RD Sharma解决方案的第8类–第14章复利–练习14.4 |套装1

问题1.城镇的现有人口为28000。如果以每年5％的速度增长，那么两年后的人口将是多少?

解决方案：

We have,

Present population of town is = 28000

Rate of increase in population is = 5% per annum

Number of years = 2

By using the formula,

A = P (1 + $\frac{R}{100}$)ⁿ

Substituting the values, we have 

Population of town after 2 years = 28000 (1 + $\frac{5}{100}$)²

= 28000 (1.05)²

= 30870

Therefore,

Population of town after 2 years will be 30870.

问题2.一个城市的人口为125,000。如果年出生率和死亡率分别为5.5％和3.5％，则计算3年后的城市人口。

解决方案：

We have,

Population of city (P) = 125000

Annual birth rate R1= 5.5 %

Annual death rate R2 = 3.5 %

Annual increasing rate R = (R1 – R2) = 5.5 – 3.5 = 2 %

Number of years = 3

By using the formula,

A = P (1 + $\frac{R}{100}$)ⁿ

Substituting the values, we have 

So, population after two years is = 125000 (1 + $\frac{2}{100}$)³

= 125000 (1.02)³

= 132651

Therefore,

Population after 3 years will be 132651.

问题3.城镇的现有人口为25000。第一年，第二年和第三年分别增长4％，5％和8％。 3年后查找其人口。

解决方案：

We have,

Present population is = 25000

First year growth R1 = 4%

Second year growth R2 = 5%

Third year growth R3 = 8%

Number of years = 3

By using the formula,

A = P (1 + $\frac{R}{100}$)ⁿ

Substituting the values, we have 

So, population after three years = P (1 + $\frac{R_1}{100}$) (1 + $\frac{R_2}{100}$) (1 + $\frac{R_3}{100}$)

= 25000(1 + $\frac{4}{100}$) (1 + $\frac{5}{100}$) (1 + $\frac{8}{100}$0)

= 25000 (1.04) (1.05) (1.08)

= 29484

Therefore,

Population after 3 years will be 29484.

问题4：三年前，一个城镇的人口为50000。如果连续三年的年增长率分别为4％，5％和3％，则求出当前人口。

解决方案：

We have,

Three years ago population of town was = 50000

Annual increasing in 3 years = 4%,5%, 3% respectively

So, let present population be = x

By using the formula,

A = P (1 + $\frac{R}{100}$)ⁿ

Substituting the values, we have 

x = 50000 (1 + $\frac{4}{100}$) (1 + $\frac{5}{100}$) (1 + $\frac{3}{100}$)

= 50000 (1.04) (1.05) (1.03)

= 56238

Therefore,

Present population of the town is 56238.

问题5：一个村庄的人口以每年5％的速度持续增长。如果它的当前人口是9261，那么3年之前是什么?

解决方案：

We have,

Present population of town is = 9261

Continuous growth of population is = 5%

So, let population three years ago be = x

By using the formula,

A = P (1 + $\frac{R}{100}$)ⁿ

Substituting the values, we have 

9261 = x (1 + $\frac{5}{100}$) (1 + $\frac{5}{100}$) (1 + $\frac{5}{100}$)

9261 = x (1.05) (1.05) (1.05)

= 8000

Therefore,

Present population of the town is 8000.

问题6.在一家工厂中，踏板车的产量在3年内从40000上升到46305。求出踏板车产量的年增长率。

解决方案：

We have,

Initial production of scooters is = 40000

Final production of scooters is = 46305

Time = 3 years

Let annual growth rate be = R%

By using the formula,

A = P (1 + $\frac{R}{100}$)ⁿ

Substituting the values, we have 

46305 = 40000 (1 + $\frac{R}{100}$) (1 + $\frac{R}{100}$) (1 + $\frac{R}{100}$)

46305 = 40000 (1 + $\frac{R}{100}$)³

(1 + $\frac{R}{100}$)³ = 46305/40000

= 9261/8000

= (21/20)³

1 + $\frac{R}{100}$ = 21/20

$\frac{R}{100}$ = 21/20 – 1

$\frac{R}{100}$ = (21-20)/20

= 1/20

R = 100/20

= 5

Therefore,

Annual rate of growth of the production of scooters is 5%.

问题7：某城市人口的年增长率为8％。如果它的当前人口是196830，那三年前是什么?

解决方案：

We have,

Annual growth rate of population of city is = 8%

Present population of city is = 196830

Let population of city 3 years ago be = x

By using the formula,

A = P (1 + $\frac{R}{100}$)ⁿ

Substituting the values, we have 

196830 = x (1 + $\frac{8}{100}$) (1 + $\frac{8}{100}$) (1 + $\frac{8}{100}$)

196830 = x (27/25) (27/25) (27/25)

196830 = x (1.08) (1.08) (1.08)

196830 = 1.259712x

x = 196830/1.259712

= 156250

Therefore,

Population 3 years ago was 156250.

问题8.城镇人口以每千人50的速度增长。 2年后它的人口将是22050。找到它的当前人口。

解决方案：

We have,

Growth rate of population of town is = 50/1000×100 = 5%

Population after 2 years is = 22050

So, let present population of town be = x

By using the formula,

A = P (1 + $\frac{R}{100}$)ⁿ

Substituting the values, we have 

22050 = x (1 + $\frac{5}{100}$) (1 + $\frac{5}{100}$)

22050 = x ($\frac{105}{100}$) ($\frac{105}{100}$)

22050 = x (1.05) (1.05)

22050 = 1.1025x

x = 22050/1.1025

= 20000

Therefore,

Present population of the town is 20000.

问题9.培养物中的细菌数量在第一小时内增长10％，在第二小时内下降8％，在第三小时内再次增长12％。如果样品中的细菌数为13125000，3小时后细菌数是多少?

解决方案：

Given details are,

Count of bacteria in sample is = 13125000

The increase and decrease of growth rates are = 10%, -8%, 12%

So, let the count of bacteria after 3 hours be = x

By using the formula,

A = P (1 + $\frac{R}{100}$)ⁿ

Substituting the values, we have 

x = 13125000 (1 + $\frac{10}{100}$) (1 – $\frac{8}{100}$) (1 + $\frac{12}{100}$)

x = 13125000 (110/100) (92/100) (112/100)

x = 13125000 (1.1) (0.92) (1.12)

= 14876400

Therefore,

Count of bacteria after three hours will be 14876400.

问题10：某个城市的人口在1998年的最后一天为72,000。在第二年，该人口增加了7％，但由于流行病，在第二年减少了10％。到2000年年底，它的人口是多少?

解决方案：

We have,

Population of city on last day of year 1998 = 72000

Increasing rate (R) in 1999 = 7%

Decreasing rate (R) in 2000 = 10 %

By using the formula,

A = P (1 + $\frac{R}{100}$)ⁿ

Substituting the values, we have 

x = 72000 (1 + $\frac{7}{100}$) (1 – $\frac{10}{100}$)

= 72000 (107/100) (90/100)

= 72000 (1.07) (0.9)

= 69336

Therefore,

Population at the end of the year 2000 will be 69336.

第14章复利–练习14.4 |套装2