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Mathematics

Case Study II

Raman Lal runs a stationery shop in Pune. The analysis of his sales, expenditures and profits showed that for x number of notebooks sold, the weekly profit (in ₹) was P(x) = - 2x2 + 88x - 680. Raman Lal found that:

  • He has a loss if he does not sell any notebook in a week.
  • There is no profit no loss for a certain value x0 of x.
  • The profit goes on increasing with an increase in x i.e. the number of notebooks sold. But he gets a maximum profit at a sale of 22 notebooks in a week.

Now answer the following questions :

1. What will be Raman Lal’s profit if he sold 20 notebooks in a week?

  1. ₹ 144
  2. ₹ 280
  3. ₹ 340
  4. ₹ 560

2. What is the maximum profit that Raman Lal can earn in a week?

  1. ₹ 144
  2. ₹ 288
  3. ₹ 340
  4. ₹ 680

3. What is Raman Lal’s loss if he does not sell any notebooks in a particular week?

  1. ₹ 0
  2. ₹ 340
  3. ₹ 680
  4. ₹ 960

4. Write a quadratic equation for the condition when Raman Lal does not have any profit or loss during a week.

  1. 2x2 - 44x + 340 = 0
  2. x2 + 44x - 340 = 0
  3. x2 - 88x + 340 = 0
  4. x2 - 44x + 340 = 0

5. What is the minimum number of notebooks x0 that Raman Lal should sell in a week so that he does not incur any loss?

  1. 0
  2. 10
  3. 11
  4. 12

Quadratic Equations

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Answer

1. Given,

⇒ P(x) = -2x2 + 88x - 680

Books sold by Raman lal is 20.Therefore x = 20

⇒ P(20) = −2(20)2 + 88(20) − 680

⇒ P(20) = −2(400) + 1760 − 680

⇒ P(20) = −800 + 1080

⇒ P(20) = 280.

Hence, option (b) is the correct option.

2. Given,

⇒ P(x) = - 2x2 + 88x - 680

Maximum profit occurs when 22 books are sold, thus x = 22.

⇒ P(22) = −2(22)2 + 88(22) − 680

= −2(484) + 1936 − 680

= −968 + 1256

= ₹ 288.

Hence, option (b) is the correct option.

3. Given,

⇒ Number of notebooks sold in a particular week = 0

Thus, x = 0

The weekly profit P(x) = -2x2 + 88x - 680.

Raman Lal’s loss if no notebooks sold in a week,

⇒ P(0) = -2(0)2 + 88(0) - 680

= −680

= loss of ₹ 680.

Hence, option (c) is the correct option.

4. Given,

⇒ P(x) = -2x2 + 88x - 680

If there is no profit/loss, then profit = ₹ 0.

⇒ P(x) = 0

⇒ −2x2 + 88x − 680 = 0

⇒ −2(x2 - 44x + 340) = 0

⇒ x2 − 44x + 340 = 0.

Hence, option (d) is the correct option.

5. When Raman lal has no profit/loss we get the equation x2 − 44x + 340 = 0.

⇒ x2 − 44x + 340 = 0

⇒ x2 − 10x - 34x + 340 = 0

⇒ x(x − 10) - 34(x - 10) = 0

⇒ (x − 10)(x - 34) = 0

⇒ (x − 10) = 0 or (x - 34) = 0      [Using Zero-product rule]

⇒ x = 10 or x = 34.

The minimum value of x with no loss is x = 10. Raman lal has to sell 10 books so that he does not incur any loss.

Hence, option (b) is the correct option.

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