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Published on: 13/05/2022
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Questions + Answers key
Take MCQ Maths Test1.
Represent the solutions of 3x + 9 ≤ 12 in a number line, where x is an integer.
2.
Represent the solutions −8 < 2x < 10 in a number line, where x is a natural number.
3.
If one worker earns Rs.200 per day, how many workers can be employed with in a monthly budget of Rs.3 Lakh?
4.
Solve: 5 − 7x ≥ 33, where x is an integer.
5.
Solve: 2x + 4 < 18, where x is a natural number.
1.
3x + 9 ≤ 12
\(\cfrac { 3x }{ 3 } +\cfrac { 9 }{ 3 } \le \cfrac { 12 }{ 3 } \) [Dividing the inequation by 3 on both sides]
x + 3 ≤ 4
x + 3 – 3 ≤ 4–3 [Subtracting 3 from both sides]
x ≤ 1
Since the solution belongs to integers, the solutions are 1, 0, −1, −2, …. It’s graph on the number line is shown below:
2.
−8 < 2x < 10
\(\cfrac { -8 }{ 2 } <\cfrac { 2x }{ 2 } <\cfrac { 10 }{ 2 } \) [Dividing the inequation by 2]
−4 < x < 5
3.
Let the number of workers be x.
Then, the wages that x workers will earn per day = Rs.200x
The wages that x workers will earn per month = Rs.(200x x 30) = Rs.6000x
Given that, this amount cannot exceed Rs.300000.
Otherwise, it can be written as 6000 x ≤ 300000
\(\frac {6000x}{6000 } \le \frac {300000 }{6000 } \)
\(x\le 50\)
Thus, up to 50 workers can be employed on a monthly budget of Rs.300000.
4.
5 − 7x ≥ 33
5 − 5 − 7x ≥ 33 − 5 [Subtracting 5 from both sides]
−7x ≥ 28
\(\cfrac { -7x }{ -7 } \ge \cfrac { 28 }{ 7 } \) [Dividing both sides by –7]
\(x\le -4\) [since, it is divided by a negative number, the inequality is reversed]
Since, solution belongs to the set of integers, that are less than −4, we take the values of x as –4, –5, –6, ...
Therefore, the solutions are –4, –5, –6,...
5.
2x + 4 < 18
2x + 4 – 4 < 18 − 4 [Subtracting 4 from both sides]
2x < 14 [Divide by 2 on both sides]
x < 7
Since the solution belongs to natural numbers, that are less than 7, we take the values of the x as 1, 2, 3, 4, 5 and 6.
Therefore, the solutions are 1, 2, 3, 4, 5 and 6.
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