Statistics & Applications

Financial Math

Medium

applied

Ram invested Rs.20,000 in a mutual fund in the year 2012. The value of the mutual fund increased to Rs.32,000 in the year 2017, then the compound annual growth rate of his investment is (Given that $(1.6)^{\frac{1}{5}} = 1.098$ )

10.8%

7.8%

9.8%

6.8%

✅ Correct Option: 3

The initial investment in 2012 is Rs. 20,000.

The value in 2017 is Rs. 32,000.

Time period = 2017 - 2012 = 5 years

Given: $(1.6)^{\frac{1}{5}} = 1.098$

The compound growth formula is:

$\text{Final Amount} = \text{Initial Amount} \times (1 + r)^n$

where $r$ is the growth rate and $n$ is the number of years.

Substituting the values:

$32,000 = 20,000 \times (1 + r)^5$

$\frac{32,000}{20,000} = (1 + r)^5$

$1.6 = (1 + r)^5$

Taking the 5th root of both sides:

$(1.6)^{\frac{1}{5}} = 1 + r$

$1.098 = 1 + r$

$r = 1.098 - 1$

$r = 0.098$

Converting to percentage:

$r = 0.098 \times 100$

$r = 9.8\%$

Therefore, the compound annual growth rate is 9.8%.

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Ram invested Rs.20,000 in a mutual fund in the year 2012. The value of the mutual fund increased to Rs.32,000 in the year 2017, then the compound annual growth rate of his investment is (Given that (1.6)15=1.098(1.6)^{\frac{1}{5}} = 1.098(1.6)51​=1.098)

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CUET Mathematics Past Year Questions (Topic-Wise)

Ram invested Rs.20,000 in a mutual fund in the year 2012. The value of the mutual fund increased to Rs.32,000 in the year 2017, then the compound annual growth rate of his investment is (Given that $(1.6)^{\frac{1}{5}} = 1.098$ )