1 Marks Question

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.

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Question 21 Mark

Fill in the blanks:
(Amount) - (Principal) = _______.

Answer

(Amount) - (Principal) = Compound interest.

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MCQ 31 Mark

The compound interest on $Rs.5000$ at $10\%$ per annum for $2$ years is:

A
$Rs. 550$
✓
$Rs. 1050$
C
$Rs. 950$
D
$Rs. 825$

Answer

Correct option: B.

$Rs. 1050$

$P = Rs. 5000$
$R = 10\%$
$n = 2$ years
Now, $\text{A}=\text{Rs}.\ \text{P}\times\Big(1+\frac{\text{R}}{100}\Big)^\text{n}$
$=\text{Rs}.\ 5000\times\Big(1+\frac{10}{100}\Big)^2$
$=\text{Rs}.\ 5000\times\Big(\frac{110}{100}\Big)^2$
$=\text{Rs}.\ 5000\times\Big(\frac{11}{10}\Big)\times\Big(\frac{11}{10}\Big)$
$=\text{Rs}.\ (50\times11\times11)$
$=\text{Rs}.\ 6050$
$\therefore CI = A - P = Rs. (6050 - 5000) = Rs. 1050$

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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.

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Question 51 Mark

Fill in the blanks: $\text{A}=\text{P}\Big(1+\frac{}{100}\Big)^\text{n}.$

Answer

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

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