Warning: Undefined array key "https://nctbsolution.com/maths-wiz-class-7-solutions/" in /home/862143.cloudwaysapps.com/hpawmczmfj/public_html/wp-content/plugins/wpa-seo-auto-linker/wpa-seo-auto-linker.php on line 192
Maths Wiz Class 7 Solutions Chapter 7 Algebra
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students for the Solutions of Maths Wiz Class 7 Math Book, Chapter 7, Algebra. Here students can easily find step by step solutions of all the problems for Algebra, Exercise 7A, 7B, 7C, 7D, 7E, 7F, 7G and 7H 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.
Algebra Exercise 7A Solution :
Question no – (1)
Solution :
(a) 8 more than x
= x + 8
(b) The sum of a and 9
= a + 9
(c) The sum of m and n
= m + n
(d) 3 less than r
= r – 3
(e) 6 times p
= 6 p
(f) One-third of
= 1/3 x
(g) P divided by q
= p/q
(h) The cube of the length l
= l³
(i) m is greater than 7
= m > 7
(j) 5 is less than t
= 5 < t
Question no – (2)
Solution :
(a) 7 more than 4 times r
= 4r + 7
(b) 4 times the sum of y and 7
= 4 (y + 7)
(c) Thrice the number x increased by y
= 3x + y
(d) 5 times the number n minus four
= 5n – 4
(e) 25 decreased by 4 times q
= 25 – 4q
(f) 6 times a less than the quotient of b by 4
= 6/4 – 6a
(g) The square of the sum of a and b
= (a + b)2
Question no – (3)
Solution :
The expression will be,
= 30 × d
= 30d
Thus, she will read 30d pages.
Question no – (4)
Solution :
The expression will be,
= (x + 4) cm
∴ Madhuri’s height in centimetres will be (x + 4) cm.
Question no – (5)
Solution :
The correct option will be – (d)
300 – 7 wn
Algebra Exercise 7B Solution :
Question no – (1)
Solution :
(a) x + 7 for x = 2
= x + 7
= 2 + 7 …[where x = 2]
= 9
(b) y – 7 for y = 10
= y – 7
= 10 – 7
= 3
[∵ y = 7]
(c) 5a for a = 2
= 5a
= 5 × 2 [∵ a = 2]
= 10
(d) 7a – 3 for a = 1
= 7a – 3
= (7 × 1) – 3 [∵ a = 1]
= 7 – 3
= 4
(e) 9d for d = 0
= 9d
= 9 × 0 [∵ d = 0]
= 0
(f) 16 – 4y for y = 4
= 16 – 4y
= 16 – (4.4) [∵ y = 4]
= 16 – 16
= 0
Question no – (2)
Solution :
(a) y – 2x for y = 25, x = 7
= Y – 2x
= 25 – 2 × 7 [∵ x = 7; Y = 25]
= 25 – 14
= 11
(b) 3a + 4b for a = 1, b = 2
= 3a + 4b
= 3 × (-1) + 4 × 2 [∵ a = – 1; B = 2]
= – 3 + 8
= 5
(c) 5 (y – x) for x = 1.5, y = 3.5
= 5 (3.5 – 1.5) [∵ x = 1.5; Y = 3.5]
= 5 (2.0)
= 5 × 2
= 10
(d) 10x + z – y for x = 3, y = – 5, z = 7
= 10x + z – y
= 10 × 3 + 7 – (-5) [∵ x = 3;Y = – 5; Z = 7]
= 30 + 7 + 5
= 42
Question no – (3)
Solution :
(a) 4x2 for x = 3
= 4x2
= 4 × 32 [∵ x = 3]
= 4 × 9
= 36
(b) a2 + 3b for a = 2, b = – 1
= a2 + 3b
= 22 + 3 × (- 1) [ ∵ x = 2; Y = – 2]
= 4 – 3
= 1
(c) 36 – 2x2 for x = 4
= 36 – 2x2
= 36 – 2 (4)2 [∵ x = 4]
= 36 – (2 × 16)
= 36 – 32
= 4
(d) x2 + y2 for x = – 1, y = – 2
= x2 + y2
= (-1)2 + (-2)2 [∵ x = – 1; Y = – 2]
= 1
Question no – (4)
Solution :
According to the given question,
A = 18
B = 0
C = -3
∴ A = 3a + 5b – 8c – 29
Or, A = 3 (18) + 5.0 – 8 (-3) – 29
Or, A = 54 + 0 + 24 – 29
or, A = 49
Therefore, the value of A will be 49.
Algebra Exercise 7C Solution :
Question no – (1)
Solution :
(a) 5 + x = 8,
Than x,
= 8 – 5
= 3
(b) 15 – q = 8,
Then q,
= 15 – 8
= 7
So, q = 7
(c) n – 6 = 3,
Then n,
= 3 + 6
= 9
Therefore, n = 9
(d) 7m = 42,
Then m,
= 42/7
= 6
Hence, m = 6
(e) 1/2x = 10,
Then x,
= 10 × 2
= 20
Thus, x = 20
Question no – (2)
Solution :
(a) B + 6 = 13
b + 6 = 13
or b = 13 – 6 = 7
Thus, the correct option is (B) b = 7
(b) 20 + w = 28
∴ 20 + w = 28
or, w = 28 – 20 = 8
So, the correct option is (C) w = 8
(c) x – 7 = 2
∴ x – 7 = 2
or, x = 7 – 5
= 2
Therefore, the correct option is (B) x = 5
(d) 16 – y = 5
∴ 16 – y = 5
or, y = 16 – 5 = 11
Hence, the correct option is (B) y = 11
(e) 8q = 56
∴ 8q = 56
or, r = 56/8
= 7
Thus, the correct option is (B) r = 7
(f) d/4 = 5
∴ d/4 = 5
or, d = 20/4
= 5
Therefore, the correct option is (A) d = 20
Algebra Exercise 7D Solution :
Question no – (1)
Solution :
(a) 7 = a – 10 → add 10
(b) -2 + x = 15 → add 2
(c) -20 = m + 5 → subtract 5
(d) 3n = 12 → divided by 3
(e) 8 = 2y → divided by 2
(f) 1/7 q = 3 → multiple by 7
Question no – (2)
Solution :
(a) 7 = a – 10
Or a = 10 – 7
= 17
(b) – 2 + x = 15
Or, x = 15 + 2
= 17
(c) – 20 = m + 5
Or m = – 20 – 5
= – 25
(d) 3n = 12
Or, n = 12/3
= 4
(e) 8 = 2y
Or, y = 8/2
= 4
(f) 1/7 = 3
= or q = 3 × 7
= 21
Question no – (3)
Solution :
(a) – (i) x + 2 = 8
= x = 8 + 2
= 10
(a) – (ii) 6 = p + 5
or, p = 6 + 5
= 11
(a) – (iii) x + 5 1/2 = 7 1/2
= or, x + 11/2 = 10/2
Or, x = 10/2 – 11/2
Or, x = 10 – 11/2
= 1/2
(a) – (iv) 6.5 = x + 4.3
Or, x = 6.5 – 4.3
= 2.2
(b) – (i) y – 3 = 5
Or, y = 5 + 3
= 8
(b) – (ii) 8 = c – 3
Or, c = 8 + 3
= 11
(b) – (iii) 9/2 = m – 7/2
Or, m = 9/2 + 7/2
Or, m = 9 + 7/2
= 16/2
= or, m = 8
(b) – (iv) z – 10.7 = 1.3
Or, y = 5 + 3
= 12
(c) – (i) 5x = 10
Or, x = 10/5
= 2
(c) – (ii) 45 = 9t
Or, t = 45/9
= 5
(c) – (iii) 0.2y = 3
Or, y = 2/10 × 3
= 2/30
(c) – (iv) 7n = 1.05
Or, n = 1.05/7
= .15
(d) – (i) a/2 = 8
= or, a = 8 × 2
= 16
(d) – (ii) 15 = d/2
Or, d = 15 × 2
= 30
(d) – (iii) 0.2y = 3
Or, n = 3 × (- 2)
Or, n = – 6
(d) – (iv) 1.8m = 36
Or, m = 36/1.8
= 36 × 10/18
= 20
Algebra Exercise 7E Solution :
Question no – (1)
Solution :
Given, x + 2 = 6
Or, x = 6 – 2
= x = 4
∴ 4 + 2 = 6 …(Solved)
Question no – (2)
Solution :
Given, y – 7 = 13
Or, y = 13 + 7
= y = 20
∴ 20 – 7 = 13 …(Solved)
Question no – (3)
Solution :
Given, 5 = p – 2
Or, p = 5 + 2
= 7
∴ 5 + 2 = 7 …(Solved)
Question no – (4)
Solution :
Given, 16y = 48
Or, y = 48/16
= y = 3 …(Solved)
Question no – (5)
Solution :
Given, 16 = m + 9
Or, m = 16 – 9
= 7
∴ 9 + 7 = 16 …(Solved)
Question no – (6)
Solution :
Given, 1/9 x = 5
Or, x = 5 × 9
∴ x = 45 …(Solved)
Question no – (7)
Solution :
x/7.5 = 1/2.5
Or, x = 7.5/2.5
∴ x = 3 …(Solved)
Question no – (8)
Solution :
Given, 1.7 = 2d
Or, d = 1.7/2
= d = 0.85 …(Solved)
Question no – (9)
Solution :
Given, c – 8 = – 13
Or, c = – 13 + 8
= c = – 5 …(Solved)
Question no – (10)
Solution :
Given, 3.8 = x/5
Or, x = 3.8 × 5
= x = 19.0 …(Solved)
Question no – (11)
Solution :
Given, z – 2 = – 10
Or, z = – 10 + 2
= z = – 8 …(Solved)
Question no – (12)
Solution :
Given, 9t = 3 3/5
Or, 9t = 18/5
Or, t = 18/5 × 9
= t = 2/5 …(Solved)
Algebra Exercise 7F Solution :
Question no – (1)
Solution :
(a) The sum of 5 and y is equal to 18.
∴ 5 + y = 18
Or, y = 18 – 5
∴ y = 13
(b) The sum of -3 and t is 5
∴ -3 + t = 5
Or, t = 5 + 3
∴ t = 8
(c) The difference of x and 7 is 3
∴ x – 7 = 3
Or, x = 3 + 7
∴ x = 10
(d) 8 more than a is -4
∴ 8 + a = – 4
Or, a = – 4 – 8
∴ a = – 12
(e) 28 less than p equal -7
∴ P – 28 = – 7
Or, p = 28 – 7
∴ P = 21
(f) The product of 6 and m is 24
∴ m × 6 = 24
or, m = 24/6
∴ m = 4
(g) The product of -5 and n is 45
∴ – 5 × n = 45
Or, n = 45/ – 5
= – 9
(h) r divided by 3 equal 6
∴ r/3 = 6
or, r = 6 × 3
= 18
(i) Given, z divided by -9 equal 7
∴ z/- 9 = 7
or, z = 7 × – 9
= – 63
(j) 25 less than b equal -8.
∴ B – 25 = – 8
Or, b = – 8 + 25
= 17
Question no – (2)
Solution :
Let, Person’s weight on earth = w
Person’s weight on Moon,
= w × 1/6
= w/6
Now, Parson’s weight on Moon = 10 kg
∴ According to question,
w/6 = 10
or, w = 60 kg
Therefore, Parson’s weight on earth will be 60 kg.
Question no – (3)
Solution :
Let, Each length of pentagon = p
Pentagon’s perimeter = 5p
Pentagon’s perimeter = 35
∴ According to question,
5p = 35
Or, p = 35/5
= 7 cm
Therefore, the length of each of a pentagon is 7 cm.
Question no – (4)
Solution :
Let, Speed = s
∴ According to equal,
5s = 300
Or, s = 300/5
= 60 km/hour
Therefore, the speed of the train will be 60 km/hour.
Question no – (5)
Solution :
Ravi’s monthly salary = 6000
∴ According to question,
M + 1500 = 6000
Or, m = 6000 – 15000
= 4500 Rs.
Therefore, his monthly expenditure will be 4500 Rs.
Question no – (6)
Solution :
Let, Each side = a
∴ According to the question,
4a = 100
Or, a = 100/4
= 25 m
Therefore, one side of the field will be 25 m.
Algebra Exercise 7G Solution :
Question no – (1)
Solution :
(a) 3a + 2 = 11
Or, 3a = 11 – 2
∴ a = 9
(b) 5p – 1 = 9
Or, 5p = 9 + 1
∴ p = 10
Or, p = 10/5
= 2
(c) 45 = – 7d – 4
Or, – 7d = 45 + 4
= 49
Or, d = 49/7
∴ d = – 7
(d) x/2 + 10 = 13
Or, x/2 = 13 + 10
= 23
Or, x = 23 × 2
∴ x = 46
(e) 9 (x – 3) = 54
Or, 9x – 27 = 54
Or, 9x = 54 + 27
= 81
Or, 9x = 81/9
∴ x = 9
(f) 17 = 3(p – 5) + 8
Or, 17= 3p – 15 +8
or, 17 = 3p-7
or, 3p = 17 + 17 = 24
or, p = 24/3
∴ p = 8
Question no – (2)
Solution :
As per the question,
(a) The number f less than 28 is 15.
∴ 28 – f = 15
Or, – f = 15 – 28
= – 7
Or, f = 7
(b) Let, The number = n
∴ According to question,
7n – 18 = 31
Or, 7n = 31 + 18
= 49
Or, n = 49/7
= 7
Therefore, the number will be 7.
Algebra Exercise 7H Solution :
Question no – (1)
Solution :
(a) Write down the first 6 terms of the pattern.
Tn = 2n + 1
T1 = 2 × 1 + 1
= 2+ 1
= 3
T2 = 2 × 2 + 1
= 4 + 1
= 5
T3 = 2 × 3 + 1
= 6 + 1
= 7
T4 = 2 × 4 + 1
= 8 + 1
= 9
T5 = 2 × 5 + 1
= 10 + 1
= 11
T6 = 2 × 6 + 1
= 12 + 1
= 13
(b) Write down the 1000th term of the pattern.
Tn = 2n + 1
Or, T1000 = 2 × 1000 + 1
= 2000 + 1
= 20001
Question no – (2)
Solution :
(a) The first 4 terms
Tn = 20n – 9
Or, T1 = 20 × 1 – 9
= 20 – 9
= 11
Or, T2 = 20 × 2 – 9
= 40 – 9
= 31
T3 = 20 × 3 – 9
= 60 – 9
= 51
Or, T4 = 20 × 4 – 9
= 80 – 9
= 71
(b) The 10th term
Tn = 20n – 9
Or, T10 = 20 × 10 – 9
= 2000 – 9
= 191
(c) The 300th term
Tn = 20n – 9
= 600 – 9
= 5991
Question no – (4)
Solution :
(a) The first 3 terms
Tn = 300 – 7n
T1 = 300 – 7 × 1
= 300 – 7
= 293
T2 = 300 – 7 × 2
= 300 – 14
= 286
T3 = 300 – 7 × 3
= 300 – 21
= 279
(b) The 20th term
Tn = 300 – 7n
Or, T120 = 300 – 7 × 20
= 300 – 140
= 60
(c) The 45th term
Tn = 300 – 7n
Or, T45 = 300 – 7 × 45
= 300 – 315
= – 15
(d) Tn = 300 – 7n
Or, T120 = 300 – 7 × 120
= 300 – 840
= – 540
Question no – (5)
Solution :
(a) y = 3x – 2
Y = 3x – 2
Y = 3 × 1 – 2
= 3 – 2
= 1
Y = 3 × 2 – 1
= 6 – 2
= 4
Y = 3 × 3 – 1
= 9 – 2
= 7
Y = 3 × 4 – 2
= 12 – 2
= 10
Y = 3 × 5 – 2
= 15 – 2
= 13
Complete Table :
| x | 1 | 2 | 3 | 4 | 5 |
| y | 1 | 4 | 7 | 10 | 13 |
(b) n = 7m – 3
n = 7m – 3
n = 7 × 1 – 3
= 7 – 3
= 4
n = 7 × 2 – 3
= 14 – 3
= 11
n = 7 × 10 – 3
= 70 – 3
= 67
n = 7 × 100 – 3
= 700 – 3
= 697
n = 7 × 250 – 3
= 1750 – 3
= 1747
Complete Table :
| m | 1 | 2 | 10 | 100 | 250 |
| n | 4 | 11 | 67 | 697 | 1747 |
Question no – (6)
Solution :
(a)
| 15 | 14 | 13 | 12 | 11 | 10 | 0 | – 3 | n |
| 10 | 9 | 8 | 7 | 6 | 5 | – 5 | – 8 | n – 5 |
Rules = (n – 5)
(b)
| 1 | 2 | 3 | 4 | 21 | 90 | 1/3 | 0 | n |
| 4 | 7 | 10 | 13 | 64 | 271 | 2 | 1 | 3n + 1 |
Rules = (3n + 1)
Question no – (7)
Solution :
(a) 5, 10, 15, 20….
(i) Multiplied by 5
(ii) 5n
(iii) 5 × 100 = 500
(b) 1, 5, 9, 13….
(i) Adding with 4
(ii) 4n – 3
(iii) 4 × 100 – 3 = 397
Next Chapter Solution :
👉 Chapter 8 👈
