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 .

Answer:

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 .

Answer:

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 .

Answer:

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.

Answer:

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 .

Answer:

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?

Answer:

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.

Answer:

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?

Answer:

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.

Answer:

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?

Answer:

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?

Answer:

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.

Answer:

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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