问题1.以每年5％的利率计算，两年复利的年复利金额为164卢比。
解决方案：
We have,
Rate = 5 % per annum
Compound Interest (CI) = ₹164
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
164 = P (1 + R/100) n – P
Substituting the values, we have
= P [(1 + R/100)n – 1]
= x [(1 + 5/100)2 – 1]
= x [(105/100)2 – 1]
164 = x ((1.05)2 – 1)
x = 164 / ((1.05)2 – 1)
= 164/0.1025
= ₹1600
Therefore
The required sum is ₹1600.
问题2。如果本金以10％的利率在两年中每年复利为₹ 210，则请求本金。
解决方案：
We have,
Rate = 10 % per annum
Compound Interest (CI) = ₹210
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
210 = P (1 + R/100)n – P
Substituting the values, we have
= P [(1 + R/100)n – 1]
= x [(1 + 10/100)2 – 1]
= x [(110/100)2 – 1]
210 = x ((1.1)2 – 1)
x = 210 / ((1.1)2 – 1)
= 210/0.21
= ₹1000
Therefore,
The required sum is ₹1000.
问题3.金额为₹756.25，两年内每年10％，每年复利。求和。
解决方案：
We have,
Rate = 10 % per annum
Amount = ₹756.25
Time (t) = 2 years
By using the formula,
A = P (1 + R/100)n
Substituting the values, we have
756.25 = P (1 + 10/100)2
P = 756.25/(1 + 10/100)2
= 756.25/1.21
= 625
Therefore,
The principal amount is ₹625.
问题4.如果利率为每年12.5％，每半年复利一次，则18个月后的总金额为₹4913?
解决方案：
We have,
Rate = 12 ½% per annum = 25/2% = 25/2/2 = 25/4% half yearly
Amount = ₹4913
Time (t) = 18months = 18/12years = 3/2 × 2 = 3 half years
By using the formula,
A = P (1 + R/100)n
Substituting the values, we have
4913 = P (1 + 25/4 ×100)3
P = 4913 / (1 + 25/400)3
= 4913/1.19946
= 4096
Therefore,
The principal amount is ₹4096.
问题5.某年利率为15％的一定金额的复利和单利之间的三年期之差为₹283.50。求和。
解决方案：
We have,
Rate = 15 % per annum
Compound Interest (CI) – Simple Interest (SI)= ₹283.50
Time (t) = 3 years
By using the formula,
CI – SI = 283.50
P [(1 + R/100)n – 1] – (PTR)/100 = 283.50
Substituting the values, we have
P [(1 + 15/100)3 – 1] – (P(3)(15))/100 = 283.50
P[1.520 – 1] – (45P)/100 = 283.50
0.52P – 0.45P = 283.50
0.07P = 283.50
P = 283.50/0.07
= 4000
Therefore,
The sum is ₹4000.
问题6.拉查纳(Rachana)以每年15％的利率借入一定金额。如果她在两年末支付了每年1290卢比的利息，请找到她的借贷金额。
解决方案：
We have,
Rate = 15 % per annum
Time = 2 years
CI = Rs 1290
By using the formula,
CI = P [(1 + R/100)n – 1]
Substituting the values, we have
1290 = P [(1 + 15/100)2 – 1]
1290 = P [0.3225]
P = 1290/0.3225
= 4000
Therefore,
The sum is ₹4000.
问题7. 2000卢比的利息每年以4％的年利率递增。找出复利为₹163.20的期间。
解决方案：
We have,
Rate = 4 % per annum
CI = ₹163.20
Principal (P) = Rs 2000
By using the formula,
CI = P [(1 + R/100)n – 1]
Substituting the values, we have
163.20 = 2000[(1 + 4/100)n – 1]
163.20 = 2000[(1.04)n -1]
163.20 = 2000 × (1.04)n – 2000
163.20 + 2000 = 2000 × (1.04)n
2163.2 = 2000 × (1.04)n
(1.04)n = 2163.2/2000
(1.04)n = 1.0816
(1.04)n = (1.04)2
So on comparing both the sides, n = 2
Therefore,
Time required is 2 years.
问题8.以每年10％的复利率计算， 5000卢布在多长时间内等于6655卢比?
解决方案：
We have,
Rate = 10% per annum
A = ₹6655
Principal (P) = ₹5000
By using the formula,
A = P (1 + R/100)n
Substituting the values, we have
6655 = 5000 (1 + 10/100)n
6655 = 5000 (11/10)n
(11/10)n = 6655/5000
(11/10)n = 1331/1000
(11/10)n = (11/10)3
So on comparing both the sides, n = 3
Therefore,
Time required is 3 years.
问题9.以每年8％的利率每半年复利一次， 4400卢比会变成4576卢比吗?
解决方案：
We have,
Rate = 8% per annum = 8/2 = 4% (half yearly)
A = Rs 4576
Principal (P) = ₹4400
Let n be ‘2T’
By using the formula,
A = P (1 + R/100)n
Substituting the values, we have
4576 = 4400 (1 + 4/100)2T
4576 = 4400 (104/100)2T
(104/100)2T = 4576/4400
(104/100)2T= 26/25
(26/25)2T = (26/25)1
So on comparing both the sides, n = 2T = 1
Therefore,
Time required is 1/2 year.
问题10. SI和CI在某笔金额为2年的年利率为每年4％时的差额为₹20。找到这笔款项。
解决方案：
We have,
Rate = 4 % per annum
Time = 2 years
Compound Interest (CI) – Simple Interest (SI)= ₹20
By using the formula,
CI – SI = 20
P [(1 + R/100)n – 1] – (PTR)/100 = 20
Substituting the values, we have
P [(1 + 4/100)2 – 1] – (P(2)(4))/100 = 20
P[51/625] – (2P)/25 = 20
51/625P – 2/25P = 20
(51P-50P)/625 = 20
P = 20 × 625
P = 20/7.918
= 12500
Therefore
The sum is ₹12500.
问题11。 1000卢比。 1331年利率为10％，复利?
解决方案：
We have,
Principal = Rs 1000
Amount = Rs 1331
Rate = 10% per annum
Let time = T years
By using the formula,
A = P (1 + R/100)n
Substituting the values, we have
1331 = 1000 (1 + 10/100)T
1331 = 1000 (110/100)T
(11/10)T = 1331/1000
(11/10)T = (11/10)3
So on comparing both the sides, n = T = 3
Therefore,
Time required is 3 years.
问题12：卢比以每年多少利率计复利。 640卢比。 2年后达到774.40?
解决方案：
We have,
Principal = Rs 640
Amount = Rs 774.40
Time = 2 years
Let rate = R%
By using the formula,
A = P (1 + R/100)n
Substituting the values, we have
774.40 = 640 (1 + R/100)2
(1 + R/100)2 = 774.40/640
(1 + R/100)2 = 484/400
(1 + R/100)2 = (22/20)2
By canceling the powers on both sides,
(1 + R/100) = (22/20)
R/100 = 22/20 – 1
= (22-20)/20
= 2/20
= 1/10
R = 100/10
= 10%
Therefore,
Required Rate is 10% per annum.
问题13：如果Rs，则求出年利率百分比。 2000卢比。在1½年内达到2662，是否每半年复利一次?
解决方案：
We have,
Principal = Rs 2000
Amount = Rs 2662
Time = 1 ½ years = 3/2 × 2 = 3 half years
Let rate be = R% per annum = R/2 % half yearly
By using the formula,
A = P (1 + R/100)n
Substituting the values, we have
2662 = 2000 (1 + R/2×100)3
(1 + R/200)3 = 1331/1000
(1 + R/100)3 = (11/10)3
By canceling the powers on both sides,
(1 + R/200) = (11/10)
R/200 = 11/10 – 1
= (11-10)/10
= 1/10
R = 200/10
= 20%
Therefore,
Required Rate is 20% per annum.
问题14.卡马拉从拉丹以一定利率向拉丹借了一笔两年的单利。她以同样的比率将这笔贷款借给了Hari，为期两年，计有复利。两年结束时，她收到了卢比。 210作为复利，但支付了卢比。 200只作为简单的兴趣。找到总和和利率。
解决方案：
We have,
C.I that Kamala receives = Rs 210
S.I that Kamala paid = Rs 200
Time = 2 years
So,
We know, SI = PTR/100
= P×2×R/100
P×R = 10000 ………….. Equation 1
CI = A – P
CI = P [(1 + R/100)n – 1]
Substituting the values, we have
210 = P [(1 + R/100)2 – 1]
210 = P (12 + R2/1002 + 2(1)(R/100) – 1) (by using the formula (a+b)2)
210 = P (1 + R2/10000 + R/50 – 1)
210 = P (R2/10000 + R/50)
210 = PR2/10000 + PR/50
We know PR = 10000 from Equation 1
210 = 10000R/10000 + 10000/50
210 = R + 200
R = 210 – 200
= 10%
In Equation 1, PR = 10000
P = 10000/R
= 10000/10
= 1000
Therefore,
Required sum is Rs 1000.