Show that the solution set of the following system of linear inequalities is an unbounded region:
Question: Show that the solution set of the following system of linear inequalities is an unbounded region: $2 x+y \geq 8, x+2 y \geq 10, x \geq 0, y \geq 0$. Solution: We have: $2 x+y \geq 8$ ...(i) $x+2 y \geq 10$ ...(ii) $x \geq 0$ ....(iii) $y \geq 0$ ...(iv) As, the solutions of the equation 2x+y= 8 are: As, the solutions of the equationx+ 2y = 10 are: Now, the graph represented by the inequalities (i), (ii), (iii) and (iv) is as follows:Since, the common shaded region is the solution set o...
Read More →The difference of squares of two numbers is 180.
Question: The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find two numbers. Solution: Let the larger numbers be $x$ Then according to question, Square of the smaller number be $=8 x$ then $x^{2}-8 x=180$ $x^{2}-8 x-180=0$ $x^{2}-18 x+10 x-180=0$ $x(x-18)+10(x-18)=0$ $(x-18)(x+10)=0$ $(x-18)=0$ $x=18$ Or $(x+10)=0$ $x=-10$ Since,xbeing a positive integer so, x cannot be negative, Therefore, When $x=18$ then smaller number be $\sqrt{8...
Read More →Factorize:
Question: Factorize: $x^{3}+27$ Solution: $x^{3}+27=(x)^{3}+(3)^{3}$ $=(x+3)\left(x^{2}-3 x+3^{2}\right)$ $=(x+3)\left(x^{2}-3 x+9\right)$...
Read More →The difference of squares of two number is 88.
Question: The difference of squares of two number is 88. If the larger number is 5 less than twice the smaller number, then find the two numbers. Solution: Let the smaller numbers be $x$ Then according to question, The larger number be $=2 x-5$, then $(2 x-5)^{2}-x^{2}=88$ $4 x^{2}-20 x+25-x^{2}-88=0$ $3 x^{2}-20 x-63=0$ $3 x^{2}-20 x-63=0$ $3 x^{2}-27 x+7 x-63=0$ $3 x(x-9)+7(x-9)=0$ $(x-9)(3 x+7)=0$ $(x-9)=0$ $x=9$ Or $(3 x+7)=0$ $x=\frac{-7}{3}$ Since,xbeing a positive integer so,xcannot be ne...
Read More →Show that the following system of linear equations has no solution:
Question: Show that the following system of linear equations has no solution: $x+2 y \leq 3,3 x+4 y \geq 12, x \geq 0, y \geq 1$ Solution: We have: $x+2 y \leq 3 \quad \ldots \ldots(\mathrm{i})$ $3 x+4 y \geq 12 \quad \ldots$ (ii) $x \geq 0 \quad \ldots$ (iii) $y \geq 1 \quad \ldots \ldots$ (iv) As, the points satisfyingx+ 2y= 3 are: Also, the points satisfying 3x+ 4y= 12 are: Now, the region representing the given inequalities is as follows:Since, there is no common region. So, the given system...
Read More →Three consecutive positive integers are such that the sum
Question: Three consecutive positive integers are such that the sum of the square of the first and the product of other two is 46, find the integers. Solution: Let three consecutive integer be $x,(x+1)$ and $(x+2)$ Then according to question $x^{2}+(x+1)(x+2)=46$ $x^{2}+x^{2}+3 x+2=46$ $2 x^{2}+3 x+2-46=0$ $2 x^{2}+3 x-44=0$ $2 x^{2}+3 x-44=0$ $2 x^{2}-8 x+11 x-44=0$ $2 x(x-4)+11(x-4)=0$ $(x-4)(2 x+11)=0$ $(x-4)=0$ $x=4$ Or $(2 x+11)=0$ $x=\frac{-11}{2}$ Since,xbeing a positive number, soxcannot...
Read More →Solve the following systems of inequations graphically:
Question: Solve the following systems of inequations graphically: (i) 2x+y 8,x+ 2y 8,x+y 6 (ii) 12x+ 12y 840, 3x+ 6y 300, 8x+ 4y 480,x 0,y 0 (iii)x+ 2y 40, 3x+y 30, 4x+ 3y 60,x 0,y 0 (iv) 5x+y 10, 2x+ 2y 12,x+ 4y 12,x 0,y 0 Solution: (i) Converting the inequations to equations, we obtain: 2x+y= 8,x+ 2y= 8,x+y= 6 2x+y= 8: This line meets thex-axis at (4, 0) andy-axis at (0, 8). Draw a thick line through these points. Now, we see that the origin $(0,0)$ does not satisfy the inequation $2 x+y \geq ...
Read More →Evaluate
Question: Evaluate (i) $(103)^{3}$ (ii) $(99)^{3}$ Solution: (i) $(103)^{3}=(100+3)^{3}$ $=(100)^{3}+(3)^{3}+3(100)^{2}(3)+3(100)(3)^{2}$ $=1000000+27+90000+2700$ $=1092727$ (ii) $(99)^{3}=(100-1)^{3}$ $=(100)^{3}-(1)^{3}-3(100)^{2}(1)+3(100)(1)^{2}$ $=1000000-1-30000+300$ $=1000300-30001$ $=970299$...
Read More →The sum of two numbers is 9.
Question: The sum of two numbers is 9. The sum of their reciprocals is 1/2. Find the numbers. Solution: Let one numbers bexthen other (9 x). Then according to question $\frac{1}{x}+\frac{1}{(9-x)}=\frac{1}{2}$ $\Rightarrow \frac{(9-x)+x}{x(9-x)}=\frac{1}{2}$ $\Rightarrow \frac{9}{x(9-x)}=\frac{1}{2}$ By cross multiplication $\Rightarrow 18=x(9-x)$ $\Rightarrow x^{2}-9 x+18=0$ $\Rightarrow x^{2}-6 x-3 x+18=0$ $\Rightarrow(x-6) x-3(x-6)=0$ $\Rightarrow(x-6)(x-3)=0$ $\Rightarrow x=6,3$ Since,xbeing...
Read More →Solve the following systems of inequations graphically:
Question: Solve the following systems of inequations graphically: (i) 2x+y 8,x+ 2y 8,x+y 6 (ii) 12x+ 12y 840, 3x+ 6y 300, 8x+ 4y 480,x 0,y 0 (iii)x+ 2y 40, 3x+y 30, 4x+ 3y 60,x 0,y 0 (iv) 5x+y 10, 2x+ 2y 12,x+ 4y 12,x 0,y 0 Solution: (i) Converting the inequations to equations, we obtain: 2x+y= 8,x+ 2y= 8,x+y= 6 2x+y= 8: This line meets thex-axis at (4, 0) andy-axis at (0, 8). Draw a thick line through these points. Now, we see that the origin $(0,0)$ does not satisfy the inequation $2 x+y \geq ...
Read More →Factorise
Question: Factorise $a^{3}-12 a(a-4)-64$ Solution: $a^{3}-12 a(a-4)-64=a^{3}-12 a^{2}+48 a-64$ $=(a)^{3}-(4)^{3}-3(a)^{2}(4)+3(a)(4)^{2}$ $=(a-4)^{3}$ Hence, factorisation of $a^{3}-12 a(a-4)-64$ is $(a-4)^{3}$....
Read More →Show that the solution set of the following linear in equations is an unbounded set:
Question: Show that the solution set of the following linear in equations is an unbounded set: x+y 9 3x+y 12 x 0,y 0 Solution: Converting the inequations to equations, we obtain: x+y= 9, 3x+y= 12 ,x= 0,y= 0 x+y= 9: This line meets thex-axis at (9, 0) andy-axis at (0, 9). Draw a thick line through these points. Now, we see that the origin $(0,0)$ does not satisfy the inequation $x+y \geq 9$ Therefore, the potion that does not contain the origin is the solution set to the inequaltion $x+y \geq 9$ ...
Read More →Find the linear inequations for which the solution set is the shaded region given in Fig. 15.42
Question: Find the linear iAlso the shaded region is in the first quadrant. Therefore, we must have $x \geq 0$ and $y \geq 0$nequations for which the solution set is the shaded region given in Fig. 15.42 Solution: Considering the line $x+y=4$, we find that the shaded region and the origin $(0,0)$ are on the same side of this line and $(0,0)$ does not satisfy the inequation $x+y \leq 4$ So, the first inequation is $x+y \leq 4$ Considering the line $y=3$, we find that the shaded region and the ori...
Read More →Find the linear inequations for which the shaded area in Fig. 15.41 is the solution set.
Question: Find the linear inequations for which the shaded area in Fig. 15.41 is the solution set. Draw the diagram of the solution set of the linear inequations: Solution: Considering the line $2 x+3 y=6$, we find that the shaded region and the origin $(0,0)$ are on the opposite side of this line and $(0,0)$ does not satisfy the inequation $2 x+3 y \geq 6$ So, the first inequation is $2 x+3 y \geq 6$ Considering the line $4 x+6 y=24$, we find that the shaded region and the origin $(0,0)$ are on...
Read More →Factorise
Question: Factorise $\frac{64}{125} a^{3}-\frac{96}{25} a^{2}+\frac{48}{5} a-8$ Solution: $\frac{64}{125} a^{3}-\frac{96}{25} a^{2}+\frac{48}{5} a-8=\left(\frac{4}{5} a\right)^{3}-(2)^{3}-3\left(\frac{4}{5} a\right)^{2}(2)+3\left(\frac{4}{5} a\right)(2)^{2}$ $=\left(\frac{4}{5} a-2\right)^{3}$ Hence, factorisation of $\frac{64}{125} a^{3}-\frac{96}{25} a^{2}+\frac{48}{5} a-8$ is $\left(\frac{4}{5} a-2\right)^{3}$....
Read More →Factorise
Question: Factorise $a^{3} x^{3}-3 a^{2} b x^{2}+3 a b^{2} x-b^{3}$ Solution: $a^{3} x^{3}-3 a^{2} b x^{2}+3 a b^{2} x-b^{3}=(a x)^{3}-(b)^{3}-3(a x)^{2}(b)+3(a x)(b)^{2}$ $=(a x-b)^{3}$ Hence, factorisation of $a^{3} x^{3}-3 a^{2} b x^{2}+3 a b^{2} x-b^{3}$ is $(a x-b)^{3}$....
Read More →Factorise
Question: Factorise $125 x^{3}-27 y^{3}-225 x^{2} y+135 x y^{2}$ Solution: $125 x^{3}-27 y^{3}-225 x^{2} y+135 x y^{2}=(5 x)^{3}-(3 y)^{3}-3(5 x)^{2}(3 y)+3(5 x)(3 y)^{2}$ $=(5 x-3 y)^{3}$ Hence, factorisation of $125 x^{3}-27 y^{3}-225 x^{2} y+135 x y^{2}$ is $(5 x-3 y)^{3}$....
Read More →The sum of two number a and b is 15,
Question: The sum of two number $a$ and $b$ is 15 , and the sum of their reciprocals $\frac{1}{a}$ and $\frac{1}{b}$ is $3 / 10$. Find the numbers $a$ and $b$. Solution: Given that $a$ and $b$ be two numbers in such a way that $b=(15-a)$. Then according to question $\frac{1}{a}+\frac{1}{b}=\frac{3}{10}$ $\frac{(b+a)}{a b}=\frac{3}{10}$ $\frac{(a+b)}{a b}=\frac{3}{10}$ By cross multiplication $10 a+10 b=3 a b$....(1) Now putting the value ofbin equation (1) $10 a+10(15-a)=3 a(15-a)$ $3 a^{2}-45 a...
Read More →Show that the solution set of the following linear inequations is empty set:
Question: Show that the solution set of the following linear inequations is empty set: (i)x 2y 0, 2xy 2,x 0,y 0 (ii)x+ 2y 3, 3x+ 4y 12,y 1,x 0,y 0 Solution: (i) Converting the inequations to equations, we obtain: $x-2 y=0,2 x-y=-2, x=0, y=0$ $x-2 y=0$ : This line meets the $x$-axis at $(0,0)$ and $y$-axis at $(0,0)$. If $x=1$, then $y=1 / 2$, so we have another point $(1,1 / 2)$. Draw a thick line through $(0,0)$ and $(1,1 / 2)$. We see that the origin $(1,0)$ satisfies the inequation $x+2 y \le...
Read More →Let f be an injective map with domain {x, y, z} and range {1, 2, 3},
Question: Letfbe an injective map with domain {x,y,z} and range {1, 2, 3}, such that exactly one of the following statements is correct and the remaining are false. $f(x)=1, f(y) \neq 1, f(z) \neq 2$ The value of $f^{-1}$ (1) is (a) $x$ (b) $y$ (c) $z$ (d) none of these Solution: Case-1: Let $f(x)=1$ be true. Then, $f(y) \neq 1$ and $f(z) \neq 2$ are false. So, $f(y)=1$ and $f(z)=2$ $\Rightarrow f(x)=1, f(y)=1$ $\Rightarrow x$ and $y$ have the same images. This contradicts the fact that $f$ is o...
Read More →Show that the solution set of the following linear inequations is empty set:
Question: Show that the solution set of the following linear inequations is empty set: (i)x 2y 0, 2xy 2,x 0,y 0 (ii)x+ 2y 3, 3x+ 4y 12,y 1,x 0,y 0 Solution: (i) Converting the inequations to equations, we obtain: $x-2 y=0,2 x-y=-2, x=0, y=0$ $x-2 y=0$ : This line meets the $x$-axis at $(0,0)$ and $y$-axis at $(0,0)$. If $x=1$, then $y=1 / 2$, so we have another point $(1,1 / 2)$. Draw a thick line through $(0,0)$ and $(1,1 / 2)$. We see that the origin $(1,0)$ satisfies the inequation $x+2 y \le...
Read More →Factorise
Question: Factorise $1+\frac{27}{125} a^{3}+\frac{9 a}{5}+\frac{27 a^{2}}{24}$ Solution: $1+\frac{27}{125} a^{3}+\frac{9 a}{5}+\frac{27 a^{2}}{25}=(1)^{3}+\left(\frac{3}{5} a\right)^{3}+3(1)^{2}\left(\frac{3}{5} a\right)+3(1)\left(\frac{3}{5} a\right)^{2}$ $=\left(1+\frac{3}{5} a\right)^{3}$ Hence, factorisation of $1+\frac{27}{125} a^{3}+\frac{9 a}{5}+\frac{27 a^{2}}{25}$ is $\left(1+\frac{3}{5} a\right)^{3}$....
Read More →Factorise
Question: Factorise $64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}$ Solution: $64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}=(4 a)^{3}-(3 b)^{3}-3(4 a)^{2}(3 b)+3(4 a)(3 b)^{2}$ $=(4 a-3 b)^{3}$ Hence, factorisation of $64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}$ is $(4 a-3 b)^{3}$....
Read More →The sum of two numbers is 18.
Question: The sum of two numbers is 18. The sum of their reciprocals is 1/4. Find the numbers. Solution: Let one of the numberbexthen other number is (18x). Then according to question, $\frac{1}{x}+\frac{1}{18-x}=\frac{1}{4}$ $\Rightarrow \frac{18-x+x}{x(18-x)}=\frac{1}{4}$ $\Rightarrow 18 \times 4=18 x-x^{2}$ $\Rightarrow 72=18 x-x^{2}$ $\Rightarrow x^{2}-18 x+72=0$ $\Rightarrow x^{2}-12 x-6 x+72=0$ $\Rightarrow x(x-12)-6(x-12)=0$ $\Rightarrow(x-6)(x-12)=0$ $\Rightarrow x-6=0$ or $x-12=0$ $\Rig...
Read More →Solve the following systems of linear inequations graphically:
Question: Solve the following systems of linear inequations graphically: (i) 2x+ 3y 6, 3x+ 2y 6,x 0,y 0 (ii) 2x+ 3y 6,x+ 4y 4,x 0,y 0 (iii)xy 1,x+ 2y 8, 2x+y 2,x 0,y 0 (iv)x+y 1, 7x+ 9y 63,x 6,y 5,x 0,y 0 (v) 2x+ 3y 35,y 3,x 2,x 0,y 0 Solution: (i) Converting the inequations to equations, we obtain: 2x+ 3y= 6, 3x+ 2y= 6,x= 0,y= 0 2x+ 3y=6: This line meets thex-axis at (3,0) and the y-axis at (0, 2). Draw a thick line joining these points. We see that the origin (0, 0) satisfies the inequation 2x...
Read More →