Question

The difference between the compound interest and the simple interest on a certain sum of money for 2 years at 11% per annum is Rs. 363. Find the sum.
1.33,000
2.31,000
3.30,000
4.32,000

Answer 1

Answer:

The sum of money = Rs 30,000

Step-by-step explanation:

➤ Given that:

• The difference between the compound interest and simple interest = Rs 363
• Time = 2 years
• Rate of interest = 11% per annum

Let us assume that the certain sum of money be x.

➤ Formula to find compound interest:

C.I. = A - p

C.I. = p(1 + 0.01r)ᵗ - p

Where,

• C.I. = Compound interest
• A = Amount
• p = Principal/Sum of money
• r = Rate of interest
• t = Time
• 0.01 = 1/100

➤ Formula to find simple interest:

S.I. = (p × r × t × 0.01)

Where,

• S.I. = Simple interest
• p = Principal/Sum of money
• r = Rate of interest
• t = Time
• 0.01 = 1/100

★ Finding the Sum of money/Principal:

C.I. - S.I. = 363

→ p(1 + 0.01r)ᵗ - p - (p × r × t × 0.01) = 363

→ x(1 + 0.11)² - x - (x × 11 × 2 × 0.01) = 363

→ x(1.11)² - x - 0.22x = 363

→ 1.2321x - 1.22x = 363

→ 0.0121x = 363

→ x = 363/0.0121

→ x = 30000

∴ Sum of money = Rs 30,000

Answer 2

$\huge\mathrm{Answer}$

★ Given :-

• Difference between the CI and SI = ₹363
• Years = 2 years
• Rate of interest = 11% pa

★ Notations :-

• p is principal
• r is the rate of interest
• t is the time period

★ Formula :-

• Difference between the compound interest and simple interest =
• p(1+(r/100))^t - p - (p×r×t)/100 = 363

★ To find :-

• Let the sum be :- p

★ Solution :-

p(1+(r/100))^t - p - (p×r×t)/100 = 363

➡ p( (1+11/100)^2 - 1 - 11×2/100 ) = 363

➡ p( (111/100)^2 - 1 - 22/100 ) = 363

➡ p ( 12321 / 10000 - 1 - 22 / 100 ) = 363

➡ p ( (1.2321 - 1.22 ) = 363

➡ p ( 0.0121 ) = 363

➡ p = 363 / 0.0121

➡ p = 3630000 / 121

➡ p = ₹ 30,000

Ans) The sum is ₹30,000