In the following question, two statements are numbered as A and B. On

1.	In the following question, two statements are numbered as A and B. On solving these statements, we get quantities A and B respectively. Solve both quantities and choose the correct option.Quantity A: If a sum of money amount to Rs. 10,070 at the end of 2 years and Rs. 10,355 at the end of 3 years. Find a sum.Quantity B: Rs. 90001. Quantity A ≥ Quantity B2. Quantity A ≤ Quantity B3. Quantity A < Quantity B4. Quantity A > Quantity B5. Quantity A = Quantity B or No relation
Answer» Correct Answer - Option 4 : Quantity A > Quantity B Quantity A: S.I = ( P × R × T)/100; S.I = A₂ – A₁; Amount = Principal + simple interest Calculation: Simple interest for 1 year = Rs. 10,355 – Rs. 10,070 = Rs. 285 285 = ( x × r × 1)/100 285 = xr/100 Let the principal be Rs x ⇒ 10,070 = x + (x × r × 2)/100 ⇒ 10,070 = x + (2 r x)/100 ⇒ 10,070 = x + 2 × 285 ⇒ 10,070 – 570 = x ⇒ x = Rs. 9500 Principal is Rs. 9500 Quantity B: Rs. 9000 ∴ Quantity A > Quantity B

In the following question, two statements are numbered as A and B. On solving these statements, we get quantities A and B respectively. Solve both quantities and choose the correct option.Quantity A: If a sum of money amount to Rs. 10,070 at the end of 2 years and Rs. 10,355 at the end of 3 years. Find a sum.Quantity B: Rs. 90001. Quantity A ≥ Quantity B2. Quantity A ≤ Quantity B3. Quantity A < Quantity B4. Quantity A > Quantity B5. Quantity A = Quantity B or No relation

Answer» Correct Answer - Option 4 : Quantity A > Quantity B

Quantity A:

S.I = ( P × R × T)/100; S.I = A₂ – A₁; Amount = Principal + simple interest

Calculation:

Simple interest for 1 year = Rs. 10,355 – Rs. 10,070 = Rs. 285

285 = ( x × r × 1)/100

285 = xr/100

Let the principal be Rs x

⇒ 10,070 = x + (x × r × 2)/100

⇒ 10,070 = x + (2 r x)/100

⇒ 10,070 = x + 2 × 285

⇒ 10,070 – 570 = x

⇒ x = Rs. 9500

Principal is Rs. 9500

Quantity B: Rs. 9000

∴ Quantity A > Quantity B