# Mathematics Questions and Answers – Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation

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This set of Mathematics Questions & Answers for Exams 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

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

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

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.
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4. If x>7 then -x>-7 is ___________
a) possible
b) certainly false
c) certainly true
d) depend on x

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}

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}

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}

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

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

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

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)

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

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.

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