This set of Class 8 Maths Chapter 2 Multiple Choice Questions & Answers (MCQs) focuses on “Solving Linear Equations with Variables on Both the Sides”.
1. If Malik has 3 coins and his sister has 4 coins each of 2 rupees, what will be the sum of amount both have?
a) 14 rupees
b) 13 rupees
c) 12 rupees
d) 7 rupees
View Answer
Explanation: There are 7 coins in total. When we multiply 2 × 7 we get 14. Hence, the total amount with them is 14 rupees.
2. The perimeter of a rectangle is 12 cm and it’s breadth is 2 cm. What will be it’s length?
a) 2 cm
b) 3 cm
c) 4 cm
d) 5 cm
View Answer
Explanation: For any rectangle the formula for perimeter is 2 × (Length + Breadth). Here we already know the breadth of the rectangle. When we substitute the known values in the formula for perimeter of a rectangle we get,
12 = 2 × (Length + 2)
∴ 6 = Length + 2
∴ Length = 4 cm.
3. At present Disha’s mother is three times the age of Disha. After 5 years their ages will sum up to 70 years. Find the present age.
a) Disha 10 and her mother 30
b) Disha 12 and her mother 36
c) Disha 13 and her mother 39
d) Disha 15 and her mother 45
View Answer
Explanation: Let Disha’s age be x years then her mother would be 3x years old. After 5 years, their ages are x + 5 (Disha) and 3x + 5 (Disha’s Mother) respectively.
(x + 5) + (3x + 5) = 70
∴ x + 5 + 3x + 5 = 70
∴ 4x + 10 = 70
∴ 4x = 60
∴ x = 15
Since x = 15 we know that Disha’s present age is 15 years, to find her mother’s age we have (3×15) i.e. 45 years.
4. Bansal has 7 times as many two-rupee coins as he has five-rupee coins. If he has in all a sum of rupees 95, how many coins of each denomination does he have?
a) 5 two-rupee and 25 five-rupee
b) 35 two-rupee and 5 five-rupee
c) 5 two-rupee and 35 five-rupee
d) 25 two-rupee and 5 five-rupee
View Answer
Explanation: Let the number of five-rupee coins with Bansal be x. Then the number of two-rupee coins will be 7x. The amount from
(i) Two-rupee coins = 7x × 2 = 14x
(ii) Five-rupee coins = x × 5 = 5x
Hence the total amount with Bansal is 19x.
∴ 19x = 95
∴ x = 5
Thus, the number of five-rupee coin = x = 5
and number of two-rupees coin = 7x = 7(5) = 35.
5. The sum of three consecutive numbers is 789, what are those consecutive numbers?
a) 262, 263 and 264
b) 263, 264 and 265
c) 264, 265 and 266
d) 265, 266 and 267
View Answer
Explanation: Let the first number be x. Since they are consecutive numbers the next number would be (x + 1) and the third number would be (x + 2). Now as the sum of the numbers is given as 789 we solve the equation by adding the three consecutive variables.
x + (x + 1) + (x + 2) = 789
∴ 3x + 3 = 789
∴ 3x = 789 – 3
∴ 3x = 786
∴ x = 262
Hence the three consecutive numbers are 262, 263 and 264.
6. The sum of two natural numbers is 5, the numbers are in a ratio 2:3. Find the numbers.
a) 2 and 3
b) 1 and 4
c) 0 and 5
d) 2 and 4
View Answer
Explanation: Let the common multiple be x.
∴ 2x + 3x = 5
∴ 5x = 5
∴ x = 1
Since, x = 1 the natural numbers are 2 and 3.
7. If the perimeter of square is 28 m. Find the length of the side.
a) 7 cm
b) 7 m
c) 14 m
d) 11 m
View Answer
Explanation: The formula for perimeter of a square is 4 × (side)
Perimeter = 4 × (side)
∴ 28 = 4 × (side)
∴ side = 7
Hence, the length of the side of the square is 7 m.
8. Sum of consecutive multiples of 23 is 1656. Find the multiples.
a) 529, 552, 575
b) 629, 662, 675
c) 189, 222, 275
d) 389, 332, 375
View Answer
Explanation: Let (x – 23), x and (x + 23) be the three consecutive multiples. Now, the sum of these numbers is 1656.
(x – 23) + x + (x + 23) = 1656
∴ x – 23 + x + x + 23 = 1656
∴ 3x = 1656
∴ x = 552
Hence, x – 23 = 529 and x + 23 = 575
The three consecutive multiples of 23 which sum up to 1656 are 529, 552 and 575.
9. When two integers are added the sum is -52. If the integers are in the ratio 6:7, then find the integers.
a) -24 and 28
b) 24 and -28
c) -24 and -28
d) 24 and 28
View Answer
Explanation: Let the common multiple be x. We know that the integers sum up to -52.
6x + 7x = -52
∴ 13x = -52
∴ x = -4
Since, x = -4 the integers are -24 and -28.
10. If Raj scores 27 marks less than the highest scorer and the highest scorer has 2 marks less than the maximum achievable score, then find the score that Raj scored, if the maximum achievable score is 100.
a) 70
b) 71
c) 72
d) 73
View Answer
Explanation: The highest scorer scores 2 marks less then maximum achievable score.
Score of highest scorer = Maximum achievable score – 2
∴ Score of highest scorer = 100 -2
∴ Score of highest scorer = 98
Now, Raj’s score can be calculated by,
Raj’s score= Score of highest scorer – 27
∴ Raj’s score = 98 – 27
∴ Raj’s score = 71.
Sanfoundry Global Education & Learning Series – Mathematics – Class 8.
To practice all chapters and topics of class 8 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]
- Practice Class 10 Mathematics MCQs
- Practice Class 9 Mathematics MCQs
- Check Class 8 - Mathematics Books
