This set of Class 11 Maths Chapter 6 Multiple Choice Questions & Answers (MCQs) focuses on “Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation”.
1. If x>7 then x+2>9 is true?
a) True
b) False
View Answer
Explanation: We can add equal number on both sides of inequality so
if x>7 then x+2>7+2 => x+2>9
2. If x>7 then which is impossible?
a) x>4
b) x<6
c) x>9
d) x<14
View Answer
Explanation: x>7 and 7>4 => x>7>4 => x>4.
If x>7 then x cannot be less than 6.
If x=11 then x>7 and x>9.
If x=11 then x>7 and x<14.
3. If x>5 then 2x>10 is true or not?
a) True
b) False
View Answer
Explanation: We can multiply a positive number on both sides of inequality without any change in inequality sign. So, if x>5 multiplying 2 on both sides 2x>10.
4. If x>7 then -x>-7 is ___________
a) possible
b) certainly false
c) certainly true
d) depend on x
View Answer
Explanation: If we multiply by negative number on both sides of inequality then sign of inequality will change i.e. if x>7 then (-1) x < (-1)7 => -x<-7.
5. If x is a positive integer and 20x<100 then find solution set of x.
a) {0,1,2,3,4,5}
b) {1,2,3,4,5}
c) {1,2,3,4}
d) {0,1,2,3,4}
View Answer
Explanation: 20x<100
Dividing by 20 on both sides, x< (100/20) => x<5
Since x is a positive integer so x = 1,2,3,4.
6. If x is a natural number and 20x≤100 then find solution set of x.
a) {0,1,2,3,4,5}
b) {1,2,3,4,5}
c) {1,2,3,4}
d) {0,1,2,3,4}
View Answer
Explanation: 20x≤100
Dividing by 20 on both sides, x ≤ (100/20) => x≤5
Since x is a natural number so x = 1,2,3,4,5.
7. If x is a whole number and 10x≤50 then find solution set of x.
a) {0,1,2,3,4,5}
b) {1,2,3,4,5}
c) {1,2,3,4}
d) {0,1,2,3,4}
View Answer
Explanation: 10x≤50
Dividing by 10 on both sides, x ≤ (50/10) => x≤5
Since x is a whole number so x = 0,1,2,3,4,5.
8. If 2x+1 > 5 then which is true?
a) x>4
b) x<4
c) x>2
d) x<2
View Answer
Explanation: 2x+1>5
=>2x>5-1
=>2x>4 => x>2.
9. If x-1>-x+7 then which is true?
a) x>4
b) x<4
c) x>2
d) x<2
View Answer
Explanation: x-1>-x+7
=>2x>8 => x>4.
10. Rahul obtained 20 and 25 marks in first two tests. Find the minimum marks he should get in the third test to have an average of at least 30 marks.
a) 60
b) 35
c) 180
d) 45
View Answer
Explanation: Average is at least 30 marks.
Let x be the marks in 3rd test.
Average = (20+25+x)/3 ≥30
=>45+x≥90 => x≥90-45 => x≥45.
Minimum marks in 3rd test should be 45.
11. Find all pairs of consecutive odd positive integers both of which are smaller than 8 such that their sum is more than 10.
a) (5,7)
b) (3,5), (5,7)
c) (3,5), (5,7), (7,9)
d) (5,7), (7,9)
View Answer
Explanation: Let two numbers be x and x+2.
x + x+2 >10 => 2x>8 => x>4
and x<8
and x+2<8 => x<6.
4<x<6 => x can be 5.
For x =5, x+2=7
So, Pairs of odd consecutive positive integers are (5,7).
12. The longest side of a triangle is 2 times the shortest side and the third side is 4 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.
a) 7
b) 9
c) 11
d) 13
View Answer
Explanation: Let shortest side be x. Then longest side = 2x.
Third side = 2x-4.
Given, perimeter of triangle is at least 61 cm
=>x+2x+2x-4 ≥ 61 => 5x≥65 = x≥13.
Minimum length of the shortest side is 13 cm.
Sanfoundry Global Education & Learning Series – Mathematics – Class 11.
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