RD Sharma解决方案的第8类–第14章复利–练习14.4 |套装2

第14章复利–练习14.4 |套装1

问题11：四年中雇用了6400名工人建造了一座河桥。在第一年年底，裁员了25％。在第二年年底，裁员了25％。但是，为了及时完成该项目，在第三年末，工人人数增加了25％。第四年有多少工人在工作?

解决方案：

We have,

Initial number of workers are = 6400

At the end of first year = 25% retrenched

At the end of second year = 25% retrenched

At the end of third year = 25% increased

By using the formula,

A = P (1 + $\frac{R}{100}$)

Substituting the values, we have

x = 6400 (1 – $\frac{25}{100}$) (1 – $\frac{25}{100}$) (1 + $\frac{25}{100}$)

= 6400 ($\frac{75}{100}$) ($\frac{75}{100}$) ($\frac{125}{100}$)

= 6400 (0.75) (0.75) (1.25)

= 4500

Therefore,

Number of workers working during the fourth year is 4500.

问题12.阿曼开办了一家工厂，初期投资为100000卢比。第一年，他蒙受了5％的损失。但是，在第二年，他获得了10％的利润，而在第三年，该利润上升到了12％。计算他整个三年的净利润。

解决方案：

We have,

Initial investment by Aman = Rs.100000

In first year = incurred a loss of 5%

In second year = earned a profit of 10%

In third year = earned a profit of 12 %

By using the formula,

A = P (1 + $\frac{R}{100}$)

Substituting the values, we have

x = 100000 (1 – $\frac{5}{100}$) (1 + $\frac{10}{100}$) (1 + $\frac{12}{100}$)

= 100000 ($\frac{95}{100}$) ($\frac{110}{100}$) ($\frac{112}{100}$)

= 100000 (0.95) (1.1) (1.12)

= 117040

Therefore,

Aman’s net profit for entire three years is 117040 – 100000 = Rs 17040.

问题13：一个城镇的人口以每年每千人40个的速度增长。如果现在的人口是175760，那么三年前的人口是多少。

解决方案：

We have,

Annul increase rate of population of town = 40/1000× 100 = 4%

Present population of town = 175760

So, let the population of town 3 years ago be = x

By using the formula,

A = P (1 + $\frac{R}{100}$)

Substituting the values, we have

175760 = x (1 + $\frac{4}{100}$) (1 + $\frac{4}{100}$) (1 + $\frac{4}{100}$)

175760 = x ($\frac{104}{100}$) ($\frac{104}{100}$ ($\frac{104}{100}$

175760 = x (1.04) (1.04) (1.04)

175760 = 1.124864x

x = 175760/1.124864

= 156250

Therefore,

Population 3 years ago was 156250.

问题14. 1996年，一家mixi公司的人口为8000个mixix。由于需求增加，它在接下来的两年中将其产量提高了15％，并且在两年后，其需求量减少了5％。 3年后将生产什么?

解决方案：

We have,

Population of mixi company in 1996 was = 8000 mixies

Production growth rate in next 2 years is = 15 %

Decrease rate in 3rd year is = 5%

By using the formula,

A = P (1 + $\frac{R}{100}$)

Substituting the values, we have

x = 8000 (1 + $\frac{15}{100}$) (1 + $\frac{15}{100}$) (1 – $\frac{5}{100}$)

= 8000 ($\frac{115}{100}$0) ($\frac{115}{100}$) ($\frac{95}{100}$)

= 8000 (1.15) (1.15) (0.95)

= 10051

Therefore,

Production after three years will be 10051 mixies.

问题15.一个城市的人口每年以每年年初增加4％的速度增长。如果1999年的人口为6760000，请在(1)2001(ii)1997中找到城市的人口。

解决方案：

We have,

Annual increase rate of population of city is = 4%

Population in 1999 was = 6760000

(i) Population of the city in 2001 (2 years after)

By using the formula,

A = P (1 + $\frac{R}{100}$)

Substituting the values, we have

x = 6760000 (1 + $\frac{4}{100}$) (1 + $\frac{4}{100}$)

= 6760000 ($\frac{104}{100}$) ($\frac{104}{100}$)

= 6760000 (1.04) (1.04)

= 7311616

Therefore,

Population in the year 2001 is 7311616

(ii) Population of city in 1997 (2 years ago)

By using the formula,

A = P (1 + $\frac{R}{100}$)

Substituting the values, we have

x = 6760000 (1 – $\frac{4}{100}$) (1 – $\frac{4}{100}$)

= 6760000 ($\frac{96}{100}$) ($\frac{96}{100}$)

= 6760000 (0.96) (0.96)

= 6230016

Therefore,

Population in the year 1997 was 6230016.

问题16. Jitendra通过投资卢比建立了一家工厂。 250万。连续两年的利润分别为5％和10％。如果每年的利润都来自上一年的资本，请计算他的总利润。

解决方案：

We have,

Initial investment by Jitendra was = Rs 2500000

Profit in first 2 successive years were = 5% and 10%

Final investment after two successive profits = 2500000 (1 + $\frac{5}{100}$) (1 + $\frac{10}{100}$)

= 2500000 ($\frac{105}{100}$) ($\frac{110}{100}$)

= 2500000 (1.05) (1.1)

= 2887500

Therefore,

Jitendra total profit is = 2887500 – 2500000 = Rs 387500.