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Linear Equation In One variable MCQ Test 1

Question 1. If 4x + 3/5 = 5, then x =
Answer: A)11/10
Solution:- 4x + 3/5 = 5
=> 4x = 5- 3/5
=> 4x = 22/5
=> x = 22/5 ÷ 4
x = 11/10
Question 2. If x/3 - 5/2 = 6, then x = ?




Answer : 51/2
Solution:- x/3 - 5/2 = 6
=>x/3 = 6 + 5/2
=>x/3 = 17/2
=> x = (17 * 3)/2
=> x = 51/2
Question 3. The age of father is 3 times the age of the age of son. If sum of their ages is 48 yeras, then the age of father ans son are (in years):




Answer : A) Father=36, Son=12
Solution:- let the age of son be x,
Given, x + 3x = 48
=> 4x = 48
=> x = 12
Son age = 12
∴ Fathers age = 36
Question 4.Length of a rectangle is 8m less than twice of breadth. If the perimeter of the rectangle is 56m. Find its length and breadth




Answer: A) Length = 16m and breadth = 12m
Solution:- 2{(2m-8)+2m}=56
=>2(2m-8)+4m=56
=>4m-8+4m=56
=>8m-8=56
=>8m=48
=>m=6
Question 5. The sum of a two digit and the number obtained by interchanging the digits of the number is 121.If the digis differ by 5, Then find the number.




Answer: A) 38,83
Solution:- Let a number have x at ten's place and y at one's place.
∴ Number = 10x + y
On interchanging the digits 10y + x
=>10x + y + 10y + x = 121 (Equation 1)
Since, digits differ by 5
∴ x = (y +5)
Now putting value of x in equation 1
=>10(y+5) + y + 10y + y + 5 = 121
=>10y + 50 + y + 10y + (y + 5) = 121
=>22y + 55 = 121
=>22y = 66
=>y = 3
∴ x = 3 + 5 = 8
∴ Number is = 83
Question 6. The age of Mira, Tina and Saina are in the ratio 6:4:7 respectively, if the sum of their ages is 34 years, what is Sania's age?




Answer: D) 14 years
Solution:- Let the age of Mira, Tina and Saina are in the ratio 6x, 4x, 7x respectively,
Given, 6x + 4x + 7x = 34
=>17x = 34
=>x = 2
∴ Mira's age = 6*2 = 12 years
∴ Tina's age = 4*2 = 8 years
∴ Sania's age = 7*2 = 14 years
Question 7. (3/4)th of a number is 20 more than half of the same number. The required number is ______.




Answer: D) 80
Solution:- Let the number be x.
Given, (3/4)x = (x/2) + 20
=> 3x/4 - x/2 = 20
=> 3x - 2x/4 = 20
=> x/4 = 20
=> x = 20*4
=> x = 80
Question 8. In a class of 100 students, 55 students have passed in Mathematics and and 67 students have passed in Physics. Then, the number of students who have passed in Physics only is.




Answer: D) 45
Solution:- Let the number of students passed by Physics be x,
Total number of students = 100
Total number of students passed in mathematics = 55
No of students who only passed by Physics is = x
=> x = 100-55
=> x = 45
Question 9. If, 8x + 3 = 27 + 2x. Then x =?




Answer: B) 4
Solution:- 8x + 3 = 27 + 2x
=> 6x + 3 = 27
=> 6x = 24
=> x = 4
Question 10. If 2x - 3 = x + 2, then x =?




Answer: C) 5
Solution:- 2x - 3 = x +2
=> (2x-x) - 3 = 2
=> x - 3 = 2
=> x = 5

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