Test: Inequalities- 3 - CA Foundation MCQ

An employer recruits experienced (x) and fresh workmen (y) for his firm under the condition that he cannot employ more than 9 people, it means he has to recriut less than or equal to 9.
∴ x + y ≤ 9.

On the average experienced person does 5 units of work while a fresh one 3 units of work daily but the employer has to maintain an output of at least 30 units of work per day.
∴ 5x+3y ≥30 , x ≥ 0, y ≥ 0

The rules and regulations demand that the employer should employ not more than 5 experienced hands to 1 fresh one.
so 5y ≥ x .

The union however forbids him to employ less than 2 experienced person to each fresh person. 2y ≤ x. or x ≥ 2y.

To plot the graph of x + ≤ 9 , first plote the graph of x + y = 9.
It will be a straight line passing through (0 ,9) and (9, 0) .
The graph of inequality is the region satisfying the inequality.
So it is on side of a line.
Now we will check any one point of one side , if it satisfy the inequality then the shading is done on that side or on the other side.
Let us check (0, 0) . 0 + 0 = 0 < 9, so the shading has to be done towards the origin.

To plot the graph of 5x + 3y ≥ 30 , first plote the graph of 5x + 3y = 30.
It will be a straight line passing through (0 ,10) and (6, 0) .
The graph of inequality is the region satisfying the inequality.
So it is one side of a line.
Now we will check any one point of one side , if it satisfy the inequality then the shading is done on that side or on the other side.
Let us check (0, 0) . 0 + 0 = 0 <30, . So the origin does not satisfy the inequality. so the shading has to be done opposite to the origin.

To plot the graph of y ≤ (½)x, first plote the graph of y =(½)x
It will be a straight line passing through (0 ,0) and (2, 1) .
The graph of inequality is the region satisfying the inequality.
So it is on one side of a line.
Now we will check any one point of one side , if it satisfy the inequality then the shading is done on that side or on the other side.
As the line passes through origin we can not check (0,0).
Let us check (2,0) , When x = 2, then y = 1. 0 < 1 , So (2,0) satisfy the inequality.
So shading must be towards the point . that is towards x axis.

The diagram shows the region which is common of the equations of inequality.
We will select one point from the shaded region and write the equations of inequalities.
Lets the consider the point (6, 2.5) and substitute in equations .
5x + 3y = 5(6) + 3(2.5) = 30 + 7.5 = 37.5 > 30
x + y = 6 + 2.5 = 8.5 < 9
x / 3 = 6/3 = 2 < 2.5 < y
So the answer is none of these.

Step 1: Check the shaded region on the graph

From the figure, the shaded region lies above all four lines (towards the top right corner).

That means for each line, instead of "=", we will have "≥".

So, the inequalities should be:

• 2x + y ≥ 9

• x + y ≥ 7

• x + 2y ≥ 10

• x + 3y ≥ 12

Step 2: Verify with a test point

Take any point from the shaded region, for example (6, 2).

• For 2x + y: 2(6) + 2 = 12 + 2 = 14, which is greater than 9.

• For x + y: 6 + 2 = 8, which is greater than 7. 

• For x + 2y: 6 + 4 = 10, equal to 10. 

• For x + 3y: 6 + 6 = 12, equal to 12. 

So, point (6, 2) satisfies all inequalities.

Step 3: Also check non-negativity

Since the shaded region is in the first quadrant, x ≥ 0 and y ≥ 0 are also valid.

Final Answer

The common region is given by:

• 2x + y ≥ 9

• x + y ≥ 7

• x + 2y ≥ 10

• x + 3y ≥ 12

• x ≥ 0, y ≥ 0

This matches option (c).