Linear InequalitiesMCQMTP June 24 Series IIQuestion 1168 of 146
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A small manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry then sent to the machine shop for finishing. The number of man-hours of labour required in each shop for the production of each unit of A and B, and the number of man-hours the firm has available per week are as follows:GadgetFoundryMachine-shopA105B64Weekly capacity1000600Let the firm manufactures x units of A and y units of B. The constraints are:

Options

A10x+6y1000,5x+4y600,x0,y0\displaystyle 10x + 6y \le 1000, 5x + 4y \le 600, x \ge 0, y \ge 0
B10x+6y1000,5x+4y600,x0,y0\displaystyle 10x + 6y \ge 1000, 5x + 4y \ge 600, x \ge 0, y \ge 0
C10x+6y1000,5x+4y600,x0,y0\displaystyle 10x + 6y \ge 1000, 5x + 4y \le 600, x \ge 0, y \le 0
D10x+6y1000,5x+4y600,x0,y0\displaystyle 10x + 6y \le 1000, 5x + 4y \ge 600, x \le 0, y \le 0
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Correct Answer

Option a10x+6y1000,5x+4y600,x0,y0\displaystyle 10x + 6y \le 1000, 5x + 4y \le 600, x \ge 0, y \ge 0

All Options:

  • A10x+6y1000,5x+4y600,x0,y0\displaystyle 10x + 6y \le 1000, 5x + 4y \le 600, x \ge 0, y \ge 0
  • B10x+6y1000,5x+4y600,x0,y0\displaystyle 10x + 6y \ge 1000, 5x + 4y \ge 600, x \ge 0, y \ge 0
  • C10x+6y1000,5x+4y600,x0,y0\displaystyle 10x + 6y \ge 1000, 5x + 4y \le 600, x \ge 0, y \le 0
  • D10x+6y1000,5x+4y600,x0,y0\displaystyle 10x + 6y \le 1000, 5x + 4y \ge 600, x \le 0, y \le 0

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Detailed Solution & Explanation

Let x\displaystyle x be the number of units of Gadget A produced and y\displaystyle y be the number of units of Gadget B produced.

Let's write the constraints for each shop:
1) **Foundry Constraint:** Production of A requires 10\displaystyle 10 man-hours and B requires 6\displaystyle 6 man-hours. The total hours required in the foundry is 10x+6y\displaystyle 10x + 6y. The weekly capacity of the foundry is 1000\displaystyle 1000 hours:
10x+6y100010x + 6y \le 1000
2) **Machine Shop Constraint:** Production of A requires 5\displaystyle 5 man-hours and B requires 4\displaystyle 4 man-hours. The total hours required in the machine shop is 5x+4y\displaystyle 5x + 4y. The weekly capacity of the machine shop is 600\displaystyle 600 hours:
5x+4y6005x + 4y \le 600
3) **Non-negativity Constraints:** The counts of gadgets produced cannot be negative:
x0,y0x \ge 0, \quad y \ge 0

Combining all these constraints, we get:
10x+6y1000,5x+4y600,x0,y010x + 6y \le 1000, \quad 5x + 4y \le 600, \quad x \ge 0, \quad y \ge 0

This matches Option A.

Hence, **Option A** is the correct answer.

About This Chapter: Linear Inequalities

Paper

Paper 3: Quantitative Aptitude

Weightage

1-3 Marks

Key Topics

Linear Inequalities in one & two variables

This chapter covers Linear Inequalities in one & two variables and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 1-3 Marks weightage. Focus on understanding core concepts rather than memorizing.

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