Download App

Maths Solve The Following Question.1MARKS

Compound Interest · Maharashtra Board STD 8 (MSBSHSE - Semi English Medium). 5 questions.

Questions

Solve The Following Question.1MARKS

Take a timed test

5 questions · self-marked practice — reveal the answer and mark yourself.

Question 11 Mark
If the value of a machine is Rs. P and it depreciates at R% per annum, then its value after 2 years is _____.
Answer
$\text{A}=\text{P}\Big(1-\frac{\text{R}}{100}\Big)^2,$ where A is the value of the machine after 2 years.
View full question & answer→
Question 31 Mark
The compound interest on Rs 5000 at 10% per annum for 2 years is:
  1. Rs. 550
  2. Rs. 1050
  3. Rs. 950
  4. Rs. 825
Answer
  1. Rs. 1050
Solution:

P = Rs. 5000

R = 10%

n = 2 years

Now, $\text{A}=\text{Rs}.\ \text{P}\times\Big(1+\frac{\text{R}}{100}\Big)^\text{n}$

$=\text{Rs}.\ 5000\times\Big(1+\frac{10}{100}\Big)^2$

$=\text{Rs}.\ 5000\times\Big(\frac{110}{100}\Big)^2$

$=\text{Rs}.\ 5000\times\Big(\frac{11}{10}\Big)\times\Big(\frac{11}{10}\Big)$

$=\text{Rs}.\ (50\times11\times11)$

$=\text{Rs}.\ 6050$

$\therefore$ CI = A - P = Rs. (6050 - 5000) = Rs. 1050
View full question & answer→
Question 41 Mark
If the population P of a town increases at R% per annum, then its population after 5 years is ______.
Answer
$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^5,$ where A is the population of the town after 5 years.
View full question & answer→
Question 51 Mark
Fill in the blanks:
$\text{A}=\text{P}\Big(1+\frac{}{100}\Big)^\text{n}.$
Answer
$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^\text{n}.$
View full question & answer→
Generate Paper FreeAll Compound Interest topics