Compound Interest
Suppose, Deepak borrows Rs. 1000 at 10% interest from Luna. The simple interest on this sum at the end of one year will be \(Rs. \frac {1000 \times 1 \times 10} {100}\) = Rs. 100. If Deepak pays this interest to Luna, Luna can get back Rs. 1100. In case, Deepak pays this interest to Luna then Luna has right to charge interest on Rs. 1100 for next year. The compound interest is infact a simple interest computed on the previous simple amount. When Deepak calculates the simple interest for 2 years then,
\(I = \frac {1000\times 2\times 10} {100}\) = Rs. 200 but whenhe calculates the compound interest for 2 years then compound interest for 2 years = simple interest for second year
\(=\frac{1000 \times 1 \times 10} {100} + \frac{1100\times 1 \times 10} {100}\) (Principal for 2^{nd }year = Rs. 1000 + interest of 1^{st} year)
=100 + 110
= Rs. 210
In this process we get Rs. 10 profit by the way of compound interest.
The following points should be remembered before calculating the compound interest.
 The compound interest for every succeeding year is always greater than the compound interest for the previous year.
 The amount of the previous year becomes the principal for the coming year.
 The final amount is equal to the sum of the original principal and the interest for all the years.
 The compound interest for the entire period is the sum of the interest for all the years that is the difference between the final amount and the original principal.
The installment is the regular interval of time in which the compound interest is calculated. The payment might be yearly, halfyearly, quarterly, monthly, daily etc. Here we use only yearly and halfyearly installments.
Derivation of yearly compound interest
Year  Principal  Time  Rate  Interest  Amount 
1^{st}  P  1 year  R%  \(\frac {PR}{100}\)  \(P +\frac {PR}{100} = P((1 + \frac {R} {100})\) 
2^{nd}  \(P (1 + \frac {R}{100})\)  1year  R%  \(P\times(1 + \frac {R} {100})\times \frac {R}{100}\)  \(P\times(1 + \frac {R} {100}) + P\times(1 + \frac {R} {100})\times \frac {R}{100}\) \(= P\times(1 + \frac {R} {100})\times (1 + \frac {R} {100})\) \(= P\times(1 + \frac {R} {100})^2\) 
3^{rd}  \(P (1 + \frac {R}{100})^2\)  1year  R%  \(P\times(1 + \frac {R} {100})^2\times \frac {R}{100}\)  \(P\times(1 + \frac {R} {100})^2 + P\times(1 + \frac {R} {100})^2 \times\frac {R}{100}\) \(=P\times(1 + \frac {R} {100})^2 \times(1 + \frac {R} {100}) \) \(=P\times(1 + \frac {R} {100})^3\) 
4^{th}  \(P\times(1 + \frac {R} {100})^{T1} \)  1year  R%  \(P\times(1 + \frac {R} {100})^{T1} \times \frac {R} {100}\)  \(P\times(1 + \frac {R} {100})^{T + 11}\) \(=P\times(1 + \frac {R} {100})^T \) 
So, the yearly compound amount for T years at R% p.a. = \(P\times \left (1 + \frac {R} {100}\right)^T \)
Compound interest for T years = Compound amount for T years  original principal
$$ = P \times \left(1 + \frac {R} {100}\right)^T  P $$
$$ =P \left \lbrace \left (1 + \frac {R} {100} \right )^T  1 \right \rbrace$$
Halfyearly compound interest
In case of halfyearly compoundinterest, time will be double and rate will be halved.
Since yearly compound interest = \( P \left (1 + \frac {R} {100} \right ) ^ T  P \)
$$ = P \times \left(1 + \frac {R} {100}\right)^2T  P $$
$$ =P \left \lbrace \left (1 + \frac {R} {100} \right )^2T  1 \right \rbrace$$
a. The amount of the previous year becomes the principal for the coming year.
b. The compound interest for every succeeding year is always greater than compound interest for the previous year.
c. The final amount is equal to the sum of the original principal and the interest for all the years.
d. The compound interest for the entire period is the sum of the interest for all the year that is a difference between the final amount and the original principal.
Solution:
Principal(P) = Rs 5000,
Rate (R) = 12%
Time (T) = 2 years
Compound interest (C.I) = ?
Now,
\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T  1\right]\\ &= 5000 \left[ \left( 1 + \frac{12}{100}\right)^2  1\right]\\ &= 5000[(1.12)^2 1]\\ &= 5000[1.2544  1] \\ &= 5000 \times 0.2544\\ &= Rs \: 1272\: \: _{Ans.}\end{align*}
Solution:
Principal (P) = Rs 700
Compound interest (A) = Rs 847
Time (T) = 2 years
Rate of interest (R) = ?
\begin{align*} Compound \: amount \: (CA) &= P \left(1 + \frac{R}{100}\right)^T \\ or, 847 &= 700\left(1 + \frac{R}{100}\right)^2 \\ or, \frac{847}{700} &= \left(1 + \frac{R}{100}\right)^2 \\ or, 1.21 &= \left(1 + \frac{R}{100}\right)^2\\ or,(1.1)^2 &= \left(1 + \frac{R}{100}\right)^2 \\ or, 1.1 &= 1 + \frac{R}{100}\\ or, \frac{R}{100} &= 1.1  1\\ or, R &= 0.1 \times 100 \\ \therefore R &= 10\% \:\:\:_{Ans.}\end{align*}
Solution:
Principal(P) = Rs 50,000
Time (T) = 2 years
Rate (R) = 10%
Compound interest (C.I) = ?
\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^2T  1\right]\\ &= 50,000\left[ \left( 1 + \frac{10}{100}\right)^{2 \times 2}  1\right] \\ &= 50,000 [(1 + 0.05)^4  1]\\ &= 50,000 [1.2155 1] \\ &= 50,000 \times 0.2155 \\ &= Rs \: 10775 \:\: \: _{Ans.} \end{align*}
Solution:
Principle (P) = Rs 12,000
Time (T) = 1 year
Month(M)=6 month
Rate (R) = 10%
Compound interest (C.I) = ?
\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T \left( 1 + \frac{MR}{1200}\right) 1\right]\\ &= 12,000 \left[ \left( 1 + \frac{10}{100}\right)^1 \left( 1 + \frac{6 \times 10}{1200}\right)  1 \right] \\ &= 12,000 [(1.1) (1.05)  1]\\ &= 12,000\: [1.155  1] \\ &= 12,000\times 0.155 \\&= Rs 930\\ \end{align*} \(\therefore Compound \: interest = Rs \: 930 \:\: _{Ans.}\)
Solution:
Principle (P) = Rs 25,000
Time(T) = 2 years
Rate (R_{1}) = 4%
Rate (R_{2}) = 5%
Compound Amount (CA) = ?
\begin{align*} Compound \: amount \: (CA) &= P \left(1 + \frac{R_1}{100}\right) \left(1 + \frac{R_2}{100}\right)
\\ &= 25000\left(1 + \frac{4}{100}\right) \left(1 + \frac{5}{100}\right) \\ &= 2500 \times 1.04 \times 1.05 \\ &= Rs \: 27300 \:\:\: _{Ans.} \end{align*}
Solution:
Compound Amount (CA) = Rs 5191.68
Time (T) = 1 year
Rate (R) = 8%
Principal (P) = ?
\begin{align*} Compound \: amount \: (CA) &= P \left(1 + \frac{R}{200}\right)^{2T} \\ or, 5191..68 &= P \left(1 + \frac{8}{200}\right)^2\\ or, 5191.68 &= P(1.04)^2\\ or, P &= \frac{5191.68}{1.0816}\\ &= \: Rs \: 4800 \end{align*}
\(\therefore\) The principal (P) = Rs \: 4800 \(_{Ans.}\)
Solution:
First compound amount (CA_{1}) = Rs 1323
Time (T_{1}) = 2 years
Second compound amount (CA_{2}) = Rs 1389.15
Time (T_{2}) = 3 years
We know that,
\begin{align*} CA &= P \left(1 + \frac{R}{100}\right)^T \\ 1323 &= P \left(1 + \frac{R}{100}\right)^2 \: \: \: \: \: ........ (1)\\ or, 1389.15 &= P \left(1 + \frac{R}{100}\right)^3 \:\:\:\:\: ........(2) \\ \end{align*}
\(\text{The equation (2 )is divided by equation (1) } \)
\begin{align*} \frac{1389.15}{1323} &= \frac{P \left(1 + \frac{R}{100}\right)^3 }{P \left(1 + \frac{R}{100}\right)^2 }\\ or, 1.05 &= 1 + \frac{R}{100}\\ or, \frac{R}{100} &= 1.05  1\\ or, R &= 0.05 \times 100\%\\ R &= 5\% \end{align*}
\(\text{Putting Value of R in}\:\: eq^n (1)\)
\begin{align*}P \left(1 + \frac{R}{100}\right)^2 &= 1323 \\ or, P \left(1 + \frac{5}{100}\right)^2 &=1323 \\ or, P \times 1.1025 &= 1323\\ P &= \frac{1323}{1.1025}\\ \therefore P &= 1200\\ \end{align*} \(\therefore Principle = Rs \: 1200 \:\: and \: rate = 5\% \)
Solution:
First compound interest (CA_{1}) = Rs 7260
Time (T_{1}) = 2 years
Second compound interest (CA_{2}) = Rs 7986
Time (T_{2}) = 3 years
We know that,
\begin{align*} Compound \: amount \: (CA) &= P \left(1 + \frac{R}{100}\right)^T \\ 7260 &= P\left( 1 + \frac{R}{100}\right)^2 \: \: \: \: ....... (1) \\ 7986 &= P\left( 1 + \frac{R}{100}\right)^3 \:\:\:\: ....... (2)\\ \end{align*}
\( \text{The equation (2) is divide d by equation (1)}\)
\begin{align*} \frac{7986}{7260} &= \frac{P\left( 1 + \frac{R}{100}\right)^3 }{P\left( 1 + \frac{R}{100}\right)^2}\\ or, 1.1 &= 1 + \frac{R}{100}\\ or, \frac{R}{100} &= 1.1  1 \\ or, \frac{R}{100} &= 1.1 1 \\ or, R &= 0.1 \times 100 \\ R &= 10\% \end{align*}
Solution:
Let, Principle (P) = Rs x
Time (T) = 2 years
Rate (R) = 5%
\begin{align*} Simple \: Interest \: (S.I) &= \frac{P\:T\:R}{100} \\ &= \frac{x \times 2 \times 5}{100} \\ &= x \times 0.1 \end{align*}
\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T  1\right]\\ &= x \left[ \left( 1 + \frac{5}{100}\right)^2  1\right] \\ &= x [(1.05)^2  1] \\ &= x[1.1025  1]\\ &= x \times 0.1025 \end{align*}
From question,
\begin{align*} S.I + C.I &= Rs \: 202.50 \\ x \times 0.1 + x \times 0.1025 &= Rs \: 202.50\\ or, x(0.1 + 0.1025) &= Rs \: 202.50\\ x &= \frac{202.50}{0.2025} \\ \therefore x &= Rs \: 10000 \end{align*}
\(\therefore\) Principle (P) = Rs 10000 \(_{Ans.}\)
Solution:
First compound interest (CI_{1}) = Rs 450
Time (T_{1}) = 1 year
Second compound interest (C_{2}) = Rs 945
Time (T_{2}) = 2 years
Let Principal = P and rate = R
We know that,
\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T  1\right]\\ C.I_1&= P \left[ \left( 1 + \frac{R}{100}\right)^T  1\right] \\ 450 &= P \left[ \left( 1 + \frac{R}{100}\right)^1  1\right] \: \: ....... (1) \\ C.I_2&= P \left[ \left( 1 + \frac{R}{100}\right)^T  1\right] \\ 945 &= P \left[ \left( 1 + \frac{R}{100}\right)^2  1\right] \:\:\: ....... (2)\end{align*}
Equation (2) is divided by equation (1)
\begin{align*} \frac{ P \left[ \left( 1 + \frac{R}{100}\right)^2  1\right]}{ P \left[ \left( 1 + \frac{R}{100}\right)^1  1\right]} &= \frac{945}{450}\\ or, \frac{\left[ \left( 1 + \frac{R}{100}\right) + 1\right] \left[ \left( 1 + \frac{R}{100}\right)  1\right]}{\left[ \left( 1 + \frac{R}{100}\right)^1  1\right]} &= 2.1 \\ or, 2 + \frac{10}{100} &= 2.1\\ or, \frac{R}{100} &= 2.1  2 \\ or, R &= 0.1 \times 100\% \\ \therefore R &= 10\%\end{align*}
Putting the value of R in equation (1)
\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)  1\right] \\ or, P \times \frac{10}{100} &= 450\\ or, P &= 450 \times 10\\ P &= 4500 \\ \\ \therefore The \: sum = 4500, rate (R) = 10\% \end{align*}
Solution:
Principal (P) = Rs. 46875
Compound interest (C.I.) = Rs. 5853
In 1year compound interest of Rs 1 is 4 paisa.
In 1year compound interest of Rs 100 is 4 × 100 paisa.
In 1year compound interest of Rs 100 is Rs. \( \frac{4 \times 100}{100} \) = Rs 4
Interest rate (R) = 4%
Time (T) =?
We know that,
\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T  1\right]\\ 5853 &= 46875 \left[ \left( 1 + \frac{4}{100}\right)^T  1\right] \\ \frac{5853}{46875} &=\left( 1 + \frac{4}{100}\right)^T  1\\ or, 0.124864 + 1 &=(1.04)^T\\ or, (1.04)^3 &= (1.04)^T \\ \therefore Time &= 3 \: years \end{align*}
Solution:
First compound amount (CA_{1}) = Rs 8820
Time (T_{1}) = 2 years
Second compounded amount (CA_{2}) = Rs 9261
Time (T_{2}) = 3 years
Let principal = P , Rate = R
\begin{align*} Compound \: amount \: (CA) &= P \left(1 + \frac{R}{100}\right)^T \\ 8820 &= P \left(1 + \frac{R}{100}\right)^2\:\:\: ...... (1)\\ 9261 &= P \left(1 + \frac{R}{100}\right)^3 \:\:\: ..... (2) \end{align*}
Equation (2) is divided by equation (1)
\begin{align*} \frac{P \left(1 + \frac{R}{100}\right)^3}{ P \left(1 + \frac{R}{100}\right)^2} &= \frac{9261}{8820} \\ or, 1 + \frac{R}{100} &= 1.05\\ or, \frac{R}{100} &= 1.05  1 \\ or, R &= 0.05 \times 100\% \\ R &= 5\% \end{align*}
Putting value of R in equation (1)
\begin{align*} P \left(1 + \frac{R}{100}\right)^T &= 8820 \\ or, P \left(1 + \frac{5}{100}\right) &= 8820\\ or, P \times (1.05)^2 &= 8820 \\ or, P \times 1.1025 &= 8820\\or,P &= \frac{8820}{1.1025}\\ P &= Rs \: 8000 \end{align*}
\(\therefore \) The principal = Rs 8000 and Rate = 5% \(_{Ans.} \)
Solution:
Time (T) = 3 years
Rate (R) = 20%
Principal (P) = ?
Compound Interest (C.I)  Simple Interest (S.I) = Rs. 384
We know that,
\begin{align*} S.I &= \frac{P \: T \: R}{100}\\ &= \frac{P \times 3 \times 20}{100}\\ &= P \times 0.6 \end{align*}
\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T  1\right]\\ &= P \left[ \left( 1 + \frac{20}{100}\right)^3  1\right]\\ &=P\: [(1.2)^3  1]\\ &= P[1.728  1]\\ &= P \times 0.728 \end{align*}
From question,
\begin{align*} C.I  S.I &= Rs\: 384\\ P \times 0.728  P \times 0.6 &= Rs\: 384 \\P\:(0.728  0.6)&= 384\\ or, P \times 0.128 &= 384\\ or, P &= \frac{384}{0.128}\\ \therefore P &= Rs\:3000 \:\:\:\: _{Ans.} \end{align*}
Solution:
For Anima,
Let, Principal (P) = Rs x,
Time (T) = 2 years
Rate (R) = 10%
\begin{align*} Simple \: Interest \: (S.I) &= \frac{P \: T \: R}{100} \\ &= \frac{x \times2 \times10 }{100}\\ &= x \times 0.2 \end{align*}
For Eline,
Principal (P) = Rs 6000  x
Time (T) = 2 years
Rate (R) = 5%
\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T  1\right]\\ &= (6000  x)\left[ \left( 1 + \frac{5}{100}\right)^2  1\right]\\ &= (6000  x ) [(1.08)^2  1] \\ &=Rs \: 998.4 \times Rs \: 0.1664x \end{align*}
From question,
\begin{align*} C.I.  S.I &= Rs \: 50\\ or, 998.4  x \times 0.1664  x \times 0.2 &=50\\ or, 998.4  x \times 0.3664&=50\\ or, x \times 0.3664&=50  998.4\\ or, x &= \frac{948.4}{0.3664}\\ &= Rs\: 2588.43\end{align*}
For Anima sum = Rs 2588.43
For Eline sum = (6000  2588.43) = Rs 3411.57
Solution:
Principal (P) = Rs 170000
Rate (R) = 21%
Time (T) = 1 year 6 month = \( 1 + \frac{6}{12} = \frac{3}{2} \: years \)
\begin{align*} Simple \: Interest \: (S.I) &= \frac{P \: T \: R}{100} \\ &= \frac{170000\times3 \times21 }{100\times 2}\\ &= Rs\: 53550 \:\:\:\:_{Ans.} \end{align*}
\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T  1\right]\\ C.I &= 170000 \left[ \left( 1 + \frac{21}{100}\right)^{\frac{3}{2}}  1\right] \\&= 170000 \left[ \left( \frac{121}{100}\right)^{\frac{3}{2}}  1\right]\\ &= 170000\left[ \left( \frac{11}{10}\right)^{\frac{2 \times3}{2}}  1\right] \\&= 170000[(1.1)^3  1]\\ &=170000[1.331  1]\\ &= Rs\: 170000 \times 0.331\\ &= Rs\: 56270 \:\:\:_{Ans.} \end{align*}
Solution:
Principle (P) = Rs 5120
Time (T) = 3 years
Rate (R) = 12.5%
\begin{align*} Simple \: Interest \: (S.I) &= \frac{P \: T \: R}{100} \\ &= \frac{5120 \times3 \times12.5 }{100} \\ &= Rs\:1920\end{align*}
\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T  1\right]\\ &= 5120 \left[ \left( 1 + \frac{12.5}{100}\right)^3  1\right]\\ &=5120 [1.4238281  1] \\ &= 5120 \times 0.4238281 \\&= Rs\: 2170 \end{align*}
\begin{align*}Difference &= C.I  S.I\\ &= 21701920\\ &= Rs\: 250 \end{align*}
Solution:
Principal (P) = Rs 4250
Rate (R) = 12%
Time (T) = 1 year
\begin{align*} Simple \: Interest \: (S.I) &= \frac{P \: T \: R}{100} \\ &= \frac{4250 \times 1 \times12 }{100}\\ &=Rs\: 510 \end{align*}
For half yearly,
\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{200}\right)^{2T}  1\right]\\ &= 4250\left[ \left( 1 + \frac{12}{200}\right)^{2\times 1}  1\right] \\ &= 4250[(1.06)^2 1]\\ &= 4250 [1.12361]\\ &=4250 \times 0.1236\\ &= Rs \:525.30 \:\:_{Ans}\end{align*}
Solution:
Principal (P) = Rs 5000
Time (T) = 2 years
Rate (R) = 20%
\begin{align*} Compound\:Interest\:(C.I) &= P \left[ \left( 1 + \frac{R}{100}\right)^T  1\right]\\ &= 5000\left[ \left( 1 + \frac{20}{100}\right)^2  1\right]\\ &= 5000[(1.2)^2  1]\\ &= 5000 \times 0.44 \\ &= Rs\:2200 \end{align*}
Let, Simple interest (S.I) = Rs 2200
Principle (P) = Rs 5000
Rate (R) = 20%
Time (T) = ?
\begin{align*} T &= \frac{I \times 100}{PR}\\ &= \frac{2200 \times 100}{5000 \times 20}\\ &= 2.2 \: years \end{align*}
Solution:
Principal (P) = Rs 5000
Rate (R) = 8%
Time (T) = 2 years
Compound amount (CA) = ?
\begin{align*} For\: yearly,\\ CA &= P \left(1 + \frac{R}{100}\right)^T \\ &= 5000 \left(1 + \frac{8}{100}\right)^2\\ &=5000 \times \frac{108}{100} \times \frac{108}{100} \\ &= Rs \: 5832 \:\:\: _{Ans.} \end{align*}
\begin{align*} For\: half \:yearly \: \\CA &= P \left(1 + \frac{R}{200}\right)^{2T} \\ &= 5000 \left(1 + \frac{8}{200}\right)^{2\times 2}\\ &= 5000 (1.04)^4\\ &= 5849.29 \end{align*}
\begin{align*} Difference \: between &= 5849.29  5832\\ &= Rs \:17.29 \:\:\: _{Ans.} \end{align*}
Solution:
For first part
First part (P_{1}) = Rs 25200  x
Time (T_{1}) = 2 years
Rate (R_{1}) = 10%
For second part
Second principal (P_{2}) = Rs x
Time (T_{2}) = 3 years
Rate (R_{3}) = `10%
\begin{align*} Compound \: amount \: (CA_1) &= P \left(1 + \frac{R}{100}\right)^T \\ &=(25200  x) \left(1 + \frac{10}{100}\right)^2\\ &= (25200  x) (1.1)^2\\ &= (25200x) \times 1.21 \end{align*}
\begin{align*} Compound \: amount \: (CA_2) &= P \left(1 + \frac{R}{100}\right)^T \\ &= x \left(1 + \frac{10}{100}\right)^3\\ &=x (1.1)^3 \\ &= x \times 1.331 \end{align*}
From Question,
\begin{align*} CA_1 &=CA_2 \\ (25200x) \times 1.21 &= x\times 1.331 \\ or, 30492  1.21x &= x \times 1.331\\ or, 1.331x+1.21x&=30492\\ or, 2.541x &= 30492\\ x&= \frac{30492}{2.541}\\ \therefore x&=12000 \end{align*}
\begin{align*} 1^{st} \: principle &= 25200 x\\&= 2520012000\\&= Rs\:13200 \:\:\:_{Ans.} \end{align*}
\(2^{nd }\: principal = x= Rs 12000 \:\:\:_{Ans.}\)

Santosh borrowed Rs130000 from Suresh at the rate of 21% per annum. Find the simple interest at the end of 2 years and interest compounded yearly.
Rs 60500, Rs 70000
Rs 54600,Rs 60333
Rs 40000, Rs45000
Rs 46000, Rs 48222

Ram borrowed Rs. 150000 from Sita at the rate of 21% per year. At the end of nine months how much compound interest should he pay in nine months compounded half yearly?
Rs 24451.87
Rs 20000
Rs 22678
Rs 21745

A man deposited Rs 2000 in the fixed deposit account of the bank for two years at the rate of 12% per annum. The interest is compounded is semiannually.How much will be the amount and the compound interest at the end of 2 years?
Rs 2289,489.3
Rs 2524.95, Rs 524.95
Rs 2222.22, Rs 421.67
Rs 2124.44,Rs 505.65

The compound interest and simple interest on Rs. 7500 in 2 years at 10% pa.
Rs 75
Rs 65
RS 70
Rs 60

Simple interest and annual compound interest on Rs 18000 for 2 yrs. at the rate of 15% pa.
Rs 505
Rs 205
Rs 405
Rs 305

The difference between the compound interest and the simple interest on a certain sum at 10% per annum for 3 years is Rs 155.
Rs 4660
Rs 5000
Rs 48000
Rs 45000

The compound interest on the sum of money in two years at the rate of10% per annum will be Rs 420 more than simple interest.Find the sum.
Rs 4000
Rs35000
Rs 42000
Rs 38000

Bishal deposited Rs.10,000 in 'A' bank and Rs 10,000 in 'B' bank at the rate of 6% per annum.'A' bank pays interest compounded halfyearly whereas 'B' bank pays interest compounded yearly. Calculate the interest which bank had paid more and much at the end of 2years.
Rs 15
Rs 19
Rs 20
Rs 18

If the difference between the compound interest compounded half yearly and yearly on a sum of money for 2 years at the rate of 20% per year is Rs 289.20, find the sum.
Rs 13500
Rs 13000
Rs 12000
Rs 11500

The difference between the annual and semiannual compound interest on a sum of money is Rs 482 at the rate of 20% per annum for 2 years.Find the sum.
Rs 20000
Rs 16000
Rs 12000
Rs 14000

The compound interest on a certain sum for 2 years at 8% per annum compounded annually is Rs 312 Find the simple interest for the same time and the same rate .
Rs 150
Rs 275
Rs 300
Rs 200

Accordingly to the yearly compound interest in what will the compound interest Rs 400000 at the rate of interest 6.5% per annum be Rs 53690?
6 years
2 years
5 years
1 years

Accordingly to the yearly compound interest in what time will the compound interest on Rs 800000 at the rate of interest 5% annum be Rs 126100?
6 years
10 years
2 years
3years

At what percent rate of compound interest per annum will the compound interest on Rs 125000 be Rs 91000 in 3 years?
15%
10%
17%
20%

The simple interest on a sum of money in 2 yrs is Rs 36 less than the compound interest compounded annually. if the rate of interest is 12% p.a find the sum.
Rs 1500
Rs 1200
Rs 2500
Rs 2000

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PRATIK ACHARYAif SI=1000 and CI=2050 and T=2years find principal and rate 
Jan 24, 2017 
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suraj gautamThe simple interest and the compound interest of a sum of money in 2 years are Rs 1000 and Rs 1050 respectively. Find annual rate of interest and the principal 
Jan 21, 2017 
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