CAT 2024 Slot 2 QA Question 6

Multiple choice (+3 / −1) · Arithmetic · Simple & Compound Interest · Try it, then check the answer and solution below.

CAT 2024 Slot 2QAArithmetic • Simple & Compound InterestModerate

Anil invests Rs 22000 for 6 years in a scheme with 4% interest per annum, compounded half-yearly. Separately, Sunil invests a certain amount in the same scheme for 5 years, and then reinvests the entire amount he receives at the end of 5 years, for one year at 10% simple interest. If the amounts received by both at the end of 6 years are equal, then the initial investment, in rupees, made by Sunil is

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Official Convenor Answer Key: B

Official Correct Answer: B. First, the problem says Anil invests Rs 22000 for 6 years at a 4% annual interest rate, compounded half-yearly. On the other hand, Sunil invests a certain amount in the same scheme for 5 years, then takes that amount and reinvests it for one year at 10% simple interest. At the end of 6 years, both Anil and Sunil have the same amount. We find out how much Sunil initially invested. Okay, let's break this down. compound interest can be calculated using the formula: $$ A = P \left(1 + \frac{r}{n}\right)^{nt} $$ where: - $ A $ is the amount after time $ t $, - $ P $ is the principal amount, - $ r $ is the annual interest rate (in decimal), - $ n $ is the number of times interest is compounded per year, - $ t $ is the time in years. Since Anil's investment is compounded half-yearly, $ n = 2 $. His principal $ P $ is 22000, rate $ r = 4\% = 0.04 $, and time $ t = 6 $ years. Compute Anil's amount first. So, plugging into the formula: $$ A_{Anil} = 22000 \left(1 + \frac{0.04}{2}\right)^{2 \times 6} $$ Simplify the inside of the brackets: $$ 1 + \frac{0.04}{2} = 1 + 0.02 = 1.02 $$ And the exponent: $$ 2 \times 6 = 12 $$ So, $$ A_{Anil} = 22000 \times (1.02)^{12} $$ I think I can calculate $ (1.02)^{12} $ using logarithms or maybe $ (1.02)^{12} $ is approximately 1.2682. Hmm, 1.02^12 is indeed around 1.2682. So, $$ A_{Anil} = 22000 \times 1.2682 $$ Calculating that: 22000 * 1.2682. Compute 22000 * 1.2682. First, 22000 * 1 = 22000. 22000 * 0.2682 = ? 0.2682 * 22000: 0.2 * 22000 = 4400 0.06 * 22000 = 1320 0.0082 * 22000 = 180.4 Adding those up: 4400 + 1320 = 5720; 5720 + 180.4 = 5900.4 So total amount is 22000 + 5900.4 = 27900.4 Wait, that seems a bit high. Alternatively, 22000 * 1.2682: 22000 * 1 = 22000 22000 * 0.2 = 4400 22000 * 0.06 = 1320 22000 * 0.0082 = 180.4 Adding all together: 22000 + 4400 = 26400; 26400 + 1320 = 27720; 27720 + 180.4 = 27900.4 Yes, that's correct. So Anil's amount after 6 years is Rs 27,900.40. Now, moving on to Sunil. He invests a certain amount, let's call it $ P $, for 5 years at the same 4% annual interest rate, compounded half-yearly. Then he takes that amount and invests it at 10% simple interest for one year. First, Computing the amount Sunil gets after 5 years from the first investment. Using the same compound interest formula: $$ A_{Sunil1} = P \left(1 + \frac{0.04}{2}\right)^{2 \times 5} $$ Simplify: $$ 1 + 0.02 = 1.02 $$ Exponent: $$ 2 \times 5 = 10 $$ So, $$ A_{Sunil1} = P \times (1.02)^{10} $$ $ (1.02)^{10} $ is approximately 1.21899. Let me confirm that. Yes, 1.02^10 ≈ 1.21899. So, $$ A_{Sunil1} = P \times 1.21899 $$ Then, Sunil reinvests this amount at 10% simple interest for one year. The simple interest formula is: $$ A = P' + P' \times r \times t $$ where $ P' $ is the principal for the simple interest investment, $ r $ is the rate, and $ t $ is the time. In this case, $ P' = A_{Sunil1} = 1.21899P $, $ r = 10\% = 0.10 $, $ t = 1 $ year. So, $$ A_{Sunil2} = 1.21899P + (1.21899P \times 0.10 \times 1) $$ Simplify: $$ A_{Sunil2} = 1.21899P + 0.121899P = 1.340889P $$ So, after 6 years, Sunil's total amount is $ 1.340889P $. According to the problem, this amount is equal to Anil's amount, which is 27,900.4

Answer and solution

Answer: B) 20808

First, the problem says Anil invests Rs 22000 for 6 years at a 4% annual interest rate, compounded half-yearly. On the other hand, Sunil invests a certain amount in the same scheme for 5 years, then takes that amount and reinvests it for one year at 10% simple interest. At the end of 6 years, both Anil and Sunil have the same amount. We find out how much Sunil initially invested.

Okay, let's break this down. compound interest can be calculated using the formula:

A = P \left(1 + \frac{r}{n}\right)^{nt}

where:

A

is the amount after time

t

P

is the principal amount,

r

is the annual interest rate (in decimal),

n

is the number of times interest is compounded per year,

t

is the time in years.

Since Anil's investment is compounded half-yearly,

n = 2

. His principal

P

is 22000, rate

r = 4\% = 0.04 , and time

t = 6$ years.

Compute Anil's amount first.

So, plugging into the formula:

A_{Anil} = 22000 \left(1 + \frac{0.04}{2}\right)^{2 \times 6}

Simplify the inside of the brackets:

1 + \frac{0.04}{2} = 1 + 0.02 = 1.02

And the exponent:

2 \times 6 = 12

So,

A_{Anil} = 22000 \times (1.02)^{12}

I think I can calculate

(1.02)^{12}

using logarithms or maybe

(1.02)^{12}

is approximately 1.2682. Hmm, 1.02^12 is indeed around 1.2682. So,

A_{Anil} = 22000 \times 1.2682

Calculating that:

22000 1.2682. Compute 22000 1.2682.

First, 22000 * 1 = 22000.

22000 * 0.2682 = ?

0.2682 * 22000:

0.2 * 22000 = 4400

0.06 * 22000 = 1320

0.0082 * 22000 = 180.4

Adding those up: 4400 + 1320 = 5720; 5720 + 180.4 = 5900.4

So total amount is 22000 + 5900.4 = 27900.4

Wait, that seems a bit high.

Alternatively, 22000 * 1.2682:

22000 * 1 = 22000

22000 * 0.2 = 4400

22000 * 0.06 = 1320

22000 * 0.0082 = 180.4

Adding all together: 22000 + 4400 = 26400; 26400 + 1320 = 27720; 27720 + 180.4 = 27900.4

Yes, that's correct. So Anil's amount after 6 years is Rs 27,900.40.

Now, moving on to Sunil. He invests a certain amount, let's call it $P , for 5 years at the same 4\% annual interest rate, compounded half-yearly. Then he takes that amount and invests it at 10\% simple interest for one year.

First, Computing the amount Sunil gets after 5 years from the first investment.

Using the same compound interest formula:

A_{Sunil1} = P \left(1 + \frac{0.04}{2}\right)^{2 \times 5}

Simplify:

1 + 0.02 = 1.02

Exponent:

2 \times 5 = 10

So,

A_{Sunil1} = P \times (1.02)^{10}

(1.02)^{10}

is approximately 1.21899. Let me confirm that. Yes, 1.02^10 ≈ 1.21899.

So,

A_{Sunil1} = P \times 1.21899

Then, Sunil reinvests this amount at 10\% simple interest for one year. The simple interest formula is:

A = P' + P' \times r \times t

where

P'

is the principal for the simple interest investment,

r

is the rate, and

t

is the time.

In this case,

P' = A_{Sunil1} = 1.21899P , r = 10\% = 0.10 , t = 1

year.

So,

A_{Sunil2} = 1.21899P + (1.21899P \times 0.10 \times 1)

Simplify:

A_{Sunil2} = 1.21899P + 0.121899P = 1.340889P

So, after 6 years, Sunil's total amount is

1.340889P

According to the problem, this amount is equal to Anil's amount, which is 27,900.4

CAT 2024 Slot 2 QA Question 6

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