1st PUC Basic Maths Question Bank Chapter 12 Linear Functions

Students can Download Basic Maths Chapter 12 Linear Functions Questions and Answers, Notes Pdf, 1st PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Basic Maths Question Bank Chapter 12 Linear Functions

Question 1.
A manufacturer produces and sells pens ’10 per unit. His fixed costs are ‘ 600 and variable cost per pen is ‘ 3.50. Calculate
(i) Revenue function
(ii) Cost function
(iii) Profit function
(iv) Breakeven point.
Answer:
(i) Revenue function R(x) = p.x. = lQx
(ii) Cost function C(x) = ax + b
(iii) Profit function p(x) = Rx – C(x) = 8x – [3.50x + 6500]
(iv) Breakeven point at BEP ⇒ TR = TC ⇒ R(x) = R(x) ⇒ P(x) = 0
∴ 4.50x – 6500 = 0 ⇒ x = \(\frac{6500}{4.50}\) = 1445. units
Breakeven point revenue is RS = 1445 × 10 = Rs. 14450

1st PUC Basic Maths Question Bank Chapter 12 Linear Functions

Question 2.
The daily cost of production ‘C’ in R for ‘x unit of an assembly is C(x) = 12.5x + 6400. If each unit is sold for ’25/- then determine the minimum number of units that should be produced and sold to ensure no loss. If the selling price is reduced by Rs. 2.5 unit, What would be the break even point.
Answer:
By date C(x) = 12.5x + 6400, S(x) = 25x
Breakeven point ⇒ C(x) = S(x)
⇒ 12.5x + 6400 = 25x ⇒ 6400 = 12.5x
⇒ x = \(\frac{6400}{12.5}\) = 512
Selling price is reduced by 2.5 unit
S(x) = (25 – 2.5) = (22.5)x
Breakeven point ⇒ C(x) = S(x)
12.5x + 640 = 22.5x
⇒ 6400 = 10x = x = \(\frac{6400}{10}\) = 640