Linear Inequalities exercise 3A CA foundation maths solutions
CA foundation maths solutions for chapter 3 exercise 3A Linear Inequalities are given.
Part A – Business maths, Linear Inequalities chapter 3 CA foundation maths solutions.
You should study the textbook lesson Linear Inequalities very well.
You must practice all example problems and solutions which are given in the textbook.
You can observe the solutions and will be try them in your own way.
Exercise 3A solutions CA foundation maths chapter 3
CA foundation maths solutions
Chapter 3
Exercise 3A
Linear Inequalities
1. i. An employer recruits experienced (X) and fresh workmen (Y) for his firm under the condition that he cannot employ more than 9 people X and Y can be related by the inquality.
a. X + Y not equal to 9
b. X + Y less than or equal to 9
c. X + Y greater than or equal to 9
d. None of these
Solution:
An employer recruits experienced (X) and fresh workmen (Y) for his firm under the condition that he cannot employ more than 9 people.
So, he has to recruit less than or equal to 9.
Answer is b.
1. ii. 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. This situation can be expressed as
Solution:
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.
1. iii. The rules and regulations demand that the employer should employ not more than 5 experienced hands to 1 fresh one and this fact can be expressed as
Solution:
The rules and regulations demand that the employer should employ not more than 5 experienced hands to 1 fresh one.
1. iv. The union however forbids him to employ less than 2 experienced person to each fresh person. This situation can be expressed as
Solution:
The union however forbids him to employ less than 2 experienced person to each fresh person.
a.
b.
c.
d. None of these.
Solution:
The graph of X + Y = 9. It is a straight line passing through (0, 9) and (9, 0).
Therefore, the graph of inequality is the region satisfying the inequality. It is one side of a line. The shading has done towards the origin.
Answer is (a).
a.
b.
c.
d. None of these.
a.
b.
c.
d.
1. viii.
2. A dietitian wishes to mix together two kinds of food so that the vitamin content of the mixture is at least 9 units of vitamin A, 7 units of vitamin B, 10 units of vitamin C and 12 units of vitamin D. The vitamin content per Kg of each food is shown below:
A B C D
Food I 2 1 1 2
Food II 1 1 2 3
Assuming x units of food I is to be mixed with y units of food II the situation can be expressed as
3. Graphs of the inequalities are drawn below :
L1 : 2x + y = 9 L2 : x + y = 7
L3 : x + 2y = 10 L4 : x + 3y = 12
The common region ( shaded part) indicated on the diagram is expressed by the set of inequalities
Solution:
The diagram shows the region which is common of the equations of inequality.
Let us take the point (6,5)
Then 2x + y = 12 + 5 = 17 > 9
x + y = 6 + 5 = 11 > 7
x + 2y = 6 + 10 = 16 > 10
x + 3y = 6 + 15 = 21 > 12
x > 0, y = 5 > 0
Answer is (c)
a.
b.
c.
d. None of these.
Solution:
Let us take the point (0, 0)
L1 : 0 + 0 = 0 > 6 L2 : 0 + 0 = 0 > 4
L3 : 0 + 0 = 0 > 6 L4 : 0 + 0 = 0 > 6
By check we get the answer (a).
Answer is (a).
Q 5. The region indicated by the shading in the graph is expressed by inequalities
Solution:
From the Information, we take point from the common region and check the equations.
Answer is (a)
a.
b.
c
d
6. ii. The region is expressed as
a.
b.
c.
d. None of these.
Solution:
Answer is (a)
Q 7.
The common region on the graph is expressed by the set of five inequalities
8. A firm makes two types of products. Type A and Type B. The profit on product A is Rs. 20 each and that on product B is Rs. 30 each. Both types are processed on three machines M1, M2 and M3. The time required in hours by each product and total time available in hours per week on each machine are as follows
Chapter 3 exercise 3A Linear inequalities CA foundation maths solutions
a.
b.
c.
d. None of thses.
Solution:
Plot the graph of equations. According the region for inequality, sade the common region.
Answer is (b)
a.
b.
c.
d. None if these.
Solution:
Plot the graph of equations. According the region for inequality, sade the common region.
Answer is (a)
Additional Questions
Solution:
Let we plot the inequality. Then find the common region. Then add the corner points of the common region.
We get the corner points as follows
(0,0), (4,0), (0,4) and (20/11), (36/11)
Answer is (a)
Solution:
Let we plot the inequality. Then find the common region. Then add the corner points of the common region.
We get the corner points as follows
(0, 18), (12, 0), (4, 2) and (2, 6)
Answer is (a)
Note: You can observe the solutions and try them in your own method.