## Wednesday, January 17, 2024

### Arithmetic+ 03: UNIZOR.COM - Math+ & Problems - Arithmetic

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 P0=\$100.
After a month the price was lower by ρ=10%.
Another month later the price went up by the same percentage ρ=10% and I sold this share.
Have I lost or gain money, or got even?

Solution A

After the first month the price dropped from P0=100 by ρ=10%, that is by P0·0.1 and was
P1 = P0 − P0·ρ =
= 100 − 100·0.1 = 90

After the second month the price went up from P1=90 by ρ=10%, that is by P1·0.1 and was
P2 = P1 + P1·ρ =
= 90 + 90·0.1 = 99

Therefore, since I bought a share for P0=\$100 and sold it for P2=\$99, my net result is
ΔP = P2 − P0 = −\$1
which 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 P0=\$100.
After a month the price was higher by ρ=10%.
Another month later the price went down by the same percentage ρ=10% and I sold this share.
Have I lost or gain money, or got even?

Solution B

After the first month the price went up from P0=100 by ρ=10%, that is by P0·0.1 and was
P1 = P0 + P0·ρ =
= 100 + 100·0.1 = 110

After the second month the price went down from P1=110 by ρ=10%, that is by P1·0.1 and was
P2 = P1 − P1·ρ =
= 110 − 110·0.1 = 99

Therefore, since I bought a share for P0=\$100 and sold it for P2=\$99, my net result is
ΔP = P2 − P0 = −\$1
which 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 ρ=10%, the price became
P1 = P0 − P0·ρ =
= 100 − 100·0.1 = 90

To return back to \$100 from \$90, the growth in price σ should satisfy the equation
100 = 90 + 90·σ
from which
σ = 1/9 ≅ 0.1111
Therefore, if we add approximately 11.11% to P1=90, the final price will be
P2 = 90 + 90·0.1111 =
= 99.999 ≅ 100

(b) If we gain ρ=10%, the price became
P1 = P0 + P0·ρ =
= 100 + 100·0.1 = 110

To return back to \$100 from \$110, the diminishing in price σ should satisfy the equation
100 = 110 − 110·σ
from which
σ = 10/11 ≅ 0.0909
Therefore, if we subtract approximately 9.09% from P1=110, the final price will be
P2 = 110 − 110·0.0909 =
= 99.999 ≅ 100

(a) To neutralize a loss of 10% we need a larger (in %) gain of approximately 11.11%.
(b) To neutralize a gain of 10% we need a smaller (in %) loss of approximately 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 of solution of salt in water with the concentration of salt 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 100 gram of solution contained 1% of salt, the mass of salt was
msalt = 100·0.01 = 1 (gram)
Hence, the mass of water was
mwater = 100 − 1 = 99 (gram)

msalt = 1 (gram)
mwater = 99 (gram)

After some water has evaporated, the glass contained the same amount salt, that is 1 gram.
Since we know that it constitutes 2% of a solution, the total mass of solution must be 50 gram because
1/0.02 = 50 (gram)

To make 50 gram of solution with 1 gram of salt, we need 49 gram of pure water.