Question
Obtain trend values for data in Problem 13 using 4-yearly moving averages.

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Use the method of least squares to fit a trend line to the data in problem 1 above. Also, obtain the trend value for the year 1975.
Property value $=\text{₹} 12,50,000$
Rate of premium, $r=\text{₹} 3 \%$
If property is fully insured, the policy value is same as property value therefore policy value =
Premium $=\frac{ r }{100} \times$ policy value
$
=\frac{\square}{100} \times 12,50,000
$
$
=\square
$
Evalute : $\int \frac{5 x^2-6 x+3}{2 x-3} d x$
Three cars were sold through an agent for $\text{₹}$ 2,40,000, $\text{₹}$ 2,22,000 and $\text{₹}$ 2,25,000 respectively. The rates of commission were $17.5 \%$ on the first, $12.5 \%$ on the second. If the agent overall received $14 \%$ commission on the total sales, find the rate of commission paid on the third car.
Solution: Total selling Price of three cars $=2,40,000+2,22,000+2,25,000$ $=\square$
Commision on total sale $=14 \%$
$=\frac{14}{100} \times \square$
Selling price of First car $=\text{₹} 2,40,000$
Rate of commission $=17.5 \%$
$=\frac{17.5}{100} \times 2,40,000=\square$
$\therefore$ Commission on first car $=\text{₹} \square$
Selling price of Second car $=\text{₹} 2,22,000$
Rate of commission $=12.5 \%$
$=\frac{12.5}{100} \times 2,22,000=\square$
$\therefore$ Commission on second car $=\text{₹} \square$
Selling price of third $\operatorname{car}=\text{₹} 2,25,000$
Let the rate of commission be $x$
Commission on third car $=\frac{x}{100} \times 2,25,000$
$\therefore$ Commission on third car $=$ Total commission - (commission on first car + commission on second car)
$\therefore \frac{x}{100} \times 2,25,000=\square-\{\square+\square\}$
$\therefore x =\square$
For the demand function $D=100-\frac{p^2}{2}$, find the elasticity of demand at (i) $p=10$ (ii) $p=6$ and comment on the results.