Explanation
Given:
Present value $$= ₹\ 10000$$
Interest rate $$= 10 \%$$ per annum
Amount $$(A) = P (1+(r/100))^n$$
Now substituting the values in above formula we get,
$$\Rightarrow A = 10000 (11/10)^3$$
$$\Rightarrow A = 121 (10) (11)$$
∴ Compound interest $$= A – P$$
$$= 13310 – 10000= ₹\ 3310$$
Let the population be $$P$$.
So, the increased population will be $$P + 1200$$.
The population after $$11\% $$ decrease in population becomes $$P - 32$$.
Therefore,
$${{P}} - 32 = \left( {{{P}} + 1200} \right) - \dfrac{{11}}{{100}}\left( {{{P}} + 1200} \right)$$
$${{P}} = 10000$$
So, the total population is $$10000$$.
Given : $$83.33$$% of $$36$$ coins are $$20$$ paisa coins
$$83.33$$% of $$36 =\left( \cfrac { 83.33 }{ 100 } \right) \times 36=30$$
So, $$30$$ coins are $$20$$ paise coins and remaining $$(36-30)= 6$$ coins are $$10$$ paise coins.
Value $$=30\times 20+6\times 10=660$$ Paise $$=Rs. 6.60$$, which is the given amount.
$$\therefore $$ Number of $$10$$ paise coins are $$6$$.
Let the price of toys be $${{Rs}}\;x$$ and the number of toys be $$y$$. So,
$$xy = 360$$
$$y = \dfrac{{360}}{x}$$
According to the question,
$$\left( {x - 2} \right)\left( {y + 2} \right) = 360$$
$$xy - 2y + 2x - 4 = 360$$
$$360 - 2\left( {\dfrac{{360}}{x}} \right) + 2x - 4 = 360$$
$${x^2} - 2x - 360 = 0$$
$${x^2} - 20x + 18x - 360 = 0$$
$$\left( {x - 20} \right)\left( {x + 18} \right) = 0$$
$$x = 20,x = - 18$$
As the price cannot be negative, so the original price of the toy is $${{Rs}}\;20$$.
Hence option (C) is correct
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