Question 1: Find all pairs of consecutive odd positive integers, both of which are smaller than $10$, such that their sum is more than $11$.

Let the smaller odd positive number be $x$.

Therefore the other odd positive number would be $(x+2)$.

It is given that all pairs of consecutive odd positive integers, both of which are smaller than $10$, such that their sum is more than $11$.

$\therefore x + 2 < 10$  and $x + ( x+2) > 11$

$\Rightarrow x < 8$ and $2x > 9 \Rightarrow x >$ $\frac{9}{2}$

$\Rightarrow$ $\frac{9}{2}$ $< x < 8$

$\therefore x \in \{ 5, 7 \}$

Hence the required pars of odd integers are $( 5, 7)$ and $( 7, 9)$.

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Question 2: Find all pairs of consecutive odd natural number, both of which are larger than $10$, such that their sum is less than $40$.

Let the smaller odd natural number be $x$.

Therefore the other odd natural number would be $(x+2)$.

It is given that all pairs of consecutive odd natural number, both of which are larger than $10$, such that their sum is less than $40$.

$\therefore x > 10$  and $x + ( x+2) < 40$

$\Rightarrow x > 10$ and $x < 19$

$\Rightarrow$ $10 < x < 19$

$\therefore x \in \{ 11, 13, 15, 17 \}$

Hence the required pars of odd integers are $( 11, 13), ( 13, 15), ( 15, 17)$ and $( 17, 19)$.

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Question 3: Find all pairs of consecutive even positive integers, both of which are larger than $5$, such that their sum is less than $23$.

Let the smaller even integer be $x$.

Then, the other even integer will be $(x+2)$.

Given all pairs of consecutive even positive integers, both of which are larger than $5$, such that their sum is less than $23$.

$\therefore x > 5$  and $x + ( x+2) < 23$

$\Rightarrow x > 5$ and $x <$ $\frac{21}{2}$

$\Rightarrow$ $5 < x <$ $\frac{21}{2}$

$\therefore x \in \{ 6, 8, 10 \}$

Hence the required pars of odd integers are $( 6,8), ( 8, 10)$ and $( 10, 12)$.

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Question 4: The marks scored by Rohit in two tests were $65$ and $70$. Find the minimum marks he should score in the third test to have an average of at least $65$ marks.

Let $x$ be the minimum marks Rohit needs to score in third test to have an average of at least $65$.

$\therefore$ $\frac{65+70+x}{3}$ $\geq 65$

$\Rightarrow 135 + x \geq 195$

$\Rightarrow x \geq 60$

Hence the minimum marks Rohit should score in the third test should be $60$.

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Question 5: A solution is to be kept between $86^{\circ}$ and $95^{\circ} F$. What is the range of temperature in degree Celsius, if the Celsius (C) / Fahrenheit (F) conversion formula is given by $F =$ $\frac{9}{5}$ $C + 32$.

Let the temperature of the solution in ${\ }^{\circ}C$ be $x$

$\therefore 86 <$ $\frac{9}{5}$ $x +32 < 95$

$\Rightarrow 86-32 <$ $\frac{9}{5}$ $x < 95-32$

$\Rightarrow 54 <$ $\frac{9}{5}$ $x < 63$

$\Rightarrow 6 <$ $\frac{1}{5}$ $x < 7$

$\Rightarrow 30 < x < 35$

Hence the range of temperature in degree Celsius is $30^{\circ}\ C$ and $35^{\circ}\ C$.

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Question 6: A solution is to be kept between $30^{\circ} C$ and $35^{\circ} C$. What is the range of temperature in degree Fahrenheit?

Let the temperature of solution in ${\ }^{\circ}\ F$ to be $x$

$\therefore 30 < x < 35$

$\Rightarrow$ $\frac{9}{5}$ $(30) + 32 < x <$ $\frac{9}{5}$ $(35) + 32$

$\Rightarrow 54 + 32 < x < 63 + 32$

$\Rightarrow 86 < x < 95$

Hence the range of temperature in degree Fahrenheit is $86^{\circ}\ F$ and $95^{\circ}\ F$

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Question 7: To receive grade ‘A’ in a course, one must obtain an average of $90$ marks or more in five papers each of $100$ marks. If Shikha scored $87, 95, 92$ and $94$ marks in first four papers, find the minimum marks that she must score in the last paper to get grade ‘A’ in the course.

Let $x$ be the minimum marks to be scored to get $\text{ 'A' }$ in course.

$\therefore 90 \leq$ $\frac{87+95+92+94+x}{5}$ $\leq 100$

$\Rightarrow 90 \leq$ $\frac{368+x}{5}$ $\leq 100$

$\Rightarrow 450 \leq 368 + x \leq 500$

$\Rightarrow 82 \leq x \leq 132$

But you cannot score more than $100$ marks

$\therefore 82 \leq x \leq 100$

Hence Shikha must score at least $82$ marks in the 5th paper to get $\text{ 'A' }$ grade in the course.

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Question 8: A company manufactures cassettes and its cost and revenue functions for a week are $C = 300 +$ $\frac{3}{2}$ $x$  and $R = 2x$ respectively, where $r$ is the number of cassettes produced and sold in a week. How many cassettes must be sold for the company to realize a profit?

To earn profit, revenue must be greater then cost.

$\therefore 2x > 300 +$ $\frac{3}{2}$ $x$

$\Rightarrow$ $\frac{1}{2}$ $x > 300$

$\Rightarrow x > 600$

Therefore the company must sell more than $600$ cassettes in a week to make profit.

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Question 9: The longest side of a triangle is three times the shortest side and the third side is $2$ cm shorter than the longest side if the perimeter of the triangles at least $61$ cm, Find the minimum length of the shortest-side.

Let the shortest side of triangle be $x$.

Therefore longest side is $3x$ and the third side $= 3x - 2$

Given Perimeter $\geq 61$

$\Rightarrow x + 3x + ( 3x-2) \geq 61$

$\Rightarrow 7x \geq 61 + 2$

$\Rightarrow x \geq 9$

Hence the minimum length of the shortest side is $9$ cm.

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Question 10: 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?

Let $x$ liters of water is added to $1125$ litres of $45\%$ solution.

Therefore the total quantity of solution $= ( 1125 + x)$

Total acid content in $1125$ litres of mixture $= 45\% of 1125$

Given resulting mixture will contain more than $25\%$ but less than $30\%$ acid content

$\Rightarrow$ $\frac{25}{100}$ $(1125 + x) <$ $\frac{45}{100}$ $(1125) <$ $\frac{30}{100}$ $( 1125 + x)$

$\Rightarrow 28125 + 25 x < 50625 < 33750 + 30x$

$\Rightarrow x < 900$ and $x > 562.5$

Hence the water to be added should be greater than $562.5$ litres and less than $900$ litres.

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Question 11: A solution of $8\%$ boric acid is to be diluted by adding $2\%$ boric acid solution to it. The resulting mixture is to be more than $4\%$ but less than $6\%$ boric acid. If there are $640$ litres of the $8\%$ solution, how many litres of $2\%$ solution will have to be added?

Let $x$ litres of $2\%$ solution is added in the existing solution of $8\%$ of Boric acid.

Given resulting mixture is to be more than $4\%$ but less than $6\%$ boric acid.

$\frac{4}{100}$ $( 640+x) <$ $\frac{8}{100}$ $\times 640 +$ $\frac{2}{100}$ $\times x <$ $\frac{6}{100}$ $( 640 + x)$

$\Rightarrow 2560 + 4x < 5120 + 2x < 3840 + 6x$

$\Rightarrow 2560 + 4x < 5120 + 2x$ and $5120 + 2x < 3840 + 6x$

$\Rightarrow 2x < 2560$ and $4x > 1280$

$\Rightarrow x < 1280$ and $x > 320$

$\therefore 320 < x < 1280$

Therefore the amount of $2\%$ boric acid solution that needs to be added needs to be more than $320$ litres and less than $1280$ litres.

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Question 12: The water acidity in a pool is considered normal when the average pH reading of three daily measurements is between $7.2$ and $7.8$. If the first two pH reading are $7.48$ and $7.85$, find the range of pH value for the third reading that will result in the acidity level being normal.

Let the third pH value be $x$

$\therefore 7.2 <$ $\frac{7.48+7.85+x}{3}$ $< 7.8$

$\Rightarrow 7.2 <$ $\frac{15.33+x}{3}$ $< 7.8$

$\Rightarrow 21.6 < 15.33+x < 23.4$

$\Rightarrow 6.27 < x < 8.07$

Hence the range for the pH value for the third reading must be between $6.27$ and $8.07$.

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