Guest

☰ Topics

P is the Principal (Amount borrowed or lent)

N is the number of years for which the sum is borrowed or lent (N = 5 in this example)

R is the annual rate of interest

\begin{align} SI &= \frac{25000 × 5 × 10}{100} \\ &= \frac{25000}{2} \\ &= 12500 \end{align}

P is the Principal (Amount borrowed or lent)

N is the number of years for which the sum is borrowed or lent (N = $\frac{6}{12}$ = 0.5 in this example)

R is the annual rate of interest

\begin{align} SI &= \frac{60000 × 0.5 × 5}{100} \\ &= \frac{60000 × 2.5}{100} \\ &= 1500 \end{align}

(Simple interest on $x$ for 3 years at 7% per annum) - (Simple interest on $x$ for 2 years at 6% per annum) = 900 \begin{align} ∴ \frac{x × 3 × 7}{100} - \frac{x × 2 × 6}{100} &= 900\\ \frac{21x}{100} - \frac{12x}{100} &= 900 \\ \frac{9x}{100} &= 900 \\ x &= 10000 \end{align}

∴ the sum invested in FD2 = $x + \frac{100}{100} × x = 2x$

and the sum invested in FD3 = $x + \frac{400}{100} × x = 5x$

(Simple interest on FD1) + (Simple interest on FD2) + (Simple interest on FD3) = 10400

\begin{align} \frac{x × 1 × 5}{100} + \frac{2x × 1 × 6}{100} + \frac{5x × 1 × 7}{100} &= 10400 \\ \frac{5x}{100} + \frac{12x}{100} + \frac{35x}{100} &= 10400 \\ \frac{52x}{100} &= 10400 \\ x &= 20000 \end{align}

∴ the sum invested in second share = ₹ 50000 - $x$

(Simple interest on share 1) + (Simple interest on share 2) = (Simple interest on ₹ 50000)

\begin{align} \frac{x × 1 × 11}{100} + \frac{(50000 - x) × 1 × 14}{100} &= \frac{50000 × 1 × 12.8}{100} \\ \frac{11x}{100} + \frac{(50000 - x) × 14}{100} &= 6400 \\ \frac{11x}{100} + \frac{(700000 - 14x)}{100} &= 6400 \\ \frac{11x - 14x + 700000}{100} &= 6400 \\ 11x - 14x &= 640000 - 700000 \\ -3x &= -60000 \\ x &= 20000 \end{align} ∴ The sum invested in the first share is ₹ 20000 and the sum invested in second share = ₹ (50000 - 20000) = ₹ 30000

P is the Principal (Amount borrowed or lent)

n is the number of times interest is compounded per year (n = 1 in this example)

t is the number of years the money is borrowed or lent (t = 5 in this example)

r is the annual rate of interest

\begin{align} CI &= 25000 × (1 + \frac{10}{100})^{1 × 5} - 25000 \\ &= 25000 × (1 + 0.1)^{1 × 5} - 25000 \\ &= 25000 × 1.61051 - 25000 \\ &= 15262.75 \end{align}

A is the amount at the end of period t

P is the Principal (Amount borrowed or lent)

n is the number of times interest is compounded per year (n = 1 in this example)

t is the number of years the money is borrowed or lent (t = 3 in this example)

r is the annual rate of interest

\begin{align} A &= 64000 × (1 + \frac{7 \frac{1}{2}}{100})^{1 × 3} \\ &= 64000 × (1 + 0.075)^{1 × 3}\\ &= 64000 × 1.2423 \\ &= 79507 \end{align}

P is the Principal (Amount borrowed or lent)

n is the number of times interest is compounded per year (n = 1 in this example)

t is the number of complete years (not partial) the money is borrowed or lent (t = 3 in this example)

r is the annual rate of interest

\begin{align} CI & = 30000 × (1 + \frac{8}{100})^{3} × (1 + \frac{\frac{3}{4} × 8}{100}) \\ &= 30000 × (\frac{108}{100})^{3} × (1 + \frac{6}{100}) \\ &= 30000 × (\frac{108}{100})^{3} × (\frac{106}{100}) \\ &= 30000 × (\frac{108}{100}) × (\frac{108}{100}) × (\frac{108}{100}) × (\frac{106}{100}) \\ &= 40058.84 \end{align}

P is the Principal (Amount borrowed or lent)

p is the annual rate of interest during the first year (p = 8% in this example)

q is the annual rate of interest during the second year (q = 9% in this example)

\begin{align} A &= 50000 × (1 + \frac{8}{100}) × (1 + \frac{9}{100}) \\ &= 25000 × (\frac{108}{100}) × (\frac{109}{100}) \\ &= 58860 \end{align}

A is the amount at the end of period t

P is the Principal (Amount borrowed or lent)

n is the number of times interest is compounded per year (n = 1 in this example)

t is the number of years the money is borrowed or lent (t = 3 in this example)

r is the annual rate of interest

\begin{align} 7290 &= 6250 × (1 + \frac{8}{100})^{1 × t} \\ &= 6250 × (\frac{108}{100})^{1 × t}\\ \frac{7290}{6250} &= (\frac{108}{100})^{1 × t}\\ (\frac{27}{25})^{2} &= (\frac{27}{25})^{t}\\ ∴ t &= 2 \end{align}

A is the amount at the end of period t

P is the Principal (Amount borrowed or lent)

t is the number of years the money is borrowed or lent (t = 3 in this example)

r is the annual rate of interest

Please note that n is treated as 1 here i.e. it is assumed that compounding is happening on an annual basis

\begin{align} A &= 125000 × (1 + \frac{2}{100})^{3} \\ &= 125000 × (\frac{102}{100})^{3}\\ &= 125000 × (1.02)^{3}\\ &= 125000 × 1.061208\\ &= 132651 \end{align}

A is the population at the end of 2 years period

P is the initial population (in year 2011, in this example, 120000)

p is the annual rate of increase during the first year (p = 6% in this example)

q is the annual rate of decrease during the second year (q = 5% in this example)

\begin{align} A &= 120000 × (1 + \frac{6}{100}) × (1 - \frac{5}{100}) \\ &= 120000 × (\frac{106}{100}) × (\frac{95}{100}) \\ &= 120000 × 1.06 × 0.95 \\ &= 120840 \end{align}

A is the amount at the end of period t

P is the Principal (Amount borrowed or lent)

n is the number of times interest is compounded per year (n = 4 in this example)

t is the number of years the money is borrowed or lent (t = 1 in this example)

r is the annual rate of interest

\begin{align} A &= 390625 × (1 + \frac{16}{4 × 100})^{4 × 1} \\ &= 390625 × (1 + \frac{4}{100})^{4}\\ &= 390625 × (1.04)^{4}\\ &= 390625 × 1.16986\\ &= 456976.5 \end{align}