ধরি, বাড়ির দাম = x টাকা
তাহলে বিনিয়োগের জন্য বাকি টাকা = (১,০০,০০০ - x) টাকা

প্রথম বিনিয়োগে,
পরিমাণ = (১,০০,০০০ - x) × ১/৩
সুদের হার = ৪%
এক বছরের সুদ = (১,০০,০০০ - x) × ১/ৃ × ৪/১০০

দ্বিতীয় বিনিয়োগে,
পরিমাণ = (১,০০,০০০ - x) × ২/৩
সুদের হার = ৬%
এক বছরের সুদ = (১,০০,০০০ - x) × ২/৩ × ৬/১০০

এখন,
মোট সুদ = ৩২০ টাকা
(১,০০,০০০ - x) × ১/৩ × ৪/১০০ + (১,০০,০০০ - x) × ২/৩ × ৬/১০০ = ৩২০
বা, (১,০০,০০০ - x) × [১/৩ × ৪/১০০ + ২/৩ × ৬/১০০] = ৩২০
বা, (১,০০,০০০ - x) × [৪/৩০০ + ১২/৩০০] = ৩২০
বা, (১,০০,০০০ - x) × [১৬/৩০০] = ৩২০
বা, (১,০০,০০০ - x) × ১৬/৩০০ = ৩২০
বা, ১,০০,০০০ - x = ৩২০ × ৩০০/১৬
বা, ১,০০,০০০ - x = ৬,০০০
বা, x = ১,০০,০০০ - ৬,০০০
বা, x = ৯৪,০০০

মনেকরি
আসল P = 100 টাকা
মুনফা- আসল A = 100 এর 7/5 = 140 টাকা
মুনাফা I = 140 - 100 = 40 টাকা
সময় n = 4 বছর
মুনাফার হার r = ?

আমরাজানি
I = Pnr
r = I/Pn
  = (40 × 100)/(100 × 4)
  = 10%

Here,
P = M
r = M%
I = M
n = 4

We know that,
I = Pnr
M = M × 4 × (M/100)
⇒ 1 = M/25
∴ M = 25

n = (S.I x 100)/P x R
   = (420 x 100)/(3000 x 7)
   = 2 years.

interest= 1200 x (9/100) x (18/12)
           = 162 tk

C = P(1+r)ⁿ
   = 5000{1+(2/100)²}
   = 5000{1+(1/50)²}
   = 5000(51/50)²
   = 5202

মূলধন, P = 50000
সুদ, I = 1000
সময় = 2 বছর
সুদের হার, r = ?
⇒r=(I×100)/Pn
= (1000×100)/(50000 x 2)
= 1%

বিক্রয় মুল্য (SP₂) = {SP₁ x (100±X₂)}÷(100±X₁)
                       = {540 x (100+20)}÷(100-10)
                       = 720     

প্রশ্ন-পরপর দুই বছরের জন্য নির্দিষ্ট টাকার চক্রবৃদ্ধি সুদ 225 টাকা এবং 238.50টাকা হলে , বার্ষিক সুদের হার কত?

225  টাকার 1 বছরের সরল সুদ =  238.50 - 225 = 13.50  টাকা
 
আমরা জানি,
হার = সুদ / (আসল × সময়)
        = 13.50/(225 × 1) × 100%
         = 6%

(প্রশ্ন- রিফাত ২১২৫০ টাকা ৬ বছরের জন্য বিনিয়োগ করল। ৬ বছর শেষে সুদসহ ২৬৩৫০ টাকা পেতে হলে তাকে শতকরা কত হার সরল সুদে বিনিয়োগ করতে হবে)

রিফাত বিনিয়োগ করে =২১২৫০ টাকা
রিফাত ৬ বছরে বিনিয়োগ করে সুদ পায় =(২৬৩৫০ - ২১২৫০) টাকা
                        =৫১০০ টাকা

২১২০০ টাকার ৬ বছরের সুদ =৫১০০ টাকা
১ টাকার ১ বছরের সুদ ৫১০০/(২১২০০×৬) টাকা
১০০ টাকার ১ বছরের সুদ =(৫১০০×১০০)/(২১২০০×৬) =৪ টাকা

অতএব শতকরা সরল সুদ =৪%

(প্রশ্ন- ইমরান বার্ষিক ১৩% সরল সুদে রহিমের কাছ থেকে ৫২০ টাকা ঋণ নিল। ৬ মাস পর রাম গৌরবকে কত টাকা পরিশোধ করবে?)

বার্ষিক ১৩% হারে ৬ মাসে সুদ
=৫২০×১/২×১৩/১০০ টাকা    [ যেহেতু I=Pnr/100 ;৬ মাস =১/২ বছর ]
= ৩৩.৮০ টাকা

আসলসহ ৬ মাস পর পরিশোধ করতে হবে =(৫২০+৩৩.৮০) =৫৫৩.৮০ টাকা

মনে করি, আসল = a টাকা 

Time period = 2 years 73 days = 2 73/365 years = 2 1/5 years

When the interest is compounded annually, but the time is in fraction, then

\(\begin{array}{l} {\rm{A}} = {\rm{P}}{\left( {1 + {\rm{\;}}\frac{{\rm{R}}}{{100}}} \right)^2} \times \left( {1 + {\rm{\;R}} \times \frac{{\frac{1}{5}}}{{100}}} \right)\\ \Rightarrow {\rm{A}} = 30000{\left( {1 + {\rm{\;}}\frac{{10}}{{100}}} \right)^2} \times \left( {1 + 10 \times \frac{{\frac{1}{5}}}{{100}}} \right)\\ \Rightarrow {\rm{A}} = 30000{\left( {\frac{{11}}{{10}}} \right)^2} \times \left( {1 + {\rm{\;}}\frac{2}{{100}}} \right)\\ \Rightarrow {\rm{A}} = 30000{\left( {\frac{{11}}{{10}}} \right)^2} \times \left( {\frac{{102}}{{100}}} \right)\\ \Rightarrow {\rm{A}} = 30000{\rm{\;}} \times {\rm{\;}}\frac{{121}}{{100}}{\rm{\;}} \times {\rm{\;}}\left( {\frac{{102}}{{100}}} \right) \end{array}\)

⇒ A = 37026

∴ C.I. = A – P = Rs. (37026 – 30000) = Rs. 7026

Let x, y and z be the amounts invested in schemes A, B and C respectively. Then,

As we know:

Simple interest (S.I.) = principal amount (P) x interest rate (r) x time (t) /100

(x × 10 × 1/100) + (y × 12 × 1/100) + (z × 15 × 1/100) = 3200

= 10x + 12y + 15z = 320000 …(i)

Now, z = 240% of y = 12y/5 …(ii)

And, z = 150% of x =  3x/2 => x = 2/3 z = (2/3) × (12/5) × y  = 8y/5 …(iii)

From (i), (ii) and (iii), we have

16y + 12y + 36y = 320000 => 64y = 320000 => y = 5000

∴ Sum invested in Scheme B = Rs.5000

Let rate of interest in case of simple interest be M and that in compound interest be N.

We know, Simple interest = (Principal × Rate × Time)/100

And, Compound Interest = Principal \(\times \left[ {{{\left( {1 + \frac{{\rm{R}}}{{100}}} \right)}^{\rm{T}}} - {\rm{\;}}1} \right]\)

If interest earned is same in 2 years,

⇒  (Principal × M × 2)/100 = Principal × (1 + N/100)2

⇒  M/50 = 1 + N2/10000 + N/50

With this equation, the ratio of M and N cannot be determined, as sum of degrees of M and N in each term is not same.

Two equal parts of Rs. 14000 will be Rs. 7000 and Rs. 7000.

Kate invested the first part in compound interest at 15% per annum rate of interest for two years.

We know, amount = Principal × \(\left[ {{{\left( {1{\rm{\;}} + {\rm{\;}}\frac{{\rm{R}}}{{100}}} \right)}^{\rm{T}}}} \right]\)

Amount that Kate has after two years = 7000 + 7000 × \(\left[ {{{\left( {1\; + \;\frac{{15}}{{100}}} \right)}^2}} \right]\) 

= 7000 + (7000 × 1.3225) = 16257.5

After two years, she withdrew the whole amount and invested it and second part in simple interest at 12% per annum rate of interest.

Simple interest = (Principal × Rate × Time)/100

Amount = Principal + Simple Interest

Amount that Kate has after 4 years = 16257.5 + (16257.5 × 12 × 2)/100 = 16257.5 + 3901.8 = 20159.3

∴ Kate will have Rs. 20159.3 after 4 years.

We know, Simple Interest = (Principal Amount × Rate of Interest × Time Period)/100

Let rate of interest be R% per annum.

On investing Rs. 20000 in simple interest for two years, the interest earned is Rs. 190 more than the interest earned when Rs. 13000 is invested in simple interest for three years, at the same rate.

⇒ (20000 × R × 2)/100 = 190 + (13000 × R × 3)/100

⇒ 400R = 190 + 390R

⇒ 10R = 190

⇒ R = 19

∴ Rate of interest is 19% per annum.

Let the sum be Rs. P

We know that, C. I. 

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

And SI = (P × r × t)/100

Where, P = principal, r = % rate of interest, t = time in years, SI = Simple interest, CI = Compound interest

\({\rm{C}}.{\rm{I\;}} = P\left[ {{{\left( {1 + \frac{20}{{100}}} \right)}^2} - 1} \right]\)

= 44P/100

S.I = P × 20 × 2/100 = 40P/100

According to question,

C.I – S.I = Rs. 760

\(\begin{array}{l} \Rightarrow \frac{{44P}}{{100}} - \frac{{40P}}{{100}} = 760\\ \Rightarrow \frac{{4P}}{{100}} = 760 \end{array}\)

⇒ P = 760 × 100/4 = 19000

Now, we have to find simple interest on this sum at 10% rate in 5 years.

SI = (P × r × t)/100

SI = (19000 × 10 × 5)/100

SI = 9500

We know that, Simple Interest = (P × R × T)/100

Where, P = principal Amount, R = % Rate of interest, T = time period in years

Let the principle amount be Rs. x

∵ the interest rate is 5% per annum for first 2 years,

Simple Interest for first 2 years = (x × 5 × 2)/100 = x/10

∵ the interest rate is 7.5% per annum for next 2 years,

Simple Interest for next 2 years = (x × 7.5 × 2)/100 = 3x/20

Now, the time period remaining out of total time = 7 – (2 + 2) = 3 years

∵ the interest rate is 10% per annum for remaining 3 years,

Simple Interest for remaining 3 years = (x × 10 × 3)/100 = 3x/10

According to given information, total interest of 7 years = Rs. 4400

∴ 4400 = (x/10) + (3x/20) + (3x/10)

⇒ 4400 = 11x/20

⇒ x = (4400 × 20)/11

⇒ x = 8000

∴ the principal amount is Rs. 8000.

We know that,  for compound interest is given by

\(\begin{array}{l} CI = P\;{\left( {1 + \frac{R}{{100}}} \right)^T} - P\\ \Rightarrow \;CI = 8000 \times {\left( {1 + \frac{{10}}{{100}}} \right)^2} - 8000\\ \Rightarrow CI = \;8000 \times \left( {\frac{{121}}{{100}} - 1} \right)\\ \Rightarrow CI = 8000 \times \frac{{21}}{{100}} = 1680 \end{array}\)

Difference in amount after 3rd year and 6th year, i.e., for a tenure of 3 years = Rs.(12167 – 8000) = Rs. 4167

Thus, Compound Interest for 3 years = Rs. 4167

Let the principal be Rs. 8000 and CI be Rs. 4167 for a time period of 3 years.

We know the formula for compound interest

\(\Rightarrow {\rm{CI}} = \left[ {{\rm{P}}\left\{ {{{\left( {1 + \frac{{\rm{R}}}{{100}}} \right)}^t} - 1} \right\}} \right]\)

Where,

CI = Compound interest

P = Principal

R = Rate of interest

T = Time period

\(\begin{array}{l} \Rightarrow 4167 = \left[ {8000\left\{ {{{\left( {1 + \frac{{\rm{R}}}{{100}}} \right)}^3} - 1} \right\}} \right]\\ \Rightarrow \frac{{4167}}{{8000}} = \left[ {\left\{ {{{\left( {1 + \frac{{\rm{R}}}{{100}}} \right)}^3} - 1} \right\}} \right]\\ \Rightarrow \frac{{4167}}{{8000}} + 1 = \left[ {\left\{ {{{\left( {1 + \frac{{\rm{R}}}{{100}}} \right)}^3}} \right\}} \right]\\ \Rightarrow \frac{{12167}}{{8000}} = \left[ {\left\{ {{{\left( {1 + \frac{{\rm{R}}}{{100}}} \right)}^3}} \right\}} \right]\\ \Rightarrow {\left( {\frac{{23}}{{20}}} \right)^3} = {\left( {1 + \frac{{\rm{R}}}{{100}}} \right)^3}\\ \Rightarrow \frac{{23}}{{20}} = \;1 + \frac{{\rm{R}}}{{100}}\; \end{array}\)

⇒ R = (3 × 100)/20 = 15% p.a.

Let the original sum be Rs. x

Amount after 3 years = Rs. 8000

We know the formula for amount-

\(\Rightarrow {\rm{A}} = \left[ {{\rm{P}}\left\{ {{{\left( {1 + \frac{{\rm{r}}}{{100}}} \right)}^t}} \right\}} \right]\)

Where,

A =Amount

P = Principal

R = Rate of interest

T = Time period

\(\begin{array}{l} \Rightarrow 8000 = \left[ {{\rm{x}}\left\{ {{{\left( {1 + \frac{{15}}{{100}}} \right)}^3}} \right\}} \right]\;\\ \Rightarrow 8000 = \left[ {{\rm{x}}\left\{ {{{\left( {1 + \frac{3}{{20}}} \right)}^3}} \right\}} \right]\\ \Rightarrow 8000 = \left[ {{\rm{x}}\left\{ {{{\left( {\frac{{23}}{{20}}} \right)}^3}} \right\}} \right]\\ \Rightarrow 8000 = \left[ {{\rm{x\;}} \times {\rm{\;}}\frac{{12167}}{{8000}}} \right] \end{array}\)

⇒ x = (8000 × 8000)/ 12167

⇒ x = 5260 (approx)

Thus, the original sum is approximately Rs. 5,260.

Money invested by George is Rs. 13000 for 2 years at 20% per annum in compound interest.

We know, in case of compound interest,

\(Amount = Principal \times {\left( {1 + \;\frac{{Rate}}{{100}}} \right)^{\;Time}}\)

Amount George will have after 2 years = 13000 × (1 + (20/100))2 = Rs. 18720

After 2 years, it was notified that the scheme will now work on the principle of simple interest. After further 2 years, George withdrew all the money from investment.

⇒ For next 2 years, George has invested Rs. 18720 at 20% per annum in simple interest.

We know, in case of simple interest,

\(Amount = Principal + \;\frac{{\left( {Principal \times Rate \times Time} \right)}}{{100}}\)

Amount George will have after next 2 years = 18720 + [(18720 × 20 × 2)/100] = Rs. 26208

Due to pre mature withdrawal, a penalty of Rs. 1000 was charged.

Amount of money that George will get = Rs. 26208 – Rs. 1000 = Rs. 25208

We know that, SI = (P × R × T)/100

Where, SI = Simple interest, P = principal, R = % rate of interest, T = time in years

Let the principal invested by Anil be p.

Anil invests 3/5th of p at 8% per annum for 2 years.

⇒ SI1 \(= \;p\; \times \;\frac{3}{5} \times \frac{8}{{100}} \times 2 = 0.096p\) ……….. (1)

Anil invests 1/4th of p at 15% per annum for 2 years.

⇒ SI2 \(= p\; \times \;\frac{1}{4} \times \frac{{15}}{{100}} \times 2 = 0.075p\) …………. (2)

Remaining principle \(= \;p\; - \;\frac{{3p}}{5} - \frac{p}{4} = \frac{{3p}}{{20}}\)

Anil invests 3/20th of p at 10% per annum for 2 years.

⇒ SI3 \(= \;p\; \times \;\frac{3}{{20}} \times \frac{{10}}{{100}} \times 2 = 0.03p\) ……….(3)

∴ As per given information:

SI1 + SI2 + SI3 = 2 × 4824 = 9648

⇒ 0.096p + 0.075p + 0.03p = 9648

⇒ 0.201p = 9648

⇒ p = 48000

Hence, Anil have a principal amount is Rs. 48000

The value of 2275 after 3 years = the value of the first installment after 2 year + the value of second installment after 1 year + the value of third installment

Let the amount of installment be x

\(\Rightarrow \;2275\;{\left( {1 + \frac{{20}}{{100}}} \right)^3}\; = \;x\;{\left( {1 + \frac{{20}}{{100}}} \right)^2}\; + \;x\;{\left( {1 + \frac{{20}}{{100}}} \right)^1}\; + \;x\)

\(\Rightarrow \;2275\;\left( {\frac{{216}}{{125}}} \right)\; = \;x\;\left( {\frac{{36}}{{25}}} \right)\; + \;x\left( {\frac{6}{5}} \right)\; + \;x\)

\(\Rightarrow \;2275\;\left( {\frac{{216}}{{125}}} \right)\; = \left( {\frac{{\;36 + 30 + 25}}{{25}}} \right)\;x\)

\(\Rightarrow \;2275\;\left( {\frac{{216}}{{125}}} \right)\; = \;\frac{{91}}{{25}}\;\;x\)

\(\Rightarrow \;x\; = \;\frac{{\;2275 \times 216 \times 25}}{{125 \times 91}}\)

⇒ x = 1080

Hence, Amount of each investment is Rs. 1080

Given that Mr. Kailash borrowed Rs. 40000 from a bank

He takes loan at 10% compound interest per annum.

When interest is compounded annually:

Amount after t years \(= P{\left( {1 + \frac{R}{{100}}} \right)^t}\) (where P = Principal, R = Rate% and t = years)

Amount at the end of 1st year \(= 40000{\left( {1 + \frac{{10}}{{100}}} \right)^1}\)

= 40000 × 11/10 = 44000

Amount to be paid after paying 1st installment = 44000 – 10000 = 34000

Amount at the end of the 2nd year \(= 34000{\left( {1 + \frac{{10}}{{100}}} \right)^1}\)

= 34000 × 11/10 = 37400

Amount to be paid after paying 2nd installment = 37400 – 10000 = 27400

Amount at the end of 3rd year \(= 27400{\left( {1 + \frac{{10}}{{100}}} \right)^1}\)

= 27400 × 11/10 = 30140

Amount to be paid after paying 3rd installment = 30140 – 10000 = 20140

Amount at the end of 4th year \(= 20140{\left( {1 + \frac{{10}}{{100}}} \right)^1}\)

= 20140 × 11/10 = 22154

Amount to be paid at the end of 4th year to clear the lone = 22154

Let the Principal be ‘a’.

∴ Interest = a/2

Let the time be ‘t’ years. Since the rate is twice the time,

∴ Rate of interest = 2 × t = 2t%

We know that,

Simple Interest = (P × R × T)/100

Where, P = principal Amount, R = % Rate of interest, T = time period in years

Here, P = a; R = 2t; T = t; SI = a/2

\(\therefore \frac{a}{2}\; = \;\frac{{a\; \times \;t\; \times \;2t}}{{100}}\)

⇒ 2t2 = 50

⇒ t2 = 25

⇒ t = 5

∴ Time = t = 5 years

⇒ Rate of interest = 2t = 2 × 5 = 10 %

{Formula :
If P is the principle kept at Compound Interest @ r% p.a , amount after n years will be P(1+ r/100)n}

Farida invests an amount of Rs. 4,800 for 3 years at the rate of 8 p.c.p.a.

So amount of the principle after 3 years will be \(= P{\left( {1 + \frac{r}{{100}}} \right)^n}\)

\(= 4800{\left( {1 + \frac{8}{{100}}} \right)^3}\)

\(= 4800{\left( {\frac{{108}}{{100}}} \right)^3}\)

= 6046.61

Amount of compound interest after 3 years = Rs (6046.61 – 4800) = Rs 1246.61 ≈ Rs 1247

Hence the answer is Rs 1247

Let the principal sum be Rs. x

We know the formula for compound interest and simple interest-

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

\(\Rightarrow {\rm{CI}} = \left[ {{\rm{P}}\left\{ {{{\left( {1 + \frac{{\rm{r}}}{{100}}} \right)}^t} - 1} \right\}} \right]\) 

Where,

CI = Compound interest

SI = Simple Interest

P = Principal

R = Rate of interest

T = Time period

Given: Rate of Interest = 15% p.a.

At the end of 3 years, difference between C.I. and S.I = Rs. 595.35

\(\therefore \;\left[ {x\left\{ {{{\left( {1 + \frac{{15}}{{100}}} \right)}^3} - 1} \right\}} \right] - \;\frac{{x\; \times 15\; \times 3}}{{100}}\; = \;Rs.\;595.35\)

\(\Rightarrow \left[ {x\left\{ {{{\left( {\frac{{115}}{{100}}} \right)}^3} - 1} \right\}} \right] - \frac{{45x}}{{100}} = 595.35\)

\(\Rightarrow \left[ {\frac{{4168}}{{8000}}x} \right] - \;\frac{{45x}}{{100}} = \;595.35\)

\(\Rightarrow \;\frac{{567x}}{{8000}} = 595.35\)

⇒ x = Rs. 8,400

Thus, the principal sum =Rs. 8,400.

Let the simple interest be Rs. x

Principal sum = Rs. 4,800

Rate of interest = 8.5% p.a. = 17/2 % p.a.

Time period = 2 years 3 months = \(2\frac{3}{{12}}\) year = 9/4 year

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

\(\Rightarrow {\rm{S}}.{\rm{I}}.{\rm{\;}} = \frac{{4800\; \times \frac{{17}}{2}\; \times \frac{9}{4}}}{{100}}\)

Thus, S.I.= Rs. 918

Therefore, the simple interest = Rs. 918.

Simple interest accrued on a certain principal is rupees 6,00,000 in 5 years at a rate of 12% per annum.

We know that, S.I. = (P × R × T)/100    where, P = principal, R = rate % per annum, T = time in years

Here, S.I. = 6,00,000, R = 12%, T = 5 years

⇒ 6,00,000 = (P × 12 × 5)/100

\(\Rightarrow {\rm{\;P\;}} = {\rm{\;}}\frac{{6,00,000 \times 100}}{{5 \times 12}}\; = {\rm{\;}}10,00,000\)

Now let’s find the compound interest on principal amount P = 10,00,000, R = 3%, T = 2 years

Compound interest \(= {\rm{\;}}P{\left( {1 + \frac{r}{{100}}} \right)^{n\;}}-\;P\)

Compound interest

\(\begin{array}{l} = {\rm{P}}\left[ {{{\left( {1 + \frac{r}{{100}}} \right)}^n} - 1} \right] = 1000000\left[ {{{\left( {1 + \frac{3}{{100}}} \right)}^2} - 1} \right]\\ = 10,00,000\left[ {{{\left( {1 + \frac{3}{{100}}} \right)}^2} - 1} \right] = \;1000000\left[ {{{\left( {\frac{{103}}{{100}}} \right)}^2} - 1} \right]\\ = {\rm{\;}}10,00,000\left[ {\frac{{10609}}{{10000}} - 1} \right] = \;1000000\left[ {\frac{{609}}{{10000}}} \right] = 60900 \end{array}\)

C.I = Amount – Principal (P)

Amount \(= \;P\;{\left( {1 + \frac{r}{{100}}} \right)^n}\) where n = time period; r = rate % p. a

\(\begin{array}{l} \therefore \;1680\; = \;8000\;{\left( {1\; + \frac{r}{{100}}} \right)^2}\;-\;8000\\ \Rightarrow \frac{{9680}}{{8000}}\; = \;{\left( {1\; + \frac{r}{{100}}} \right)^2}\\ \Rightarrow \frac{{121}}{{100}}\; = \;{\left( {1\; + \frac{r}{{100}}} \right)^2}\\ \Rightarrow \frac{{11}}{{10}}\; = \;1\; + \frac{r}{{100}} \end{array}\)

⇒ r = 10% p. a                                 ---(1)

Now, S.I = Pnr/100

Where, P = principal amount, n = time period; r = rate % p. a

For this case, r = 2 × 10 = 20% p.a.

And n = 2/2 = 1 year

⇒ S.I = (8,000 × 20 × 1)/100 (from 1)

Thus, the simple interest is Rs. 1600