Test: Simple And Compound Interest Including Annuity - 1 - CA Foundation MCQ

The formula for Simple Interest (SI) is: SI = (P x R x T) / 100

Where, P = Principal amount (Rs. 3500) R = Rate of interest (12%) T = Time (3 years)

So, SI = (3500 x 12 x 3) / 100 = 1260

Therefore, the simple interest on Rs. 3500 for 3 years at 12% per annum is Rs. 1260. So the answer is 2. Rs. 1260

Calculation of Rate Percent per Annum

- Given:
- Principal amount (P) = Rs. 12000
- Amount (A) = Rs. 16500
- Time (T) = 2 ½ years

- We know that:
- Simple Interest (SI) = A - P
- SI = P * R * T / 100

- Substituting the given values:
- A - P = P * R * T / 100
- 16500 - 12000 = 12000 * R * 2.5 / 100
- 4500 = 300 * R
- R = 4500 / 300
- R = 15%

Therefore, the rate percent per annum is 15%, which corresponds to option A.

⇒ Hence, r = 5/100 = 0.05.

⇒ So, A = 100(1 + 0.05/1)^(1×20).

⇒ This yields A = 100(1 + 0.05)^20 = 100(1.05)^20.

⇒ This comes out to A =~ Rs. 265.33.

Therefore, Rs. 100 will become approximately Rs. 265.33 after 20 years at 5% per annum compound interest.

Hence, the correct answer is approximately 265.50.

The amount after 1 year at 8% per annum when interest is compounded half yearly is

=100(1+3/2/100)^2×1=100(101.5/100)²=103.0225
CI for 1 year = 103.0225 – 100 = 3.0225
The effective annual rate of interest is = 3.0225%

Cost of machine (P) = Rs 1,00,000

Scrap value (A) = Rs 30,000

Rate of Depreciation = 20% per annum on reducing value

The effective life of the machine in years is the number of years in which P (Rs 1,00,000) would reduce to A (scrap value Rs 30,000) reducing at the rate of 20% per annum of the value at the start of that year year.

Value of the machine at time t= 0 years = P

The depreciated cost at end of one year = P[1 — 20%] = P[1 — 0.2] = P × 0.8

At the end of second year = P × 0.8²

At the end of 3rd year = P × 0.8³

And so on.

Let after n years the value depreciate to scrap value. We are required to find n.

P(0.8)^n = A

1,00000 (0.8)^n = 30,000

=> (0.8)^n = (30,000)/(1,00,000) = 0.3

Taking log of both sides

n log (0.8) = log (0.3)

=>n × (-0.09691) = (-0.52288)

=> n = (-0.52288)/(-0.09691)= 5.396 year ~5.4 years

To determine the number of years it takes for the population to increase by 40% given a 2% annual increase, we can use the compound interest formula:

Thus, the number of years it takes for the population to increase by 40% is approximately 17 years.

Let the principal be P. The rate of interest is 6% per annum, and the time period is 3 years.

So, the sum of money is Rs. 10,000.

• The nominal interest rate is 6% per annum, compounded half-yearly.
• This means the interest is applied twice a year at a rate of 3% (6% / 2).
• To find the effective annual rate (EAR), use the formula: EAR = (1 + r/n)^(nt) - 1, where r is the nominal rate, n is the number of compounding periods per year, and t is the number of years.
• Plugging in the values: EAR = (1 + 0.03)^2 - 1 = 0.0609 or 6.09%.

Given: b( birth rate)=39.4/1000, d(death rate) =19.4/1000

b-d= 39.4–19.4=20/1000 i.e.0.02

P( population) = 2P( population)

To find: t=?

Solution :Let's assume P=1000

applying formula:

An= p(1+i)ⁿ

Putting values:

2000=1000(1+0.02)ⁿ

2=1.02ⁿ

n=35 years.

Calculation of Compound Interest
 

• Principal amount (P) = Rs. 4000


• Rate of interest (R) = 12% per annum


• Time period (T) = 6 months = 0.5 years


• Interest compounded quarterly, so n = 4


• Using the formula for compound interest: A = P(1 + R/n)^(nt)


• Substitute the values: A = 4000(1 + 0.12/4)^(4*0.5)


• Calculating further, A = 4000(1 + 0.03)^2


• A = 4000(1.03)^2 = 4000(1.0609) = Rs. 4243.60


• Compound Interest (CI) = A - P = 4243.60 - 4000 = Rs. 243.60


 

Therefore, the compound interest on Rs. 4000 for 6 months at 12% per annum payable quarterly is Rs. 243.60. So, option A is the correct answer.

Given:

• Future value (amount accumulated): Rs. 50,000
• Interest rate (compound interest rate): 5% per annum (0.05 in decimal)
• Number of years (periods): 25 years