A two-digit number is such that the product of its digits is 8.
Question: A two-digit number is such that the product of its digits is 8. When 18 is subtracted from the number, the digits interchange their places. Find the number. Solution: Let the tens digit be $x$, then, the unit digits $=\frac{8}{x}$ Therefore, number $=\left(10 x+\frac{8}{x}\right)$ And number obtained by interchanging the digits $=\left(10 \times \frac{8}{x}+x\right)$ Then according to question $\left(10 x+\frac{8}{x}\right)-\left(10 \times \frac{8}{x}+x\right)=18$ $\left(10 x+\frac{8}{...
Read More →Represent to solution set of each of the following inequations graphically in two dimensional plane:
Question: Represent to solution set of each of the following inequations graphically in two dimensional plane: 4.x 2y 0 Solution: Converting the inequation to equation, we obtainx-2y= 0 Puttingy= 0 andx= 0 in this equation, we obtainx= 0 andy= 0 respectively. So, this line meets thex-axis at (0,0) and they- axis at (0,0). Ifx= 1, theny= 1/2. So, we have another point (1,1/2). We plot these points and join them by a thin line. This divides thexyplane into two parts. We take a point (0, 2) and it ...
Read More →Let f : R→R be given by
Question: Let $f: R \rightarrow R$ be given by $f(x)=\left[x^{2}\right]+[x+1]-3$ where $[x]$ denotes the greatest integer less than or equal to $x$. Then, $f(x)$ is (a) many-one and onto (b) many-one and into(c) one-one and into(d) one-one and onto Solution: (b) many-one and into $f: R \rightarrow R$ $f(x)=\left[x^{2}\right]+[x+1]-3$ It is many one function because in this case for two different values ofxwe would get the same value off(x) . For $x=1.1,1.2 \in R$ $f(1.1)=\left[(1.1)^{2}\right]+[...
Read More →Factorize:
Question: Factorize: $4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z$ Solution: We have : $4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z$ $=(2 x)^{2}+(3 y)^{2}+(-4 z)^{2}+2(2 x)(3 y)+2(3 y)(-4 z)+2(-4 z)(2 x)$ $=[(2 x)+(3 y)+(-4 z)]^{2}$ $=(2 x+3 y-4 z)^{2}$...
Read More →Represent to solution set of each of the following inequations graphically in two dimensional plane:
Question: Represent to solution set of each of the following inequations graphically in two dimensional plane: 3.x+ 2 0 Solution: Converting the inequation to equation, we obtainx+ 2 = 0, i.ex=-2. Clearly, it is a parallel line toy-axis at a distance of-2 units from it. This line divides thexyplane into two parts, viz LHS ofx=-2 and RHS ofx=-2. To determine the region represented by the given inequality, consider pointO(0,0). Clearly, (0,0) does not satisfy the inequality. So, the region that do...
Read More →The sum of a number and its reciprocal is 17/4.
Question: The sum of a number and its reciprocal is 17/4. Find the number. Solution: Let a numbers be $x$ and its reciprocal is $\frac{1}{x}$ Then according to question $x+\frac{1}{x}=\frac{17}{4}$ $\frac{x^{2}+1}{x}=\frac{17}{4}$ By cross multiplication $4 x^{2}+4=17 x$ $4 x^{2}-17 x+4=0$ $4 x^{2}-17 x+4=0$ $4 x^{2}-x-16 x+4=0$ $x(4 x-1)-4(4 x-1)=0$ $(4 x-1)(x-4)=0$ $(4 x-1)=0$ $x=\frac{1}{4}$ Or $(x-4)=0$ $x=4$ Thus, two consecutive number be either 4 or $\frac{1}{4}$...
Read More →Expand:
Question: Expand: (i) $(2 a-5 b-7 c)^{2}$ (ii) $(-3 a+4 b-5 c)^{2}$ (iii) $\left(\frac{1}{2} a-\frac{1}{4} a+2\right)^{2}$ Solution: (i) $(2 a-5 b-7 c)^{2}=[(2 a)+(-5 b)+(-7 c)]^{2}$ $=(2 a)^{2}+(-5 b)^{2}+(-7 c)^{2}+2(2 a)(-5 b)+2(-5 b)(-7 c)+2(2 a)(-7 c)$ $=4 a^{2}+25 b^{2}+49 c^{2}-20 a b+70 b c-28 a c$ (ii) $(-3 a+4 b-5 c)^{2}=[(-3 a)+(4 b)+(-5 c)]^{2}$ $=(-3 a)^{2}+(4 b)^{2}+(-5 c)^{2}+2(-3 a)(4 b)+2(4 b)(-5 c)+2(-3 a)(-5 c)$ $=9 a^{2}+16 b^{2}+25 c^{2}-24 a b-40 b c+30 a c$ (iii) $\left(\f...
Read More →Represent to solution set of each of the following in equations graphically in two dimensional plane:
Question: Represent to solution set of each of the following in equations graphically in two dimensional plane: 2.x+ 2y 6 Solution: Converting the in equation to equation, we obtainx+ 2y= 6, i.ex+ 2y-6 = 0. Puttingy= 0 andx= 0 in this equation, we obtainx= 6 andy= 3. So, this line meetsx-axis at (6,0) andy-axis at (0,3). We plot these points and join them by a thick line. This divides thexyplane into two parts. To determine the region represented by the given inequality, consider point O(0,0). C...
Read More →Determine two consecutive multiples
Question: Determine two consecutive multiples of 3 whose product is 270. Solution: Let the required number be $3 x$ and $(3 x+3)$ Then according to question $(3 x)(3 x+3)=270$ $9 x^{2}+9 x-270=0$ $9\left(x^{2}+x-30\right)=0$ $x^{2}+x-30=0$ $x^{2}+x-30=0$ $x^{2}-5 x+6 x-30=0$ $x(x-5)+6(x-5)=0$ $(x-5)(x+6)=0$ $(x-5)=0$ $x=5$ Or $(x+6)=0$ $x=-6$ Since,xbeing a positive number, soxcannot be negative. Therefore, When $x=5$ then positive integer $3 x=3 \times 5$ $=15$ And $3 x+3=3 \times 5+3$ $=18$ Th...
Read More →Let A={x ∈ R : −1≤x≤1}=B. Then, the mapping
Question: Let $A=\{x \in R:-1 \leq x \leq 1\}=B$. Then, the mapping $f: A \rightarrow B$ given by $f(x)=x|x|$ is (a) injective but not surjective(b) surjective but not injective(c) bijective(d) none of these Solution: Injectivity:Letxandybe any two elements in the domainA. Case-1: Letxandybe two positive numbers, such that $f(x)=f(y)$ $\Rightarrow x|x|=y|y|$ $\Rightarrow x(x)=y(y)$ $\Rightarrow x^{2}=y^{2}$ $\Rightarrow x=y$ Case-2: Letxandybe two negative numbers, such that $f(x)=f(y)$ $\Righta...
Read More →Represent to solution set of each of the following inequations graphically in two dimensional plane:
Question: Represent to solution set of each of the following inequations graphically in two dimensional plane: 1.x+ 2yy 0 Solution: We have: 1.x+ 2yy 0 Converting the given inequation to equation, we obtainx+ 2y-y= 0, i.ex+y= 0 Puttingy= 0 andx= 0 in this equation, we obtainx= 0 andy= 0. So, this line intersects thex-axis and they-axis at (0,0). We draw the line of the equationx+y= 0 Now we take a point (1, 1) ( any point which does not lie on the linex+y= 0 ) (1, 1) does not satisfy the inequal...
Read More →Expand:
Question: Expand: (i) $(a+2 b+5 c)^{2}$ (ii) $(2 b-b+c)^{2}$ (iii) $(a-2 b-3 c)^{2}$ Solution: (i) $(a+2 b+5 c)^{2}=(a)^{2}+(2 b)^{2}+(5 c)^{2}+2(a)(2 b)+2(2 b)(5 c)+2(5 c)(a)$ $=a^{2}+4 b^{2}+25 c^{2}+4 a b+20 b c+10 a c$ (ii) $(2 a-b+c)^{2}=[(2 a)+(-b)+(c)]^{2}$ $=(2 a)^{2}+(-b)^{2}+(c)^{2}+2(2 a)(-b)+2(-b)(c)+4(a)(c)$ $=4 a^{2}+b^{2}+c^{2}-4 a b-2 b c+4 a c$ (iii) $(a-2 b-3 c)^{2}=[a+(-2 b)+(-3 c)]^{2}$ $=(a)^{2}+(-2 b)^{2}+(-3 c)^{2}+2(a)(-2 b)+2(-2 b)(-3 c)+2(a)(-3 c)$ $=a^{2}+4 b^{2}+9 c^{...
Read More →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.
Question: 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 Solution: Letxbe the third pH value. Then, $7.2\frac{7.48+7.85+x}{3}7.8$ $\Rightarrow 7.2\frac{15.33+x}{3}7.8$ $\Rightarrow 21.615.33+x23.4 \quad$ (Multiplying throughout by 3 ) $\Rightarrow 21.6-15.3315.33+x-x23.4...
Read More →Evaluate
Question: Evaluate $\{(999) 2-1\}$ Solution: $\left\{(999)^{2}-1\right\}=\left\{(999)^{2}-1^{2}\right\}$ $=(999-1)(999+1)$ $=(998)(1000)$ $=998000$ Hence, $\left\{(999)^{2}-1\right\}=998000$....
Read More →The function f : A→B defined by
Question: The function $f: A \rightarrow B$ defined by $f(x)=-x^{2}+6 x-8$ is a bijection if (a) $A=(-\infty, 3]$ and $B=(-\infty, 1]$ (b) $A=[-3, \infty)$ and $B=(-\infty, 1]$ (c) $A=(-\infty, 3]$ and $B=[1, \infty)$ (d) $A=[3, \infty)$ and $B=[1, \infty)$ Solution: (a) $A=(-\infty, 3]$ and $B=(-\infty, 1]$ $f(x)=-x^{2}+6 x-8$, is a polynomial function And the domain of polynomial function is real number.\ $\therefore x \in R$ $f(x)=-x^{2}+6 x-8$ $=-\left(x^{2}-6 x+8\right)$ $=-\left(x^{2}-6 x+...
Read More →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.
Question: 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 Solution: Letxbe the third pH value. Then, $7.2\frac{7.48+7.85+x}{3}7.8$ $\Rightarrow 7.2\frac{15.33+x}{3}7.8$ $\Rightarrow 21.615.33+x23.4 \quad$ (Multiplying throughout by 3 ) $\Rightarrow 21.6-15.3315.33+x-x23.4...
Read More →Factorise:
Question: Factorise: $4 x^{4}+7 x^{2}-2$ Solution: $4 x^{4}+7 x^{2}-2=4 x^{4}+8 x^{2}-x^{2}-2$ $=4 x^{2}\left(x^{2}+2\right)-1\left(x^{2}+2\right)$ $=\left(4 x^{2}-1\right)\left(x^{2}+2\right)$ Hence, factorisation of $4 x^{4}+7 x^{2}-2$ is $\left(4 x^{2}-1\right)\left(x^{2}+2\right)$....
Read More →The sum of two numbers is 16.
Question: The sum of two numbers is 16. The sum of their reciprocals is 1/3. Find the numbers. Solution: Let one numbers be $x$ then other $(16-x)$. Then according to question $\frac{1}{x}+\frac{1}{(16-x)}=\frac{1}{3}$ $\frac{16}{\left(16 x-x^{2}\right)}=\frac{1}{3}$ By cross multiplication $16 x-x^{2}=48$ $x^{2}-16 x+48=0$ $x^{2}-12 x-4 x-48=0$ $x(x-12)-4(x-12)=0$ $(x-12)(x-4)=0$ $(x-12)=0$ $x=12$ Or $(x-4)=0$ $x=4$ Since,xbeing a number, Therefore, When $x=12$ then $16-x=16-12$ $=4$ Thus, two ...
Read More →Factorise:
Question: Factorise: $(a+2 b)^{2}+101(a+2 b)+100$ Solution: $(a+2 b)^{2}+101(a+2 b)+100=(a+2 b)^{2}+100(a+2 b)+1(a+2 b)+100$ $=(a+2 b)[(a+2 b)+100]+1[(a+2 b)+100]$ $=[(a+2 b)+1][(a+2 b)+100]$ $=(a+2 b+1)(a+2 b+100)$ Hence, factorisation of $(a+2 b)^{2}+101(a+2 b)+100$ is $(a+2 b+1)(a+2 b+100)$...
Read More →A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it.
Question: 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? Solution: Supposexlitres of 2% solution is added in the existing solution of 8% of boric acid. Resulting mixture = (640 +x) L Therefore, as per given conditions: $4 \%$ of $(640+x)8 \%$ of $640+2 \%$ of $x6 \%$ of $(640+x)$ $\Righta...
Read More →The sum of the squares of three consecutive natural number is 149.
Question: The sum of the squares of three consecutive natural number is 149. Find the numbers. Solution: Let three consecutive integer be $x,(x+1)$ and $(x+2)$ Then according to question $x^{2}+(x+1)^{2}+(x+2)^{2}=149$ $x^{2}+x^{2}+2 x+1+x^{2}+4 x+4=149$ $3 x^{2}+6 x+5-149=0$ $3 x^{2}+6 x-144=0$ $3 x^{2}+6 x-144=0$ $3\left(x^{2}+2 x-48\right)=0$ $x^{2}+2 x-48=0$ $x^{2}+8 x-6 x-48=0$ $x(x+8)-6(x+8)=0$ $(x+8)(x-6)=0$ $(x+8)=0$ $x=-8$ Or $(x-6)=0$ $x=6$ Since,xbeing a positive number, soxcannot be ...
Read More →A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it.
Question: 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? Solution: Supposexlitres of 2% solution is added in the existing solution of 8% of boric acid. Resulting mixture = (640 +x) L Therefore, as per given conditions: $4 \%$ of $(640+x)8 \%$ of $640+2 \%$ of $x6 \%$ of $(640+x)$ $\Righta...
Read More →Factorise:
Question: Factorise: $6\left(2 x-\frac{3}{x}\right)^{2}+7\left(2 x-\frac{3}{x}\right)-20$ Solution: $6\left(2 x-\frac{3}{x}\right)^{2}+7\left(2 x-\frac{3}{x}\right)-20=6\left(2 x-\frac{3}{x}\right)^{2}+15\left(2 x-\frac{3}{x}\right)-8\left(2 x-\frac{3}{x}\right)-20$ $=\left[3\left(2 x-\frac{3}{x}\right)\right]\left[2\left(2 x-\frac{3}{x}\right)+5\right]-4\left[2\left(2 x-\frac{3}{x}\right)+5\right]$ $=\left[2\left(2 x-\frac{3}{x}\right)+5\right]\left[3\left(2 x-\frac{3}{x}\right)-4\right]$ $=\le...
Read More →Find two natural numbers which differ by 3
Question: Find two natural numbers which differ by 3 and whose squares have the sum 17. Solution: Let one natural number be $x$ and other $(x-3)$. Then according to question $(x)^{2}+(x-3)^{2}=117$ $x^{2}+x^{2}-6 x+9=117$ $2 x^{2}-6 x+9-117=0$ $2 x^{2}-6 x-108=0$ $2 x^{2}-6 x-108=0$ $2\left(x^{2}-3 x-54\right)=0$ $\left(x^{2}-3 x-54\right)=0$ $x^{2}-9 x+6 x-54=0$ $x(x-9)+6(x-9)=0$ $(x-9)(x+6)=0$ $(x-9)=0$ $x=9$ $(x+6)=0$ $x=-6$ Since,xbeing a natural number, soxcannot be negative. Therefore, Whe...
Read More →Factorise:
Question: Factorise: $10\left(3 x+\frac{1}{x}\right)^{2}-\left(3 x+\frac{1}{x}\right)-3$ Solution: $10\left(3 x+\frac{1}{x}\right)^{2}-\left(3 x+\frac{1}{x}\right)-3=10\left(3 x+\frac{1}{x}\right)^{2}-6\left(3 x+\frac{1}{x}\right)+5\left(3 x+\frac{1}{x}\right)-3$ $=\left[2\left(3 x+\frac{1}{x}\right)\right]\left[5\left(3 x+\frac{1}{x}\right)-3\right]+1\left[5\left(3 x+\frac{1}{x}\right)-3\right]$ $=\left[5\left(3 x+\frac{1}{x}\right)-3\right]\left[2\left(3 x+\frac{1}{x}\right)+1\right]$ $=\left(...
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