5 Marks Questions

Question 15 Marks

IQ of a person is given by the formula
$\mathrm{IQ}=\frac{\mathrm{MA}}{\mathrm{CA}} \times 100$
where MA is mental age and CA is chronological age. If 80 $\le$ IQ $\le$ 140 for a group of 12 years old children, find the range of their mental age.

Answer

It is given that $80 \leqslant Q \leqslant 140$ and CA = 12
We have $IQ = \frac{{MA}}{{CA}} \times 100$
$\therefore 80 \leqslant \frac{{MA}}{{12}} \times 100 \leqslant 140$
$ \Rightarrow 960 \leqslant MA \times 100 \leqslant 1680$
$ \Rightarrow 9.6 \leqslant MA \leqslant 16.8$
Thus minimum MA is 9.6 and maximum 16.8

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Question 25 Marks

How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?

Answer

Let x litres of water be added to 1125 litres of 45% acid solution.
Then total quantity of mixture = (1125 +x ) litres
$\frac{{45}}{{100}} \times 1125 + 0 \times \frac{x}{{100}} > $$\frac{{25}}{{100}} \times (1125 + x)$
and $\frac{{45}}{{100}} \times 1125 + 0 \times \frac{x}{{100}} < \frac{{30}}{{100}}$$ \times (1125 + x)$
Combining the above inequations, we get
$\frac{{25}}{{100}} \times 100 \leqslant \frac{{2025 \times 100}}{{4(1125 + x)}} \leqslant \frac{{30}}{{100}} \times 100$
$ \Rightarrow 25 \leqslant \frac{{50625}}{{1125 + x}} \leqslant 30$
$ \Rightarrow 25 \leqslant \frac{{50625}}{{1125 + x}}$ and $\frac{{50625}}{{1125 + x}} \leqslant 30$
$ \Rightarrow 28125 + 25x \leqslant 50625$ and $50625 \leqslant 33750 + 30x$
$ \Rightarrow 25x \leqslant 22500$ and $30x \geqslant 1687.5$
$ \Rightarrow x \leqslant 900$ and $x \geqslant 562.5$
$ \Rightarrow 562.5 \leqslant x \leqslant 900$
Thus minimum 562.5 litres and maximum 900 litres of water need to be added.

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Question 35 Marks

A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?

Answer

Let x litre of 2% boric acid solution be added to 640 litres of 8% boric acid solution. Then
Total quality of mixture = (640 + x) litres
Total boric acid in (640 + x) litres of mixture $ = \frac{{2x}}{{100}} + \frac{8}{{100}} \times 640$
$ = \frac{x}{{50}} + \frac{{256}}{5}$
It is given that the resulting mixture must be more than 4% but less than 6% boric acid.
$\therefore \frac{4}{{100}}(640 + x) < \frac{x}{{50}} + \frac{{256}}{5}$$ < \frac{6}{{100}}(640 + x)$
$ \Rightarrow \frac{{640 + x}}{{25}} < \frac{{x + 2560}}{{50}} < \frac{{1920 + 3x}}{{50}}$
$ \Rightarrow $ 1280 + 2x < x + 2560 < 1920 + 3x
$ \Rightarrow $ 1280 + 2x < x + 2560 and x + 2560 < 1920 + 3x
$ \Rightarrow $ x < 1280 and
$ \Rightarrow $ x < 1280 and x > 320
$ \Rightarrow $ 320 < x < 1280
Thus 2% boric acid solution must be more than 320 litres but less than 1280 litres.

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