Find ten rational numbers between
Question: Find ten rational numbers between $\frac{-2}{5}$ and $\frac{1}{2}$. Solution: L. C.M of the denominator $s(2$ and 5$)$ is 10 . We can write: $\frac{-2}{5}=\frac{-2 \times 2}{5 \times 2}=\frac{-4}{10}$ and $\frac{1}{2}=\frac{1 \times 5}{2 \times 5}=\frac{5}{10}$ Since the integers between the numerators $(-4$ and 5$)$ of both the fractions are not sufficient, we will multiply the fractions by 2 . $\therefore \frac{-4}{10}=\frac{-4 \times 2}{10 \times 2}=\frac{-8}{20}$ $\frac{5}{10}=\fra...
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Question: If $\sin \theta=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$, find the values of all T-ratios of $\theta$. Solution: We have $\sin \theta=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$, As, $\cos ^{2} \theta=1-\sin ^{2} \theta$ $=1-\left(\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\right)^{2}$ $=\frac{1}{1}-\frac{\left(a^{2}-b^{2}\right)^{2}}{\left(a^{2}+b^{2}\right)^{2}}$ $=\frac{\left(a^{2}+b^{2}\right)^{2}-\left(a^{2}-b^{2}\right)^{2}}{\left(a^{2}+b^{2}\right)^{2}}$ $=\frac{\left[\left(a^{2}+b^{2}\right)-\left(a^{2}-b^{2...
Read More →Prove that:
Question: Prove that: $\frac{\sec \theta+\tan \theta-1}{\tan \theta-\sec \theta+1}=\frac{\cos \theta}{1-\sin \theta}$ Solution: Here we have to prove that, $\left(\frac{\sec \theta+\tan \theta-1}{\tan \theta-\sec \theta+1}\right)=\frac{\cos \theta}{1-\sin \theta}$ Firs we take the left hand side of the given equation $L H S=\left(\frac{\sec \theta+\tan \theta-1}{\tan \theta-\sec \theta+1}\right)$ Now we are using the trigonometric identity $1+\tan ^{2} \theta=\sec ^{2} \theta$ $\Rightarrow 1=\se...
Read More →Find ten rational numbers between
Question: Find ten rational numbers between $\frac{1}{4}$ and $\frac{1}{2}$. Solution: The L.C.M of the denominators $(2$ and 4$)$ is 4 . So, we can write $\frac{1}{4}$ as it is. Also, $\frac{1}{2}=\frac{1 \times 2}{2 \times 2}=\frac{2}{4}$ As the integers between the numerators 1 and 2 of both the fractions are not sufficient, we will multiply the fractions by 20 . $\therefore \frac{1}{4}=\frac{1 \times 20}{4 \times 20}=\frac{20}{80}$ $\frac{2}{4}=\frac{2 \times 20}{4 \times 20}=\frac{40}{80}$ ...
Read More →Prove that the ratio of the areas of two similar triangles
Question: Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. Solution: Here we have to prove that the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides. We have two trianglesandin which In the given figure AD is perpendicular to BC and PM is perpendicular to QR The areas of triangle ABC and triangle PQR are given by Area $(\triangle \mathrm{ABC})=\frac{1}{2} \times$ ...
Read More →Find two rational numbers between
Question: Find two rational numbers between $\frac{1}{5}$ and $\frac{1}{2}$. Solution: $R$ ational number between $\frac{1}{5}$ and $\frac{1}{2}=\frac{\left(\frac{1}{5}+\frac{1}{2}\right)}{2}=\frac{\frac{2+5}{10}}{2}=\frac{7}{20}$ $R$ ational number between $\frac{1}{5}$ and $\frac{7}{20}=\frac{\left(\frac{1}{5}+\frac{7}{20}\right)}{2}=\frac{\frac{4+7}{20}}{2}=\frac{11}{40}$ Therefore, two rational numbers between $\frac{1}{5}$ and $\frac{1}{2}$ are $\frac{7}{20}$ and $\frac{11}{40}$....
Read More →Find two rational numbers between
Question: Find two rational numbers between $\frac{-2}{9}$ and $\frac{5}{9}$. Solution: Since both the fractions $\left(\frac{-2}{9}\right.$ and $\left.\frac{5}{9}\right)$ have the same denominator, the integers between the numerators $(-2$ and 5$)$ are $-1,0,1,2,3,4$. Hence, two rational numbers between $\frac{-2}{9}$ and $\frac{5}{9}$ are $\frac{0}{9}$ or 0 and $\frac{1}{9}$....
Read More →Prove that
Question: If $\operatorname{cosec} \theta=\sqrt{10}$, the find the values of all T-ratios of $\theta$. Solution: Let us first draw a right $\triangle \mathrm{ABC}$, right angled at $\mathrm{B}$ and $\angle C=\theta$. Now, we know that $\operatorname{cosec} \theta=\frac{\text { Hypotenuse }}{\text { Perpendicular }}=\frac{A C}{A B}=\frac{\sqrt{10}}{1}$. So, if $A C=(\sqrt{10}) k$, then $A B=k$, where $k$ is a positive number. Now, by using Pythagoras theorem, we have:AC2= AB2+ BC2 $\Rightarrow B ...
Read More →Find other zeroes of the polynomial p(x) =
Question: Find other zeroes of the polynomial $p(x)=2 x^{4}+7 x^{3}-19 x^{2}-14 x+30$ if two of its zeroes are $\sqrt{2}$ and $\sqrt{-2}$. Solution: The given polynomial is $p(x)=2 x^{4}+7 x^{3}-19 x^{2}-14 x+30$ and two zeroes of this polynomial are $-\sqrt{2}$ and $\sqrt{2}$, then we have to find the two other zeroes of $p(x)$ If two zeroes of the polynomial $p(x)$ are $-\sqrt{2}$ and $\sqrt{2}$, then $(x+\sqrt{2})(x-\sqrt{2})=x^{2}-2$ is a factor of $p(x)$ Thereforecan be written as $p(x)=2 x...
Read More →Find any five rational numbers less than 2.
Question: Find any five rational numbers less than 2. Solution: We can write : $2=\frac{2}{1}=\frac{2 \times 5}{1 \times 5}=\frac{10}{5}$ Integers less than 10 are $0,1,2,3,4,5 \ldots 9$ Hence, five rational numbers less than 2 are $\frac{0}{5}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}$ and $\frac{4}{5}$....
Read More →Find the median of the following data
Question: Find the median of the following data Solution: The given distribution is We have to find the median of the above distribution. Now we have to find the cumulative frequency as From the above distribution median class is 60-70 and $l=60$ $N=53$ $h=10$ $f=7$ $C=22$ Now we are using the following relation median $=l+\frac{\frac{N}{2}-C}{f} \times h$ $=60+\frac{\frac{53}{2}-22}{7} \times 10$ $=66.43$ Hence the median of the given distribution is 66.43...
Read More →Find a rational number between −3 and 1.
Question: Find a rational number between 3 and 1. Solution: $R$ ational number between $-3$ and $1=\frac{-3+1}{2}=\frac{-2}{2}=-1$...
Read More →If 24 trousers of equal size can be prepared in 54 metres of cloth,
Question: If 24 trousers of equal size can be prepared in 54 metres of cloth, what length of cloth is required for each trouser? Solution: Cloth needed to prepare 24 trousers $=54 \mathrm{~m}$ $\therefore L$ ength of the cloth required for each trousers $=54 \div 24=\frac{54}{24}=\frac{9}{4} \mathrm{~m}=2 \frac{1}{4}$ metres...
Read More →Divide the sum of
Question: Divide the sum of $\frac{65}{12}$ and $\frac{12}{7}$ by their difference. Solution: $\left(\frac{65}{12}+\frac{12}{7}\right) \div\left(\frac{65}{12}-\frac{12}{7}\right)$ $=\frac{65 \times 7+12 \times 12}{84} \div \frac{65 \times 7-12 \times 12}{84}$ $=\frac{455+144}{84} \div \frac{455-144}{84}$ $=\frac{599}{84} \div \frac{311}{84}$ $=\frac{599}{84} \times \frac{84}{311}$ $=\frac{599}{311}$...
Read More →Find the mean of the following frequency distribution,
Question: Find the mean of the following frequency distribution, using step-deviation method: Solution: The following frequency distribution is given We have to find the mean of the above frequency distribution using step deviation method. By using step deviation method, we have From the above distribution, we have $h=10$ $a=25$ $\sum f_{i}=50$ $\sum f_{i} u_{i}=0$ We know the mean of a given distribution is given by $\operatorname{Mean}(\bar{x})=a+\frac{\sum f_{t} u_{i}}{\sum f_{i}} \times h$ $...
Read More →If cot θ = 2, find the value of all T-ratios of θ.
Question: If cot = 2, find the value of all T-ratios of . Solution: Let us first draw a right $\triangle \mathrm{ABC}$, right angled at $\mathrm{B}$ and $\angle C=\theta$. Now, we know that $\cot \theta=\frac{\text { base }}{\text { Perpendicular }}=\frac{B C}{A B}=2$. So, if BC = 2k, then AB =k, wherekis a positive number.Now, using Pythagoras theorem, we have:AC2= AB2+ BC2= (2k)2+ (k)2⇒ AC2= 4k2+k2= 5k2 $\Rightarrow \mathrm{AC}=\sqrt{5} \mathrm{k}$ Now, finding the other T-ratios using their d...
Read More →Divide the sum of
Question: Divide the sum of $\frac{-13}{5}$ and $\frac{12}{7}$ by the product of $\frac{-31}{7}$ and $\frac{-1}{2}$. Solution: $\left(\frac{-13}{5}+\frac{12}{7}\right) \div\left(\frac{-31}{7} \times \frac{-1}{2}\right)$ $=\frac{-13 \times 7+12 \times 5}{35} \div \frac{31}{14}$ $=\frac{-91+60}{35} \div \frac{31}{14}$ $=\frac{-31}{35} \times \frac{14}{31}$ $=\frac{-2}{5}$...
Read More →By what number should
Question: By what number should $\frac{-33}{16}$ be divided to get $\frac{-11}{4} ?$ Solution: Let the number be $\mathrm{x}$. $\therefore \frac{-33}{16} \div \mathrm{x}=\frac{-11}{4}$ Or $\frac{-33}{16} \times \frac{1}{\mathrm{x}}=\frac{-11}{4}$ or $\frac{1}{\mathrm{x}}=\frac{-11}{4} \times \frac{16}{-33}$ or $\frac{1}{\mathrm{x}}=\frac{4}{3}$ or $\mathrm{x}=\frac{3}{4}$ Thus, the number is $\frac{3}{4}$....
Read More →The cost of
Question: The cost of $2 \frac{1}{3}$ metres of cloth is Rs $75 \frac{1}{4}$. Find the cost of cloth per metre. Solution: The cost of $2 \frac{1}{3}$ metres of cloth is Rs $75 \frac{1}{4}$. $\therefore C$ ost per metre $=75 \frac{1}{4} \div 2 \frac{1}{3}$ $=\frac{301}{4} \div \frac{7}{3}$ $=\frac{301}{4} \times \frac{3}{7}$ $=\frac{129}{4}$ $=\operatorname{Rs} 32 \frac{1}{4}$ Thus, Rs $32 \frac{1}{4}$ or Rs $32.25$ is the cost of cloth per metre....
Read More →Solve this
Question: If $\tan \theta=\frac{15}{8}$ find the values of all T-ratios of $\theta$. Solution: Let us first draw a right $\triangle \mathrm{ABC}$, right angled at $\mathrm{B}$ and $\angle C=\theta$. Now, we know that $\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}=\frac{A B}{B C}=\frac{15}{8}$. So, if BC = 8k, then AB = 15k, wherekis a positive number.Now, using Pythagoras theorem, we have:AC2= AB2+ BC2= (15k)2+ (8k)2⇒ AC2= 225k2+ 64k2= 289k2⇒ AC = 17kNow, finding the other T-ratios ...
Read More →In Fig. 8, and ΔDBC are on the same base BC and on
Question: In Fig. 8, and ΔDBCare on the same baseBCand on opposite sides ofBCadQis the point of intersection ofADandBC. Prove that $\frac{\operatorname{area}(\Delta A B C)}{\operatorname{area}(\Delta D B C)}=\frac{A O}{D O}$ Solution: We have given the diagram in which We have to prove that $\frac{\text { Area of } \triangle A B C}{\text { Area of } \triangle D B C}=\frac{A O}{D O}$ Firstly, we draw a line fromAperpendicular to lineBCand after that we draw a line fromDperpendicular toBC. From th...
Read More →The cost of
Question: The cost of $7 \frac{2}{3}$ metres of rope is Rs $12 \frac{3}{4}$. Find its cost per metre. Solution: The cost of $7 \frac{2}{3}$ metres of rope is Rs $12 \frac{3}{4}$. $\therefore C$ ost per metre $=12 \frac{3}{4} \div 7 \frac{2}{3}$ $=\frac{51}{4} \div \frac{23}{3}$ $=\frac{51}{4} \times \frac{3}{23}$ $=\frac{153}{92}$ $=\operatorname{Rs} 1 \frac{61}{92}$...
Read More →Solve this
Question: If $\cos \theta=\frac{7}{25}$ find the values of all T-ratios of $\theta$. Solution: Let us first draw a right $\triangle \mathrm{ABC}$, right angled at $\mathrm{B}$ and $\angle C=\theta$. Now, we know that $\cos \theta=\frac{\text { Base }}{\text { hypotenuse }}=\frac{B C}{A C}=\frac{7}{25}$. So, if BC = 7k, then AC = 25k, wherekis a positive number.Now, using Pythagoras theorem, we have:AC2= AB2+ BC2⇒ AB2= AC2-BC2= (25k)2-(7k)2.⇒ AB2= 625k2-49k2= 576k2⇒ AB = 24kNow, finding the other...
Read More →In Fig. 7, AB ⊥ BC, FG ⊥ BC and DE ⊥ AC.
Question: In Fig. 7, $A B \perp B C, F G \perp B C$ and $D E \perp A C$. Prove that $\triangle A D E \sim \triangle G C F$ Solution: In the given figure, we have $A B \perp B C$ $F G \perp B C$, and $D E \perp A C$ Then we have to prove that The following diagram is given In, we have $\angle A+\angle D=90^{\circ}$.......(1) In $\triangle A B C$, we have $\angle A+\angle C=90^{\circ} \ldots \ldots(2)$ From equation (1) and equation (2), we have $\angle A+\angle C=\angle A+\angle D$ $\Rightarrow \...
Read More →Find (x + y) ÷ (x − y), if
Question: Find (x+y) (xy), if (i) $x=\frac{2}{3}, y=\frac{3}{2}$ (ii) $x=\frac{2}{5}, y=\frac{1}{2}$ (iii) $x=\frac{5}{4}, y=\frac{-1}{3}$ (iv) $x=\frac{2}{7}, y=\frac{4}{3}$ (v) $x=\frac{1}{4}, y=\frac{3}{2}$ Solution:...
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