A sum of money invested at compound interest amounts to \u20b9 400 in 2 years and to \u20b9 420 in 3 years. The rate of interest per annum is<\/p>\n
[amp_mcq option1=”2.5%” option2=”5.5%” option3=”4%” option4=”5%” correct=”option4″]<\/p>\n
So, we have two equations:
\n1) $400 = P(1+r)^2$
\n2) $420 = P(1+r)^3$<\/p>\n
To find the rate r, divide equation (2) by equation (1):
\n$\\frac{420}{400} = \\frac{P(1+r)^3}{P(1+r)^2}$
\n$\\frac{42}{40} = (1+r)^{3-2}$
\n$\\frac{21}{20} = 1+r$<\/p>\n
Now, solve for r:
\n$r = \\frac{21}{20} – 1$
\n$r = \\frac{21 – 20}{20}$
\n$r = \\frac{1}{20}$<\/p>\n
To express the rate as a percentage, multiply by 100: A sum of money invested at compound interest amounts to \u20b9 400 in 2 years and to \u20b9 420 in 3 years. The rate of interest per annum is [amp_mcq option1=”2.5%” option2=”5.5%” option3=”4%” option4=”5%” correct=”option4″] This question was previously asked in UPSC CBI DSP LDCE – 2023 Download PDFAttempt Online Let the principal amount be … <\/p>\n
\nRate = $\\frac{1}{20} \\times 100\\% = 5\\%$.
\n<\/section>\n
\n$\\frac{A_{t+1}}{A_t} = 1+r$
\n<\/section>\n
\nThe interest rate can be calculated as the interest earned in the 3rd year divided by the principal at the beginning of the 3rd year (which is the amount after 2 years).
\nInterest rate = $\\frac{\\text{Interest earned in 3rd year}}{\\text{Amount after 2 years}} = \\frac{20}{400} = \\frac{1}{20}$.
\nAs a percentage, this is $\\frac{1}{20} \\times 100\\% = 5\\%$. This method is valid only because the time difference is exactly one year.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"