Question 13 Marks
Solve the following linear inequations in R:
$\text{x}-2\leq\frac{5\text{x}+8}{3}$
Answer$\text{x}-2\leq\frac{5\text{x}+8}{3}$
$3(\text{x}-2)\leq5\text{x}+8$
$3\text{x}-6\leq5\text{x}+8$
$-2\text{x}\leq14$
$2\text{x}\geq-14$
$\text{x}\geq-7$
$\therefore$ The solution set is $[-7,\infty)$
View full question & answer→Question 23 Marks
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.
AnswerLet the pH value of third reading be x.
$\therefore7.2<\frac{7.48+7.85+\text{x}}{3}<7.8$
⇒ 21.6 < 7.48 + 7.85 + x < 23.4
⇒ 21.6 < 15.33 + x < 23.4
⇒ 21.6 - 15.33 < x < 23.4 - 15.33
⇒ 6.27 < x < 8.07
$\therefore$ The range of pH value for the third reading is lies between 6.27 and 8.07.
View full question & answer→Question 33 Marks
The longest side of a triangle is three times the shortest side and 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.
AnswerLet the length of the shortest side be x.
Then, the length of te longest side and third side of the triangle are 3x and 3x - 2 respectively.
According to question, perimeter of triangle > 61
perimeter of triangle $\geq61$
$\Rightarrow\text{x}+3\text{x}-2+3\text{x}\geq61$
$\Rightarrow7\text{x}\geq61+2$
$\Rightarrow7\text{x}\geq63$
$\Rightarrow\text{x}\geq\frac{63}{7}$
$\Rightarrow\text{x}\geq9$
$\therefore$ The maximum length of the shortest side is 9cm.
View full question & answer→Question 43 Marks
Solve the following linear inequations in R:
$\frac{3\text{x}-2}{5}\leq\frac{4\text{x}-3}{2}$
Answer$\frac{3\text{x}-2}{5}\leq\frac{4\text{x}-3}{2}$
$\Rightarrow\frac{3\text{x}}{5}-\frac{2}{5}\leq\frac{4\text{x}}{2}-\frac{3}{2}$
$\Rightarrow\frac{3\text{x}}{5}-\frac{4\text{x}}{2}\leq\frac{-3}{2}+\frac{2}{5}$
$\Rightarrow\frac{3\text{x}}{5}-\frac{4\text{x}}{2}\leq\frac{-3}{2}+\frac{2}{5}$
$\Rightarrow\frac{6\text{x}-20\text{x}}{10}\leq\frac{-15+4}{10}$
$\Rightarrow-14\text{x}\leq-11$
$\Rightarrow14\text{x}\geq11$
$\Rightarrow\text{x}\geq\frac{11}{14}$
$\Big[\frac{11}{14},\infty\Big)$ is the solution set
View full question & answer→Question 53 Marks
Solve the following linear inequations in R:
$\frac{5\text{x}-6}{\text{x}+6}<1$
Answer$\frac{5\text{x}-6}{\text{x}+6}<1$
$\frac{5\text{x}-6}{\text{x}+6}-1<0$
$\frac{5\text{x}-6-(\text{x}+6)}{\text{x}+6}<0$
$\frac{5\text{x}-6-\text{x}-6}{\text{x}+6}<0$
$\frac{4\text{x}-12}{\text{x}+6}<0$
Case 1: $4\text{x}-2>0$ and $\text{x}+6<0$
$\Rightarrow\text{x}>3$ and $\text{x}<-6$
Case 2: $4\text{x}-2<0$ and $\text{x}+6>0$
$\Rightarrow\text{x}<3$ and $\text{x}>-6$
Hence the solution set is (-6, 3)
View full question & answer→Question 63 Marks
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 there are 640 litres of the 8% solution, how many litres of 2% solution will have to be added?
AnswerLet x liters of 2% solution will have to be added to 640 liters of the 8% solution of acid.
Total quantity of mixture = (640 + x)
Total acid in the (640+x) liters of mixture
$\frac{2}{100}\text{x}+\frac{8}{100}640$
It is given that acid content in the resulting mixture must be more than 4% but less than 6%.
$\frac{4}{100}[640+\text{x}]<\Big(\frac{2}{100}\text{x}+\frac{8}{100}640\Big)<\frac{6}{100}[640+\text{x}]$
⇒ 4 [640 + x] < ( 2x + 8640) < 6 [640 + x]
⇒ 2560 + 4x < 2x + 8640 and 2x + 8640 < 3840 + 6x
⇒ 2650 - 8640 < 2x - 4x and 2x - 6x < 3840 - 8640
⇒ x < 1280 and x > 320
More than 320 litres but less than 1280 litres of 2% is to be added.
View full question & answer→Question 73 Marks
A solution is to be kept between $30^\circ C$ and $35^\circ C$. What is the range of temperature in degree Fahrenheit?
AnswerWe have,
$C_1 = 30^\circ C$
$\therefore\text{F}_1=\frac{9}{5}\text{C}_1+32$$\Big[\because\text{F}=\frac{9}{5}\text{C}+32\Big]$
$\Rightarrow\text{F}_1=\frac{9}{5}\times30+32$
$\Rightarrow\text{F}_1=9\times6+32$
$\Rightarrow\text{F}_1=54+32$
$\Rightarrow\text{F}_1=86^\circ\text{F}$
Now,
$C^2 = 35^\circ C$
$\therefore\text{F}_2=\frac{9}{5}\text{C}_2+32$
$\Rightarrow\text{F}_2=\frac{9}{5}\times35+32$
$\Rightarrow\text{F}_2=9\times7+32$
$\Rightarrow\text{F}_2=63+32$
$\Rightarrow\text{F}_2=95^\circ\text{F}$
Hence, the temperature of the solution lies between 86ºF to 95ºF.
View full question & answer→Question 83 Marks
Solve the following system of equations in R.
$\frac{1}{|\text{x}|-3}\leq\frac{1}{2}$
Answer$\frac{1}{|\text{x}|-3}\leq\frac{1}{2}$
$\Rightarrow\frac{1}{|\text{x}|-3}-\frac{1}{2}\leq0$
$\Rightarrow\frac{2-|\text{x}|+3}{2(|\text{x}|-3)}\leq0$
$\Rightarrow\frac{5-|\text{x}|}{2(|\text{x}|-3)}\leq0$
$\Rightarrow\frac{|\text{x}|-5}{2(|\text{x}|-3)}\geq0$
$\Rightarrow|\text{x}|\geq5\ \text{or}|\text{x}|<3$
$\Rightarrow\text{x}\in(-\infty,-5)\text{ or }\text{x}\in(-3,-3)$
$\Rightarrow\text{x}\in(-\infty,-5)\cup(-3,-3)\cup[5,\infty)$
The solution set is $(-\infty,-5]\cup(-3,-3)\cup[5,\infty).$
View full question & answer→Question 93 Marks
Solve the following linear inequations in R:
$\frac{2\text{x}+3}{5}-2<\frac{3(\text{x}-2)}{5}$
Answer$\frac{2\text{x}+3}{5}-2<\frac{3(\text{x}-2)}{5}$
$\frac{2\text{x}+3-10}{5}<\frac{3\text{x}-6}{5}$
2x - 7 < 3x - 6
2x - 3x < -6 + 7
-x < 1
x > -1
View full question & answer→Question 103 Marks
Solve the following system of equations in R.
$\frac{7\text{x}-1}{2}<-3,\frac{3\text{x}+8}{5}+11<0$
AnswerConsider the first inequation,
$\frac{7\text{x}-1}{2}<-3$
7x - 1 < -6
7x < -6 + 1
7x < -5
Consider the second inequation,
$\frac{3\text{x}+8}{5}+11<0$
$\frac{3\text{x}+8+55}{5}<0$
$\frac{3\text{x}+63}{5}<\frac{0}{1}$
$3\text{x}+63<0$
$3\text{x}<-63$
$\text{x}<-21\ ...(\text{ii})$
From (i) and (ii), $(-\infty,-21)$ is the solution set of the simultaneous equations.
View full question & answer→Question 113 Marks
Solve the following linear inequations in R:
$\frac{\text{x}}{\text{x}-5}>\frac{1}{2}$
Answer$\frac{\text{x}}{\text{x}-5}>\frac{1}{2}$
$\frac{\text{x}}{\text{x}-5}-\frac{1}{2}>0$
$\frac{2\text{x}-(\text{x}-5)}{2(\text{x}-5)}>0$
$\frac{2\text{x}-\text{x}+5}{2\text{x}-10}>0$
$\frac{\text{x}+5}{2\text{x}-10}>0$
Case 1: $\text{x}+5>0$ and $2\text{x}-10>0$
$\Rightarrow\text{x}>-5$ and $\text{x}>5$
$\Rightarrow\text{x}>5$
Case 2: $\text{x}+5<0$ and $2\text{x}-10<0$
$\Rightarrow\text{x}<-5$ and $\text{x}<5$
$\Rightarrow\text{x}<-5$
Hence the solution set is $(-\infty,-5)\cup(5,\infty)$
View full question & answer→Question 123 Marks
Find all pairs of consecutive odd positive integers, both of which are smaller than 10, such that their sum is more than 11.
AnswerLet x be the smaller of the two consecutive odd positive intgers. Then the other odd integer is x + 2. It is given that both the integers are smaller than 10 and their sum is more than 11.
$\therefore$ x + 2 < 10 and, x + (x + 2) > 11
⇒ x < 10 - 2 and 2x + 2 > 11
⇒ x < 8 and 2x > 9
⇒ x < 8 and $\text{x}>\frac{9}{2}$
$\Rightarrow\frac{9}{2}<\text{x}<8$
⇒ x = 5, 7 [$\because$ is an odd integer]
Hence, the required pairs of odd integers are (5, 7) and (7, 9).
View full question & answer→Question 133 Marks
Solve the following system of equations in R.
$\frac{|\text{x}-2|}{\text{x}-2}>0$
AnswerWe have
$\frac{|\text{x}-2|}{\text{x}-2}>0\ ...(\text{i})$
Case1: when $|\text{x}-2|\geq0$
$\text{x}\geq2$
$\Rightarrow\frac{\text{x}-2}{\text{x}-2}\geq0$
$\Rightarrow\text{x}-2\geq0$
$\text{x}\geq2\ ..(\text{ii})$
Case 2: When |x - 2| < 0
x < 2
$\Rightarrow-\frac{(\text{x}-2)}{\text{x}-2}>0$
⇒ -(x - 2) > 0
⇒ -x + 2 < 0
⇒ x < -2
⇒ x > 2 ...(iii)
Combining (ii) and (iii), we get $(2,\infty)$ as the solution set.
View full question & answer→Question 143 Marks
Solve the following linear inequations in R:
$\frac{\text{x}-1}{3}+4<\frac{\text{x}-5}{5}-2$
Answer$\frac{\text{x}-1}{3}+4<\frac{\text{x}-5}{5}-2$
$\frac{\text{x}-1+12}{3}<\frac{\text{x}-5-10}{5}$
5 (x - 1 + 12) < 3 (x - 5 - 10)
5(x + 11) < 3 (x - 15)
5x + 55 < 3x - 45
5x - 3x < -45 -55
2x < -100
x < -50
$\therefore$ The solution set is $(-\infty,-50)$
View full question & answer→Question 153 Marks
A company manufactures cassettes and its cost and revenue functions for a week are $\text{C}=300+\frac{3}{2}\text{x}$ R = 2x respectively, where x is the number of cassettes produced and sold in a week. How many cassettes must be sold for the company to realize a profit?
AnswerWe have,
profit = Revenue - cost
therefore, to ear some profit, we must have
Revenue > Cost
$\Rightarrow2\text{x}>300+\frac{3}{2}\text{x}$
$\Rightarrow2\text{x}-\frac{3}{2}\text{x}>300$
$\Rightarrow\frac{4\text{x}-3\text{x}}{2}>300$
$\Rightarrow\text{x}>300\times2$
$\Rightarrow\text{x}>600$
Hence, then manufacture must sell more that 600 cassettes to realize some profit.
View full question & answer→Question 163 Marks
Solve the following system of equations in R.
$\Big|\text{x}+\frac{1}{3}\Big|>\frac{8}{3}$
AnswerConsider the first inequation, $\text{x}+\frac{1}{3}\geq0$ $\therefore\text{ex}\geq\frac{-1}{3}.$ $\Big|\text{x}+\frac{1}{3}\Big|>\frac{8}{3}>0$ $\frac{3\text{x}-7}{3}>0$ $3\text{x}-7>0$ $\text{x}>\frac{7}{3}\ ..(\text{i})$ Consider the second inequation, $\text{x}+\frac{1}{3}<0\therefore\text{e}.\text{x}<-\frac{1}{3}$ $\Big|\text{x}+\frac{1}{3}-\frac{8}{3}>0\Big|$ $-\text{x}-\frac{1}{3}-\frac{8}{3}>0$ $-3\text{x}-9>0$ $-3\text{x}>9$ $3\text{x}<-9$ $\text{x}<\frac{-9}{3}$ $\text{x}<-3\ ..(\text{ii})$From (i) and (ii), $(-\infty,-3)\cup\Big(\frac{7}{3},\infty\Big)$is the solution set of the simultaneous equations.
View full question & answer→Question 173 Marks
Solve the following system of equations in R.
$\frac{4}{\text{x}+1}\leq3\leq\frac{6}{\text{x}+1},\text{x}>0$
AnswerConsider the first inequation,
$\frac{4}{\text{x}+1}\leq3\leq\frac{6}{\text{x}+1}$
$\Rightarrow4\leq3(\text{x}+1)\leq6$
$\Rightarrow\frac{4}{3}\leq(\text{x}+1)\leq6$
$\Rightarrow\frac{4}{3}-1\leq\text{x}\leq2-1$
$\Rightarrow\frac{1}{3}\leq\text{x}\leq1$
Solution set for given inequation is $\Big[\frac{1}{3},1\Big].$
View full question & answer→Question 183 Marks
Find all pairs of consecutive odd natural number, both of which are larger than 10, such that their sum is less than 40.
AnswerLet x be the smaller of the two consecutive odd natural numbers. Then the other odd integer İS x + 2
It is given that both the natural number are greater than 10 and their sum is less than 40.
$\therefore$ x > 10 and x + x + 2 < 40
⇒ x > 10 and 2x < 38
⇒ x > 10 and x < 19
⇒ 10< x < 19
⇒ x = 11, 13, 15, 17 [ $\because$ is an odd number]
Hence, the required pairs of odd natural numbers are (11, 13), (13, 15) (15, 17) and (17, 19).
View full question & answer→Question 193 Marks
Solve the following linear inequations in R:
$\frac{2\text{x}-3}{3\text{x}-7}>0$
Answer$\frac{2\text{x}-3}{3\text{x}-7}>0$
Case 1: 2x - 3 > 0 and 3x - 7 > 0
$\Rightarrow\text{x}>\frac{3}{2}$ and $\text{x}<\frac{7}{3}$
$\Rightarrow\text{x}>\frac{7}{3}$
Case 2: 2x - 3 < 0 and 3x - 1 < 0
$\Rightarrow\text{x}<\frac{3}{2}$ and $\text{x}<\frac{7}{3}$
$\Rightarrow\text{x}<\frac{3}{2}$
$\therefore$ $\Big(-\infty,\frac{3}{2}\Big)\cup\Big(\frac{7}{3},\infty\Big)$Solution set
View full question & answer→Question 203 Marks
Solve the following linear inequations in R:
$\frac{5\text{x}+8}{4-\text{x}}<2$
Answer$\frac{5\text{x}+8}{4-\text{x}}<2$
$\frac{5\text{x}+8}{4-\text{x}}-2<0$
$\frac{5\text{x}+8-2(4-\text {x})}{4-\text{x}}<0$
$\frac{5\text{x}+8-8+2\text{x}}{4-\text{x}}<0$
$\frac{7\text{x}}{4-\text{x}}<0$
Case 1: $7\text{x}>0$ and $4-\text{x}<0$
$\Rightarrow\text{x}>0$ and $4<\text{x}$
$\Rightarrow\text{x}<0$
Hence solution set is $(-\infty,0)\cup(4,\infty)$
View full question & answer→Question 213 Marks
Solve the following linear inequations in R:
$\frac{2\text{x}+3}{4}-3<\frac{\text{x}-4}{3}-2$
Answer$\frac{2\text{x}+3}{4}-3<\frac{\text{x}-4}{3}-2$
$\frac{2\text{x}+3-12}{4}<\frac{\text{x}-4-6}{3}$
3 (2x + 3 - 12) < 4 (x - 4 - 6)
3(2x - 9) < 4 (x - 10)
6x - 27 < 4x - 40
6x - 4x < -40 + 27
2x < -13
$\text{x}<-\frac{13}{2}$
$\therefore$ The solution set is $\Big(-\infty,-\frac{13}{2}\Big)$
View full question & answer→Question 223 Marks
Solve the following system of equations in R.
$\frac{2\text{x}-3}{4}-2\geq\frac{4\text{x}}{3}-6,2(2\text{x}+3)<6(\text{x}-2)+10$
AnswerConsider the first inequation,
$\frac{2\text{x}-3}{4}-2\geq\frac{4\text{x}}{3}-6$
$\frac{2\text{x}-3-8}{4}\geq\frac{4\text{x}-18}{3}$
$3(2\text{x}-11)\geq4(4\text{x}-18)$
$6\text{x}-33\geq16\text{x}-72$
$6\text{x}-16\text{x}\geq-72+33$
$-10\text{x}\geq-39$
$\text{x}\leq\frac{39}{10}\ ...(\text{i})$
Consider the second inequation,
2 (2x + 3) < 6 (x - 2) + 10
4x + 6 < 6x - 12 + 10
4x - 6x < -12 - 6 + 10
-2x < -8
x > 8 ..(ii)
From (i) and (ii), There is no solution set of the simultaneous equations.
View full question & answer→Question 233 Marks
Solve the following linear inequations in R:
$\frac{4\text{x}+3}{2\text{x}-5}<6$
Answer$\frac{4\text{x}+3}{2\text{x}-5}<6$
$\frac{4\text{x}+3}{2\text{x}-5}-6<0$
$\frac{4\text{x}+3-12\text{x}+30}{2\text{x}-5}<0$
$\frac{-8\text{x}+33}{2\text{x}-5}<0$
$\frac{8\text{x}-33}{2\text{x}-5}>0$
Case: 1 $8\text{x}-33>0$ and $2\text{x}-5>0$
$\Rightarrow\text{x}>\frac{33}{8}$ and $\text{x}>\frac{5}{2}$
$\Rightarrow\text{x}>\frac{33}{8}$
Case 2: $8\text{x}-33<0$ and $2\text{x}-5<0$
$\Rightarrow\text{x}>\frac{33}{8}$ and $\text{x}<\frac{5}{2}$
$\Rightarrow\text{x}<\frac{5}{2}$
Hence the solution set is $\Big(-\infty,\frac{5}{2}\Big)\cup\Big(\frac{33}{8},\infty\Big)$
View full question & answer→Question 243 Marks
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 $F =\frac{9}{5} C +32$.
AnswerWe have,
$F_1 = 86^\circ F$
$\therefore\text{F}_1=\frac{9}{5}\text{C}_1+32$ $\Big[\because\text{F}=\frac{9}{5}\text{C}+32\Big]$
$\Rightarrow86=\frac{9}{5}\text{C}_1+32$
$\Rightarrow86-32=\frac{9}{5}\text{C}_1$
$\Rightarrow54=\frac{9}{5}\text{C}_1$
$\Rightarrow9\text{C}_1=5\times54$
$\Rightarrow\text{C}_1=\frac{5\times54}{9}$
$\Rightarrow\text{C}_1=5\times6=30^\circ\text{C}$
Now, $\text{F}_2=95^\circ\text{F}$
$\therefore\text{F}_2=\frac{9}{5}\text{C}_2+32$
$\Rightarrow95-32=\frac{9}{5}\text{C}_2$
$\Rightarrow63=\frac{9}{5}\text{C}_2$
$\Rightarrow9\text{C}_2=63\times5$
$\Rightarrow\text{C}_2=\frac{63\times5}{9}$
$\Rightarrow\text{C}_2=7\times5=35^\circ\text{C}$
The range of temperature of the solution is from $30^\circ C$ to $35^\circ C.$
View full question & answer→Question 253 Marks
Solve the following linear inequations in R:
$\frac{\text{x}-1}{\text{x}+3}>2$
Answer$\frac{\text{x}-1}{\text{x}+3}>2$
$\frac{\text{x}-1}{\text{x}+3}-2<0$
$\frac{\text{x}-1-2(\text {x}+3)}{\text{x}+3}>0$
$\frac{\text{x}-1-2\text{x}-6}{\text{x}+3}>0$
$\frac{-\text{x}-7}{\text{x}+3}>0$
$\frac{\text{x}+7}{\text{x}+3}<0$
Case 1: $\text{x}+7>0$ and $\text{x}+3<0$
$\Rightarrow\text{x}>-7$ and $\text{x}>-3$
Case 2: $\text{x}+7<0$ and $\text{x}+3>0$
$\Rightarrow\text{x}<-7$ and $\text{x}>-3$
This is not possible.
$\therefore$ The solution set is (-7, -3)
View full question & answer→Question 263 Marks
Solve the following system of equations in R.
$1\leq|\text{x}-2|\leq3$
Answer$1\leq|\text{x}-2|\leq3$
$\Rightarrow\text{x}\in[-3+2,-1+2]\cup[1+2,3+2]$
$\Rightarrow\text{x}\in[-1,1]\cup[3,5]$
$\therefore$ The solution set is for given inequality is $[-1,1]\cup[3,5].$
View full question & answer→Question 273 Marks
Find all pairs of consecutive even positive integers, both of which are larger than 5, such that their sum is less than 23.
AnswerLet x be the smaller of the two consecutive even positive integers.
Then the other even integer is x + 2.
It is given that both the even integers are greater than 5 and their sum is less than 23.
$\therefore$ x > 5 and x + x + 2 < 23
⇒ x > 5 and 2x < 21
$\Rightarrow\text{x}>5$ and $\text{x}<\frac{21}{2}$
$\Rightarrow5<\text{x}<\frac{21}{2}=10.5$
⇒ x = 6, 8, 10 [ $\because$ is an even integer]
Hence, the required pairs of even positive integer are (6, 8),(8, 10) and (10, 12).
View full question & answer→Question 283 Marks
Solve the following system of equations in R.
$-5<2\text{x}-3<5$
AnswerConsider the first inequation,
-5 < 2x - 3
2x - 3 > -5
2x > -5 + 3
2x > -2
x > -1 ...(i)
Consider the second inequation,
2x - 3 < 5
2x < 5 + 3
2x < 8
x < 4 ...(ii)
From (i) and (ii), (-1, 4) is the solution set of the simultaneous equations.
View full question & answer→Question 293 Marks
Solve the following linear inequations in R:
$\frac{2(\text{x}-1)}{5}\leq\frac{3(2+\text{x})}{7}$
Answer$\frac{2(\text{x}-1)}{5}\leq\frac{3(2+\text{x})}{7}$
$\Rightarrow7(2(\text{x}-1))\leq5(3(2+\text{x}))$
$14(\text{x}-1)\leq15(2+\text{x})$
$\Rightarrow14\text{x}-14\leq30+15\text{x}$
$\Rightarrow14\text{x}-15\text{x}\leq30+14$
$\Rightarrow-\text{x}\leq44$
$\Rightarrow\text{x}\geq-44$
$\therefore$ The solution set is $[-44,\infty)$
View full question & answer→Question 303 Marks
Solve the following linear inequations in R:
$\frac{7\text{x}-5}{8\text{x}+3}>4$
Answer$\frac{7\text{x}-5}{8\text{x}+3}>4$
$\frac{7\text{x}-5}{8\text{x}+3}-4>0$
$\frac{7\text{x}-5-4(8\text{x}+3)}{8\text{x}+3}>0$
$\frac{7\text{x}-5-32\text{x}-12}{8\text{x}+3}>0$
$\frac{-25\text{x}-17}{8\text{x}+3}>0$
$\frac{25\text{x}+17}{8\text{x}+3}<0$
Case 1: $25\text{x}+17>0$ and $8\text{x}+3<0$
$\Rightarrow\text{x}>\frac{-17}{25}$ and $\text{x}<\frac{-3}{8}$
Case 2: $25\text{x}+17<0$ and $8\text{x}+3>0$
$\Rightarrow\text{x}<\frac{-17}{25}$ and $\text{x}>\frac{-3}{8}$
This is not possible
$\therefore$ Hence the solution set is $\Big(\frac{-17}{25},\frac{-3}{8}\Big)$
View full question & answer→Question 313 Marks
Solve the following linear inequations in R:
$\frac{3}{\text{x}-2}<1$
Answer$\frac{3}{\text{x}-2}<1$
$\frac{3}{\text{x}-2}-1<0$
$\frac{3-(\text{x}-2)}{\text{x}-2}<0$
$\frac{3-\text{x}+2}{\text{x}-2}<0$
$\frac{5-\text{x}}{\text{x}-2}<0$
$\frac{\text{x}-5}{\text{x}-2}>0$
Case1: x - 5 > 0 and x - 2 > 0
⇒ x > 5 and x > 2
⇒ x > 5
Case 2: x - 5 < 0 and x - 2 < 0
⇒ x < 5 and x < 2
⇒ x < 2
$\therefore$ solution set is $(-\infty,2)\cup(5,\infty)$
View full question & answer→Question 323 Marks
Solve the following system of equations in R.
|4 - x| + 1 < 3
AnswerWe have,
|4 - x| + 1 < 3
⇒ |4 - x| - 2 < 0 ...(i)
Case 1: when |4 - x| > 0
|4 - x| - 2 < 0
⇒ 4 - x - 2 < 0
⇒ 2 - x < 0
⇒ -x < -2
⇒ x > 2 ...(ii)
Case 2: When |4 - x| < 0
|4 - x| - 2 < 0
⇒ -(4 - x) - 2 < 0
⇒ -4 + x - 2 < 0
⇒ x - 6 < 0
⇒ x < 6 ...(iii)
Combining (ii) and (iii) we get (2, 6) as the solution set.
View full question & answer→Question 333 Marks
Solve the following system of equations in R.
$|3-4\text{x}|\geq9$
Answer$|3-4\text{x}|\geq9$
$\Rightarrow4\Big|\frac{3}{4}-\text{x}\Big|\geq9$
$\Rightarrow\Big|\frac{3}{4}-\text{x}\Big|\geq\frac{9}{4}$
Case 1: When $-\infty<\text{x}\leq-\frac{3}{4}$
$\Big|\frac{3}{4}-\text{x}\Big|=\Big(\frac{3}{4}-\text{x}\Big)$
$\therefore\Big|\frac{3}{4}-\text{x}\Big|\geq\frac{9}{4}$
$\Rightarrow\Big(\frac{3}{4}-\text{x}\Big)\geq\frac{9}{4}$
$\Rightarrow-\frac{6}{4}\geq\text{x}$
$\Rightarrow-\frac{3}{2}\geq\text{x}$
But, $-\infty<\text{x}<-1$
$\therefore$ The solution set of the given inequation is $\Big(-\infty,-\frac{3}{2}\Big]$
Case 2: When $-\frac{3}{4}<\text{x}<\infty$
$\Big|\frac{3}{4}-\text{x}\Big|=-\Big(\frac{3}{4}-\text{x}\Big)$
$\therefore\Big|\frac{3}{4}-\text{x}\Big|\geq\frac{9}{4}$
$\Rightarrow\text{x}\geq3$
But, $-\frac{3}{4}<\text{x}<\infty$
$\therefore$ The solution set of the given inequation is $[3,\infty).$
Combining case 1 and case 2,
We obtain that the solution set of given in equality is $\Big(-\infty,-\frac{3}{2}\Big]\cup(3,\infty)$
View full question & answer→Question 343 Marks
Solve the following linear inequations in R:
$\frac{5-2\text{x}}{3}<\frac{\text{x}}{6}-5$
Answer$\frac{5-2\text{x}}{3}<\frac{\text{x}}{6}-5$
$\frac{5-2\text{x}}{3}<\frac{\text{x}-30}{6}$
6(5 - 2x) < 3 (x - 30)
30 - 12x < 3x - 90
-12x - 3x < -90 - 30
- 15x < -120
15x > 120
$\text{x}>\frac{120}{15}=8$
$\therefore$ The solution set is $(8,\infty)$
View full question & answer→Question 353 Marks
Solve the following linear inequations in R:
$\frac{5\text{x}}{2}+\frac{3\text{x}}{4}\geq\frac{39}{4}$
Answer$\frac{5\text{x}}{2}+\frac{3\text{x}}{4}\geq\frac{39}{4}$
$\Rightarrow\frac{10\text{x}+3\text{x}}{4}\geq\frac{39}{4}$
$\Rightarrow13\text{x}\geq39$
$\Rightarrow\text{x}\geq\frac{39}{13}=3$
$\Rightarrow\text{x}\geq3$
$\therefore$ The solution set is $[3,\infty)$
View full question & answer→Question 363 Marks
Solve the following system of equations in R.
5x - 1 < 24, 5x + 1 > -24
Answer5x - 1 < 24
5x < 24 + 1
5x < 25
$\text{x}<\frac{25}{5}$
x < 5 ...(i)
And
5x + 1 > -24
5x > -24 - 1
5x > -25
x > -5 ...(ii)
From equation (i) and (ii),
< x < 5
⇒ (-5, 5)
View full question & answer→Question 373 Marks
Solve the following system of equations in R.
$\frac{|\text{x}+2|-\text{x}}{\text{x}}<2$
AnswerWe have,
$\frac{|\text{x}+2|-\text{x}}{\text{x}}<0$
$\frac{|\text{x}+2|-\text{x}}{\text{x}}-2<0$
$\frac{|\text{x}+2|-\text{x}-2\text{x}}{\text{x}}<0$
$\frac{|\text{x}+2|-3\text{x}}{\text{x}}<0\ ...(\text{i})$
Case 1: When $|\text{x}+2|\geq0$
$\Rightarrow\frac{\text{x}+2-3\text{x}}{\text{x}}<0$
⇒ - 2x + 2 < 0
⇒ -2x < - 2 and x > 0
⇒ x > 1 ...(ii)
case2: |x + 2|< 0
x < -2
⇒ -(x + 2)- 3x < 0
⇒ - x - 2 - 3x < 0
⇒ -4x - 2 < 0
⇒ -4x < 2
$\Rightarrow\text{x}>\frac{-1}{2}\ ...(\text{iii})$
and x < 0
Combining (ii) and (iii), we get $(-\infty,0)\cup(1,\infty)$ as the solution set.
View full question & answer→Question 383 Marks
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.
AnswerSuppose Rohit scores x mark in the third test then,
$65\leq\frac{65+70+\text{x}}{3}$
$195\leq135+\text{x}$
$\Rightarrow195-135\leq\text{x}$
$\Rightarrow60\leq\text{x}$
Hence, the minimum marks Rohit should score in the third test test is 60.
View full question & answer→Question 393 Marks
Solve the following system of equations in R.
$5\text{x} - 7 < 3(\text{x} + 3), 1-\frac{3\text{x}}{2}\geq\text{x}-4$
AnswerConsider the first inequation, 5x - 7 < 3(x + 3) 5x - 7 < 3x + 9 5x - 3x < 9 + 7 2x < 16 X < 8 ...(i) Consider the second inequation, $1-\frac{3\text{x}}{2}\geq\text{x}-4$ $\frac{-3\text{x}}{2}-\text{x}\geq-4-1$ $\frac{-3\text{x}-2\text{x}}{2}\geq-5$ $-5\text{x}>-10$ $\text{x}\leq2\ ...(\text{ii})$From (i) and (ii), $(-\infty,2)$ is the solution set of the simultaneous equations.
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