*Notes to a video lecture on http://www.unizor.com*

__Arithmetic+ 03__

*Problem A*

I have purchased a share of some company XYZ for price

*.*

**P**_{0}=$100After a month the price was lower by

*.*

**ρ=10%**Another month later the price went up by the same percentage

*and I sold this share.*

**ρ=10%**Have I lost or gain money, or got even?

*Solution A*

After the first month the price dropped from

*by*

**P**_{0}=100*, that is by*

**ρ=10%***and was*

**P**_{0}·0.1

**P**

= 100 − 100·0.1 = 90_{1}= P_{0}− P_{0}·ρ == 100 − 100·0.1 = 90

After the second month the price went up from

*by*

**P**_{1}=90*, that is by*

**ρ=10%***and was*

**P**_{1}·0.1

**P**

= 90 + 90·0.1 = 99_{2}= P_{1}+ P_{1}·ρ == 90 + 90·0.1 = 99

Therefore, since I bought a share for

*and sold it for*

**P**_{0}=$100*, my net result is*

**P**_{2}=$99Δ

**P = P**_{2}− P_{0}= −$1which is a

*.*

**loss****Notice that percentage of my loss was equal to percentage of my gain, but in dollars it was larger because the same percentage was applied to a larger amount on the way down, than to a smaller amount on the way up**.

*Problem B*

I have purchased a share of some company XYZ for price

*.*

**P**_{0}=$100After a month the price was higher by

*.*

**ρ=10%**Another month later the price went down by the same percentage

*and I sold this share.*

**ρ=10%**Have I lost or gain money, or got even?

*Solution B*

After the first month the price went up from

*by*

**P**_{0}=100*, that is by*

**ρ=10%***and was*

**P**_{0}·0.1

**P**

= 100 + 100·0.1 = 110_{1}= P_{0}+ P_{0}·ρ == 100 + 100·0.1 = 110

After the second month the price went down from

*by*

**P**_{1}=110*, that is by*

**ρ=10%***and was*

**P**_{1}·0.1

**P**

= 110 − 110·0.1 = 99_{2}= P_{1}− P_{1}·ρ == 110 − 110·0.1 = 99

Therefore, since I bought a share for

*and sold it for*

**P**_{0}=$100*, my net result is*

**P**_{2}=$99Δ

**P = P2 − P**_{0}= −$1which is a

*.*

**loss****Notice that percentage of my loss was equal to percentage of my gain, but in dollars it was larger because the same percentage was applied to a larger amount on the way down, than to a smaller amount on the way up**.

*IMPORTANT CONCLUSION*:

If market price of a stock regularly goes up and down by the same percentage, the price per share goes down, investors lose money.

To be profitable, the price should go up more frequently or/and by a greater percentage than down.

*Problem C*

(a) Within the framework of Problems A, what the percentage of gain would neutralize the loss after price moving down by

*?*

**ρ=10%**(b) Within the framework of Problems B, what the percentage of loss would neutralize the gain after price moving up by

*?*

**ρ=10%***Solution C*

(a) If we lost

*, the price became*

**ρ=10%**

**P**

= 100 − 100·0.1 = 90_{1}= P_{0}− P_{0}·ρ == 100 − 100·0.1 = 90

To return back to

*from*

**$100***, the growth in price*

**$90***should satisfy the equation*

**σ**

**100 = 90 + 90·σ**from which

**σ = 1/9 ≅ 0.1111**Therefore, if we add approximately

*to*

**11.11%***, the final price will be*

**P**_{1}=90

**P**

= 99.999 ≅ 100_{2}= 90 + 90·0.1111 == 99.999 ≅ 100

(b) If we gain

*, the price became*

**ρ=10%**

**P**

= 100 + 100·0.1 = 110_{1}= P_{0}+ P_{0}·ρ == 100 + 100·0.1 = 110

To return back to

*from*

**$100***, the diminishing in price*

**$110***should satisfy the equation*

**σ**

**100 = 110 − 110·σ**from which

**σ = 10/11 ≅ 0.0909**Therefore, if we subtract approximately

*from*

**9.09%***, the final price will be*

**P**_{1}=110

**P**

= 99.999 ≅ 100_{2}= 110 − 110·0.0909 == 99.999 ≅ 100

*Answer C*:

(a) To neutralize a loss of

*we need a larger (in %) gain of approximately*

**10%***.*

**11.11%**(b) To neutralize a gain of

*we need a smaller (in %) loss of approximately*

**10%***.*

**9.09%***Problem D*(suggested by Max)

Concentration of salt dissolved in water is measured in percentage of the mass of salt to the total mass of solution (salt + water).

Initially, a glass contained

**100**gram*.*

**1%***Question 1*: how much salt and how much water was in a glass?

This same glass was standing on a table for a week and part of water has evaporated, so the concentration of salt after a week became

*.*

**2%***Question 2*: how much salt and how much water was in a glass after this process of evaporation?

*Question 3*: how much water has evaporated in a week?

*Solution D*

If

*of solution contained*

**100**gram*of salt, the mass of salt was*

**1%**

**m**(gram)_{salt}= 100·0.01 = 1Hence, the mass of water was

**m**(gram)_{water}= 100 − 1 = 99*Answer 1*(initially):

**m**(gram)_{salt}= 1

**m**(gram)_{water}= 99After some water has evaporated, the glass contained the same amount salt, that is

*.*

**1**gramSince we know that it constitutes

*of a solution, the total mass of solution must be*

**2%***because*

**50**gram

**1/0.02 = 50**(gram)To make

*of solution with*

**50**gram*of salt, we need*

**1**gram

**49**gram*Answer 2*(after a week):

**m**(gram)_{salt}= 1

**m**(gram)_{water}= 49*Answer 3*:

That means,

**99−49=50**gram
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