These Class 7 Maths Extra Questions Chapter 4 Simple Equations will help students prepare well for the exams.
Class 7 Maths Chapter 4 Extra Questions Simple Equations
Class 7 Maths Simple Equations Extra Questions – 2 Marks
Question 1.
p + 3 = 10, p = –1
Solution:
p + 3 = 10 ∵ p = –1
– 1 + 3 = 2 ≠ 10
∴ p = –1 is Not the solution.
Question 2.
7p – 3 = 1, p = 5
Solution:
7p – 3 = 1 ∵ p = 5
7(5) – 3 = 35 – 3 = 32 ≠ 1
∴ p = 51 is Not the solution.
Question 3.
1 – 3p = 7, p = 0
Solution:
1 – 3p = 7 ∵ p = 0
1 – 3(0) = 1 – 0 = 1 ≠ 7
∴ p = 0 is not the solution.
Question 4.
7 – x = –1, x = 8
Solution:
7 – x = –1 ∵ x = 8
7 – 8 = –1
∴ x = 8 is the solution.
Question 5.
3p = –12, p = –4
Solution:
3p = –12,
∵ p = –4
3(–4) = –12
–12 = –12
∴ p = –4 is the solution.
Question 6.
\(\frac{3 n}{2}\) + 1 = 0, n = \(\frac{-2}{3}\)
Solution:
\(\frac{3 n}{2}\) + 1 = 0
∵ n = \(\frac{-2}{3}\) = \(\frac{3 \times-2}{2 \times 3}\) + 1 = – 1 + 1 = 0
∴ n = \(\frac{-2}{3}\) is the solution.
Question 7.
4l – 3 = 8l, l = 1.
Solution:
4l – 3 = 8l ∵ l = 1
4(1) – 3 = 4 – 3 = 1
8l = 8 × 1 = 8
l ≠ 8
∴ l = 1 is not the solution.
Question 8.
18x = 0, x = 0
Solution:
18x = 0 ∵ x = 0
18x = 18 × 0 = 0
∴ x = 0 is the solution.
Express the following statements in the form of simple equations.
Question 9.
i) 3 is added to twice of x, result is – 1.
ii) 8 times p is –54.
Solution:
i) 2x + 3 = –1
ii) 8p = –54
Question 10.
i) 1 is subtracted from b gives 10
ii) Three fourth of a number z is 9.
Solution:
i) b – 1 = 10
ii) \(\frac { 3 }{ 4 }\) z = 9
Question 11.
i) \(\frac { 7 }{ 2 }\) is added to thrice of p is equal to 7.
ii) 8 times of 8x is equal to 64.
Solution:
i) 3p + \(\frac { 7 }{ 2 }\) = 7
ii) 8 × 8x = 64 ⇒ 64x = 64
Question 12.
i) Twice of a is added to – 7, the result is 10.5.
ii) 8 times of ‘x plus 2 ‘ is 91 .
Solution:
i) 2a – 7 = 10.5
ii) 8(x+2) = 91
Question 13.
i) Ten times of n is equal to 1000.
ii) p is divided by 6 gives the quotient 8.
Solution:
i) 10n = 1000
ii) \(\frac { p }{ 6 }\) = 8
Question 14.
x – 3 = 5
Solution:
x – 3 = 5
Adding ‘ 3 ‘ on both sides
x – 3 + 3 = 5 + 3
x = 8
Question 15.
1 + x = –9
Solution:
1 + x = –9
Adding – 1 on both sides
1 + x – 1 = – 9 – 1
x = –10
Question 16.
l – 3 = – 1
Solution:
l – 3 = – 1
Adding 3 on both sides
l – 3 + 3 = – 1 + 3 ⇒ l = 2
Question 17.
y – 7 = – 4
Solution:
y – 7 = – 4
Adding 7 on both sides
y – 7 + 7 = – 4 + 7 ⇒ y = 3
Question 18.
3p = 1
Solution:
3p = 1
Divide both sides by 3
\(\frac{3 \mathrm{p}}{3}\) = \(\frac { 1 }{ 3 }\) ⇒ p = \(\frac { 1 }{ 3 }\)
Question 19.
4x = 20
Solution:
4x = 20
Divide both sides by 4
\(\frac { 4x }{ 4 }\) = \(\frac { 20 }{ 4 }\) – x – 5
Question 20.
\(\frac { b }{ 3 }\) = 4
Solution:
\(\frac { b }{ 3 }\) = 4
Multiply both sides by 3
\(\frac { b }{ 3 }\) × 3 = 4 × 3 ⇒ b = 12
Question 21.
Solve 2p – 5 = 5
Solution:
2p – 5 = 5
Adding 5 on both sides
2p – 5 + 5 = 5 + 5
2p = 10
p = \(\frac { 10 }{ 2 }\) (divide both sides by 2)
p = 5
Question 22.
Solve 1 – 2x = 3
Solution:
1 – 2x = 3
Subtracting 1 from both sides
1 – 2x – 1 = 3 – 1
– 2x = 2
Divide both sides by ‘–2’
\(\frac{-2 x}{-2}\) = \(\frac{2}{-2}\)
⇒ x = –1
Question 23.
2x + 5 = 1, x = – 2,1
Solution:
2x + 5 = 1, x = –2
Case (i) :
LHS = 2x + 5 = 2(– 2) + 5 = – 4 + 5 = 1
RHS = 1
LHS = RHS
∴ x = –2 is the solution.
Case (ii): x = 1
LHS = 2x + 5 = 2(1) + 5 = 2 + 5 = 7
RHS = 1
LHS ≠ RHS
x = 2 is not the solution.
Question 24.
9p = 18, p = 2,–2
Solution:
9p = 18
Let p = 2
9p = 9 × 2 = 18
Let p = –2
9p = 9 × – 2 = –18
∴ p = 2 is the solution.
Question 25.
Solve “Add 8 to eight times of a number you get 24” What is that number?
Solution:
Let the number be x
8 times of the number = 8x
According to the problem
8 + 8x = 24
8x = 24 – 8 (Transposing 8 to RHS)
8x = 16
x = \(\frac { 16 }{ 8 }\) (Divide both sides by 8 )
x = 2
∴ The required number = 2
Simple Equations Extra Questions Class 7 – 3 Marks
Question 1.
Anwar’s father’s age is 5 years more than three times Anwar’s age. Find Anwar’s age, if his father is 47 years.
Solution:
Let the age of Anwar = x years
Three times of x = 3x
According to the problem
3x + 5 = 47
3x = 47 – 5 (Transposing 5 to RHS)
3x = 42
x = \(\frac { 42 }{ 3 }\) (Divide both sides by 3)
x = 14 years
∴ Age of Anwar = 14 years.
Question 2.
Find a number such that one fifth of the number is 4 more than 10.
Solution:
Let the required number = p
One fifth of the number = \(\frac { p }{ 5 }\)
According to the problem
\(\frac { p }{ 5 }\) – 10 = 4
\(\frac { p }{ 5 }\) = 10 + 4 (Transposing 10 to RHS)
\(\frac { p }{ 5 }\) = 14
p = 14 × 5 (Multiply both sides by 5)
p = 70
∴ Required number = 70
Question 3.
Solve 4(x – 1) = 16
Solution:
4(x – 1) = 16
x – 1 = \(\frac { 16 }{ 4 }\) (divide by 4 on both sides)
x – 1 = 4
x = 4 + 1 (adding 1 on both sides)
x = 5
Question 4.
Solve 0 = – 8 + (p – 3) 2
Solution:
0 = –8 + (p – 3)2
8 = (p – 3) 2 (adding 8 on both sides)
\(\frac { 8 }{ 2 }\) = p – 3 (divide by 2 on both sides)
4 = p – 3
p = 4 + 3 (adding 3 on both sides)
p = 7
Question 5.
Solve \(\frac { a }{ 2 }\) – 3 = 2
Solution:
\(\frac { a }{ 2 }\) – 3 = 2
\(\frac { a }{ 2 }\) = 2 + 3 (adding 3 on both sides)
\(\frac { a }{ 2 }\) = 5
a = 5 × 2 (Multiply by 2 on both sides)
a = 10
Question 6.
Solve 3(5 – x) = –30
Solution:
3(5–x) = – 30
5 – x = \(\frac{-30}{5}\) (Divide by 3 on both sides)
5 – x = – 6
–x = – 6 – 5 (Transposing 5 to RHS)
–x = – 11
x = 11 (Divide both sides by –1)
Question 7.
Solve \(\frac{8x}{3}\) = \(\frac{3}{8}\)
Solution:
\(\frac{8x}{3}\) = \(\frac{3}{8}\)
Multiply both sides by 3
8x = \(\frac{3}{8}\) × 3
8x = \(\frac{9}{8}\)
x = \(\frac{9}{8 \times 8}\) (Divide both sides by 8)
x = \(\frac{9}{64}\)
Question 8.
Solve \(\frac{2 x}{5}\) =– 4
Solution:
\(\frac{2 x}{5}\) =– 4
2x = –4 × 5(Multiply both sides by 5)
2x = –20
x = \(\frac{-20}{2}\) (Divide both sides by 2)
x = –10
Question 9.
Solve 3k – 12 = – 6
Solution:
3k – 12 = – 6
3k = – 6 + 12 (adding 12 on both sides)
3k = 6
k = \(\frac{6}{3}\) (divide both sides by 3)
k = 2
Question 10.
Solve \(\frac{2}{3}\)(x + 1) = 1
Solution:
\(\frac{2}{3}\)(x + 1) = 1
x – 1 = 1 × \(\frac{3}{2}\)
x – 1 = \(\frac{3}{2}\)
x = \(\frac{3}{2}\) + 1 ⇒ x = \(\frac{3+2}{2}\) ⇒ x = \(\frac{5}{2}\)
Extra Questions of Simple Equations Class 7 – 5 Marks
Question 1.
Solve \(\frac{2 x+1}{2}\) + 1 = \(\frac{x-2}{3}\) – 1
Solution:
Question 2.
Solve 2(2x + 5) – 3(2x – 1) = 7
Solution:
2(2x + 5) – 3(2x – 1) = 7
4x + 10– 6x + 3 = 7
–2x + 13 = 7
–2x = 7 – 13
–2x = – 6
\(\frac{-2 x}{-2}\) = \(\frac{-6}{-2}\) ⇒ x = 3
Question 3.
Solve \(\frac{p+3}{10}\) – 2 = p
Solution:
\(\frac{p+3}{10}\) – 2 = p
\(\frac{p+3-20}{10}\) = p
p – 17 = 10p
p – 10p = 17
–9p = 17 ⇒ p = \(\frac{17}{-9}\)
p = \(\frac{-17}{9}\)
Question 4.
Solve \(\frac{x}{3}\) + 4 = \(\frac{x}{2}\) – 1
Solution:
\(\frac{x}{3}\) + 4 = \(\frac{x}{2}\) – 1
\(\frac{x}{3}\) – \(\frac{x}{2}\) = – 1 – 4
\(\frac{2 x-3 x}{6}\) = –5
\(\frac{-x}{6}\) = –5 ⇒ \(\frac{x}{6}\) = 5
⇒ x = 5 × 6 ⇒ x = 30
Question 5.
3(x – 1) = 12
Solution:
3(x – 1) = 12
x – 1 = \(\frac{12}{3}\)
x – 1 = 4
x = 4 + 1 ⇒ x = 5
Verification :
LHS = 3(x – 1) = 3(5 – 1) = 3 × 4 = 12
RHS = 12
∴ LHS = RHS
Question 6.
3x – \(\frac{1}{3}\) = 3
Solution:
3x – \(\frac{1}{3}\) = 3
3x = 3 + \(\frac{1}{3}\) ⇒ 3x = \(\frac{9+1}{3}\)
3x = \(\frac{10}{3}\) ⇒ x = \(\frac{10}{3}\) × \(\frac{1}{3}\) ⇒ x = \(\frac{10}{9}\)
Verification :
LHS = 3x – \(\frac{1}{3}\) = \(3\left(\frac{10}{9}\right)\) – \(\frac{1}{3}\)
= \(\frac{10}{3}\) – \(\frac{1}{3}\) = \(\frac{10-1}{3}\) = \(\frac{9}{3}\) = 3
RHS = 3
∴ LHS = RHS
Question 7.
\(\frac{2}{8x}\) = 16
Solution:
Question 8.
Find the three consecutive natural numbers so that the sum of the first and second is 15 more than the third.
Solution:
Let the first number = x
Second number = x + 1
Third number = x + 1 + 1 = x + 2
According to the problem
x + x + 1 = 15 + x + 2
2x + 1 = 17 + x
2x – x = 17 – 1 ⇒ x = 16
First number = 16
Second number = 16 + 1 = 17
Third number = 17 + 1 = 18
∴ The three consecutive number are 16,17 and 18.
Question 9.
The sum of two consecutive odd number is 68. Find the number.
Solution:
Let the first odd number = x
Second odd number = x + 2
According to the problem
x + x + 2 = 68
2x + 2 = 68