1. Solve $24x<100$, when (i) x is a natural number (ii) x is an integer
Given that, $24x<100$
$\frac{24x}{24}<\frac{100}{24}$ [Divide both sides by same number]
$x<\frac{25}{6}$
(i) x is a natural number
Natural numbers less than $\frac{25}{6}$ are 1,2,3,4
∴ solution set is {1,2,3,4}
(ii) x is an integer
Integers less than $\frac{25}{6}$ are...-3, -2, -1, 0, 1, 2, 3, 4
∴ solution set is {...-3, -2, -1, 0, 1,2,3,4}
2. Solve $-12x>30$, when (i) x is a natural number (ii) x is an integer
Given that, $-12x>30$
$\frac{-12x}{-12}<\frac{30}{-12}$ [Since divided by negative number, inequality sign is reversed]
$x<\frac{-5}{2}$
(i) x is a natural number
There is no natural number less than $\frac{-5}{2}$
∴ no solution
(ii) x is an integer
Integers less than $\frac{-5}{2}$ are...-5, -4, -3
∴ solution set is {...-5,-4,-3}
3. Solve $5x-3<7$, when (i) x is an integer (ii) x is a real number
Given that, $5x-3<7$
$5x-3+3<7+3$
$5x<10$
$\frac{5x}{5}<\frac{10}{5}$
$x<2$
(i) x is an integer
Integers less than $2$ are ...-4, -3,-2,-1,0,1
∴ solution set is {...-4, -3,-2,-1,0,1}
(ii) x is a real number
Solution set is (-∞,2)
4. Solve $3x+8>2$, when (i) x is an integer (ii) x is a real number
Given that, $3x+8>2$
$3x+8-8>2-8$
$3x>-6$
$\frac{3x}{3}>\frac{-6}{3}$
$x>-2$
(i) x is an integer
Integers greater than $-2$ are -1,0,1,2,3,4...
∴ solution set is {-1,0,1,2,3,4...}
(ii) x is a real number
Solution set is (-2,∞)
Solve the inequalities for real x.
5. $4x+3<6x+7$
$4x+3-7<6x+7-7$
$4x-4<6x$
$4x-4-4x<6x-4x$
$-4<2x$
$2x>-4$
$\frac{2x}{2}>\frac{-4}{2}$
$x>-2$
∴ Solution set is (-2,∞)
6. $3x-7>5x-1$
$3x-7+1>5x-1+1$
$3x-6>5x$
$3x-6-3x>5x-3x$
$-6>2x$
$2x<-6$
$\frac{2x}{2}<\frac{-6}{2}$
$x<-3$
∴ Solution set is (-∞,-3)
7. $3(x-1)\leq2(x-3)$
$3x-3\leq 2x-6$
$3x-3+3\leq 2x-6+3$
$3x\leq 2x-3$
$3x-2x\leq 2x-3-2x$
$x\leq -3$
∴ Solution set is (-∞,-3]
8. $3(2-x)\geq 2(1-x)$
$6-3x\geq 2-2x$
$6-3x-2\geq 2-2x-2$
$4-3x\geq -2x$
$4-3x+3x\geq -2x+3x$
$x\leq 4$
∴ Solution set is (-∞,4]
9. $x+\frac{x}{2}+\frac{x}{3}<11$
$x(1+\frac{1}{2}+\frac{1}{3})<11$
$x(\frac{11}{6})<11$
$\frac{11x}{6}<11$
$\frac{11x}{6*11}<\frac{11}{11}$
$\frac{x}{6}<1$
$x<6$
∴ Solution set is (-∞,6)
10. $\frac{x}{3}>\frac{x}{2}+1$
$\frac{x}{3}-\frac{x}{2}>1$
$\frac{2x-3x}{6})>1$
$\frac{-x}{6}>1$
$-x>6$
$x<-6$
∴ Solution set is (-∞,-6)
11. $\frac{3(x-2)}{5}\leq \frac{5(2-x)}{3}$
$\frac{3x-6}{5}\leq \frac{10-5x}{3}$
$15*(\frac{3x-6}{5})\leq 15*(\frac{10-5x}{3})$
$3(3x-6)\leq 5(10-5x)$
$9x-18\leq 50-25x$
$9x+25x\leq 50+18$
$34x\leq 68$
$\frac{34x}{34}\leq \frac{68}{34}$
$x\leq 2$
∴ Solution set is (-∞,2]
12. $\frac{1}{2}(\frac{3x}{5}+4)\geq \frac{1}{3}(x-6)$
$\frac{3x}{10}+\frac{4}{2}\geq \frac{x}{3}-\frac{6}{3}$
$\frac{3x}{10}+2\geq \frac{x}{3}-2$
$2+2\geq \frac{x}{3}-\frac{3x}{10}$
$4\geq \frac{10x-9x}{30}$
$4\geq \frac{x}{30}$
$120\geq x$
$x\leq 120$
∴ Solution set is (-∞,120]
13. $2(2x+3)-10<6(x-2)$
$4x+6-10<6x-12$
$4x-4<6x-12$
$-4+12<6x-4x$
$8<2x$
$4<x$
$x>4$
∴ Solution set is (4, ∞)
14. $37-(3x+5)\geq 9x-8(x-3)$
$37-3x-5\geq 9x-8x+24$
$32-3x\geq x+24$
$32-24\geq x+3x$
$8\geq 4x$
$2\geq x$
$x\leq 2$
∴ Solution set is (-∞,2]
15. $\frac{x}{4}<\frac{(5x-2)}{3}-\frac{(7x-3)}{5}$
$\frac{x}{4}<\frac{5(5x-2)-3(7x-3)}{15}$
$\frac{x}{4}<\frac{25x-10-21x+9}{15}$
$\frac{x}{4}<\frac{4x-1}{15}$
$15x<4(4x-1)$
$15x<16x-4$
$4<16x-15x$
$4<x$
$x>4$
∴ Solution set is (4,∞)
16. $\frac{2x-1}{3}\geq \frac{(3x-2)}{4}-\frac{(2-x)}{5}$
$\frac{2x-1}{3}\geq \frac{5(3x-2)-4(2-x)}{20}$
$\frac{2x-1}{3}\geq \frac{15x-10-8+4x}{20}$
$\frac{2x-1}{3}\geq \frac{19x-18}{20}$
$20(2x-1)\geq 3(19x-18)$
$40x-20\geq 57x-54$
$-20+54\geq 57x-40x$
$34\geq 17x$
$2\geq x$
$x\leq 2$
∴ Solution set is (-∞,2]
Solve the inequalities and show the graph of the solution on number line
17. $3x-2<2x+1$
$3x-2x<1+2$
$x<3$
solution is (-∞,3)
18. $5x-3\geq 3x-5$
$5x-3x\geq -5+3$
$2x\geq -2$
$x\geq -1$
solution is [-1,∞)
19. $3(1-x)<2(x+4)$
$3-3x<2x+8$
$3-8<2x+3x$
$-5<5x$
$-1<x$
$x>-1$
solution is (-1,∞)
20. $\frac{x}{2}<\frac{(5x-2)}{3}-\frac{(7x-3)}{5}$
$\frac{x}{2}<\frac{5(5x-2)-3(7x-3)}{15}$
$\frac{x}{2}<\frac{25x-10-21x+9}{15}$
$\frac{x}{2}<\frac{4x-1}{15}$
$15x<2(4x-1)$
$15x<8x-2$
$15x-8x<-2$
$7x<-2$
$x<\frac{-2}{7}$
∴ Solution set is $(-\infty,\frac{-2}{7})$
21. Ravi obtained 70 and 75 marks in first two unit test. Find the number if minimum marks he should get in the third test to have an average of at least 60 marks.
Let x be the marks obtained in third unit test.
Average must be atleast 60 marks.
∴ $\frac{70+75+x}{3}\geq 60$
$\frac{145+x}{3}\geq 60$
$145+x\geq 180$
$x\geq 180-145$
$x\geq 35$
Thus Ravi must obtain minimum of 35 marks to have an average of at least 60 marks.
22. To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade ‘A’ in the course.
Let x be the marks obtained in fifth examination.
Average must be 90 marks or more.
∴ $\frac{87+92+94+95+x}{5}\geq 90$
$\frac{368+x}{5}\geq 90$
$368+x\geq 450$
$x\geq 450-368$
$x\geq 82$
Thus Sunita must obtain 82 marks or more in fifth examination.
23. Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.
Let x be the smallest odd positive integer.
∴ consecutive odd integer is x+2.
Given that, both integers are smaller than 10.
∴ $x+2<10$
$x<10-2$
$x<8$ ...(i)
Also given that, their sum is more than 11
∴ $x+(x+2)>11$
$2x+2>11$
$2x>11-2$
$2x>9$
$x>\frac{9}{2}$
$x>4.5$ ...(ii)
from (i) and (ii) we have,
$4.5<x<8$
∴ values of x are 5 and 7.
Thus required consecutive pairs are (5,7) and (7,9)
24. Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.
Let x be the smallest even positive integer.
∴ consecutive even integer is x+2.
Given that, both integers are greater than 5.
$x>5$ ...(i)
Also given that, their sum is less than 23
∴ $x+(x+2)<23$
$2x+2<23$
$2x<23-2$
$2x<21$
$x<\frac{21}{2}$
$x<10.5$ ...(ii)
from (i) and (ii) we have,
$5<x<10.5$
∴ values of x are 6,8,10.
Thus required consecutive pairs are (6,8), (8,10), (10,12)
25. The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.
Let the shortest side of triangle be x cm.
given that,
longest side = 3x cm
third side = 3x-2 cm
Perimeter is at least 61 cm
∴ $x+(3x)+(3x-2)\geq 61$
$7x-2\geq 61$
$7x\geq 61+2$
$7x\geq 63$
$x\geq \frac{63}{7}$
$x\geq 9$
Thus, minimum length of shortest side is 9 cm.
26. A man wants to cut three lengths from a single piece of board of length 91cm. The second length is to be 3cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5cm longer than the second? [ Hint: If x is the length of the shortest board, then x , (x + 3) and 2x are the lengths of the second and third piece, respectively. Thus, x + (x + 3) + 2x ≤ 91 and 2x ≥ (x + 3) + 5].
Let length of the shortest piece be x cm
Given that,
second piece = x + 3
third piece = 2x
$x+(x+3)+(2x)\leq 91$
$4x+3\leq 91$
$4x\leq 91-3$
$4x\leq 88$
$x\leq 22$ ...(i)
Also given that third piece is at least 5cm longer than the second
∴ $2x\geq 5+(x+3)$
$2x\geq x+8$
$2x-x\geq 8$
$x\geq 8$ ...(ii)
from (i) and (ii) we have,
$8\leq x \leq 22$
Thus, shortest board length is greater than or equal to 8 cm but less than or equal to 22 cm.
Given that, $24x<100$
$\frac{24x}{24}<\frac{100}{24}$ [Divide both sides by same number]
$x<\frac{25}{6}$
(i) x is a natural number
Natural numbers less than $\frac{25}{6}$ are 1,2,3,4
∴ solution set is {1,2,3,4}
(ii) x is an integer
Integers less than $\frac{25}{6}$ are...-3, -2, -1, 0, 1, 2, 3, 4
∴ solution set is {...-3, -2, -1, 0, 1,2,3,4}
2. Solve $-12x>30$, when (i) x is a natural number (ii) x is an integer
Given that, $-12x>30$
$\frac{-12x}{-12}<\frac{30}{-12}$ [Since divided by negative number, inequality sign is reversed]
$x<\frac{-5}{2}$
(i) x is a natural number
There is no natural number less than $\frac{-5}{2}$
∴ no solution
(ii) x is an integer
Integers less than $\frac{-5}{2}$ are...-5, -4, -3
∴ solution set is {...-5,-4,-3}
3. Solve $5x-3<7$, when (i) x is an integer (ii) x is a real number
Given that, $5x-3<7$
$5x-3+3<7+3$
$5x<10$
$\frac{5x}{5}<\frac{10}{5}$
$x<2$
(i) x is an integer
Integers less than $2$ are ...-4, -3,-2,-1,0,1
∴ solution set is {...-4, -3,-2,-1,0,1}
(ii) x is a real number
Solution set is (-∞,2)
4. Solve $3x+8>2$, when (i) x is an integer (ii) x is a real number
Given that, $3x+8>2$
$3x+8-8>2-8$
$3x>-6$
$\frac{3x}{3}>\frac{-6}{3}$
$x>-2$
(i) x is an integer
Integers greater than $-2$ are -1,0,1,2,3,4...
∴ solution set is {-1,0,1,2,3,4...}
(ii) x is a real number
Solution set is (-2,∞)
Solve the inequalities for real x.
5. $4x+3<6x+7$
$4x+3-7<6x+7-7$
$4x-4<6x$
$4x-4-4x<6x-4x$
$-4<2x$
$2x>-4$
$\frac{2x}{2}>\frac{-4}{2}$
$x>-2$
∴ Solution set is (-2,∞)
6. $3x-7>5x-1$
$3x-7+1>5x-1+1$
$3x-6>5x$
$3x-6-3x>5x-3x$
$-6>2x$
$2x<-6$
$\frac{2x}{2}<\frac{-6}{2}$
$x<-3$
∴ Solution set is (-∞,-3)
7. $3(x-1)\leq2(x-3)$
$3x-3\leq 2x-6$
$3x-3+3\leq 2x-6+3$
$3x\leq 2x-3$
$3x-2x\leq 2x-3-2x$
$x\leq -3$
∴ Solution set is (-∞,-3]
8. $3(2-x)\geq 2(1-x)$
$6-3x\geq 2-2x$
$6-3x-2\geq 2-2x-2$
$4-3x\geq -2x$
$4-3x+3x\geq -2x+3x$
$x\leq 4$
∴ Solution set is (-∞,4]
9. $x+\frac{x}{2}+\frac{x}{3}<11$
$x(1+\frac{1}{2}+\frac{1}{3})<11$
$x(\frac{11}{6})<11$
$\frac{11x}{6}<11$
$\frac{11x}{6*11}<\frac{11}{11}$
$\frac{x}{6}<1$
$x<6$
∴ Solution set is (-∞,6)
10. $\frac{x}{3}>\frac{x}{2}+1$
$\frac{x}{3}-\frac{x}{2}>1$
$\frac{2x-3x}{6})>1$
$\frac{-x}{6}>1$
$-x>6$
$x<-6$
∴ Solution set is (-∞,-6)
11. $\frac{3(x-2)}{5}\leq \frac{5(2-x)}{3}$
$\frac{3x-6}{5}\leq \frac{10-5x}{3}$
$15*(\frac{3x-6}{5})\leq 15*(\frac{10-5x}{3})$
$3(3x-6)\leq 5(10-5x)$
$9x-18\leq 50-25x$
$9x+25x\leq 50+18$
$34x\leq 68$
$\frac{34x}{34}\leq \frac{68}{34}$
$x\leq 2$
∴ Solution set is (-∞,2]
12. $\frac{1}{2}(\frac{3x}{5}+4)\geq \frac{1}{3}(x-6)$
$\frac{3x}{10}+\frac{4}{2}\geq \frac{x}{3}-\frac{6}{3}$
$\frac{3x}{10}+2\geq \frac{x}{3}-2$
$2+2\geq \frac{x}{3}-\frac{3x}{10}$
$4\geq \frac{10x-9x}{30}$
$4\geq \frac{x}{30}$
$120\geq x$
$x\leq 120$
∴ Solution set is (-∞,120]
13. $2(2x+3)-10<6(x-2)$
$4x+6-10<6x-12$
$4x-4<6x-12$
$-4+12<6x-4x$
$8<2x$
$4<x$
$x>4$
∴ Solution set is (4, ∞)
14. $37-(3x+5)\geq 9x-8(x-3)$
$37-3x-5\geq 9x-8x+24$
$32-3x\geq x+24$
$32-24\geq x+3x$
$8\geq 4x$
$2\geq x$
$x\leq 2$
∴ Solution set is (-∞,2]
15. $\frac{x}{4}<\frac{(5x-2)}{3}-\frac{(7x-3)}{5}$
$\frac{x}{4}<\frac{5(5x-2)-3(7x-3)}{15}$
$\frac{x}{4}<\frac{25x-10-21x+9}{15}$
$\frac{x}{4}<\frac{4x-1}{15}$
$15x<4(4x-1)$
$15x<16x-4$
$4<16x-15x$
$4<x$
$x>4$
∴ Solution set is (4,∞)
16. $\frac{2x-1}{3}\geq \frac{(3x-2)}{4}-\frac{(2-x)}{5}$
$\frac{2x-1}{3}\geq \frac{5(3x-2)-4(2-x)}{20}$
$\frac{2x-1}{3}\geq \frac{15x-10-8+4x}{20}$
$\frac{2x-1}{3}\geq \frac{19x-18}{20}$
$20(2x-1)\geq 3(19x-18)$
$40x-20\geq 57x-54$
$-20+54\geq 57x-40x$
$34\geq 17x$
$2\geq x$
$x\leq 2$
∴ Solution set is (-∞,2]
Solve the inequalities and show the graph of the solution on number line
17. $3x-2<2x+1$
$3x-2x<1+2$
$x<3$
solution is (-∞,3)
18. $5x-3\geq 3x-5$
$5x-3x\geq -5+3$
$2x\geq -2$
$x\geq -1$
solution is [-1,∞)
19. $3(1-x)<2(x+4)$
$3-3x<2x+8$
$3-8<2x+3x$
$-5<5x$
$-1<x$
$x>-1$
solution is (-1,∞)
20. $\frac{x}{2}<\frac{(5x-2)}{3}-\frac{(7x-3)}{5}$
$\frac{x}{2}<\frac{5(5x-2)-3(7x-3)}{15}$
$\frac{x}{2}<\frac{25x-10-21x+9}{15}$
$\frac{x}{2}<\frac{4x-1}{15}$
$15x<2(4x-1)$
$15x<8x-2$
$15x-8x<-2$
$7x<-2$
$x<\frac{-2}{7}$
∴ Solution set is $(-\infty,\frac{-2}{7})$
21. Ravi obtained 70 and 75 marks in first two unit test. Find the number if minimum marks he should get in the third test to have an average of at least 60 marks.
Let x be the marks obtained in third unit test.
Average must be atleast 60 marks.
∴ $\frac{70+75+x}{3}\geq 60$
$\frac{145+x}{3}\geq 60$
$145+x\geq 180$
$x\geq 180-145$
$x\geq 35$
Thus Ravi must obtain minimum of 35 marks to have an average of at least 60 marks.
22. To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade ‘A’ in the course.
Let x be the marks obtained in fifth examination.
Average must be 90 marks or more.
∴ $\frac{87+92+94+95+x}{5}\geq 90$
$\frac{368+x}{5}\geq 90$
$368+x\geq 450$
$x\geq 450-368$
$x\geq 82$
Thus Sunita must obtain 82 marks or more in fifth examination.
23. Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.
Let x be the smallest odd positive integer.
∴ consecutive odd integer is x+2.
Given that, both integers are smaller than 10.
∴ $x+2<10$
$x<10-2$
$x<8$ ...(i)
Also given that, their sum is more than 11
∴ $x+(x+2)>11$
$2x+2>11$
$2x>11-2$
$2x>9$
$x>\frac{9}{2}$
$x>4.5$ ...(ii)
from (i) and (ii) we have,
$4.5<x<8$
∴ values of x are 5 and 7.
Thus required consecutive pairs are (5,7) and (7,9)
24. Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.
Let x be the smallest even positive integer.
∴ consecutive even integer is x+2.
Given that, both integers are greater than 5.
$x>5$ ...(i)
Also given that, their sum is less than 23
∴ $x+(x+2)<23$
$2x+2<23$
$2x<23-2$
$2x<21$
$x<\frac{21}{2}$
$x<10.5$ ...(ii)
from (i) and (ii) we have,
$5<x<10.5$
∴ values of x are 6,8,10.
Thus required consecutive pairs are (6,8), (8,10), (10,12)
25. The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.
Let the shortest side of triangle be x cm.
given that,
longest side = 3x cm
third side = 3x-2 cm
Perimeter is at least 61 cm
∴ $x+(3x)+(3x-2)\geq 61$
$7x-2\geq 61$
$7x\geq 61+2$
$7x\geq 63$
$x\geq \frac{63}{7}$
$x\geq 9$
Thus, minimum length of shortest side is 9 cm.
26. A man wants to cut three lengths from a single piece of board of length 91cm. The second length is to be 3cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5cm longer than the second? [ Hint: If x is the length of the shortest board, then x , (x + 3) and 2x are the lengths of the second and third piece, respectively. Thus, x + (x + 3) + 2x ≤ 91 and 2x ≥ (x + 3) + 5].
Let length of the shortest piece be x cm
Given that,
second piece = x + 3
third piece = 2x
$x+(x+3)+(2x)\leq 91$
$4x+3\leq 91$
$4x\leq 91-3$
$4x\leq 88$
$x\leq 22$ ...(i)
Also given that third piece is at least 5cm longer than the second
∴ $2x\geq 5+(x+3)$
$2x\geq x+8$
$2x-x\geq 8$
$x\geq 8$ ...(ii)
from (i) and (ii) we have,
$8\leq x \leq 22$
Thus, shortest board length is greater than or equal to 8 cm but less than or equal to 22 cm.
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