Compound Interest

Let A be the amount obtained when ₹5000 is compound annually.
CI = P(1 + (R/100))ⁿ - P
= 5000×(1 + (10/100))² - 5000
= 5000×(110/100)² - 5000
= 5000 × [(110×110)/(100×100)] - 5000
= 6050 - 5000
= ₹ 1050
Given, simple interest is half the compound interest = 1050/2 = ₹ 525
Let P be the sum of money invested on SI.
SI = PNR/100
525 = (P×5×3)/100
P = (525×100)/15
P = ₹ 3500

Lets first calculate the total rate % that Rinki will have after 3 years:
As per the question Rinki invested at rate of 12% pa simple interst
So, for 3 years tenure he will get = 12 × 3 = 36%
And the amount that Pinki invested at rate of 10% pa compound interest
By net% effect formula, we can calculate the total perecntage for 3 years tenure = 33.1% (sub details)
So, the difference between SI and CI = 36% – 33.1% = 2.9% (SI is more)
Here Rinki will get, 2.9% of 10000 = 290
So Rinki will have ₹ 290 more than Pinki.
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Sub-details:-
Net% effect = x + y = xy/100 %
For the first 2 years: Here,×= y = 10%
= 10 + 10 = 10× 10 = 21 /100 %
And for the next year: Here×= 21% and y = 10%
= 21 + 10 = 21 × 10 = 33.1/100 %
Hence, option C is correct:

Given: Principal = ₹10,000
n = 2
r = 4%
Amount = P [ 1 + ( r / 100 )]ⁿ
= 10000×[ 1 +( 4 / 100 )²]
= 10000×( 104 / 100 )²
= 10000×( 104 / 100 )×( 104 / 100 )
= 104×104
= ₹10,816
Compound Interest = Amount - Principal
= 10816 - 10000
= ₹ 816

Given: Principal = 216x; C.I = 127x
Amount = 216x + 127x = 343x
For three years ratio of amount and principal
A = P(1 + r/100)ⁿ
A/P = 343x/216x = (1 + r/100)³
1 + r/100 = ∛(343/216)
1 + r/100 = 7/6
r/100 = 7/6 - 1 = 1/6
r = 100/6 = 50/3 = 16($$\frac{2}{3}$$)%
3rd year C.I - 2nd year C.I = 245
(3 years C.I - 2 years C.I) - (2 years C.I - 1st years C.I) = 245
3 years C.I - 2 × (2 years C.I) + 1st years C.I = 245
P[(1 + r/100)³ - 1] - 2×P[(1 + r/100)² - 1] + P[(1 + r/100)¹ - 1] = 245
P[(1 + r/100)³ - 1 - 2×(1 + r/100)² + 2 + 1 + r/100 - 1] = 245
P[1 + 3r/100 + 3r²/100² + r³/100³ + 1 - 2(1 + r²/100² + 2r/100) + r/100] = 245
2 + 3r/100 + 3r²/100² + r³/100³ - 2 - 2r²/100² - 4r/100 + r/100 = 245
P(r²/100² + r³/100³) = 245
Pr²/100³ × (100 + r) = 245
P×(50/3)²/100³ × (100 + (50/3)) = 245
P/(4×100×9) × (350/3)= 245
P/(8×9) × (7/3) = 245
P = (245×8×9×3)/7 = 7560
Increased rate of interest = 100/6 × 138/100 = 23%
The difference between C.I and S.I of same sum of money for 2 years = P[(1 + r/100)² - 1] - Pnr/100
= P[1 + r²/100² + 2r/100 - 1 - 2r/100]
= P(r²/100²)
= (7560×23×23) / (100×100) = ₹ 399.9

Solution is :
Given Compound Interest = ₹2100
Rate of Interest ( r ) = 10 % p.a
No.of years ( n ) = 2
To find , amount received at the beginning ⇒ principal
Compound Interest = P [ 1 + ( r / 100 )ⁿ- 1 ]
⇒ 2100 = P[ 1 + ( 10 / 100 )²- 1 ]
⇒ 2100 = P[ 1 + ( 1 / 10 )²- 1 ]
⇒ 2100 = P[ ( 11 / 10 )²- 1 ]
⇒ 2100 = P[ ( 121 / 100 ) - 1 ]
⇒ 2100 = P[ 21 / 100 ]
⇒ 2100×( 100 / 21 ) = P
Principal = ₹ 10000
Amount invested at the beginning = ₹ 10000

Given, C.I + S.I = 5160
[(x - 1000)(1 + 30/100)² - (x - 1000)] + [(x + 1000)×20×2]/100 = 5160
[(x - 1000) × 69/100] + [(x + 1000) × 40/100] = 5160
69x - 69000 + 40x + 40000 = 516000
109x = 516000 + 29000
x = 545000/109 = 5000
Amount invested on scheme B = 5000 + 1000 = ₹ 6000

We know, Shiva invested an amount into two parts in the ratio = 4 : 3
Let the total amount invested by Shiva be ₹ 7X.
If the rate of compound interest for 2 years is then equivalent rate of interest will be
CI = r + r + [(r×r)/100]
Equivalent rate of interest for 2 yearss at 20% = 20 + 20 + [(20×20)/100]
= 20 + 20 + 4 = 44%
Equivalent rate of interest for 2 yearss at 15% = 15 + 15 + [(15×15)/100]
= 129/4 = 32.25%
We got from Q,
Amount invested in earlier interest - amount invested in exchange of interest = ₹ 705
[(4X)×44% + (3X)×32.25%] - [(4X)×32.25% + (3X)×44% ] = 705
[4X×(44/100) + 3X×(32.25/100)] - [4X×(32.25/100) + 3X×(44/100)] = 705
[1.76X + 0.9675X] - [1.29X + 1.32X] = 705
2.7275X - 2.61X = 705
0.1175X = 705
X = ₹ 6000
So, total amount invested by Shiva = 7x = 7(6000) = ₹ 42,000
∴ SI for 2yrs at 17.5% = PNR/100 = 42000×2×17.5/100
= ₹ 14,700.

CI - SI = 150
P(1 + r/100)ⁿ - Pnr/100 = 150
15000[(1 + (2x/100))² - 1] - (15000×2x×2)/100 = 150
15000[((100 + 2x)/100)² - 1] - 600x = 150
15000[(10000 + 400x + 4x²)/10000 - 1] - 600x = 150
15000[(10000 + 400x + 4x² - 10000)/10000] - 600x = 150
15000[(400x + 4x²)/10000] - 600x = 150
3/2(400x + 4x²) - 600x = 150
(3×4/2)(100x + x²) - 600x = 150
600x + 6x² - 600x = 150
6x² = 150
x² = 25
x = 5
The rate of interest per annum = 2x% = (2×5)% = 10%

Let Ram's share be '₹ x'.
∴ Shyam share = ₹ (260200 - x)
A = P[1 + (R/100)]ⁿ
we know that,amount recieved by Anitesh is same as that of Shyam,
x[1 + (4/100)]⁴ = (260200 - x)[1 + (4/100)]⁶
x = (260200 - x)[1 + (4/100)]²
x = (260200 - x)[1 + ($$\frac{1}{2}$$5)]²
x = (260200 - x)[26/25]²
x = (260200 - x) (676/625)
(625x/676) = 260200 - x
(625x/676) +×= 260200
(625x + 676x)/676 = 260200
1301x = 260200×676
x = (260200×676)/1301
x = ₹ 1,35,200
∴ Ram's share is ₹ 1,35,200:

Let the amount invested in the scheme A be '₹ X'.
∴ the amount invested in the scheme B = (40000 - X).
Given that the simple and compound interest for 1st year is same.
we know that,
15% of X = 10% of (40000 - X)
15X/100 = (400000 - 10X)/100
15X = 400000 - 10X
15X + 10X = 400000
25X = 400000
X = ₹ 16,000
Amount invested in scheme B = 40000 - X
= 40000 - 16000
= ₹ 24,000
∴ ₹ 24,000 invested in scheme B.