\displaystyle \textbf{Question 1: } \text{Find all pairs of consecutive odd positive integers, both smaller than }10,
\displaystyle \text{such that their sum is more than }11.
\displaystyle \text{Answer:}
\displaystyle \text{Let the smaller odd positive integer be }x.
\displaystyle \text{Then the next consecutive odd positive integer is }x+2.
\displaystyle \text{Since both integers are smaller than }10,
\displaystyle x+2<10\Rightarrow x<8.
\displaystyle \text{Also, their sum is more than }11.
\displaystyle x+(x+2)>11.
\displaystyle \Rightarrow 2x+2>11.
\displaystyle \Rightarrow 2x>9\Rightarrow x>\frac{9}{2}.
\displaystyle \therefore \frac{9}{2}<x<8.
\displaystyle \text{Since }x\text{ is an odd positive integer, }x\in\{5,7\}.
\displaystyle \therefore \text{The required pairs are }(5,7)\text{ and }(7,9).
\displaystyle \\

\displaystyle \textbf{Question 2: } \text{Find all pairs of consecutive odd natural numbers, both larger than }10,
\displaystyle \text{such that their sum is less than }40.
\displaystyle \text{Answer:}
\displaystyle \text{Let the smaller odd natural number be }x.
\displaystyle \text{Then the next consecutive odd natural number is }x+2.
\displaystyle \text{Since both numbers are larger than }10,
\displaystyle x>10.
\displaystyle \text{Also, their sum is less than }40.
\displaystyle x+(x+2)<40.
\displaystyle \Rightarrow 2x+2<40.
\displaystyle \Rightarrow 2x<38\Rightarrow x<19.
\displaystyle \therefore 10<x<19.
\displaystyle \text{Since }x\text{ is an odd natural number, }x\in\{11,13,15,17\}.
\displaystyle \therefore \text{The required pairs are }(11,13),(13,15),(15,17)\text{ and }(17,19).
\displaystyle \\

\displaystyle \textbf{Question 3: } \text{Find all pairs of consecutive even positive integers, both larger than }5,
\displaystyle \text{such that their sum is less than }23.
\displaystyle \text{Answer:}
\displaystyle \text{Let the smaller even positive integer be }x.
\displaystyle \text{Then the next consecutive even positive integer is }x+2.
\displaystyle \text{Since both integers are larger than }5,
\displaystyle x>5.
\displaystyle \text{Also, their sum is less than }23.
\displaystyle x+(x+2)<23.
\displaystyle \Rightarrow 2x+2<23.
\displaystyle \Rightarrow 2x<21\Rightarrow x<\frac{21}{2}.
\displaystyle \therefore 5<x<\frac{21}{2}.
\displaystyle \text{Since }x\text{ is an even positive integer, }x\in\{6,8,10\}.
\displaystyle \therefore \text{The required pairs are }(6,8),(8,10)\text{ and }(10,12).
\displaystyle \\

\displaystyle \textbf{Question 4: } \text{The marks scored by Rohit in two tests were }65\text{ and }70. \text{ Find the}
\displaystyle \text{minimum marks he should score in the third test to have an average of at least }65.
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ be the minimum marks Rohit should score in the third test.}
\displaystyle \therefore \frac{65+70+x}{3}\geq65.
\displaystyle \Rightarrow 135+x\geq195.
\displaystyle \Rightarrow x\geq60.
\displaystyle \therefore \text{The minimum marks Rohit should score in the third test is }60.
\displaystyle \\

\displaystyle \textbf{Question 5: } \text{A solution is to be kept between }86^\circ\text{F and }95^\circ\text{F}. \text{ Find the}
\displaystyle \text{range of temperature in degree Celsius if }F=\frac{9}{5}C+32.
\displaystyle \text{Answer:}
\displaystyle \text{Let the temperature of the solution be }x^\circ\text{C}.
\displaystyle \therefore 86\leq\frac{9}{5}x+32\leq95.
\displaystyle \Rightarrow 54\leq\frac{9}{5}x\leq63.
\displaystyle \Rightarrow 30\leq x\leq35.
\displaystyle \therefore \text{The required range of temperature is }30^\circ\text{C}\leq x\leq35^\circ\text{C}.
\displaystyle \\

\displaystyle \textbf{Question 6: } \text{A solution is to be kept between }30^\circ\text{C and }35^\circ\text{C}. \text{ Find the}
\displaystyle \text{corresponding range of temperature in degrees Fahrenheit.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the temperature of the solution be }x^\circ\text{C}.
\displaystyle \therefore 30\leq x\leq35.
\displaystyle \text{Using }F=\frac{9}{5}C+32,
\displaystyle \frac{9}{5}(30)+32\leq F\leq\frac{9}{5}(35)+32.
\displaystyle \Rightarrow 54+32\leq F\leq63+32.
\displaystyle \Rightarrow 86\leq F\leq95.
\displaystyle \therefore \text{The required range is }86^\circ\text{F}\leq F\leq95^\circ\text{F}.
\displaystyle \\

\displaystyle \textbf{Question 7: } \text{To receive grade `A' in a course, one must obtain an average of }90\text{ marks or}
\displaystyle \text{more in five papers, each carrying }100\text{ marks. Shikha scored }87,95,92\text{ and }94
\displaystyle \text{marks in the first four papers. Find the minimum marks she must score in the last}
\displaystyle \text{paper to receive grade `A' in the course.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ be the marks scored by Shikha in the fifth paper.}
\displaystyle \text{For grade `A', her average must be at least }90.
\displaystyle \therefore \frac{87+95+92+94+x}{5}\geq90.
\displaystyle \Rightarrow \frac{368+x}{5}\geq90.
\displaystyle \Rightarrow 368+x\geq450.
\displaystyle \Rightarrow x\geq82.
\displaystyle \therefore \text{Shikha must score at least }82\text{ marks in the fifth paper.}
\displaystyle \\

\displaystyle \textbf{Question 8: } \text{A company manufactures cassettes. Its weekly cost and revenue functions are}
\displaystyle C=300+\frac{3}{2}x\text{ and }R=2x\text{ respectively, where }x\text{ is the number of cassettes}
\displaystyle \text{produced and sold in a week. How many cassettes must be sold for the company}
\displaystyle \text{to realize a profit?}
\displaystyle \text{Answer:}
\displaystyle \text{To earn a profit, the revenue must be greater than the cost.}
\displaystyle \therefore R>C.
\displaystyle \Rightarrow 2x>300+\frac{3}{2}x.
\displaystyle \Rightarrow 2x-\frac{3}{2}x>300.
\displaystyle \Rightarrow \frac{1}{2}x>300.
\displaystyle \Rightarrow x>600.
\displaystyle \text{Since }x\text{ is a whole number, the least possible value of }x\text{ is }601.
\displaystyle \therefore \text{The company must sell at least }601\text{ cassettes in a week to make a profit.}
\displaystyle \\

\displaystyle \textbf{Question 9: } \text{The longest side of a triangle is three times the shortest side, and the third}
\displaystyle \text{side is }2\text{ cm shorter than the longest side. If the perimeter of the triangle is at}
\displaystyle \text{least }61\text{ cm, find the minimum length of the shortest side.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the shortest side of the triangle be }x\text{ cm.}
\displaystyle \text{Then the longest side is }3x\text{ cm and the third side is }(3x-2)\text{ cm.}
\displaystyle \text{Given that the perimeter is at least }61\text{ cm,}
\displaystyle x+3x+(3x-2)\geq61.
\displaystyle \Rightarrow 7x-2\geq61.
\displaystyle \Rightarrow 7x\geq63.
\displaystyle \Rightarrow x\geq9.
\displaystyle \therefore \text{The minimum length of the shortest side is }9\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 10: } \text{How many litres of water must be added to }1125\text{ litres of a }45\%
\displaystyle \text{acid solution so that the resulting mixture contains more than }25\%\text{ but less}
\displaystyle \text{than }30\%\text{ acid?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ litres of water be added to }1125\text{ litres of the solution.}
\displaystyle \text{Then the total volume of the mixture is }(1125+x)\text{ litres.}
\displaystyle \text{The quantity of acid remains }45\%\text{ of }1125\text{ litres.}
\displaystyle \therefore \frac{25}{100}(1125+x)<\frac{45}{100}(1125)<\frac{30}{100}(1125+x).
\displaystyle \Rightarrow 28125+25x<50625<33750+30x.
\displaystyle \Rightarrow 25x<22500\text{ and }16875<30x.
\displaystyle \Rightarrow x<900\text{ and }x>562.5.
\displaystyle \therefore 562.5<x<900.
\displaystyle \therefore \text{The quantity of water to be added must be greater than }562.5\text{ litres}
\displaystyle \text{and less than }900\text{ litres.}
\displaystyle \\

\displaystyle \textbf{Question 11: } \text{An }8\%\text{ boric acid solution is to be diluted by adding a }2\%\text{ boric acid}
\displaystyle \text{solution. The resulting mixture is to contain more than }4\%\text{ but less than }6\%
\displaystyle \text{boric acid. If there are }640\text{ litres of the }8\%\text{ solution, find how many}
\displaystyle \text{litres of the }2\%\text{ solution must be added.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ litres of the }2\%\text{ boric acid solution be added.}
\displaystyle \text{The total volume of the mixture is }(640+x)\text{ litres.}
\displaystyle \text{The total quantity of boric acid is }\frac{8}{100}\times640+\frac{2}{100}\times x.
\displaystyle \therefore \frac{4}{100}(640+x)<\frac{8}{100}\times640+\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.
\displaystyle \therefore \text{The quantity of }2\%\text{ boric acid solution to be added must be greater than}
\displaystyle 320\text{ litres and less than }1280\text{ litres.}
\displaystyle \\

\displaystyle \textbf{Question 12: } \text{The water acidity in a pool is considered normal when the average pH of}
\displaystyle \text{three daily measurements is between }7.2\text{ and }7.8.\text{ If the first two readings are}
\displaystyle \text{ }7.48\text{ and }7.85,\text{ find the range of pH values for the third reading.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the third pH reading be }x.
\displaystyle \therefore 7.2\leq\frac{7.48+7.85+x}{3}\leq7.8.
\displaystyle \Rightarrow 7.2\leq\frac{15.33+x}{3}\leq7.8.
\displaystyle \Rightarrow 21.6\leq15.33+x\leq23.4.
\displaystyle \Rightarrow 6.27\leq x\leq8.07.
\displaystyle \therefore \text{The third pH reading must satisfy }6.27\leq x\leq8.07.
\displaystyle \\

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