This set of Mathematics MCQs for Class 8 focuses on “Solving Linear Equations with Variables on One Side and Number on the Other”.
1. Find the solution for the equation 3x + 3 = 4.
a) 3
b) \(\frac{1}{3}\)
c) 4
d) \(\frac{1}{4}\)
View Answer
Explanation: 3x + 3 = 4
Step 1: Subtract 3 from both the sides,
3x + 3 – 3 = 4 – 3
3x = 1
Step 2: Divided both the sides by 3,
x = \(\frac{1}{3}\)
So, we have solution for the equation x = \(\frac{1}{3}\)
2. In equation 3x + 4 = 10, by transposing the variable on RHS we get ________
a) -4 = 10 – 3x
b) 4 = 3x + 10
c) 4 = -3x + 10
d) -4 = – 3x – 10
View Answer
Explanation: While shifting the variable from LHS to RHS the sign changes from positive to negative and vice versa. Here, the given equation is 3x + 4 = 10.
3. Solve equation 7x + 14 = 21 to find value of x.
a) x = 1
b) x =-1
c) x = 2
d) x = -2
View Answer
Explanation: 7x + 14 = 21
Step 1: Solving the equation, we take constant 14 on the RHS making 7x = 7
Step 2: Dividing LHS and RHS with 7 we get x = 1.
4. Solve: \(\frac{5}{2}x + \frac{3}{2} = \frac{6}{4}\).
a) x = 1
b) x = 2
c) x = 3
d) x = 0
View Answer
Explanation: \(\frac{5}{2}x + \frac{3}{2} = \frac{6}{4}\)
Step 1: Reducing the fraction we get,
\(\frac{5}{2}x + \frac{3}{2} = \frac{3}{2}\)
Step 2: Multiplying LHS and RHS by 2 we get,
5x + 3 = 3
i.e. 5x = 0
Step 3: Dividing both sides by 5 we get,
x = 0.
5. Solve: 7x – 2 = 3.
a) x = \(\frac{5}{7}\)
b) x = \(\frac{7}{5}\)
c) x = \(\frac{3}{5}\)
d) x = \(\frac{5}{3}\)
View Answer
Explanation: 7x – 2 = 3
Step 1: Transposing -2 on RHS we get,
7x = 5
Step 2: Dividing both sides by 7 we get,
x = \(\frac{5}{7}\).
6. Pick the equation from the given one’s which have solution as z = 2.
a) 2z -2 = 3
b) 3z -2 = -2
c) 3z -3 = 3
d) 4z + 3 = 3
View Answer
Explanation: Solving the 3z – 3 = 3 we get,
Step 1: Transposing -3 on RHS we get,
3z = 6
Step 2: Dividing both sides by 3 we get,
z = 2
Hence, this is the solution of our question.
7. Pick the equation which has the solution in the form of prime number.
a) 2x = 3
b) 3z = -6
c) 4y – 3 = 2
d) 2z – 2 = 2
View Answer
Explanation: A prime number is the number which can be formed by multiplying only two numbers which are 1 and the number itself.
Solving equation, 2z – 2 = 2
Step 1: Transposing – 2 on RHS we get,
2z = 4
Step 2: Dividing equation by 2 we get,
z = 2
Since 2 is greater than 0 and can be formed only by multiplying 1 and 2 it is a prime number.
8. Solve: 16 = 3m – 2.
a) m = -5
b) m = 5
c) m = 6
d) m = -6
View Answer
Explanation: Step 1: Transposing – 2 to LHS we get,
18 = 3m
Step 2: Dividing both the sides by 3 we get,
m = 6.
9. What is the solution of the equation, 7 + 2x = 0?
a) \(\frac{2}{7}\)
b) \(\frac{-2}{7}\)
c) \(\frac{7}{2}\)
d) \(\frac{-7}{2}\)
View Answer
Explanation: 7 + 2x = 0
Step 1: Transposing 7 to RHS we get,
2x = -7
Step 2: Dividing both sides by 2 we get,
x = \(\frac{-7}{2}\).
10. The solution of equation 3x + 6 = -3 is x = -3.
a) True
b) False
View Answer
Explanation: 3x + 6 = -3
Step 1: Transposing 6 to RHS we get,
3x = -9
Step 2: Dividing both sides by 3 we get
x = -3.
Sanfoundry Global Education & Learning Series – Mathematics – Class 8.
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