1991 AMC 12 第 9 题

Question

从时刻

t=0

到

t=1

，某种群增长了

i\%

；从时刻

t=1

到

t=2

，该种群又增长了

j\%

。因此，从时刻

t=0

到

t=2

，该种群增长了

From time

t=0

to time

t=1

a population increased by

i\%,

and from time

t=1

to time

t=2

the population increased by

j\%.

Therefore, from time

t=0

to time

t=2

the population increased by

(i+j)\%

ij\%

(i+ij)\%

\left(i+j+\frac{ij}{100}\right)\%

\left(i+j+\frac{i+j}{100}\right)\%

用乘法因子表示两次增长

Represent the two increases by multiplication factors

展开

\left(1+\frac{i}{100}\right)\left(1+\frac{j}{100}\right)

Expand

\left(1+\frac{i}{100}\right)\left(1+\frac{j}{100}\right)

总增长因子为

\begin{aligned} &\left(1+\frac{i}{100}\right) \left(1+\frac{j}{100}\right)\\ &\qquad=1+\frac{i+j}{100} +\frac{ij}{10000} \end{aligned}\text{。}

因此，增长的百分数为

i+j+\frac{ij}{100}

。

因此正确答案为 D。

The combined growth factor is

\begin{aligned} &\left(1+\frac{i}{100}\right) \left(1+\frac{j}{100}\right)\\ &\qquad=1+\frac{i+j}{100} +\frac{ij}{10000}. \end{aligned}

Hence the percent increase is

i+j+\frac{ij}{100}.

Thus the correct answer is D.

其他年份的第 9 题