YkUea
Per dvejus metus miestelio gyventojų skaičius padidėjo 44%. Keliais procentais padidėdavo miestelio gyventojų skaičius kiekvienais metais, jei šis procentas ir pirmaisiais, ir antraisiais metais buvo toks pat?
Sprendimas:
Sudėtinių procentų formulė $$S_{n} = S_0\cdot (1+\frac{p}{100})^{n}$$
n = 2 (du metai).
$$S_{n} = 1.44\cdot S_0$$, nes skaičius padidėjo 44 proc.
Sudarome lygtį
1.44* S_0 =
S_0* (1+
)^2
S_0* (1+
| p |
| / 100 |
|
1.44* S_0 = S_0* (1+
)^2$$1.44\cdot S_0$$ = $$S_0\cdot (1+\frac{p}{100})^{2}$$
| p |
| / 100 |
| 1.44* S_0 |
| / S_0 |
| p |
| / 100 |
$${\normalsize \frac{1.44\cdot S_0}{S_0}}$$ = $${\normalsize 1.44}$$
1.44 = (1+
)^2$$1.44$$ = $$(1+\frac{p}{100})^{2}$$
| p |
| / 100 |
saknis(1.44) = saknis( (1+
)^2)$$\sqrt {1.44}$$ = $$\sqrt {(1+\frac{p}{100})^{2}}$$
1.44) = saknis( (1+ | p |
| / 100 |
$${\normalsize \sqrt {(1+\frac{p}{100})^{2}}}$$ = $${\normalsize (1+\frac{p}{100})}$$
saknis(1.44) = 1+
$$\sqrt {1.44}$$ = $$1+\frac{p}{100}$$
1.44) = 1+ | p |
| / 100 |
$${\normalsize \sqrt {1.44}}$$ = $${\normalsize 1.2}$$
1.2 = 1+
$$1.2$$ = $$1+\frac{p}{100}$$
| p |
| / 100 |
1.2-1 =
$$1.2-1$$ = $$\frac{p}{100}$$
| p |
| / 100 |
$${\normalsize 1.2-1}$$ = $${\normalsize 0.2}$$
0.2 =
$$0.2$$ = $$\frac{p}{100}$$
| p |
| / 100 |
0.2* 100 = p$$0.2\cdot 100$$ = $$p$$
$${\normalsize 0.2\cdot 100}$$ = $${\normalsize 20}$$
20 = p$$20$$ = $$p$$
p = 20$$p$$ = $$20$$
$$1.44\cdot S_0$$ = $$S_0\cdot (1+\frac{p}{100})^{2}$$
$$1.44$$ = $$(1+\frac{p}{100})^{2}$$
$$\sqrt {1.44}$$ = $$\sqrt {(1+\frac{p}{100})^{2}}$$
$$1.2$$ = $$1+\frac{p}{100}$$
$$0.2$$ = $$\frac{p}{100}$$
$$0.2\cdot 100$$ = $$p$$
$$p$$ = $$20$$
Atsakymas: 20%