8类RD Sharma解决方案–第14章复利–练习14.3 |套装2

第14章复利–练习14.3 |套装1

问题15：如果Rs，则求出年利率百分比。 2000卢比。一年半的2315.25，每个月六个月复利一次。

解决方案：

We have,

Principal = Rs 2000

Amount = Rs 2315.25

Time = 1 ½ years = 3/2 years

Let rate be = R % per annum

By using the formula,

A = P (1 + $\frac{R}{100}$)ⁿ

Substituting the values, we have

2315.25 = 2000 (1 + $\frac{R}{100}$)^3/2

(1 + $\frac{R}{100}$)^3/2 = 2315.25/2000

(1 + $\frac{R}{100}$)^3/2 = (1.1576)

(1 + $\frac{R}{100}$) = 1.1025

$\frac{R}{100}$ = 1.1025 – 1

= 0.1025 × 100

= 10.25

Therefore,

Required Rate is 10.25% per annum.

问题16。求出如果每年加息的话，一笔钱将在3年内增加一倍的利率。

解决方案：

We have,

Time = 3 years

Let rate be = R %

Also principal be = P

So, amount becomes = 2P

By using the formula,

A = P (1 + $\frac{R}{100}$)ⁿ

Substituting the values, we have

2P = P (1 + $\frac{R}{100}$)³

(1 + $\frac{R}{100}$)³ = 2

(1 + $\frac{R}{100}$) = 2^1/3

1 + $\frac{R}{100}$ = 1.2599

$\frac{R}{100}$ = 1.2599-1

= 0.2599

R = 0.2599 × 100

= 25.99

Therefore,

Required Rate is 25.99% per annum.

问题17：如果两年期加息，则求出一笔钱在两年内将变成原始金额四倍的利率

解决方案：

We have,

Time = 2 years = 2×2 = 4 half years

Let rate = R % per annum = R/2% half years

Let principal be = P

So, Amount becomes = 4P

By using the formula,

A = P (1 + $\frac{R}{100}$)ⁿ

Substituting the values, we have

4P = P (1 + $\frac{R}{100 \times 2}$)⁴

(1 + $\frac{R}{200}$)⁴ = 4

(1 + $\frac{R}{200}$) = 4^1/4

1 + $\frac{R}{200}$ = 1.4142

$\frac{R}{200}$ = 1.4142-1

= 0.4142

R = 0.4142 × 200

= 82.84%

Therefore,

Required Rate is 82.84% per annum.

问题18.一定金额等于卢比。 2年内5832的年利率为8％。求和。

解决方案：

We have,

Amount = Rs 5832

Time = 2 years

Rate = 8%

Let principal be = P

By using the formula,

A = P (1 + $\frac{R}{100}$)ⁿ

Substituting the values, we have

5832 = P (1 + $\frac{8}{100}$)²

5832 = P (1.1664)

P = 5832/1.1664

= 5000

Therefore,

Required sum is Rs 5000.

问题19：按年利率7.5％计算的某笔总金额的复利和单利之间的差额为2年，每年为7.5卢比。 360.找到总和。

解决方案：

We have,

Time = 2 years

Rate = 7.5 % per annum

Let principal = Rs P

Compound Interest (CI) – Simple Interest (SI) = Rs 360

C.I – S.I = Rs 360

By using the formula,

P [(1 + $\frac{R}{100}$)ⁿ – 1] – (PTR)/100 = 360

Substituting the values, we have

P [(1 + $\frac{7.5}{100}$)² – 1] – (P(2)(7.5))/100 = 360

P[249/1600] – (3P)/20 = 360

249/1600P – 3/20P = 360

(249P-240P)/1600 = 360

9P = 360 × 1600

P = 576000/9

= 64000

Therefore,

The sum is Rs 64000.

问题20.一定金额的单利和复利之间的差额，以3年卢比计，年利率623％。 46.确定总和。

解决方案：

We have,

Time = 3 years

Rate = 6$\frac{2}{3}$ % per annum = 20/3%

Let principal = Rs P

Compound Interest (CI) – Simple Interest (SI) = Rs 46

C.I – S.I = Rs 46

By using the formula,

P [(1 + $\frac{R}{100}$)ⁿ – 1] – (PTR)/100 = 46

Substituting the values, we have

P [(1 + $\frac{20}{3\times100}$)³ – 1] – (P(3)(20/3))/100 = 46

P[(1 + $\frac{20}{300}$)³ – 1] – P/5 = 46

P[721/3375] – P/5 = 46

721/3375P – 1/5P = 46

(721P-675P)/3375 = 46

46P = 46 × 3375

46P = 46 × 3375/46

= 3375

Therefore, 

The sum is Rs 3375.

问题21：Ishita投资了Rs。 12000，年利率5％。她收到了一定数量的卢比。 n年后的13230。找到n的值。

解决方案：

We have,

Principal = Rs 12000

Amount = Rs 13230

Rate = 5% per annum

Let time = T years

By using the formula,

A = P (1 + $\frac{R}{100}$)ⁿ

Substituting the values, we have

13230 = 12000 (1 + $\frac{5}{100}$)^T

13230 = 12000 ($\frac{105}{100}$)^T

(21/20)^T = 13230/12000

(21/20)^T = 441/400

(21/20)^T = (21/20)²

So on comparing both the sides, n = T = 2

Therefore, 

Time required is 2 years.

问题22.以每年的百分比率计算的总和为Rs。 4000卢比的复利。 2年内410吗?

解决方案：

We have,

Principal = Rs 4000

Time = 2 years

CI = Rs 410

Rate be = R% per annum

By using the formula,

CI = P [(1 + $\frac{R}{100}$)ⁿ – 1]

Substituting the values, we have

410 = 4000 [(1 + $\frac{R}{100}$)² – 1]

410 = 4000 (1 + $\frac{R}{100}$)² – 4000

410 + 4000 = 4000 (1 + $\frac{R}{100}$)²

(1 + $\frac{R}{100}$)² = 4410/4000

(1 + $\frac{R}{100}$)² = 441/400

(1 + $\frac{R}{100}$)² = (21/20)²

By canceling the powers on both the sides,

1 + $\frac{R}{100}$ = 21/20

$\frac{R}{100}$ = 21/20 – 1

= (21-20)/20

= 1/20

R = 100/20

= 5

Therefore,

Required Rate is 5% per annum.

问题23.以每年2％的比例每年存入的资金的总和为Rs。 10年末(2年末)。找到存入的总和。

解决方案：

We have,

Time = 2years

Amount = Rs 10404

Rate be = 2% per annum

Let principal be = Rs P

By using the formula,

A = P [(1 + $\frac{R}{100}$)ⁿ

Substituting the values, we have

10404 = P [(1 + $\frac{2}{100}$)²]

10404 = P [1.0404]

P = 10404/1.0404

= 10000

Therefore, 

Required sum is Rs 10000.

问题24.多少时间总和为Rs。 1600卢比。 1852.20，年利率为5％?

解决方案：

We have,

Principal = Rs 1600

Amount = Rs 1852.20

Rate = 5% per annum

Let time = T years

By using the formula,

A = P (1 + $\frac{R}{100}$)ⁿ

Substituting the values, we have

1852.20 = 1600 (1 + $\frac{5}{100}$)^T

1852.20 = 1600 ($\frac{105}{100}$)^T

(21/20)^T = 1852.20/1600

(21/20)^T = 9261/8000

(21/20)^T = (21/20)³

So on comparing both the sides, n = T = 3

Therefore,

Time required is 3 years.

问题25.总和的卢比率为多少? 1000卢比。 1102.50在2年后以复利计算吗?

解决方案：

We have,

Principal = Rs 1000

Amount = Rs 1102.50

Rate = R% per annum

Let time = 2 years

By using the formula,

A = P (1 + $\frac{R}{100}$)ⁿ

Substituting the values, we have

1102.50 = 1000 (1 + $\frac{R}{100}$)²

(1 + $\frac{R}{100}$)² = 1102.50/1000

(1 + $\frac{R}{100}$)² = 4410/4000

(1 +$\frac{R}{100}$)² = (21/20)²

1 + $\frac{R}{100}$ = 21/20

$\frac{R}{100}$= 21/20 – 1

= (21-20)/20

= 1/20

R = 100/20

= 5

Therefore,

Required Rate is 5%.

问题26. Rs的复利。 1800卢比，在每年10％的特定时间段内为Rs。 378.以年为单位查找时间。

解决方案：

We have,

Principal = Rs 1800

CI = Rs 378

Rate = 10% per annum

Let time = T years

By using the formula,

CI = P [(1 + $\frac{R}{100}$)ⁿ – 1]

Substituting the values, we have

378 = 1800 [(1 + $\frac{10}{100}$)^T – 1]

378 = 1800 [($\frac{110}{100}$)^T – 1]

378 = 1800 [($\frac{11}{10}$)^T – 1800

378 + 1800 = 1800 [($\frac{11}{10}$)^T

(11/10)^T = 2178/1800

(11/10)^T = 726/600

(11/10)^T = 121/100

(11/10)^T = (11/10)²

So on comparing both the sides, n = T = 2

Therefore,

Time required is 2 years.

问题27.多少钱等于卢比。两年期年利率为6¾％的45582.25，每年加息

解决方案：

We have,

Time = 2years

Amount = Rs 45582.25

Rate be = 6 ¾ % per annum = 27/4%

Let principal be = Rs P

By using the formula,

A = P [(1 + $\frac{R}{100}$)ⁿ

Substituting the values, we have

45582.25 = P [(1 + 27/4×100)²]

45582.25 = P (1 + $\frac{27}{100}$)²

45582.25 = P ($\frac{427}{400}$)²

45582.25 = P × 427/400 × 427/400

P = (45582.25 × 400 × 400) / (427×427)

P = 7293160000/182329

= 40000

Therefore,

Required sum is Rs 40000.

问题28.金额为卢比。 453690在2年中，每年以6.5％的年增长率复合。求和。

解决方案：

We have,

Time = 2years

Amount = Rs 453690

Rate be = 6.5 % per annum

Let principal be = Rs P

By using the formula,

A = P [(1 + $\frac{R}{100}$)ⁿ

Substituting the values, we have

453690 = P [(1 + $\frac{6.5}{100}$)²]

453690 = P ($\frac{106.5}{100}$)²

453690 = P × 106.5/100 × 106.5/100

P = (453690 × 100 × 100) / (106.5×106.5)

P = 4536900000/11342.25

= 400000

Therefore,

Required sum is Rs 400000.