Compound Interest

Principal : P = 20,000 ₹
Rate of Interest : r = 6 %
Number of years : n = 2
Amount = P×(1 + r/100)ⁿ
⇒ Amount = 20,000×(1+6/100)²
=20000×(1+3/50)²
=20000×(53/50)×(53/50)
=561800 / 25
= 22472
Thus, the Amount that Pandey receives at the end of 2 years= ₹22472

We know, CI: P - ₹ 1200; N - 2yrs; R - 10%
C.I. = P[(1+R/100)^N] – P
CI = 1200[(1+ 10/100)²] – 1200
= 1200[(11/10)² - 1]
= 1200[(121 - 100)/100]
= ₹ 252
We got from Q, SI is half of the CI
SI = 252/2 = ₹ 126
SI = PNR/100
126 = (P×3×8)/100
P = (126×100)/(3×8)
P = ₹ 525.

Given
Principal : P = 10000 ₹ Rate of Interest : r = 4 %
Number of years : n = 2
Amount = P×(1 + r/100)ⁿ
Amount = 10000×(1+4/100)²
= 10000×(1+$$\frac{1}{2}$$5)²
= 10000×(26/25)×(26/25)
= ₹ 10816
Amount to be paid after 2 years =₹ 10816:

We know, In SI: P - ₹ A; N - 2 years; R - 6%
In CI: P - ₹1500; N - 2 years; R - 20%
CI = P{[1 + (R/100)]^N - 1}
CI = 1500{[1 + (20/100)]² - 1}
= 1500{(6/5)² - 1}
= 1500[(36 - 25)/25]
= 1500(1$$\frac{1}{2}$$5)
= ₹ 660
If the ratio of SI and CI after 2 years is 6: 11,
SI/CI = 6/11
(PNR/100)/P{[1 + (R/100)]^N - 1} = 6/11
(A×2×6)/(660×100) = 6/11
A = (6×660×100)/(2×6×11)
A = ₹ 3000.

Principal = ₹ 4000, t= 2 years , rate of percent, r = 10 %
Amount = P(1 + r/100)^t = 4000×(1 + 10/100)^2 = 4000×(11/10)×(11/10) = 40×121 = 4840 ₹
Amount = Principal + CI ⇒ 4000 + CI = 4840 ⇒ CI = 840 ₹

Total Income = 67,280
After giving 50% salary to his wife the man is left with an amount = 33,640
Let's assume the man gave ₹×to A. Therefore B will get ₹ (33640 – x).
↙ 33640 ↘
14 years A 12 years B
x (33640 – x)
Now, as per the question A & B will be getting an equal amount with CI at 5% rate per year at the 18th year.
⇒×(1 + 5/100)⁴ = (33640 – x)[1+ 5/100] ⁶
⇒ x/(33640 – x) = (1 + 5/100 ) ⁶/(1 + 5/100) ⁴
⇒ x/(33640 – x) = = (2$$\frac{1}{2}$$0× 2$$\frac{1}{2}$$0)
⇒ 400×= 33640 × 441 – 441x
⇒ 841x = 33640 ×441
-⇒ x = 33640 × 441/841 = 40 × 441 = 17640/-
∴ at the time of divison of money, B would have got a sum = (33640 – 17640) = ₹ 16000
Hence, option D is correct:

Let man invested at the rate of 4x % per annum and for the period of time is×yr
ATQ
(2400 × 4x × x)/100 = 864
4x² = 864/24
4x² = 36
x² = 9
x = 3
So, Rate of interest = 4 × 3 = 12% per annum
Time of period = 3 years.
Equivalent C.I. of two year at the rate of 12% per annum
CI = P[(1 + r/100)ⁿ - 1]
814.08 = (2400 + x) [(1 + 12/100)² - 1]
814.08 = (2400 + x) [(112/100)(112/100) - 1]
814.08 = (2400 + x) [(28/25)(28/25) - 1]
814.08 = (2400 + x) [784/625 - 1]
814.08 = (2400 + x) [(784 - 625)/625]
814.08 = (2400 + x) (159/625)
814.08×(625/159) = 2400 + x
3200 = 2400 + x
x = 3200 - 2400 = 800

Let the principal be ₹100
Amount doubles it'self in 4 years = ₹200
Amount = P [1 + (r/100 )ⁿ]
200 = 100[1 + (r/100)⁴]
2 = [1 + (r/100)⁴] .....(i)
If sum become 16 times in the time n years,
1600 = 100[1 + (r/100)ⁿ]
16 = [1 + (r/100)ⁿ]
2⁴ = [1 + (r/100)ⁿ] ....(ii)
Using (i) in (ii),
[1 + (r/100)⁴]⁴ = [1 + (r/100)ⁿ]
[1 + (r/100)]¹⁶ = [1 + (r/100)ⁿ]
Hence, n = 16 yrs.

This problem can be solved by compound interest method.
Where R = (1/5) of it's height = (1/5)×100 = 20 %
Time period n = 3 years so,
Height after 3 years = P (1 + (R/100))ⁿ
= 50 × (1 + (20/100))³
= 50 × (1 + (1/5))³
= 50 × ((6×6×6) / (5×5×5))
= 86.4 cm.

P = 1000, A = 1728, n = 3 years
The amount is given by A = P (1 + (r/100))ⁿ
(1 + (r/100))³ = A/P
(1 + (r/100))³ = (1728/1000)
(1 + (r/100))³ = (12/10)³
Here both cubes cancels each other.
1 + (r/100) = 12/10
r/100 = (12/10) - 1
r/100 = 1/5
Therefore rate of interest r = 20%.