old in next year
she will (x + 5) years old in 5 years
she was (x – 10) years old in 10 years ago.
(f) His profit was x rupees
he has tripled his yearly profit
∴ His profit at the end of the period is = 3x rupees
(g) The ones digit is x
The tens digit of a number is 4 less than
its ones digit = 10 (x – 4)
∴ The number will be 10 (x – 4) + x
(h) In (g) part if the numbers are interchanged
Then the number will be,
= 10x + (x – 4)
= 11x – 4
∴ Required number will be 11x – 4
(i) The numerator is y
The denominator is 3 more than the numerator y + 3
∴ The fraction is = y/y + 3
(j) The angle will be = 360° – x°
Question no – (2)
Solution :
As per the question,
Forty-six more than a number is = 13.
∴ The number is,
= 13 – 46
= – 33
Therefore, the required number will be -33
Question no – (3)
Solution :
Let the three consecutive odd numbers are x, x + 2, x + 4
Given, x + x + 2 + x + 4 = 33
= 3x + 6 = 33
= 3x = 33 – 6
= x = 27/3
= x = 9
∴ The numbers are 9, 9 + 2, 9 + 4
= 9, 11, 13
Hence, the numbers are 9, 11, 13
Question no – (4)
Solution :
Let the larger number be = x
∴ The small number is (87 – x)
Now, as per the given condition,
x + 2 (87 – x) = 120
= x + 2 × 87 – 2x = 120
= – x = 120 – 174
= – x = – 54
= x = 54
Therefore, the larger number will be 54
Question no – (5)
Solution :
Let the three consecutive numbers are x, (x + 1) (x + 2)
Given condition,
x + (x + 1) = (x + 2) + 15
= 2x + 1 = x + 2 + 15
= 2x – x = 17 – 1
= x = 16
Thus, the numbers are = 16, 17, 18
Question no – (6)
Solution :
Let, the third number is x
second number is (x + 10)
Let first number is y
1st, condition,
2y = x + 10 – 6
= y = x + 4/2
2nd, condition,
x + (x + 10) + x + 4/2 = 57
= 2x + 10 + x + 4/2 = 57
= 4x + 20 + x + 4/2 = 57
= 5x + 24 = 57 × 2
= 5x = 114 – 24
= x = 90/5
= x = 18
∴ The numbers are,
= 18+4/2, (18 + 10), 18
= 22/2, 28, 18
= 11, 28, 18
Hence, the number will be 11, 28, 18
Question no – (7)
Solution :
Let the original number is 10a + b
1st, condition,
b = 2a
If the numbers are reverse 10b + a
2nd, condition,
10a+b – 27 = 10b+a
= 10a + 2a + 27 = 10 × 2a + a
= 12a + 27 = 20a + a
= 12a + 27 = 21a
= 21a – 12a = 27
= 9a = 27
= a = 27/9
= a = 3
∴ b = 2 × 3 = 6
∴ The number is,
= (10 × 3) + 6
= 30 + 6
= 36
Therefore, the required number will be 36
Question no – (8)
Solution :
Let the number be x
As per the given condition,
9 + x/17 + x = 5/7
= 63 + 7x = 85 + 5x
= 7x – 5x = 85 – 63
= 2x = 22
= x = 11
Hence, the number will be 11
Question no – (9)
Solution :
Let the four consecutive odd numbers are (x + 1), (x + 3), (x + 5), (x + 7)
Given condition,
3(x + 7) + 2 (x + 1) = 93
= 3x + 21 + 2x + 2 = 93
= 5x + 23 = 93
= 5x = 93 – 23
= x = 70/5
= x = 14
∴ The numbers are
= (14 + 1), (14 + 3), (14 + 5), (14 + 7)
= 15, 17, 19, 21
17 and 19 are prime numbers.
Question no – (12)
Solution :
Let the denominator = x
∴ The numerator is = x – 2
Given condition is,
(x – 2)/x – 1 = 1/2
= x – 3/x – 1 = 1/2
= 2x – 6 = x – 1
= 2x – x = 6 – 1
= x = 5
Therefore, the rational number = 5 – 2/5 = 3/5
Question no – (13)
Solution :
The boy was p years old r years ago.
He is now = (p + r) years old
In (p – r) years time he will be (p + r) + (p – r)
= 2p years old
Question no – (14)
Solution :
Let Anushka’s present age x years
Now, Hari is twice as old Anushka = 2x 3 years ago,
Anushka’s age (x – 3) years
Then Hari was 5 years older then Anushka
∴ Now, Hari’s age,
2x – 3 = x + 2
= 2x – x = 2 + 3
= x = 5
∴ Anushka’s age 5 years
∴ Hari’s age = 2 × 5 = 10 years.
Therefore, the present age of Anushka will be 5 years and Hari will be 10 years.
Question no – (15)
Solution :
Let, Neeta’s Present age x years
∴ Ravi’s present age (x + 6) years
Three years ago,
Neeta’s age (x – 3) years
Ravi’s age (x + 6 – 3) = (x + 3) years
As per the given condition,
(x + 3) = 2 (x – 3)
= 2x – 6 = x + 3
= 2x – x = 3 + 6
= x = 9
∴ Ravi’s present age (9 + 6) = 15 years
∴ Neeta present age 9 years
Therefore, the present age of Ravi will be 15 years, and Neeta will be 9 years.
Question no – (16)
Solution :
Let, Deepa’s present is x years
Her mother’s present age (3x + 4) years
Six years,
Deepa’s age (x + 6) years
Her mother’s age
= (3x + 4 + 6)
= (3x – 10) years
Now, as per the given condition,
(3x + 10) = 2 (x + 6) + 8
= 3x + 10 = 2x + 12 + 8
= 3x – 2x = 20 – 10
= x = 10
∴ Deepa’s age 10 years.
∴ Deepa’s mother’s age
= (3 × 10 + 4) = 34 years
Therefore, the age of Deep will be 10 years, and her mother will be 34 years.
Question no – (17)
Solution :
Let the present age of two children are 2x and 3x respectively
Five years ago their ages will be (2x – 5) and (3x – 5) respectively
Now, as per the given condition,
2x – 5/3x – 5 = 1/2
= 4x – 10 = 3x – 5
= 4x – 3x = 10 – 5
= x = 5
∴ There present ages (2 × 5) and (3 × 5) = 10 and 15 years.
Therefore, the present ages of them will 10 years and 15 years.
Question no – (18)
Solution :
Let wide of the playground x m
∴ The length of the playground (x + 30) m
∴ It’s perimeter,
2 (x + 30 + x) = 300
= 2x + 30 = 300/2
= 2x = 150 – 30
= x = 120/2
= x = 60
Length is (60 + 30) = 90m
Wide is = 60m
∴ Area = (length × wide) m2
= (90 × 60) m2
= 5400 m2
Hence, the Area of the playground will be 5400 m2
Question no – (20)
Solution :
The sides of the triangle are,
5x + 3, 3x + 7, 1 + 1/3 y
Given, the triangle is equilateral
∴ 5x + 3 = 3x + 7 = 1 + 1/3 y
5x + 3 = 3x + 7
= 5x – 3x = 7 – 3
= 2x = 4
= x = 2
And,
3 × 2 + 7 = 1 + 1/3 y
= 6 + 7 = 1 + 1/3 y
= 1/3y = 13 – 1
= y = 12 × 3
= y = 36
∴ x = 2 and y = 36
Hence, the value of x will be 2 and y will be 36
∴ The perimeter of the triangle,
5x + 3 + 3x + 7 + 1 + 1/3 y
= 8x + 11 + 1/3 y
= 8 × 2 + 11 + 1/3 × 36
= 16 + 11 + 12
= 39 cm
Therefore, the perimeter of the triangle will be 39 cm.
Question no – (22)
Solution :
Let the man walking 3x km in x hours and the man travel in scooter 24y km in y hour
∴ 3x + 24y = 25 …..(1) × 2
x + y = 2 1/2
= x + y = 5/2
= 2x + 2y = 5 ……(2) × 3
∴ Multiplying (1) × 2 and (2) × 3 we get
6x + 48y = 50
6x + 6y = 15
(-) (-) (-)
—————————
42y = 35
y = 35/42 = 5/6
∴ 3x + 24 × 5/6 = 25
= 3x + 20 = 25
= 3x = 25 – 20
= 3x = 5
= x = 5/3
Therefore, the man walking 5 km in 5/3 hours.
Question no – (23)
Solution :
8 km – 1 hr
1 km – 1/8 hr
∴ x km – x/8 hr
Again 12km – 1 h
1 km – 1/12 h
x km – x/12 h
Given that double journey take 5 hr
∴ x/8 + x/12 = 5
= 3x + 2x/24 = 5
= 5x = 5 × 24
= x = 5 × 24/5
= x = 24
Therefore, the x will be 24
Question no – (24)
Solution :
48 min = 48/60 = 4/5 hr
Let, x km up stream he goes
3 km – 1h
1 km – 1/3 h
x km = x/3 h
5 km – 1 h
1 km – 1/5 h
x km = x/5 h
x/3 + x/5 = 4/5
= 5x + 3x/15 = 4/5
= 8x = 4 × 15/5
= x = 4 × 3/8
= 3/2
= 1.5 km
Therefore, he will go 1.5 km upstream
Question no – (26)
Solution :
Let, speed in still water = x
Stream flows at 3 km/h
3 hour goes to downstream = 3 (x + 3)
5 hour goes to upstream = 5 (x – 3)
∴ 5 (x – 3) = 3 (x + 3)
= 5x – 15 = 3x + 9
= 5x – 3x = 9 + 15
= 2x = 24
= x = 12
∴ Speed of the streamer upstream
= (12 – 3)
= 9 km/h.
Therefore, the speed of the streamer will be 9 km/h
Question no – (28)
Solution :
∴ 4(x + 7) + 5 (x – 20) = 180°
= 4x + 28 + 5x – 100 = 180°
= 9x – 72 = 180°
= 9x = 252
= x = 252/9
= x = 28
Therefore, the value of x will be 28
Question no – (30)
Solution :
Let, apple = A, peach = P, Banana = B
Given, A = P + 29 = P = A – 29
B = A + 13
and {2p + 2b} – 3A = 43
{2(A – 29) + 2 (A + 13)} – 3A = 43
= 2A – 58 + 2A + 26 – 3A = 43
= A – 32 = 43
= A = 43 + 32
= A = 75
Therefore, an apple have 75 calories.
Question no – (31)
Solution :
Let, he got x cards ₹ 10
Given, 10x + 2/3x × 5 = 1/3 × x × 15 = 245
= 13x + 10x/3 = 245
= 30x + 10x = 245 × 3
= 49x = 245 × 3
= x = 245 × 3/49
= 15
∴ 10 rupees card = 15
∴ 5 rupees card
= 2/3 × 15
= 10
∴ 15 rupees card
= 1/5 × 15
= 3
Next Chapter Solution :
👉 Chapter 8 👈
