By what number should
Question: By what number should $\frac{-3}{4}$ be multiplied in order to produce $\frac{2}{3} ?$ Solution: Let the other number that should be multiplied with $\frac{-3}{4}$ to produce $\frac{2}{3}$ be $\mathrm{x}$. $\therefore \mathrm{x} \times \frac{-3}{4}=\frac{2}{3}$ or $\mathrm{x}=\frac{2}{3} \div \frac{-3}{4}$ or $\mathrm{x}=\frac{2}{3} \times \frac{4}{-3}$ or $x=\frac{-8}{9}$ Thus, the number is $\frac{-8}{9}$....
Read More →Solve this
Question: If $\sin \theta=\frac{\sqrt{3}}{2}$, find the value 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 $\sin \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{A B}{A C}=\frac{\sqrt{3}}{2}$. So, if $\mathrm{AB}=\sqrt{3} k$, then $\mathrm{AC}=2 k$, where $k$ is a positive number. Now, using Pythagoras theorem, we have:AC2= AB2+ BC2 $\Rightarrow \mathrm{BC}^{2}=\ma...
Read More →If cos θ + sin θ = 2–√ cos θ,
Question: If $\cos \theta+\sin \theta=\sqrt{2} \cos \theta$, show that $\cos \theta-\sin \theta=\sqrt{2} \sin \theta .$ Solution: Given that if $\cos \theta+\sin \theta=\sqrt{2} \cos \theta$, then we have to prove that $\cos \theta-\sin \theta=\sqrt{2} \sin \theta$ We have, $\cos \theta+\sin \theta=\sqrt{2} \cos \theta$ $\Rightarrow \quad \sin \theta=\sqrt{2} \cos \theta-\cos \theta$ $\Rightarrow \quad \sin \theta=(\sqrt{2}-1) \cos \theta$ $\Rightarrow \quad \sin \theta=\frac{(\sqrt{2}-1)(\sqrt{...
Read More →By what number should we multiply
Question: By what number should we multiply $\frac{-8}{13}$ so that the product may be 24 ? Solution: Let the number be $\mathrm{x}$. $\therefore \mathrm{x} \times \frac{-8}{13}=24$ or $\mathrm{x}=24 \div \frac{-8}{13}$ or $\mathrm{x}=24 \times \frac{13}{-8}$ or $x=-39$ Thus, the number is $-39$....
Read More →Prove that (cosec θ − sin θ) (sec θ − cos θ) =
Question: Prove that $(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)=\frac{1}{\tan \theta+\cot \theta}$. Solution: Here we have to prove that $(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)=\frac{1}{\tan \theta+\cot \theta}$ Left hand side $=(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)$ $=\left(\frac{1}{\sin \theta}-\sin \theta\right)\left(\frac{1}{\cos \theta}-\cos \theta\right)$ $=\frac{\left(1-\sin ^{2} \theta\right)\left(1-\cos ^{...
Read More →By what number should we multiply
Question: By what number should we multiply $\frac{-15}{28}$ so that the product may be $\frac{-5}{7}$ ? Solution: Let the other number be $\mathrm{x}$. $\therefore \mathrm{x} \times \frac{-15}{28}=\frac{-5}{7}$ or $\mathrm{x}=\frac{-5}{7} \div \frac{-15}{28}$ or $\mathrm{x}=\frac{-5}{7} \times \frac{28}{-15}$ or $\mathrm{x}=\frac{4}{3}$ Thus, the other number is $\frac{4}{3}$....
Read More →By what number should we multiply
Question: By what number should we multiply $\frac{-1}{6}$ so that the product may be $\frac{-23}{9} ?$ Solution: Let the number $b e \mathrm{x}$. $\therefore \mathrm{x} \times \frac{-1}{6}=\frac{-23}{9}$ $\begin{aligned} \mathrm{x} =\frac{-23}{9} \div \frac{-1}{6} \\ \mathrm{x} =\frac{-23}{9} \times \frac{6}{-1}=\frac{46}{3} \end{aligned}$ Therefore, the other number is $\frac{46}{3}$....
Read More →are zeroes of the polynomial x2 − 2x − 15,
Question: If $\alpha, \beta$ are zeroes of the polynomial $x^{2}-2 x-15$, then form a quadratic polynomial whose zeroes are $(2 \alpha)$ and $(2 \beta)$. Solution: The given polynomial is $P(x)=x^{2}-2 x-15$ and $\alpha$ and $\beta$ are the zeroes of the polynomial $P(x)$, then we have to find another polynomial whose zeroes are $2 \alpha, 2 \beta$ Now we comparing the given polynomial $P(x)$ with $a x^{2}+b x+c$, we get $a=1, b=-2$, and $c=-15$ We know that $\alpha+\beta=-\frac{b}{a}$ $=-\frac{...
Read More →The product of two rational numbers is
Question: The product of two rational numbers is $\frac{-8}{9}$. If one of the numbers is $\frac{-4}{15}$, find the other. Solution: Let the other number be $\mathrm{x}$. $\therefore \mathrm{x} \times \frac{-4}{15}=\frac{-8}{9}$ or $\mathrm{x}=\frac{-8}{9} \div \frac{-4}{15}$ or $\mathrm{x}=\frac{-8}{9} \times \frac{15}{-4}$ or $\mathrm{x}=\frac{10}{3}$ Thus, the other number is $\frac{10}{3}$....
Read More →The product of two rational numbers is 15.
Question: The product of two rational numbers is 15. If one of the numbers is 10, find the other. Solution: Let the other number be $x$. $\therefore x \times(-10)=15$ or $x=\frac{15}{-10}=\frac{3}{-2}$ So, the other number is $\frac{-3}{2}$....
Read More →A person can row a boat at the rate of 5 km/hour
Question: A person can row a boat at the rate of 5 km/hour in still water. He takes thrice as much time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream. Solution: Given that, Speed of boat in still water = 5 km/hour. Distance covered by boat = 40 km Now we are assuming that the speed of the stream is, then Speed of the boat rowing upstream =, and Speed of the boat rowing downstream = According to the given condition, time taken to coverin upstream is three time...
Read More →Draw two concentric circles of radii 4 cm and 6 cm.
Question: Draw two concentric circles of radii 4 cm and 6 cm. Construct a tangent to the smaller circle from a point on the larger circle. Measure the length of this tangent. Solution: Steps of Construction :Step 1. Draw a circle with O as centre and radius 6 cm.Step 2. Draw another circle with O as centre and radius 4 cm.Step 2 . Mark a point P on the circle with radius 6 cm.Step 3. Join OP and bisect it at M.Step 4. Draw a circle with M as centre and radius equal to MP to intersect the given c...
Read More →Prove that
Question: Prove that $\frac{2 \sqrt{3}}{5}$ is irrational. Solution: Here we have to prove that the number $\frac{2 \sqrt{3}}{5}$ is an irrational number. Now let us suppose that $\frac{2 \sqrt{3}}{5}=x$, where $x$ is a rational number, then $2 \sqrt{3}=5 x$ $\Rightarrow \sqrt{3}=\frac{5 x}{2}$ ..........(1) As $x$ is rational number, therefore form equation (1), so is $\frac{5 x}{2}$ and $\sqrt{3}$ is also a rational number which is a contradiction as $\sqrt{3}$ is an irrational number. Therefo...
Read More →Draw a circle of radius 4 cm.
Question: Draw a circle of radius 4 cm. Draw a tangent to the circle, making an angle of 60 with a line passing through the centre. Solution: Steps Of construction:Step 1. Draw a circle with centre O and radius 4 cm.Step 2. Draw radius OA and produce it to B. Step 3. Make $\angle A O P=30^{\circ}$ Step 4. Draw $\mathrm{PQ} \perp O P$, meeting $\mathrm{OB}$ at $\mathrm{Q}$. Step 5. Then, $\mathrm{PQ}$ is the desired tangent, such that $\angle O Q P=60^{\circ}$....
Read More →Find the value and express as a rational number in standard form:
Question: Find the value and express as a rational number in standard form: (i) $\frac{2}{5} \div \frac{26}{15}$ (ii) $\frac{10}{3} \div \frac{-35}{12}$ (iii) $-6 \div\left(\frac{-8}{17}\right)$ (iv) $\frac{-40}{99} \div(-20)$ (v) $\frac{-22}{27} \div \frac{-110}{18}$ (vi) $\frac{-36}{125} \div \frac{-3}{75}$ Solution: (i) $\frac{2}{5} \div \frac{26}{15}=\frac{2}{5} \times \frac{15}{26}=\frac{3}{13}$ (ii) $\frac{10}{3} \div \frac{-35}{12}=\frac{10}{3} \times \frac{12}{-35}=\frac{-8}{7}$ (iii) $-...
Read More →Find the value and express as a rational number in standard form:
Question: Find the value and express as a rational number in standard form: (i) $\frac{2}{5} \div \frac{26}{15}$ (ii) $\frac{10}{3} \div \frac{-35}{12}$ (iii) $-6 \div\left(\frac{-8}{17}\right)$ (iv) $\frac{-40}{99} \div(-20)$ (v) $\frac{-22}{27} \div \frac{-110}{18}$ (vi) $\frac{-36}{125} \div \frac{-3}{75}$ Solution: (i) $\frac{2}{5} \div \frac{26}{15}=\frac{2}{5} \times \frac{15}{26}=\frac{3}{13}$ (ii) $\frac{10}{3} \div \frac{-35}{12}=\frac{10}{3} \times \frac{12}{-35}=\frac{-8}{7}$ (iii) $-...
Read More →Show that any positive odd integers is of the form
Question: Show that any positive odd integers is of the form 4q+1 or 4q+3 where q is a positive integer. Solution: Here we have to prove that for any positive integerq, the positive odd integer will be form of 4q+1 or 4q+3. Now let us suppose that the positive odd integer isathen by Euclids division rule a= 4q+r(1 ) Whereq(quotient) andr(remainder) are positive integers, and We are putting the values ofrfrom 0 to 3 in equation (1), we get But we can easily see that 4qand 4q+2 are both even numbe...
Read More →Draw a circle of radius 3.5 cm. Draw a pair of tangents to this circle,
Question: Draw a circle of radius 3.5 cm. Draw a pair of tangents to this circle, which are inclined to each other at an angle of 60. Write the steps of construction. Solution: Steps of Construction:Step 1. Draw a circle with centre O and radius = 3.5 cm.Step 2. Draw any diameter AOB of this circle. Step 3. Construct $\angle B O C=60^{\circ}$, such that the radius $O C$ meets the circle at $C$. Step 4. Draw $\mathrm{MA} \perp A B$ and $N C \perp O C$. Let AM and CN intersect at P. Then, PA and P...
Read More →Divide:
Question: Divide: (i) 1 by $\frac{1}{2}$ (ii) 5 by $\frac{-5}{7}$ (iii) $\frac{-3}{4}$ by $\frac{9}{-16}$ (iv) $\frac{-7}{8}$ by $\frac{-21}{16}$ (v) $\frac{7}{-4}$ by $\frac{63}{64}$ (vi) 0 by $\frac{-7}{5}$ (vii) $\frac{-3}{4}$ by $-6$ (viii) $\frac{2}{3}$ by $\frac{-7}{12}$ (ix) $-4$ by $\frac{-3}{5}$ (x) $\frac{-3}{13}$ by $\frac{-4}{65}$ Solution: (i) $1 \div \frac{1}{2}=1 \times \frac{2}{1}=2$ (ii) $5 \div \frac{-5}{7}=5 \times \frac{7}{-5}=-7$ (iii) $\frac{-3}{4} \div \frac{9}{-16}=\frac{...
Read More →Find the mode of the following distribution of marks obtained by 80 students:
Question: Find the mode of the following distribution of marks obtained by 80 students: Solution: We have the following distribution We have to find the mode of the above distribution From the given distribution, we can easily see that the class (20-30) has the maximum frequencyi.e.32. Lower limit of the modal class $x_{k}=30$ Class interval $h=40-30=10$ Frequency of the modal class $f_{k}=32$ Frequency of the class preceding the modal class $f_{k-1}=12$ Frequency of the class succeeding the mod...
Read More →Divide:
Question: Divide: (i) 1 by $\frac{1}{2}$ (ii) 5 by $\frac{-5}{7}$ (iii) $\frac{-3}{4}$ by $\frac{9}{-16}$ (iv) $\frac{-7}{8}$ by $\frac{-21}{16}$ (v) $\frac{7}{-4}$ by $\frac{63}{64}$ (vi) 0 by $\frac{-7}{5}$ (vii) $\frac{-3}{4}$ by $-6$ (viii) $\frac{2}{3}$ by $\frac{-7}{12}$ (ix) $-4$ by $\frac{-3}{5}$ (x) $\frac{-3}{13}$ by $\frac{-4}{65}$ Solution: (i) $1 \div \frac{1}{2}=1 \times \frac{2}{1}=2$ (ii) $5 \div \frac{-5}{7}=5 \times \frac{7}{-5}=-7$ (iii) $\frac{-3}{4} \div \frac{9}{-16}=\frac{...
Read More →Draw a circle of radius 4.8 cm. Take a point P on it.
Question: Draw a circle of radius 4.8 cm. Take a pointPon it. Without using the centre of the circle, construct a tangent at the pointP. Write the steps of construction. Solution: Steps of Construction :Step 1. Draw a circle of radius 4.8 cm.Step 2. Mark a point P on it.Step 3. Draw any chord PQ.Step 4. Take a point R on the major arc QP.Step 5. Join PR and RQ. Step 6. Draw $\angle Q P T=\angle P R Q$ Step 7. Produce TP to T', as shown in the figure. T'PT is the required tangent....
Read More →Fill in the blanks:
Question: Fill in the blanks: (i) $-4 \times \frac{7}{9}=\frac{7}{9} \times \ldots \ldots$ (ii) $\frac{5}{11} \times \frac{-3}{8}=\frac{-3}{8} \times \ldots . . .$ (iii) $\frac{1}{2} \times\left(\frac{3}{4}+\frac{-5}{12}\right)=\frac{1}{2} \times \ldots . . .+\ldots \times \frac{-5}{12}$ (iv) $\frac{-4}{5} \times\left(\frac{5}{7}+\frac{-8}{9}\right)=\left(\frac{-4}{5} \times \ldots\right) \times \frac{-8}{9}$ Solution: (i) $-4$ $\mathrm{x} \times \mathrm{y}=\mathrm{y} \times \mathrm{x}$ (commuta...
Read More →Prove that in a triangle, if the square of one side is equal
Question: Prove that in a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angles opposite to the first side is a right angle. Solution: Given: if square of one side is equal to the sum of the squares of the other two sides, then we have to prove that the angle opposite the first side is a right angle. Here, we are given a triangle ABC with. We need to prove that . Now, we construct a triangle PQR right angled at Q such thatand. We have the ...
Read More →Draw ∆ABC, right-angled at B, such that AB = 3 cm and BC = 4 cm.
Question: Draw $\triangle A B C$, right-angled at $B$, such that $A B=3 \mathrm{~cm}$ and $B C=4 \mathrm{~cm}$. Now, construct a triangle similar to $\triangle A B C$, each of whose sides is $\frac{7}{5}$ times the corresponding sides of ∆ABC. Solution: Step 1. Draw a line segment BC = 4 cm.Step 2. With B as centre, draw an angle of 90o.Step 3. With B as centre and radius equal to 3 cm, cut an arc at the right angle and name it A.Step 4. Join AB and AC.Thus, △ ABC is obtained . Step 5. Extend BC...
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