How many litres of water will have to be added to 1125 litres of the 45%
Question: 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? Solution: Letxlitres of water be added to the 1125 litres of 45% solution of the acid. Total quantity of mixture is (1125+x) litres. Total acid content in 1125 litres of mixture = 45% of 1125 It is given that the acid content in the resulting mixture must be more than $25 \%$ and less than $30 \%$. $\therefore...
Read More →How many litres of water will have to be added to 1125 litres of the 45%
Question: 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? Solution: Letxlitres of water be added to the 1125 litres of 45% solution of the acid. Total quantity of mixture is (1125+x) litres. Total acid content in 1125 litres of mixture = 45% of 1125 It is given that the acid content in the resulting mixture must be more than $25 \%$ and less than $30 \%$. $\therefore...
Read More →The difference of two numbers is 4.
Question: The difference of two numbers is 4 . If the difference of their reciprocal is $\frac{4}{21}$, find the numbers. Solution: Let one numbers be $x$ then other $(x+4)$. Then according to question $\frac{1}{x}-\frac{1}{(x+4)}=\frac{4}{21}$ $\frac{4}{\left(x^{2}+4 x\right)}=\frac{4}{21}$ By cross multiplication $4 x^{2}+16 x=84$ $4 x^{2}+16 x-84=0$ $4\left(x^{2}+4 x-21\right)=0$ $\left(x^{2}+4 x-21\right)=0$ $x^{2}+7 x-3 x-21=0$ $x(x+7)-3(x+7)=0$ $(x+7)(x-3)=0$ $(x+7)=0$ $x=-7$ Or $(x-3)=0$ ...
Read More →Factorise:
Question: Factorise: $7(x-2 y)^{2}-25(x-2 y)+12$ Solution: $7(x-2 y)^{2}-25(x-2 y)+12=7(x-2 y)^{2}-21(x-2 y)-4(x-2 y)+12$ $=[7(x-2 y)](x-2 y-3)-4(x-2 y-3)$ $=[7(x-2 y)-4](x-2 y-3)$ $=(7 x-14 y-4)(x-2 y-3)$ Hence, factorisation of $7(x-2 y)^{2}-25(x-2 y)+12$ is $(7 x-14 y-4)(x-2 y-3)$....
Read More →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,
Question: 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. Solution: Let the shortest side of the triangle bexcm. Then, the longest side will be 3xand the third side will be 3x 2. $\therefore$ Perimeter of the triangle $\geq 61$ $\Rightarrow x+3 x+3 x-261$ $\Rightarrow 7 x \geq 61+2$ $\Rightarrow x \geq 9 \quad$ (Dividing throughout b...
Read More →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,
Question: 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. Solution: Let the shortest side of the triangle bexcm. Then, the longest side will be 3xand the third side will be 3x 2. $\therefore$ Perimeter of the triangle $\geq 61$ $\Rightarrow x+3 x+3 x-261$ $\Rightarrow 7 x \geq 61+2$ $\Rightarrow x \geq 9 \quad$ (Dividing throughout b...
Read More →Factorize:
Question: Factorize: $9(2 a-b)^{2}-4(2 a-b)-13$ Solution: We have: $9(2 a-b)^{2}-4(2 a-b)-13$ Let : $(2 a-b)=p$ Thus, the given expression becomes $9 p^{2}-4 p-13$ Now, we must split $(-4)$ into two numbers such that their sum is $(-4)$ and their product is $(-117)$. Clearly, $-13+9=-4$ and $-13 \times 9=-117$. $\therefore 9 p^{2}-4 p-13=9 p^{2}+9 p-13 p-13$ $=9 p(p+1)-13(p+1)$ $=(p+1)(9 p-13)$ Putting $p=(2 a-b)$, we get: $9(2 a-b)^{2}-4(2 a-b)-13=[(2 a-b)+1][9(2 a-b)-13]$ $=(2 a-b+1)[18 a-9 b-...
Read More →Let the function f : R−{−b}→R−{1} be defined by
Question: Let the function $f: R-\{-b\} \rightarrow R-\{1\}$ be defined by $f(x)=\frac{x+a}{x+b}, a \neq b$. T hen, (a)fis one-one but not onto(b)fis onto but not one-one(c)fis both one-one and onto(d) None of these Solution: (c)fis both one-one and onto Injectivity: Let $x$ and $y$ be two elements in the domain $R-\{-b\}$, such that $f(x)=f(y)$ $\Rightarrow \frac{x+a}{x+b}=\frac{y+a}{y+b}$ $\Rightarrow(x+a)(y+b)=(x+b)(y+a)$ $\Rightarrow x y+b x+a y+a b=x y+a x+b y+a b$ $\Rightarrow b x+a y=a x+...
Read More →A company manufactures cassettes and its cost and revenue functions for a week are
Question: A company manufactures cassettes and its cost and revenue functions for a week are $C=300+\frac{3}{2} x$ and $R=2 x$ 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? Solution: To realise profit, revenue must be greater than the cost. $\therefore 2 x300+\frac{3}{2} x$ $\Rightarrow 2 x-\frac{3}{2} x300$ $\Rightarrow \frac{1}{2} x300$ $\Rightarrow x600$ Thus, the company must sell more than...
Read More →Find the consecutive even integers whose squares have the sum 340.
Question: Find the consecutive even integers whose squares have the sum 340. Solution: Let two consecutive even integer be $2 x$ and other $(2 x+2)$ Then according to question $(2 x)^{2}+(2 x+2)^{2}=340$ $4 x^{2}+4 x^{2}+8 x+4=340$ $8 x^{2}+8 x=340-4$ $8 x^{2}+8 x-336=0$ $8 x^{2}+8 x-336=0$ $8\left(x^{2}+x-42\right)=0$ $\left(x^{2}+x-42\right)=0$ $x^{2}+7 x-6 x-42=0$ $x(x+7)-6(x+7)=0$ $(x+7)(x-6)=0$ $(x+7)=0$ $x=-7$ or $(x-6)=0$ $x=6$ Since,xbeing a positive number, soxcannot be negative. Theref...
Read More →A company manufactures cassettes and its cost and revenue functions for a week are
Question: A company manufactures cassettes and its cost and revenue functions for a week are $C=300+\frac{3}{2} x$ and $R=2 x$ 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? Solution: To realise profit, revenue must be greater than the cost. $\therefore 2 x300+\frac{3}{2} x$ $\Rightarrow 2 x-\frac{3}{2} x300$ $\Rightarrow \frac{1}{2} x300$ $\Rightarrow x600$ Thus, the company must sell more than...
Read More →The function f : R → R defined by
Question: The function $f: R \rightarrow R$ defined by $f(x)=2^{x}+2^{|x|}$ is (a) one-one and onto(b) many-one and onto(c) one-one and into(d) many-one and into Solution: (d) many-one and into Graph for the given function is as follows. A line parallel to $X$ axis is cutting the graph at two different values. Therefore, for two different values ofxwe are getting the same value ofy.That means it is many one function. From the given graph we can see that the range is $[2, \infty)$ and R is the co...
Read More →Factorize:
Question: Factorize: $2(x+y)^{2}-9(x+y)-5$ Solution: We have: $2(x+y)^{2}-9(x+y)-5$ Let : $(x+y)=u$ Thus, the given expression becomes $2 u^{2}-9 u-5$ $2 u^{2}-9 u-5$ Now, we have to split $(-9)$ into two numbers such that their sum is $(-9)$ and their product is $(-10)$. Clearly, $-10+1=-9$ and $-10 \times 1=-10$ $\therefore 2 u^{2}-9 u-5=2 u^{2}-10 u+u-5$ $=2 u(u-5)+1(u-5)$ $=(u-5)(2 u+1)$ Putting $u=(x+y)$, we get: $2(x+y)^{2}-9(x+y)-5=(x+y-5)[2(x+y)+1]$ $=(x+y-5)(2 x+2 y+1)$...
Read More →To receive grade 'A' in a course, one must obtain an average of 90 marks or more in five papers each of 100 marks.
Question: To receive grade 'A' in a course, one must obtain an average of 90 marks or more in five papers each of 100 marks. If Shikha scored 87, 95, 92 and 94 marks in first four paper, find the minimum marks that she must score in the last paper to get grade 'A' in the course. Solution: Letxbe the minimum marks scored in the last paper. Then, $90 \leq \frac{87+95+92+94+x}{5} \leq 100$ $\Rightarrow 90 \leq \frac{368+x}{5} \leq 100$ $\Rightarrow 450 \leq 368+x \leq 500$ $\Rightarrow 450-368 \leq...
Read More →To receive grade 'A' in a course, one must obtain an average of 90 marks or more in five papers each of 100 marks.
Question: To receive grade 'A' in a course, one must obtain an average of 90 marks or more in five papers each of 100 marks. If Shikha scored 87, 95, 92 and 94 marks in first four paper, find the minimum marks that she must score in the last paper to get grade 'A' in the course. Solution: Letxbe the minimum marks scored in the last paper. Then, $90 \leq \frac{87+95+92+94+x}{5} \leq 100$ $\Rightarrow 90 \leq \frac{368+x}{5} \leq 100$ $\Rightarrow 450 \leq 368+x \leq 500$ $\Rightarrow 450-368 \leq...
Read More →The sum of the squares of two numbers is 233 and
Question: The sum of the squares of two numbers is 233 and one of the number is 3 less than twice the other number. Find the numbers. Solution: Let the numbers be integers. One of the numbers be $x$. So, the other will be $(2 x-3)$. Then according to question, $x^{2}+(2 x-3)^{2}=233$ $x^{2}+4 x^{2}-12 x+9=233$ $5 x^{2}-12 x+9-233=0$ $5 x^{2}-12 x-224=0$ $5 x^{2}-40 x+28 x-224=0$ $5 x(x-8)+28(x-8)=0$ $(x-8)(5 x+28)=0$ $(x-8)=0$ $x=8$ Or $(5 x+28)=0$ $x=\frac{-28}{5}$ Since, we have assumed the nu...
Read More →Factorise:
Question: Factorise: $2 x^{2}-x+\frac{1}{8}$ Solution: $2 x^{2}-x+\frac{1}{8}=2 x^{2}-\frac{1}{2} x-\frac{1}{2} x+\frac{1}{8}$ $=2 x\left(x-\frac{1}{4}\right)-\frac{1}{2}\left(x-\frac{1}{4}\right)$ $=\left(x-\frac{1}{4}\right)\left(2 x-\frac{1}{2}\right)$ Hence, factorisation of $2 x^{2}-x+\frac{1}{8}$ is $\left(x-\frac{1}{4}\right)\left(2 x-\frac{1}{2}\right)$....
Read More →A solution is to be kept between 30°C and 35°C.
Question: A solution is to be kept between 30C and 35C. What is the range of temperature in degree Fahrenheit? Solution: Letxdegree Fahrenheit be the temperature of the solution. Then, $30^{\circ} \mathrm{C}x^{\circ} \mathrm{F}35^{\circ} \mathrm{C}$ Now, $F=\frac{9}{5} C+32$ $\Rightarrow \frac{9}{5} \times 30+32x^{\circ} F\frac{9}{5} \times 35+32$ $\Rightarrow 54+32x63+32$ $\Rightarrow 86^{\circ}x^{\circ}95^{\circ}$ Hence, the range of the temperature in Fahrenheit is between $86^{\circ}$ and $9...
Read More →A solution is to be kept between 30°C and 35°C.
Question: A solution is to be kept between 30C and 35C. What is the range of temperature in degree Fahrenheit? Solution: Letxdegree Fahrenheit be the temperature of the solution. Then, $30^{\circ} \mathrm{C}x^{\circ} \mathrm{F}35^{\circ} \mathrm{C}$ Now, $F=\frac{9}{5} C+32$ $\Rightarrow \frac{9}{5} \times 30+32x^{\circ} F\frac{9}{5} \times 35+32$ $\Rightarrow 54+32x63+32$ $\Rightarrow 86^{\circ}x^{\circ}95^{\circ}$ Hence, the range of the temperature in Fahrenheit is between $86^{\circ}$ and $9...
Read More →If f : A→B given by
Question: If $f: A \rightarrow B$ given by $3^{f(x)}+2^{-x}=4$ is a bijection, then (a) $A=\{x \in R:-1x\infty\}, B=\{x \in R: 2x4\}$ (b) $A=\{x \in R:-3x\infty\}, B=\{x \in R: 2x4\}$ (c) $A=\{x \in R:-2x\infty\}, B=\{x \in R: 2x4\}$ (d) None of these Solution: (d) None of these $f: A \rightarrow B$ $3^{f(x)}+2^{-x}=4$ $\Rightarrow 3^{f(x)}=4-2^{-x}$ Taking $\log$ on both the sides, $f(x) \log 3=\log \left(4-2^{-x}\right)$ $\Rightarrow f(x)=\frac{\log \left(4-2^{-x}\right)}{\log 3}$ Logaritmic f...
Read More →The product of two successive integral multiples of 5 is 300.
Question: The product of two successive integral multiples of 5 is 300. Determine the multiples. Solution: Let the successive integer multiples of 5 be $5 x$, and $5(x+1)$ Then according to question $5 x \times 5(x+1)=300$ $25\left(x^{2}+x\right)=300$ $x^{2}+x=12$ $x^{2}+x-12=0$ $x^{2}-3 x+4 x-12=0$ $x(x-3)+4(x-3)=0$ $(x-3)(x+4)=0$ Therefore, $(x-3)=0$ $x=3$ Or $(x+4)=0$ $x=-4$ When $x=3$ then integer $5 x=5 \times 3$ $=15$ $5(x+1)=5(3+1)$ $=5 \times 4$ $=20$ And when $x=-4$ then integer $5 x=5 ...
Read More →A solution is to be kept between 86° and 95°F. What is the range of temperature in degree Celsius,
Question: A solution is to be kept between $86^{\circ}$ and $95^{\circ} \mathrm{F}$. What is the range of temperature in degree Celsius, if the Celsius (C)/ Fahrenheit (F) conversion formula is given by $F=\frac{9}{5} C+32$. Solution: Suppose the temperature of the solution is $x$ degree Celsius. $\therefore x$ in Fahrenheit $=\frac{9}{5} x+32$ Then, as per the given condition: $86\frac{9}{5} x+3295$ $\Rightarrow 86-32\frac{9}{5} x95-32 \quad$ (Subtratcting 32 throughout) $\Rightarrow 54\frac{9}...
Read More →Factorise:
Question: Factorise: $\frac{3}{5} x^{2}-\frac{19}{5} x+4$ Solution: $\frac{3}{5} x^{2}-\frac{19}{5} x+4=\frac{3}{5} x^{2}-3 x-\frac{4}{5} x+4$ $=3 x\left(\frac{1}{5} x-1\right)-4\left(\frac{1}{5} x-1\right)$ $=\left(\frac{1}{5} x-1\right)(3 x-4)$ Hence, factorisation of $\frac{3}{5} x^{2}-\frac{19}{5} x+4$ is $\left(\frac{1}{5} x-1\right)(3 x-4)$....
Read More →The marks scored by Rohit in two tests were 65 and 70.
Question: 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. Solution: Letxbe the minimum marks he scores in the third test. Then, $\frac{65+70+x}{3} \geq 65$ $\Rightarrow \frac{135+x}{3} \geq 65$ $\Rightarrow 135+x \geq 195 \quad$ (Multiplying both the sides by 3$)$ $\Rightarrow x \geq 195-135$ $\Rightarrow x \geq 60$ Hence, the minimum marks Rohit should score in the third test should be 60 ....
Read More →The marks scored by Rohit in two tests were 65 and 70.
Question: 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. Solution: Letxbe the minimum marks he scores in the third test. Then, $\frac{65+70+x}{3} \geq 65$ $\Rightarrow \frac{135+x}{3} \geq 65$ $\Rightarrow 135+x \geq 195 \quad$ (Multiplying both the sides by 3$)$ $\Rightarrow x \geq 195-135$ $\Rightarrow x \geq 60$ Hence, the minimum marks Rohit should score in the third test should be 60 ....
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