Statistics & Applications

Financial Math

Medium

applied

Mr. Mittal invested Rs. 20,000 in a mutual fund in the year 2019. The value of the mutual fund increased to Rs. 32,000 in the year 2024. The compound annual growth rate of his investment is:

[Given $(1.6)^{1/5} = 1.098$ ]

8.5%

10%

9.8%

9.1%

✅ Correct Option: 3

Mr. Mittal's investment grew from Rs. 20,000 to Rs. 32,000 over the period from 2019 to 2024.

Time period: 2024 - 2019 = 5 years

For compound growth, the relationship between initial and final values is:

Final Value = Initial Value × $(1 + r)^n$

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

Substituting the given values:

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

Dividing both sides by 20,000:

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

$1.6 = (1 + r)^5$

Taking the 5th root of both sides:

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

Given that $(1.6)^{1/5} = 1.098$ :

$(1 + r) = 1.098$

$r = 1.098 - 1$

$r = 0.098$

Converting to percentage:

$r = 0.098 \times 100 = 9.8\%$

Therefore, the compound annual growth rate is 9.8%.

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A money-lender charges interest at the rate of 10 rupees per 110 rupees semi-annually, payable in advance. Then

Match the LIST-I with LIST-II

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Choose the correct answer from the options given below:

Mr. Mittal invested Rs. 20,000 in a mutual fund in the year 2019. The value of the mutual fund increased to Rs. 32,000 in the year 2024. The compound annual growth rate of his investment is: [Given (1.6)1/5=1.098(1.6)^{1/5} = 1.098(1.6)1/5=1.098]

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

Mr. Mittal invested Rs. 20,000 in a mutual fund in the year 2019. The value of the mutual fund increased to Rs. 32,000 in the year 2024. The compound annual growth rate of his investment is:

[Given $(1.6)^{1/5} = 1.098$ ]