[2 Mark Question Answer]

Question 12 Marks

6000 workers were employed to construct a river bridge in four years. At the end of first year, 20% workers were retrenched; At the end of second year 5% of the workers at that time were retrenched. However, to complete the project in time, the number of workers was increased by 15% at the end of third year. How many workers were working during the fourth year?

Answer

The number of workers who were working during the fourth year
= 6000 $\left(1-\frac{20}{100}\right)\left(1-\frac{5}{100}\right)\left(1+\frac{15}{100}\right)$
$V = V _0\left(1-\frac{r_1}{100}\right)\left(1-\frac{r_2}{100}\right)\left(1+\frac{r_3}{100}\right)$
= 6000 $\times \frac{4}{5} \times \frac{19}{20} \times \frac{23}{20}$
= 5244

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Question 22 Marks

There is a continuous growth in population of a village at the rate of 5% per annum. If its present population is 9,261, what population was 3 years ago?

Answer

Let P be the population 3 years ago.
Then, present population = $P \times\left(1+\frac{5}{100}\right)^3$
$\begin{aligned} & \therefore 9,261=\left( P \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}\right) \\ & \text { or } P =\left(9,261 \times \frac{20}{21} \times \frac{20}{21} \times \frac{20}{21}\right)\end{aligned}$
= 8,000
Hence, 3years ago the population of village was = 8,000

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Question 32 Marks

The population of a city is 1,25,000. If the annual birth rate and death rate are 5.5% and 3.5% respectively. Calculate the population of the city after 3 years.

Answer

Here net growth = (5.5 - 3.5)% = 2%
P = ₹ 1,25,000
n = 3 years
$\therefore$ Population of the city after 3 years
$
=1,25,000\left(1+\frac{2}{100}\right)^3
$
$=1,25,000\left(\frac{102}{100}\right)^3$
$\begin{aligned} & =1,25,000 \times \frac{102 \times 102 \times 102}{100 \times 100 \times 100} \\ & =1,32,651 .\end{aligned}$

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Question 42 Marks

In how many years will Rs.15,625 amount to Rs. 17,576 at 4% p.a., compound interest?

Answer

A = ₹ 17,576, P = ₹ 15,625,
r = 4%, n = ?
$A = P \left(1+\frac{r}{100}\right)^n$
17,576 = 15,625 $\left(1+\frac{4}{100}\right)^n$
$\frac{17,576}{15,625}=\left(1+\frac{1}{25}\right)^n$
$\frac{17,576}{15,625}=\left(\frac{26}{25}\right)^n$
$\left(\frac{26}{25}\right)^3=\left(\frac{26}{25}\right)^n$
n = 3 years

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Question 52 Marks

If the interest is compounded half yearly, calculate the amount when the Principal is Rs. 7,400, the rate of interest is 5% per annum and the duration is one year.

Answer

For half yearly payable
$A=P\left(1+\frac{r}{2 \times 100}\right)^{2 \times n}$
where P is principal r is rate p.a. and n is time in years.
$A=7400\left(1+\frac{5}{200}\right)^2$
$=7400 \times \frac{205}{200} \times \frac{205}{200}$
= ₹ 7774.63

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Question 62 Marks

Find the Compound Interest on Rs. 2,000 for 3 years at 15% per annum Compounded annually.

Answer

Amount at the end of the third year
$A = P \left(1+\frac{r}{100}\right)^n$
Given P = ₹ 2,000, r = 15%, n = 3 years
A = ₹ 2,000 $\left(1+\frac{15}{100}\right)^3$
= ₹ 2,000 x $\frac{23}{20} \times \frac{23}{20} \times \frac{23}{20}$
= ₹ 3,041.75
Compound Interest = (A - P)
= ₹ (3,041.75 - 2,000)
= ₹ 1,041.75

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