Compound interest questions

Let equation 1 be

\( p \times(1 + \frac{ R}{100}) )^ 6 = 8900 \)

Let equation 2 be

\( p \times(1 + \frac{ R}{100}) )^ {12}=11200 \)

eqn 2 \( \div \) eqn1 gives

\( (1 + \frac{ R}{100}) )^ 6 = \frac{11200}{8900} =1.258 \)

sub ( 1+ (R /100 ) )^ 6 =1.258 in eqn

\( p \times1.258 =8900 \)

\( p = \frac{8900}{1.258} =7074.72 \)

\( \frac{PR^2}{100^2} = 10 \)
\( \Rightarrow R = 10 \% \)

2 years, therefore \(11^{2}\) = 121

1 \(\rightarrow\)13000(Sum) \(\rightarrow\)\(1\times13000\)=13000

2 \(\rightarrow\)\(13000\times\frac{13}{100} \) = 1690\(\rightarrow\)\(2\times1690\)=3380

1 \(\rightarrow\)\(1690\times\frac{13}{100} \) =219.70 \(\rightarrow\)\(1\times219.70\)=219.70

Total compound interest \( = 3380+219.70 = 3599.70\)

SI for 2 years = Rs 1800 =.> Si 1 year = Rs 900

In the second years Rs 72 is added in CI (1872-1800) which is 8% of 900

Hence R=8%

We know 8%=900

Hence 100%=11250

i.e. sum=11250

\[ \textrm{CI for 3 years= } P×(1+\frac{R}{100})^3-P\]

\[ =11250×[(1+\frac{8}{100})^3-P]=2921.76\]

Difference between C.I and S.I for 2 years = 36.30
S.I. for one year = 330.
S.I. on Rs 330 for one year = 36.30
R= \( \frac{36.30\times 100}{330}\) =11%

Let the difference between compound interest and simple interest for 3 years be \(D3\) and two years be \(D2\).

We know that,
\( D3 = 3\times D2 + \frac{r}{100}\times D2\)

Given,
\(D3 = D2+41\) and \(r = 5%\)

\(D2+41 = 3D2 + \frac{5}{100}\times D2\)

Solving we get,
\( D2 = 20\)
Let the principal be \(P\),
\(P\times \frac{r}{100} \times \frac{r}{100} = D2\)
\(P\times \frac{5}{100} \times \frac{5}{100} = 20\)

We get, \(P = 8000\) Rs.

2 years, therefore

\(11^{2}\)= 121

\(1\rightarrow 10000 \) Amount\(\rightarrow 1 \times 10000\) = 10000

\(2 \rightarrow 10000 \times \frac{12}{100} = 1200 \rightarrow 2 \times 1200= 2400\)

\(1\rightarrow 1200 \times \frac{12}{100} = 144 \rightarrow 1\times 144 = 144\)

Total compound interest = 2400 + 144 = 2544

Interests for one year = 24883.2 - 20736 = Rs 4147.2
\(ie.\) Interest obtained for one year will be Rs 4147.2
Compound Interest for one year = Simple Interest for one year.
\(\therefore\) Rate = \(\frac{Intrest\times 100}{P \times T}\)
Rate = \(\frac{4147.2\times 100}{20736 \times 1}\)
Rate = \(\frac{414720}{20736}\)
Rate = 20 %

S.I. for 3 years = Rs. 3000
S.I. for 2 years =\(\frac{3000}{3}\times 2 = Rs. 2000\)
C.I. – S.I. = 2050 – 2000 = Rs. 50
S.I. =\(\frac{PR \times 3}{100}\)
\(\rightarrow PR = \frac{3000 \times 100}{3} = Rs.100000\)
Difference = \(\frac{P\times R^2}{10000}\)
50 = \(\frac{P\times (100000)^2}{10000 \times P^2}\)
\(P = \frac{1000000}{50} = Rs.20000\)

SI for 1 years= \( \frac{480}{3} \)=160

(as SI is same for every year)

SI for 2 years=320

CI for w year=352; diff=32 32=20% of 160

Hence R=20%

20%=160

100%=800

\( 50 = P \times 2 \times \frac{R}{100} \)
\( \Rightarrow PR = 2500 \)
\( C.I - S.I = .80 \Rightarrow \frac{PR^2}{100^2} = .80 \Rightarrow PR^2 = 8,000 \)
Therefore, \( R = \frac{16}{5} = 3\frac{1}{5} \)

Principal = 5000, R = 10%

Compound interest= P \((1+\frac{R}{100})^3\)-5000

\( = 5000 \times(1+\frac{10}{100})^3 \)-5000

\( = 5000 \times(\frac{101}{100})^3 \)-5000

\( = 5000 \times\frac{110}{100} \times\frac{110}{100} \times\frac{110}{100} -5000= 1655\)

Simple interest \(= 5000 \times\frac{10}{100} \times 3 = 1500\)

Difference= 155

Principal = Rs. 7200

Time = 18 months = 1.5 years

Rate = 20% p.a.

if the interest is compounded half-yearly then
Time \(=1.5 \times 2=3\) years
Rate \(=\frac{20 \%}{2}=10 \%\)
Amount \(=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{T}}=7200\left(1+\frac{10}{100}\right)^{3}=7200 \times \frac{11}{10} \times\frac{11}{10} \times \frac{11}{10}=\)Rs.( 9583.2

Compound Interest \(=\)Rs.\(9583.2-\)Rs.\(7200=\) Rs. \(2383.2\)

\( P{(1+\frac R{100})}^3=\;\;37,044 \)

and \( P{(1+\frac R{100})}^2=\;35,280 \)

Divide both equations,

\( (1+\frac R{100})=\frac{37044}{35280}=\frac{21}{20} \)

So \( P{(\frac{21}{20})}^{2\;}=35280\\WE\;GET\;P\;=32000 \)\

S.I= \((\frac{8000 \times 5 \times 3}{100} )= 1200 \)Rs.

C.I= Rs\(8000 \times (1+\frac{5}{100})^{3} -8000 = 1261\)Rs.

Difference = \(1261-1200 = 61\)

P(1+\(\frac{R1}{100}\))\( \times \)(1+\(\frac{R2}{100}\))

25000 (1+\(\frac{4}{100}\))\( \times \)(1+\(\frac{5}{100}\))

25000 x (\(\frac{104}{100}\)\( \times \)\(\frac{105}{100}\))

= 27300

Calculation :

• Quarterly rate of interest \(=\frac{6}{4} \%\)

• Time \(=3\) quarter

• C.I. \(=800\left[\left(1+\frac{6}{4 \times 100}\right)^{3}-1\right]\)

  \(=800\left[\left(\frac{203}{200} \times \frac{203}{200} \times \frac{203}{200}\right)-1\right]\)

  \(=800\left(\frac{8365427-8000000}{8000000}\right)\)

  \(=\frac{800 \times 365427}{8000000}\)

  \(=\) Rs. \(36.50\) (Approximate)

Hence, the answer is option B

Interest for one year = Rs (13176 - 12960) = 216
That means a certain percentage of 12960 is 216
That percentage will be the desired Rate, R
\(12960\times\frac{R}{100} = 216\)
To find Rate, R = \(\frac{216\times100}{12960}\)
R = \(1\tfrac{2}{3}\) %

Given that,

• Amount = 2 \( \times\) Principal
• Time = 6 years

We know that,

• Simple Interest = Principal \( \times\) Time \( \times\)R/100
• Amount = Principal + Simple interest

Case 1

• Let the principal be x

\( \Rightarrow \)So, the amount is 2x

• Interest = 2x - x = x

\( \Rightarrow \) x = x \( \times\) 6 \( \times\) R/100

⇒ R = \( \frac{100}{6}= \frac{50}{3}\) %

Case 2

• Amount = 5x
• Principal = x
• Simple Interest = 5x - x = 4x

⇒\( 4x = x \times Time \times \frac{50}{3 \times 100} \)

⇒ Time = 24 years.

\(\because A _{1}= Rs 578.40\),

\( n _{1}=2\) years,

\(A _{2}=614.55\),

\(n _{2}=3\) years

\(A = P (1+ \frac{r }{ 100})^{ n }\)

\(578.40= P (1+ \frac{r }{ 100})^{2}\)

\(614.55= P (1+ \frac{r }{ 100})^{3}\)

Dividing, we get

\((1+\frac{r }{ 100})= \frac{614.55}{578.40}\)

\(\frac{r}{100}= \frac{614.55}{578.40}-1\)

\(= \frac{36.15}{578.40}\)

\( r = 100 \times \frac{ 36.15}{578.40} = 6¼ \) %

Change in amount in 1 year = 2730 – 2600 = 130

Required Rate % = (\(\frac{130}{2600}\) ) \(\times\) 100 = 5% .

Hence , the answer is option A .

Let sum of money \(= P\), rate \(= R\)

\(P\left(1+\frac{R}{100}\right)^{5}=A P\)

\(\left(1+\frac{ R }{100}\right)^{5}=4\)

For 64 times

\(\left(\left(1+\frac{ R }{100}\right)^{5}\right)^{3}=4^{3}\)

\(\left(1+\frac{R}{100}\right)^{15}=64\)

So, it will become 64 times in 15 years

Hence, option \(C\) is correct.

\begin{aligned}
&A=P\left(1+\frac{r}{100}\right)^{t} \\
&\Rightarrow 3 P=P\left(1+\frac{r}{100}\right)^{10} \\
&\Rightarrow 3=\left(1+\frac{r}{100}\right)^{10}---(1) \\
&\Rightarrow 81 P=P\left(1+\frac{r}{100}\right)^{n} \\
&\Rightarrow 3^{4}=\left(1+\frac{r}{100}\right)^{n} \quad----(2) \\
&\text { From (1) and (2) } \\
&\Rightarrow\left(1+\frac{r}{100}\right)^{10 \times 4}=\left(1+\frac{r}{100}\right)^{n}
\end{aligned} ⇒ n = 40 years

∴ A sum of money becomes 81 times in 40 years.

SI for 2 years = Rs 1000 =.> Si 1 year = Rs 500

In the second years Rs 25 is added in CI (1025-1000)

which is 5% of 500

Hence R=5%

5%=500

100%=10000

sum=10000

CI for 3 years= \( P(1+\frac{R}{100})^3 -P\)

= \( 10000(1+\frac{5}{100})^3 -10000\)

=\( 1576.25 \)

\( R=\frac{R}{2}=\frac{10}{2}=5% \)

\( A=P(1+\frac{5}{100})^n \)

\( A=16000(\frac{105}{100})^4 \)

(N=2 years =6 months + 6 months + 6 months + 6 months)

\( A =16000\times \frac{105^4}{100^4} \)

\(=19448.1 \)

Principal,P = 10000

Rate, r = 20%

Time, t = 6 months = \(\frac{1}{2} \)year

Compounded interest quarterly,

Amount \( = P(1+\frac{5}{100})^{4t}\)

= \(10000\times(1+\frac{5}{100})^{2} \)

= \( 10000\times(\frac{105}{100})^2\)

= \( 10000\times\frac{105}{100}\times\frac{105}{100}\)

= \(105\times105\)

= \(11025\)

Simple interest =

\( \frac{P\times R\times T}{100}\)

where,P = principal,R = rate,T = time period

∴ 3200=

\( \frac{P\times 5\times 2}{100}=\frac{P}{10}\)

∴ Principal = Rs.32000

Compound Interest =P

\( \left[{\left(1+\frac{r}{100}\right)}^{n}-1\right]\)

Compound Interest =

\( P\left[{\left(1+\frac{r}{100}\right)}^{n}-1\right]=32000\left[{\left(1+\frac{1}{20}\right)}^{3}-1\right]\)

Compound Interest =

\( P\left[{\left(\frac{21}{20}\right)}^{3}-1\right]=32000\times \left[\frac{21}{20}\times \frac{21}{20}\times \frac{21}{20}-1\right]\)

Compound Interest =

\( 32000\times \left[\frac{9261}{8000}-1\right]=32000\times \left[\frac{9261-8000}{8000}\right]\)

Compound Interest =32000×

\( \frac{1261}{8000}\)

Principal,P = 3500
Rate, r = 4%
Time, t = 2 years
Amount after 2 years, \( A = P ( 1+\frac{r}{100})^t\)
= \(3500(1+ \frac{4}{100})^2\)
=\(3500(1+ \frac{1}{25})^2\)
= \(3500\times \frac{26}{25}\times \frac{26}{25}\)
= \(3785.6\)
CI = 3785.6 – 3500 = Rs. 285.6 .

S.I. \(=\frac{P \times R \times T}{100}\)

\(\frac{P \times 5 \times 3}{100}=1200\)

\(P=\frac{1200 \times 100}{5 \times 3}=8000\)

\(\therefore\) Compound Interest \(=\mathbf{P}\left(1+\frac{r}{100}\right)^{t}-P\)

\(=8000\left(1+\frac{5}{100}\right)^{3}-8000\)

\(=8000\left(\frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}\right)-8000\)

\(=9261-8000=1261\) rupees.

Compound Interest \( = P(1+\frac{R}{100})^N-P \)

\( =(2000 \times(1+\frac{5}{200})^3)- 2000 \)

\( =(2000 \times(\frac{105}{200})^3)- 2000 \)

\( = 2315.25 - 2000 = Rs.315.25 \)

\(Difference = P\times\left(\frac{R}{100}\right )^{2}\times\left(\frac{R+300}{100} \right)\)
= \(Difference = 5000\times\left(\frac{4}{100}\right )^{2}\times\left(\frac{4+300}{100} \right)\)
= \(Difference = 5000\times\left(\frac{4}{100}\right )\times\left(\frac{4}{100}\right )\times\left(\frac{304}{100} \right)\)
= Rs 24.3

difference b/w 2 years =\( P(\frac{R}{100})^2 \)

\( 96=15000(\frac{R}{100})^2 \)

\( \frac{960}{15}=R^2 \)

\( 8=R \)

Interest in one year = 2622-2420= 242

Rate= \(\frac{242\times100}{2420}=10\)%

Let rate of interest be R and principal be Rs. P

After 3 year, amount (Rs.) = 13380

Therefore, \( P(1+\frac{R}{100}^3) = 13380 \)-----(1)

After 6 year, amount Rs= 26760

Therefore, \( P(1+\frac{R}{100}^6) = 26760 \)------(2)

eq(2) \( \div \) eq(1) we get \( = (1+\frac{R}{100}^3) =\frac{26760}{13380}= 2 \)

\( P(1+\frac{R}{100}^3) = 13380 \)

\( P \times 2 = 13380 \)
\( P = \frac{13380}{2} = 6690\)

\(A_{1}\) = 5436 , \(A_{2}\) = 16308

\( A_I= P(1+\frac{R}{100})^3 \)

\( A_2 = P(1+\frac{R}{100})^6 \)

\( \frac{A_2}{A_1} = (1+\frac{R}{100})^3 = \frac{16308}{5436}\)

\( A_I= P(1+\frac{R}{100})^3 = 5436 \)

\( 5436 = P ( \frac{16308}{5436}) \)

\( P = 1812 \) rupees.

\(SI = \frac{ P \times 10 \times 2}{100} = \frac{P}{5}\)
\( CI = P[ 1 + \frac{10}{100}]^{2} - P\)
\(CI = P[ \frac{110}{100} \times \frac{110}{100} - 1] \)
\(CI = P[ \frac{110 \times 110 - 10000}{10000}] = \frac{21}{100}P\)
C.I - S.I = 15
\(\frac{21}{100}P - \frac{P}{5} = 15 \)
\(21P - 20P = 1500\)
\(P = 1500 \)

\(SI = \frac{P \times N \times R}{100}\)
\(CI = P \times (1 + \frac{R}{100}) – P\)
\(1806 = P \times (1 + \frac{R}{100})^{2} – P\)
\(1806 = P \times ((1 + \frac{R}{100})^{2} – 1) -------(1)\)
\(2520 = \frac{P \times 3 \times R}{100}\)
\(PR = 84000 -----------(2)\)
Solve (1) and (2)
\(1806 = \frac{PR}{100} \times (\frac{R}{100} + 2)\)
\(1806 = \frac{84000}{100} \times (\frac{R}{100} + 2)\)
\(\frac{1806}{840} = \frac{R}{100} + 2\)
\(\frac{(1806 – 1680)}{840} = \frac{R}{100}\)
\(R = 15\%\)
\(P \times 15 = 84000\)
\(P = Rs.\:5600\)

\(S.I = \frac{P\times T\times R}{100}\)
\((880 - P) = \frac{P\times 2\times 5}{100}\)
\((880 - P) = \frac{P}{10}\)
\(10\times (880 - P) = P\)
\(8800 -10 P = P\)
\(8800 = P +10 P\)
\(8800 = 11 P\)
\(P = \frac{8800}{11}\)
\(P = 800\)
\(A = P\times\left ( 1+\frac{R}{100} \right )^{n}\)
\(A = 800\times\left ( 1+\frac{5}{100} \right )^{2}\)
\(A = 800\times\left (\frac{105}{100} \right )^{2}\)
\(A = 800\times\left (\frac{105}{100} \right )\times\left (\frac{105}{100} \right )\)
\(A = 882\)

Let the rate of interst be \(r \%\) per annum.

According to the question,

As, Amount = Principal \(\left(1+\frac{r}{100}\right)^{n}\)

so,

\(8280= P \left(1+\frac{r}{100}\right)^{2} \ldots \ldots . .( i )\)

\(\Rightarrow\) And

\(9108=P\left(1+\frac{r}{100}\right)^{3} \ldots \ldots .\) (ii)

Dividing equation (ii) by equation (i)

we get

\(1+\frac{r}{100}=\frac{9108}{8280}\)

\(\frac{r}{100}=\frac{9108}{8280}-1\)

\(\frac{r}{100}=\frac{9108-8280}{8280}\)

\(\frac{r}{100}=\frac{828}{8280}\)

\(\frac{r}{100}=\frac{1}{10} \Rightarrow r=10 \%\)

So, the rate of compound interest \(=10 \%\) per annum.

Hence, option (A) is correct.

Increase in interest \(=6750-6000=₹ 750\)

Rate of interest \(=\frac{\text { Increase in interest }}{\text { Previous Year's interest }} \times 100\)

\(=\frac{750}{6000} \times 100=12 .5 \%\)

SI for \(3\) years = Rs \(1200\) =.> Si \(1\) year = Rs \(400\)

Given that difference between second year CI and SI is Rs \(20\). which is \(5\)% of \(400\)

Hence R=\(5\)%

We know \(5\)%=\(400\)

Hence \(100\)%=\(8000\)

i.e. sum=\(8000\)

\(\textrm{CI for 2 years= } (P\times (1+\frac{R}{100})^2)-P\)
\(=(8000\times (1+\frac{5}{100})^2)-8000=820\)

Alternate method.

SI for first years = Rs \(400\)

SI for second years = Rs \(400\)

Total SI for \(2\) years = Rs \(800\)

Given that difference between second year CI and SI is Rs \(20\)

So total CI for \(2\) years = \(800+20\)= Rs \(820\) .

(D) 6 years

\(\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{T}}\)

Let P = 1, A =2

\(\Rightarrow 2=1\left(1+\frac{\mathrm{R}}{100}\right)^{3}\)

On squaring both sides.

\(4=1\left(1+\frac{\mathrm{R}}{100}\right)^{6}\)

\(\therefore \quad\) Time \(=6\) years

3 years, therefore

\(11^{3}\) = 1331

1 \(\rightarrow\)10000(Amount) \(\rightarrow\)\(1\times10000\)=10000

3 \(\rightarrow\)\(10000\times\frac{5}{100} \) = 500\(\rightarrow\)\(3\times500\)=1500

3 \(\rightarrow\)\(500\times\frac{5}{100} \) = 25\(\rightarrow\)\(3\times25\)=75

1 \(\rightarrow\)\(25\times\frac{5}{100} \) =1.25 \(\rightarrow\)\(1\times1.25\)=1.25

Total compound interest \(= 1500 + 75 + 1.25 = 1576.25 \)