Class 11: Linear Inequations – Exercise 15.4 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

Answer:

Let the smaller odd positive number be $\displaystyle x$ .

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

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

$\displaystyle \therefore x + 2 < 10 \text{ and } x + ( x+2) > 11$

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

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

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

Hence the required pars of odd integers are $\displaystyle ( 5, 7) \text{ and } ( 7, 9)$ .

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

Answer:

Let the smaller odd natural number be $\displaystyle x$ .

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

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

$\displaystyle \therefore x > 10 \text{ and } x + ( x+2) < 40$

$\displaystyle \Rightarrow x > 10 \text{ and } x < 19$

$\displaystyle \Rightarrow 10 < x < 19$

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

Hence the required pars of odd integers are $\displaystyle ( 11, 13), ( 13, 15), ( 15, 17) \text{ and } ( 17, 19)$ .

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

Answer:

Let the smaller even integer be $\displaystyle x$ .

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

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

$\displaystyle \therefore x > 5 \text{ and } x + ( x+2) < 23$

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

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

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

Hence the required pars of odd integers are $\displaystyle ( 6,8), ( 8, 10) \text{ and } ( 10, 12)$ .

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

Answer:

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

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

$\displaystyle \Rightarrow 135 + x \geq 195$

$\displaystyle \Rightarrow x \geq 60$

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

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

Answer:

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

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

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

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

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

$\displaystyle \Rightarrow 30 < x < 35$

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

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

Answer:

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

$\displaystyle \therefore 30 < x < 35$

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

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

$\displaystyle \Rightarrow 86 < x < 95$

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

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

Answer:

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

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

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

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

$\displaystyle \Rightarrow 82 \leq x \leq 132$

But you cannot score more than $\displaystyle 100$ marks

$\displaystyle \therefore 82 \leq x \leq 100$

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

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Question 8: A company manufactures cassettes and its cost and revenue functions for a week are $\displaystyle C = 300 + \frac{3}{2} x \text{ and } R = 2x$ respectively, where $\displaystyle r$ 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.

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

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

$\displaystyle \Rightarrow x > 600$

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

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

Answer:

Let the shortest side of triangle be $\displaystyle x$ .

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

Given Perimeter $\displaystyle \geq 61$

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

$\displaystyle \Rightarrow 7x \geq 61 + 2$

$\displaystyle \Rightarrow x \geq 9$

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

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Question 10: How many litres of water will have to be added to $\displaystyle 1125$ litres of the $\displaystyle 45\%$ solution of acid so that the resulting mixture will contain more than $\displaystyle 25\%$ but less than $\displaystyle 30\%$ acid content?

Answer:

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

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

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

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

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

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

$\displaystyle \Rightarrow x < 900 \text{ and } x > 562.5$

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

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

Answer:

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

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

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

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

$\displaystyle \Rightarrow 2560 + 4x < 5120 + 2x \text{ and } 5120 + 2x < 3840 + 6x$

$\displaystyle \Rightarrow 2x < 2560 \text{ and } 4x > 1280$

$\displaystyle \Rightarrow x < 1280 \text{ and } x > 320$

$\displaystyle \therefore 320 < x < 1280$

Therefore the amount of $\displaystyle 2\%$ boric acid solution that needs to be added needs to be more than $\displaystyle 320$ litres and less than $\displaystyle 1280$ 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 $\displaystyle 7.2 \text{ and } 7.8$ . If the first two pH readings are $\displaystyle 7.48 \text{ and } 7.85$ , find the range of pH value for the third reading that will result in the acidity level is normal.