Solve the inequalities
1. $2\leq 3x-4\leq 5$
$2+4\leq 3x\leq 5+4$
$6\leq 3x\leq 9$
$\frac{6}{3}\leq x\leq \frac{9}{3}$
$2\leq x\leq 3$
∴ solution set is [2,3]
2. $6\leq -3(2x-4)< 12$
$\frac{6}{-3}\geq 2x-4> \frac{12}{-3}$
$-2\geq 2x-4> -4$
$-2+4\geq 2x> -4+4$
$2\geq 2x> 0$
$\frac{2}{2}\geq x> 0$
$1\geq x> 0$
∴ solution set is (0,1]
3. $-3\leq 4-\frac{7x}{2}\leq 18$
$-3-4\leq -\frac{7x}{2}\leq 18-4$
$-7\leq -\frac{7x}{2}\leq 14$
$-7*\frac{-2}{7}\geq x\geq 14*\frac{-2}{7}$
$2\geq x\geq -4$
∴ solution set is [-4,2]
4. $-15< \frac{3(x-2)}{5}\leq 0$
$-15*\frac{5}{3}< x-2\leq 0*\frac{5}{3}$
$-25< x-2\leq 0$
$-25+2< x\leq 0+2$
$-23< x\leq 2$
∴ solution set is (-23,2]
5. $-12< 4-\frac{3x}{-5}\leq 2$
$-12-4< -\frac{3x}{-5}\leq 2-4$
$-16< \frac{3x}{5}\leq -2$
$-16*\frac{5}{3}< x\leq -2*\frac{5}{3}$
$\frac{-80}{3}< x\leq \frac{-10}{3}$
∴ solution set is $(\frac{-80}{3},\frac{-10}{3}]$
6. $7\leq \frac{3x+11}{2}\leq 11$
$7*2\leq 3x+11\leq 11*2$
$14\leq 3x+11\leq 22$
$14-11\leq 3x\leq 22-11$
$3\leq 3x\leq 11$
$\frac{3}{3}\leq x\leq \frac{11}{3}$
$1\leq x\leq \frac{11}{3}$
∴ solution set is $[1,\frac{11}{3}]$
Solve the inequalities and represent the solution graphically on number line.
7. $5x+1> -24$, $5x-1< 24$
Consider, $5x+1> -24$
$5x> -24-1$
$5x> -25$
$x> \frac{-25}{5}$
$x> -5$ ...(i)
Consider, $5x-1< 24$
$5x< 24+1$
$5x< 25$
$x< \frac{25}{5}$
$x< 5$ ...(ii)
From (i) and (ii), solution set is (-5,5)
8. $2(x-1)< x+5$, $3(x+2)> 2-x$
Consider, $2(x-1)< x+5$
$2x-2< x+5$
$2x-x< 5+2$
$x< 7$ ...(i)
Consider, $3(x+2)> 2-x$
$3x+6> 2-x$
$3x+x> 2-6$
$4x> -4$
$x> -1$ ...(ii)
From (i) and (ii), solution set is (-1,7)
9. $3x-7> 2(x-6)$, $6-x> 11-2x$
Consider, $3x-7> 2(x-6)$
$3x-7> 2x-12$
$3x-2x> -12+7$
$x> -5$ ...(i)
Consider, $6-x> 11-2x$
$-x+2x> 11-6$
$x> 5$ ...(ii)
From (i) and (ii), solution set is (5,∞)
10. $5(2x-7)-3(2x+3)\leq 0$, $2x+19\leq 6x+47$
Consider, $5(2x-7)-3(2x+3)\leq 0$
$10x-35-6x-9\leq 0$
$4x-44\leq 0$
$4x\leq 44$
$x\leq 11$ ...(i)
Consider, $2x+19\leq 6x+47$
$19-47\leq 6x-2x$
$-28\leq 4x$
$-7\leq x$ ...(ii)
From (i) and (ii), solution set is [-7,11]
11. A solution is to be kept between 68℉ and 77℉. What is the range in temperature in degree Celsius (C) if the Celsius/Fahrenheit (F) conversion formula is given by, $F=\frac{9}{5}C+32$?
Given conversion formula,
$F=\frac{9}{5}C+32$
Also, solution is to be kept between 68℉ and 77℉
∴ $68< F< 77$
substitute equation of F,
⇒ $68< \frac{9}{5}C+32< 77$
$68-32< \frac{9}{5}C< 77-32$
$36< \frac{9}{5}C< 45$
$36*\frac{5}{9}< C< 45*\frac{5}{9}$
$20< C< 25$
∴ the range in temperature in degree Celsius is 20℃ to 25℃.
12. 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 8% solution, how many litres of the 2% solution will have to be added?
Let x litres of 2% solution is to be added.
boric acid in resulting mixture = 8% of 640+2% of x
total mixture = (x+640) litres
given that, resulting mixture is to be more than 4% but less than 6% boric acid
∴ $8% of 640+2% of x> 4% of (x+640)$ and $8% of 640+2% of x< 6% of (x+640)$
Consider, $8% of 640+2% of x> 4% of (x+640)$
$\frac{8}{100}*640+\frac{2}{100}*x> \frac{4}{100}* (x+640)$
$\frac{256}{5}+\frac{x}{50}> \frac{4}{100}*640+\frac{4x}{100}$
$\frac{256}{5}+\frac{x}{50}> \frac{128}{5}+\frac{x}{25}$
$\frac{256}{5}-\frac{128}{5}> \frac{x}{25}-\frac{x}{50}$
$\frac{128}{5}> \frac{2x-x}{50}$
$\frac{128}{5}> \frac{x}{50}$
$\frac{128}{5}*50> x$
$1280> x$
Consider, $8% of 640+2% of x< 6% of (x+640)$
$\frac{8}{100}*640+\frac{2}{100}*x< \frac{6}{100}* (x+640)$
$\frac{256}{5}+\frac{2x}{50}< \frac{6x}{100}+\frac{6}{100}*640$
$\frac{256}{5}+\frac{x}{50}< \frac{3x}{50}+\frac{192}{5}$
$\frac{256}{5}-\frac{192}{5}< \frac{3x}{50}-\frac{x}{50}$
$\frac{64}{5}< \frac{3x-x}{50}$
$\frac{64}{5}< \frac{2x}{50}$
$\frac{64}{5}< \frac{x}{25}$
$\frac{64}{5}*25< x$
$320< x$
Therefore, number of litres of the 2% solution to be added must be more than 320 litres but less than 1280 litres.
13. 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 litres of water is to be added.
acid in resulting mixture = 45% of 1125
total mixture = (x+1125) litres
given that, resulting mixture will contain more than 25% but less than 30% acid content
∴ $45% of 1125> 25% of (x+1125)$ and $45% of 1125< 30% of (x+1125)$
Consider, $45% of 1125> 25% of (x+1125) $
$\frac{45}{100}*1125> \frac{25}{100}*(x+1125)$
$\frac{2025}{4}> \frac{1}{4}*(x+1125)$
$\frac{2025}{4}> \frac{x}{4}+\frac{1125}{4}$
$\frac{2025}{4}-\frac{1125}{4}> \frac{x}{4}$
$\frac{900}{4}> \frac{x}{4}$
$900> x$
Consider, $45% of 1125< 30% of (x+1125) $
$\frac{45}{100}*1125< \frac{30}{100}*(x+1125)$
$\frac{2025}{4}< \frac{3}{10}*(x+1125)$
$\frac{2025}{4}< \frac{3x}{10}+\frac{3*1125}{10}$
$\frac{2025}{4}< \frac{3x}{10}+\frac{675}{2}$
$\frac{2025}{4}-\frac{675}{2}< \frac{3x}{10}$
$\frac{2025-2*675}{4}< \frac{3x}{10}$
$\frac{2025-1350}{4}< \frac{3x}{10}$
$\frac{675}{4}< \frac{3x}{10}$
$\frac{675}{4}*\frac{10}{3}< x$
$562.5< x$
Therefore, number of litres of water to be added must be more than 562.5 litres but less than 900 litres.
14. IQ of a person is given by the formula $IQ=\frac{MA}{CA}*100$, where MA is mental age and CA is chronological age. If $80\leq IQ\leq 140$ for a group of 12 years old children, find the range of their mental age.
Given that,
$IQ=\frac{MA}{CA}*100$
and for a group of 12 years old children,
$80\leq IQ\leq 140$
Also, CA=12 years
Consider, $80\leq IQ\leq 140$
$80\leq \frac{MA}{CA}*100\leq 140$
$80\leq \frac{MA}{12}*100\leq 140$
$80\leq \frac{MA}{3}*25\leq 140$
$80*\frac{3}{25}\leq MA\leq 140*\frac{3}{25}$
$\frac{48}{5}\leq MA\leq \frac{84}{5}$
$9.6\leq MA\leq 16.8$
1. $2\leq 3x-4\leq 5$
$2+4\leq 3x\leq 5+4$
$6\leq 3x\leq 9$
$\frac{6}{3}\leq x\leq \frac{9}{3}$
$2\leq x\leq 3$
∴ solution set is [2,3]
2. $6\leq -3(2x-4)< 12$
$\frac{6}{-3}\geq 2x-4> \frac{12}{-3}$
$-2\geq 2x-4> -4$
$-2+4\geq 2x> -4+4$
$2\geq 2x> 0$
$\frac{2}{2}\geq x> 0$
$1\geq x> 0$
∴ solution set is (0,1]
3. $-3\leq 4-\frac{7x}{2}\leq 18$
$-3-4\leq -\frac{7x}{2}\leq 18-4$
$-7\leq -\frac{7x}{2}\leq 14$
$-7*\frac{-2}{7}\geq x\geq 14*\frac{-2}{7}$
$2\geq x\geq -4$
∴ solution set is [-4,2]
4. $-15< \frac{3(x-2)}{5}\leq 0$
$-15*\frac{5}{3}< x-2\leq 0*\frac{5}{3}$
$-25< x-2\leq 0$
$-25+2< x\leq 0+2$
$-23< x\leq 2$
∴ solution set is (-23,2]
5. $-12< 4-\frac{3x}{-5}\leq 2$
$-12-4< -\frac{3x}{-5}\leq 2-4$
$-16< \frac{3x}{5}\leq -2$
$-16*\frac{5}{3}< x\leq -2*\frac{5}{3}$
$\frac{-80}{3}< x\leq \frac{-10}{3}$
∴ solution set is $(\frac{-80}{3},\frac{-10}{3}]$
6. $7\leq \frac{3x+11}{2}\leq 11$
$7*2\leq 3x+11\leq 11*2$
$14\leq 3x+11\leq 22$
$14-11\leq 3x\leq 22-11$
$3\leq 3x\leq 11$
$\frac{3}{3}\leq x\leq \frac{11}{3}$
$1\leq x\leq \frac{11}{3}$
∴ solution set is $[1,\frac{11}{3}]$
Solve the inequalities and represent the solution graphically on number line.
7. $5x+1> -24$, $5x-1< 24$
Consider, $5x+1> -24$
$5x> -24-1$
$5x> -25$
$x> \frac{-25}{5}$
$x> -5$ ...(i)
Consider, $5x-1< 24$
$5x< 24+1$
$5x< 25$
$x< \frac{25}{5}$
$x< 5$ ...(ii)
From (i) and (ii), solution set is (-5,5)
8. $2(x-1)< x+5$, $3(x+2)> 2-x$
Consider, $2(x-1)< x+5$
$2x-2< x+5$
$2x-x< 5+2$
$x< 7$ ...(i)
Consider, $3(x+2)> 2-x$
$3x+6> 2-x$
$3x+x> 2-6$
$4x> -4$
$x> -1$ ...(ii)
From (i) and (ii), solution set is (-1,7)
9. $3x-7> 2(x-6)$, $6-x> 11-2x$
Consider, $3x-7> 2(x-6)$
$3x-7> 2x-12$
$3x-2x> -12+7$
$x> -5$ ...(i)
Consider, $6-x> 11-2x$
$-x+2x> 11-6$
$x> 5$ ...(ii)
From (i) and (ii), solution set is (5,∞)
10. $5(2x-7)-3(2x+3)\leq 0$, $2x+19\leq 6x+47$
Consider, $5(2x-7)-3(2x+3)\leq 0$
$10x-35-6x-9\leq 0$
$4x-44\leq 0$
$4x\leq 44$
$x\leq 11$ ...(i)
Consider, $2x+19\leq 6x+47$
$19-47\leq 6x-2x$
$-28\leq 4x$
$-7\leq x$ ...(ii)
From (i) and (ii), solution set is [-7,11]
11. A solution is to be kept between 68℉ and 77℉. What is the range in temperature in degree Celsius (C) if the Celsius/Fahrenheit (F) conversion formula is given by, $F=\frac{9}{5}C+32$?
Given conversion formula,
$F=\frac{9}{5}C+32$
Also, solution is to be kept between 68℉ and 77℉
∴ $68< F< 77$
substitute equation of F,
⇒ $68< \frac{9}{5}C+32< 77$
$68-32< \frac{9}{5}C< 77-32$
$36< \frac{9}{5}C< 45$
$36*\frac{5}{9}< C< 45*\frac{5}{9}$
$20< C< 25$
∴ the range in temperature in degree Celsius is 20℃ to 25℃.
12. 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 8% solution, how many litres of the 2% solution will have to be added?
Let x litres of 2% solution is to be added.
boric acid in resulting mixture = 8% of 640+2% of x
total mixture = (x+640) litres
given that, resulting mixture is to be more than 4% but less than 6% boric acid
∴ $8% of 640+2% of x> 4% of (x+640)$ and $8% of 640+2% of x< 6% of (x+640)$
Consider, $8% of 640+2% of x> 4% of (x+640)$
$\frac{8}{100}*640+\frac{2}{100}*x> \frac{4}{100}* (x+640)$
$\frac{256}{5}+\frac{x}{50}> \frac{4}{100}*640+\frac{4x}{100}$
$\frac{256}{5}+\frac{x}{50}> \frac{128}{5}+\frac{x}{25}$
$\frac{256}{5}-\frac{128}{5}> \frac{x}{25}-\frac{x}{50}$
$\frac{128}{5}> \frac{2x-x}{50}$
$\frac{128}{5}> \frac{x}{50}$
$\frac{128}{5}*50> x$
$1280> x$
Consider, $8% of 640+2% of x< 6% of (x+640)$
$\frac{8}{100}*640+\frac{2}{100}*x< \frac{6}{100}* (x+640)$
$\frac{256}{5}+\frac{2x}{50}< \frac{6x}{100}+\frac{6}{100}*640$
$\frac{256}{5}+\frac{x}{50}< \frac{3x}{50}+\frac{192}{5}$
$\frac{256}{5}-\frac{192}{5}< \frac{3x}{50}-\frac{x}{50}$
$\frac{64}{5}< \frac{3x-x}{50}$
$\frac{64}{5}< \frac{2x}{50}$
$\frac{64}{5}< \frac{x}{25}$
$\frac{64}{5}*25< x$
$320< x$
Therefore, number of litres of the 2% solution to be added must be more than 320 litres but less than 1280 litres.
13. 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 litres of water is to be added.
acid in resulting mixture = 45% of 1125
total mixture = (x+1125) litres
given that, resulting mixture will contain more than 25% but less than 30% acid content
∴ $45% of 1125> 25% of (x+1125)$ and $45% of 1125< 30% of (x+1125)$
Consider, $45% of 1125> 25% of (x+1125) $
$\frac{45}{100}*1125> \frac{25}{100}*(x+1125)$
$\frac{2025}{4}> \frac{1}{4}*(x+1125)$
$\frac{2025}{4}> \frac{x}{4}+\frac{1125}{4}$
$\frac{2025}{4}-\frac{1125}{4}> \frac{x}{4}$
$\frac{900}{4}> \frac{x}{4}$
$900> x$
Consider, $45% of 1125< 30% of (x+1125) $
$\frac{45}{100}*1125< \frac{30}{100}*(x+1125)$
$\frac{2025}{4}< \frac{3}{10}*(x+1125)$
$\frac{2025}{4}< \frac{3x}{10}+\frac{3*1125}{10}$
$\frac{2025}{4}< \frac{3x}{10}+\frac{675}{2}$
$\frac{2025}{4}-\frac{675}{2}< \frac{3x}{10}$
$\frac{2025-2*675}{4}< \frac{3x}{10}$
$\frac{2025-1350}{4}< \frac{3x}{10}$
$\frac{675}{4}< \frac{3x}{10}$
$\frac{675}{4}*\frac{10}{3}< x$
$562.5< x$
Therefore, number of litres of water to be added must be more than 562.5 litres but less than 900 litres.
14. IQ of a person is given by the formula $IQ=\frac{MA}{CA}*100$, where MA is mental age and CA is chronological age. If $80\leq IQ\leq 140$ for a group of 12 years old children, find the range of their mental age.
Given that,
$IQ=\frac{MA}{CA}*100$
and for a group of 12 years old children,
$80\leq IQ\leq 140$
Also, CA=12 years
Consider, $80\leq IQ\leq 140$
$80\leq \frac{MA}{CA}*100\leq 140$
$80\leq \frac{MA}{12}*100\leq 140$
$80\leq \frac{MA}{3}*25\leq 140$
$80*\frac{3}{25}\leq MA\leq 140*\frac{3}{25}$
$\frac{48}{5}\leq MA\leq \frac{84}{5}$
$9.6\leq MA\leq 16.8$
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