Compound Interest

Let us take sum be y
A = P (1 + (rn)/100)
19500 = y (1 + 30/100)
19500 = y (130/100)
y = 19500 × (10/13) = 15000
SI = A - P = 19500 - 15000
= 4500
Given,
CI = SI + 2100 = 4500 + 2100
= 6600
Total amount = P + I = 15000 + 6600
= 21600
A = P (1 + r/100)ⁿ
21600 = 15000 (1 + x/100)²
(1 + x/100)² = (6/5)²
1 + x/100 = 6/5
x/100 = 1/5
x = 20%

Here, sum = A₁(A₁/A₂)ⁿ
Where, A₁ = ₹ 8000, A₂ = ₹ 10000, n = 3 years
Sum = 8000(8000/10000)³
= ₹ 4096

Let×= equal instalment at the end of each year
Now 1st year,
P = ₹ 1100
Interest at the end of 1st year = (1100 × 20 × 1)/100 = ₹ 22
Now, at the beginning of 2nd year,
P = ₹ (1100 + 220 - x) = ₹ (1320 - x)
Interest at the end of 2nd year
= ((1320 - x) × 20 × 1)/100 = 264 - x/5
Now amount remaining after 2 years
= (1320 - x) + (264 - x/5) -×= 0
⇒ 2x + x/5 = 1320 + 264
⇒ 11x/5 = 1584
⇒ x = 720

We know, Sum of money amounts in 2 yearss = ₹ 2420
Sum of money amounts in 3 yearss = ₹ 2662
So, interest on ₹2420 for 1 yr = 2662 - 2420
= ₹ 242
SI = PNR/100
242 = (2420×1×R)/100
R = (242×100)/2420
R = 10%:

Rate = r = 8% p.a = 4% per half-year
Time = n = 3/2 years = 3 half years
Formula for amount, A = P(1+(r/100))ⁿ
= 37500 (1+(4/100))³
= 37500 (1+($$\frac{1}{2}$$5))³
= 37500 (26/25)³
= ₹ 42812.4
∴ amount A = ₹ 42812.4

Given
Here, P = ₹10,00,000, R1 = 11 , R2 = 12, R3 = 13.
Each rate of interest is calculated for one year.
Hence, N = 1 year.
Amount after 3 years,
= P(1 + R1/100) (1 + R2/100) (1 + R3/100)
= 10,00,000 × (1 + 11/100) × (1 + 12/100) × (1 + 13/100)
= 10,00,000 × (111/100) × (112/100) × (113/100)
= 111×112×113
= 14,04,816
Hence the total amount after 3 years is ₹14,04,816

Given: P = 220, R = 5%
Amount = P[1 + (R/100)]ⁿ
Amount = [200(1 + (5/100))³] + [200(1 + (5/100))²] + [200(1 + (5/100)]
= [200(1 + ($$\frac{1}{2}$$0))³] + [200(1 + ($$\frac{1}{2}$$0))²] + [200(1 + ($$\frac{1}{2}$$0)]
= [200(2$$\frac{1}{2}$$0)³] + [200(2$$\frac{1}{2}$$0)²] + [200(2$$\frac{1}{2}$$0)]
= [(200×9261)/8000] + [(200× 441)/400] + [210]
= 231.525 + 220.5 + 210
= ₹ 662.025
Therefore amount at the end of 3 years be ₹ 662.025.

Given, r = 10%
n = 2 years
Compound interest, C.I = ₹ 210
Compound Interest = Amount – Principal
⇒ C.I = P {[1 + (r/100)]^n - 1}
⇒ 210 = P {[1 + (10/100)]^2 - 1}
⇒ 210 = P {[1 + (1/10)]^2 - 1}
⇒ 210 = P {[(10 + 1)/10]^2 - 1}
⇒ 210 = P {[11/10]^2 - 1}
⇒ 210 = P {[121 / 100] – 1}
⇒ 210 = P {(121 – 100) / 100}
⇒ 210 = P {21 / 100}
⇒ P = (210 × 100) / 21
⇒ P = 1000 ₹
Thus, Principal = ₹ 1000

Let the rate of interest be r% per annum.
Amount = P[1 + (R/100)]ⁿ
Sum of money amounts to ₹ 4840 in 2 years,
4840 = P[1 + (r/100)]² ....(i)
Sum of money amounts to ₹ 5324 in 3 years,
5324 = P[1 + (r/100)]³ .....(ii)
Dividing equation (ii) by (i), we get
5324/4840 = 1 + (r/100)
(5324/4840) - 1 = (r/100)
(5324 - 4840)/4840 = (r/100)
484/4840 = r/100
r = (484/4840)×100
r = 10%
∴ rate of interest per annum is 10%.

Given
Principal : P = 20000 ₹
Rate of Interest : r = 6 %
Number of years : n = 2
Amount = P×(1 + r/100)^n
⇒ Amount = 20000×(1+6/100)^2
= 20000×(1+3/50)^2
= 20000×(53/50)×(53/50)
= 22472
∴ Amount received by Neel at the end of two years = ₹ 22472