If a line is drawn parallel to one side of a triangle
Question: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio. Solution: Given that ain which a line parallel to BC is drawn which meet the other two sides at the point D and E, then we have to prove that $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$ We have the following diagram with some additional construction. In the above figure, we can see that EF is perpe...
Read More →If a line is drawn parallel to one side of a triangle
Question: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio. Solution: Given that ain which a line parallel to BC is drawn which meet the other two sides at the point D and E, then we have to prove that $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$ We have the following diagram with some additional construction. In the above figure, we can see that EF is perpe...
Read More →A stationary observer receives sound from two identical tuning forks,
Question: A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears 2 beats/sec. The oscillation frequency of each tuning fork is $v_{0}=1400 \mathrm{~Hz}$ and the velocity of sound in air is $350 \mathrm{~m} / \mathrm{s}$. The speed of each tuning fork is close to:$\frac{1}{2} \mathrm{~m} / \mathrm{s}$$1 \mathrm{~m} / \mathrm{s}$$\frac{1}{4} \mathrm{~m} / \ma...
Read More →What should be subtracted from
Question: What should be subtracted from $\left(\frac{3}{4}-\frac{2}{3}\right)$ to get $\frac{-1}{6} ?$ Solution: Let $x$ be subtracted. $\therefore\left(\frac{3}{4}-\frac{2}{3}\right)-x=\frac{-1}{6}$ $\Rightarrow\left(\frac{9}{12}-\frac{8}{12}\right)-x=\frac{-1}{6}$ $\Rightarrow\left(\frac{9-8}{12}\right)-x=\frac{-1}{6}$ $\Rightarrow x=\frac{1}{12}-\frac{-1}{6}$ $\Rightarrow x=\frac{1}{12}-\frac{-2}{12}$ $\Rightarrow x=\frac{1-(-2)}{12}$ $\Rightarrow x=\frac{1+2}{12}$ $\Rightarrow x=\frac{3}{12...
Read More →Divide 30x4 + 11x3 − 82x2 − 12x + 48
Question: Divide 30x4+ 11x3 82x2 12x+ 48 by (3x2+ 2x 4) and verify the result by division algorithm. Solution: Here we have to divide $30 x^{4}+11 x^{3}-82 x^{2}-12 x+48$ by $3 x^{2}+2 x-4$. $30 x^{4}+20 x^{3}-40 x^{2}$ $-9 x^{3}-42 x^{2}-12 x+48$ $-9 x^{3}-6 x^{2}+12 x$ $-36 x^{2}-24 x+48$ $-36 x^{2}-24 x+48$ According to division algorithm, Dividend = Divisor Quotient + Remainder. This can be verified as, Divisor Quotient + Remainder $\left(3 x^{2}+2 x-4\right) \times\left(10 x^{2}-3 x-12\righ...
Read More →Two concentric circles are of radii 5 cm and 3 cm, respectively.
Question: Two concentric circles are of radii 5 cm and 3 cm, respectively. Find the length of the chord of the larger circle that touches the smaller circle. Solution: Given: Two circles have the same centreOandABis a chord of the larger circle touching thesmaller circle at C; also, OA = 5 cm and OC = 3 cm. In $\Delta \mathrm{OAC}, \mathrm{OA}^{2}=\mathrm{OC}^{2}+\mathrm{AC}^{2}$ $\therefore \mathrm{AC}^{2}=\mathrm{OA}^{2}-\mathrm{OC}^{2}$ $\Rightarrow \mathrm{AC}^{2}=5^{2}-3^{2}$ $\Rightarrow \...
Read More →Speed of a transverse wave on a straight wire mass 6.0 g,
Question: Speed of a transverse wave on a straight wire (mass $6.0$ $\mathrm{g}$, length $60 \mathrm{~cm}$ and area of cross-section $1.0 \mathrm{~mm}^{2}$ ) is $90 \mathrm{~ms}^{-1}$. If the Young's modulus of wire is $16 \times 10^{11} \mathrm{Nm}^{-2}$ the extension of wire over its natural length is:$0.03 \mathrm{~mm}$$0.02 \mathrm{~mm}$$0.04 \mathrm{~mm}$$0.01 \mathrm{~mm}$Correct Option: 1 Solution: (1) Given, $l=60 \mathrm{~cm}, m=6 \mathrm{~g}, A=1 \mathrm{~mm}^{2}, v=90 \mathrm{~m} / \m...
Read More →What should be added to
Question: What should be added to $\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}\right)$ to get $3 ?$ Solution: Let $x$ be added. $\therefore x+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}\right)=3$ $\Rightarrow x+\left(\frac{15}{30}+\frac{10}{30}+\frac{6}{30}\right)=3$ $\Rightarrow x+\left(\frac{15+10+6}{30}\right)=3$ $\Rightarrow x+\frac{31}{30}=3$ $\Rightarrow x=\frac{3}{1}-\frac{31}{30}$ $\Rightarrow x=\frac{90}{30}-\frac{31}{30}$ $\Rightarrow x=\frac{90-31}{30}$ $\Rightarrow x=\frac{59}{30}$...
Read More →Prove that the parallelogram circumscribing a circle is a rhombus.
Question: Prove that the parallelogram circumscribing a circle is a rhombus. Solution: Given, a parallelogram ABCD circumscribes a circle with centre O. $A B=B C=C D=A D$ We know that the lengths of tangents drawn from an exterior point to a circleare equal. $\therefore A P=A S \ldots \ldots(i) \quad[$ tangents from $A]$ $B P=B Q \ldots . \ldots \ldots(i i) \quad[$ tangents from $B]$ $C R=C Q \ldots \ldots \ldots(i i i) \quad[$ tangents from $C]$ $D R=D S \ldots \ldots \ldots \ldots(i v) \quad[$...
Read More →A sound source $S$ is moving along a straight track with speed $v$,
Question: A sound source $S$ is moving along a straight track with speed $v$, and is emitting sound of frequency $v_{\mathrm{o}}$ (see figure). An observer is standing at a finite distance, at the point $O$, from the track. The time variation of frequency heard by the observer is best represented by: $\left(t_{0}\right.$ represents the instant when the distance between the source and observer is minimum)Correct Option: , 2 Solution: (2) Frequency heard by the observer $v_{\text {observed }}=\lef...
Read More →What should be added to
Question: What should be added to $\left(\frac{2}{3}+\frac{3}{5}\right)$ to get $\frac{-2}{15} ?$ Solution: Let $x$ be added. $\therefore x+\left(\frac{2}{3}+\frac{3}{5}\right)=\frac{-2}{15}$ $\Rightarrow x+\left(\frac{10}{15}+\frac{9}{15}\right)=\frac{-2}{15}$ $\Rightarrow x+\left(\frac{10+9}{15}\right)=\frac{-2}{15}$ $\Rightarrow x=\frac{-2}{15}-\frac{19}{15}$ $\Rightarrow x=\frac{-2-19}{15}$ $\Rightarrow x=\frac{-21}{15}$ $\Rightarrow x=\frac{-7}{5}$...
Read More →Find the missing frequencies in the following
Question: Find the missing frequencies in the following frequency distribution table, ifN= 100 and median is 32. Solution: We have to find the missing term of the following distribution table if $N=100$ and median is 32 . Suppose the missing term arexandy. Now we have to find the cumulative frequency as From the above distribution median class is 30-40 and $l=30$ $N=100$ $h=10$ $f=30$ $C=35+x$ Therefore, $N=100=75+x+y$ $\Rightarrow x+y=25$.........(1) Now we are using the following relation medi...
Read More →The mean of the following frequency
Question: The mean of the following frequency distribution is 50. Find the value ofp. Solution: Given that the mean value of the following distribution is 50. We have to find the value ofp. To find the value ofp, we have following procedure Thus, we have $\sum f_{i}=96+p$ $\sum f_{i} x_{i}=4320+70 p$ We know that the formula of mean is given by mean $=\frac{\sum f_{i} x_{i}}{\sum f_{i}}$ $\Rightarrow \quad 50=\frac{4320+70 p}{96+p}$ $\Rightarrow 4800+50 p=4320+70 p$ $\Rightarrow \quad 20 p=480$ ...
Read More →What number should be subtracted from
Question: What number should be subtracted from $\frac{3}{7}$ to get $\frac{5}{4} ?$ Solution: Let, $x$ be subtracted. $\therefore \frac{3}{7}-x=\frac{5}{4}$ $\Rightarrow x=\frac{3}{7}-\frac{5}{4}$ $\Rightarrow x=\frac{12}{28}-\frac{35}{28}$ $\Rightarrow x=\frac{12-35}{28}=\frac{-23}{28}$...
Read More →A driver in a car, approaching a vertical wall notices that the frequency of his car horn,
Question: A driver in a car, approaching a vertical wall notices that the frequency of his car horn, has changed from $440 \mathrm{~Hz}$ to $480 \mathrm{~Hz}$, when it gets reflected from the wall. If the speed of sound in air is $345 \mathrm{~m} / \mathrm{s}$, then the speed of the car is :$54 \mathrm{~km} / \mathrm{hr}$$36 \mathrm{~km} / \mathrm{hr}$$18 \mathrm{~km} / \mathrm{hr}$$24 \mathrm{~km} / \mathrm{hr}$Correct Option: 1 Solution: (1) Let $f_{1}$ be the frequency heard by wall, $f_{1}=\...
Read More →A driver in a car, approaching a vertical wall notices that the frequency of his car horn,
Question: A driver in a car, approaching a vertical wall notices that the frequency of his car horn, has changed from $440 \mathrm{~Hz}$ to $480 \mathrm{~Hz}$, when it gets reflected from the wall. If the speed of sound in air is $345 \mathrm{~m} / \mathrm{s}$, then the speed of the car is :54 \mathrm{~km} / \mathrm{hr}$36 \mathrm{~km} / \mathrm{hr}$$18 \mathrm{~km} / \mathrm{hr}$$24 \mathrm{~km} / \mathrm{hr}$Correct Option: 1 Solution: (1) Let $f_{1}$ be the frequency heard by wall, $f_{1}=\le...
Read More →What number should be subtracted from
Question: What number should be subtracted from $\frac{-5}{3}$ to get $\frac{5}{6}$ ? Solution: Let $x$ be subtracted. $\therefore \frac{-5}{3}-x=\frac{5}{6}$ $\Rightarrow x=\frac{-5}{3}-\frac{5}{6}$ $\Rightarrow x=\frac{-10}{6}-\frac{5}{6}$ $\Rightarrow x=\frac{-10-5}{6}=\frac{-15}{6}=\frac{-5}{2}$...
Read More →The diagonals of a trapezium ABCD with AB || DC
Question: The diagonals of a trapezium ABCD with AB || DC intersect each other at point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD. Solution: Given that the diagonals of trapeziumABCDwithintersect each other at pointOand if, then we have to find the ratio of areas of triangleAOBand triangleCOD. We have the following diagram. From the above figure, in trianglesAOBandCOD, we have $\angle A O B=\angle C O D$ $\angle O A B=\angle O C D$ $\Rightarrow \triangle A O B \sim \t...
Read More →Prove that the tangents drawn at the ends
Question: Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord. Solution: Let $R A$ and $R B$ be two tangents to the circle with centre $O$ and let $A B$ be $a$ chord of the circle. We have to prove that $\angle R A B=\angle R B A$. $\therefore$ Now, $R A=R B$ (Since tangents drawn from an external point to a circle are equal) $\therefore$ In $\Delta R A B, \angle R A B=\angle R B A$ (S ince opposite sides are equal, their base angles are also equal)...
Read More →What number should be added to
Question: What number should be added to $\frac{-5}{7}$ to get $\frac{-2}{3} ?$ Solution: Let $x$ be added. $\therefore x+\frac{-5}{7}=\frac{-2}{3}$ $\Rightarrow x=\frac{-2}{3}-\frac{-5}{7}$ $\Rightarrow x=\frac{-14}{21}-\frac{-15}{21}$ $\Rightarrow x=\frac{-14-(-15)}{21}$ $\Rightarrow x=\frac{-14+15}{21}$ $\Rightarrow x=\frac{1}{21}$...
Read More →In Figure 6, P and Q are the midpoints of the sides CA and CB
Question: In Figure 6, P and Q are the midpoints of the sides CA and CB respectively of ABC right angled at C. Prove that 4(AQ2+ BP2) = 5 AB2. Solution: In the given figurePandQare the mid points ofACandBC, the we have to prove that The following figure is given. Using Pythagoras theorem in $\triangle A B C$, we get AB2=AC2+BC2 (1) Similarly, by using Pythagoras theorem in $\triangle A C Q$ and $\triangle B C P$, we get AQ2=CQ2+AC2 (2) BP2=CP2+BC2 (3) Now adding equation (2) and equation (3), we...
Read More →Two coherent sources of sound,
Question: Two coherent sources of sound, $S_{1}$ and $S_{2}$, produce sound waves of the same wavelength, $\lambda=1 \mathrm{~m}$, in phase. $S_{1}$ and $S_{2}$ are placed $1.5 \mathrm{~m}$ apart (see fig.). A listener, located at $L$, directly in front of $S_{2}$ finds that the intensity is at a minimum when he is $2 \mathrm{~m}$ away from $S_{2}$. The listener moves away from $S_{1}$, keeping his distance from $S_{2}$ fixed. The adjacent maximum of intensity is observed when the listener is at...
Read More →What number should be added to
Question: What number should be added to $\frac{-5}{11}$ so as to get $\frac{26}{33} ?$ Solution: Let $x$ be added. $\therefore x+\frac{-5}{11}=\frac{26}{33}$ $\Rightarrow x=\frac{26}{33}-\frac{-5}{11}$ $\Rightarrow x=\frac{26}{33}-\frac{-15}{33}$ $\Rightarrow x=\frac{26-(-15)}{33}$ $\Rightarrow x=\frac{26+15}{33}$ $\Rightarrow x=\frac{41}{33}$...
Read More →If two tangents are drawn to a circle from an external point,show
Question: If two tangents are drawn to a circle from an external point,show that they subtend equal angles at the centre. Solution: Given : A circle with centre O and a pointAoutside it. Also,APandAQare the two tangents to the circle. :To prove: $\angle A O P=\angle A O Q$. Proof : In $\Delta A O P$ and $\Delta A O Q$, we have : $A P=A Q \quad$ [tangents from an external point are equal] $O P=O Q \quad$ [radii of the same circle] $O A=O A \quad[$ common side $]$ $\therefore \Delta A O P \cong \D...
Read More →What should be added to
Question: What should be added to $\frac{-7}{8}$ so as to get $\frac{5}{9} ?$ Solution: Let $x$ be added to $\frac{-7}{8}$ so as to get $\frac{5}{9}$. $\therefore x+\frac{-7}{8}=\frac{5}{9}$ $\Rightarrow x=\frac{5}{9}-\frac{-7}{8}$ $\Rightarrow x=\frac{40}{72}-\frac{-63}{72}$ $\Rightarrow x=\frac{40-(-63)}{72}$ $\Rightarrow x=\frac{40+63}{72}=\frac{103}{72}$...
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