A bag contains tickets numbered 11, 12, 13, ..., 30.
Question: A bag contains tickets numbered 11, 12, 13, ..., 30. A ticket is taken out from the bag at random. Find the probability that the number on the drawn ticket (i) is a multiple of 7 (ii) is greater than 15 and a multiple of 5. Solution: GIVEN: Tickets are marked with one of the numbers 11, 12, 1330 are placed in a bag and mixed thoroughly. One ticket is picked at random. TO FIND: Probability of getting (i) multiple of 7 (ii) greater than15 and multiple of 5 Total number of cards is (i) Nu...
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Question: If $A=\left[\begin{array}{rrr}2 -3 -5 \\ -1 4 5 \\ 1 -3 -4\end{array}\right]$ and $B=\left[\begin{array}{rrr}-1 3 5 \\ 1 -3 -5 \\ -1 3 5\end{array}\right]$, show that $A B=B A=O_{3 \times 3}$. Solution: Here, $A B=\left[\begin{array}{ccc}2 -3 -5 \\ -1 4 5 \\ 1 -3 -4\end{array}\right]\left[\begin{array}{ccc}-1 3 5 \\ 1 -3 -5 \\ -1 3 5\end{array}\right]$ $\Rightarrow A B=\left[\begin{array}{ccc}-2-3+5 6+9-15 10+15-25 \\ 1+4-5 -3-12+15 -5-20+25 \\ -1-3+4 3+9-12 5+15-20\end{array}\right]$ ...
Read More →A bag contains 6 red balls and some blue balls.
Question: A bag contains 6 red balls and some blue balls. if the probability of a drawing a blue ball from the bag is twice that of a red ball, find the number of blue balls in the bag. Solution: GIVEN: A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball from the bag is twice that of a red ball, TO FIND: the number of blue balls in the bag. Let the probability of getting a red ball be The probability of not getting a red ball or getting a blue ball be We kno...
Read More →Cards marked with numbers 13, 14, 15, ..., 60 are placed in a box and mixed thoroughly
Question: Cards marked with numbers 13, 14, 15, ..., 60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that number on the card drawn is(i) divisible by 5(ii) a number is a perfect square Solution: GIVEN: Cards are marked with numbers 13 to 60 are placed in a box and mixed thoroughly. If one card is drawn at random from the box TO FIND: Probability that it bears (i) A number divisible by 5 (ii) A perfect square Total number of cards is (i)...
Read More →In ∆ABC, if D is the midpoint of BC and E is the midpoint of AD, then ar(∆BED) = ?
Question: In∆ABC, ifDis the midpoint ofBCandEis the midpoint ofAD, then ar(∆BED) = ? (a) $\frac{1}{2} \operatorname{ar}(\Delta A B C)$ (b) $\frac{1}{3} \operatorname{ar}(\Delta A B C)$ (c) $\frac{1}{4} \operatorname{ar}(\Delta A B C)$ (d) $\frac{2}{3} \operatorname{ar}(\Delta A B C)$ Solution: (c) $\frac{1}{4} \operatorname{ar}(\triangle A B C)$ SinceDis the mid point ofBC,ADis a median of ∆ABCandBEis the median of ∆ABD.We know that a median of a triangle divides it into two triangles of equal a...
Read More →A group consists of 12 persons, of which 3 are extremely patient,
Question: A group consists of 12 persons, of which 3 are extremely patient, other 6 are extremely honest and rest are extremely kind. A person form the group is selected at random. Assuming that each person is equally likely to be selected, find the probability of selecting a person who is (i) extremely patient (ii) extremely kind or honest. Which of the above you prefer more. [CBSE 2013] Solution: Number of persons in the group = 12 Total number of outcomes = 12(i) Number of persons who are ext...
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Question: If $A=\left[\begin{array}{rr}\cos 2 \theta \sin 2 \theta \\ -\sin 2 \theta \cos 2 \theta\end{array}\right]$, find $A^{2}$ Solution: Given : $A=\left[\begin{array}{cc}\cos 2 \theta \sin 2 \theta \\ -\sin 2 \theta \cos 2 \theta\end{array}\right]$ Now, $A^{2}=A A$ $\Rightarrow A^{2}=\left[\begin{array}{cc}\cos 2 \theta \sin 2 \theta \\ -\sin 2 \theta \cos 2 \theta\end{array}\right]\left[\begin{array}{cc}\cos 2 \theta \sin 2 \theta \\ -\sin 2 \theta \cos 2 \theta\end{array}\right]$ $\Right...
Read More →The sum of three numbers in G.P. is 21 and the sum of their squares is 189.
Question: The sum of three numbers in G.P. is 21 and the sum of their squares is 189. Find the numbers. Solution: Let the required numbers be $a, a r$ and $a r^{2}$. Sum of the numbers = 21 $\Rightarrow a+a r+a r^{2}=21$ $\Rightarrow a\left(1+r+r^{2}\right)=21$ ...(1) Sum of the squares of the numbers = 189 $\Rightarrow a^{2}+(a r)^{2}+\left(a r^{2}\right)^{2}=189$ $\Rightarrow a^{2}+(a r)^{2}+\left(a r^{2}\right)^{2}=189$ $\Rightarrow a^{2}\left(1+r^{2}+r^{4}\right)=189$ ....(2) Now, $a\left(1+...
Read More →A box contains cards numbered 3, 5, 7, 9, ..., 35, 37. A card is drawn at random form the box.
Question: A box contains cards numbered 3, 5, 7, 9, ..., 35, 37. A card is drawn at random form the box. Find the probability that the number on the drawn card is a prime number. Solution: The numbers 3, 5, 7, 9, ..., 35, 37 are in AP.Here,a= 3 andd= 5 3 = 2Suppose there arenterms in the AP. $\therefore a_{n}=37$ $\Rightarrow 3+(n-1) \times 2=37 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow 2 n+1=37$ $\Rightarrow 2 n=37-1=36$ $\Rightarrow n=18$ Total number of outcomes = 18 Let E be the event ...
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Question: If $A=\left[\begin{array}{cc}a b b^{2} \\ -a^{2} -a b\end{array}\right]$, show that $A^{2}=0$ Solution: Given : $A=\left[\begin{array}{cc}a b b^{2} \\ -a^{2} -a b\end{array}\right]$ Now, $A^{2}=A A$ $\Rightarrow A^{2}=\left[\begin{array}{cc}a b b^{2} \\ -a^{2} -a b\end{array}\right]\left[\begin{array}{cc}a b b^{2} \\ -a^{2} -a b\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{cc}a^{2} b^{2}-a^{2} b^{2} a b^{3}-a b^{3} \\ -a^{3} b+a^{3} b -a^{2} b^{2}+a^{2} b^{2}\end{array}\ri...
Read More →A number is selected at random from first 50 natural numbers.
Question: A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4. Solution: Total number of outcomes = 50Let E be the event of getting a number which is a multiple of 3 and 4.Now, the common multiples of 3 and 4 among first 50 natural numbers are 12, 24, 36 and 48.So, the favourable number of outcomes are 4. $\therefore$ Required probability $=\mathrm{P}(\mathrm{E})=\frac{\text { Favourable number of outcomes }}{\text { Total number of...
Read More →A box contains 100 red cards, 200 yellow cards and 50 blue cards.
Question: A box contains 100 red cards, 200 yellow cards and 50 blue cards. If a card is drawn at random from the box, then find the probability that it will be (i) a blue card (ii) not a yellow card (iii) neither yellow nor a blue card. [CBSE 2012] Solution: Total number of cards = 100 + 200 + 50 = 350 Total number of outcomes = 350(i) Number of blue cards = 50So, the number of favourable outcomes are 50. $\therefore P($ drawing a blue card $)=\frac{\text { Favourable number of outcomes }}{\tex...
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Question: If $A=\left[\begin{array}{ll}1 1 \\ 0 1\end{array}\right]$, show that $A^{2}=\left[\begin{array}{ll}1 2 \\ 0 1\end{array}\right]$ and $A^{3}=\left[\begin{array}{ll}1 3 \\ 0 1\end{array}\right]$. Solution: Given : $A=\left[\begin{array}{ll}1 1 \\ 0 1\end{array}\right]$ Now, $A^{2}=A A$ $\Rightarrow A^{2}=\left[\begin{array}{ll}1 1 \\ 0 1\end{array}\right]\left[\begin{array}{ll}1 1 \\ 0 1\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{ll}1+0 1+1 \\ 0+0 0+1\end{array}\right]$ $...
Read More →The figure formed by joining the midpoints of the adjacent sides of a rectangle of sides 8 cm and 6 cm is a
Question: The figure formed by joining the midpoints of the adjacent sides of a rectangle of sides 8 cm and 6 cm is a (a) rectangle of area $24 \mathrm{~cm}^{2}$ (b) square of area $24 \mathrm{~cm}^{2}$ (c) trapezium of area $24 \mathrm{~cm}^{2}$ (d) rhombus of area $24 \mathrm{~cm}^{2}$ Solution: (d) rhombus of $24 \mathrm{~cm}^{2}$ We know that the figure formed by joining the midpoints of the adjacent sides of a rectangle is a rhombus.So,PQRSis a rhombus andSQandPRare its diagonals.i.e.,SQ= 8...
Read More →Two dice, one blue and one grey, are thrown at the same time. Complete the following table:
Question: Two dice, one blue and one grey, are thrown at the same time. Complete the following table: From the above table a student argues that there are 11 possible outcomes $2,3,4,5,6,7,8,9,10,11$ and 12 . Therefore, each of them has a probability $\frac{1}{11}$. Do you agree with this argument? Solution: GIVEN: Two dice are thrown TO FIND: Probability of the following: Let us first write the all possible events that can occur (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,...
Read More →Find three numbers in G.P. whose product is 729 and the sum of their products in pairs is 819.
Question: Find three numbers in G.P. whose product is 729 and the sum of their products in pairs is 819. Solution: Let the required numbers be $\frac{a}{r}, a$ and $a r$. Product of the G.P. = 729 $\Rightarrow a^{3}=729$ $\Rightarrow a=9$ Sum of the products in pairs = 819 $\Rightarrow \frac{a}{r} \times a+a \times a r+a r \times \frac{a}{r}=819$ $\Rightarrow a^{2}\left(\frac{1}{r}+r+1\right)=819$ $\Rightarrow 81\left(\frac{1+r^{2}+r}{r}\right)=819$ $\Rightarrow 9\left(r^{2}+r+1\right)=91 r$ $\R...
Read More →Harpreet tosses two different coins simultaneously
Question: Harpreet tosses two different coins simultaneously (say, one is of Re 1 and other of Rs 2). What is the probability that he gets at least one head? Solution: GIVEN: Harpreet tosses two different coins simultaneously (Re1 and Rs2) TO FIND: Probability of getting at least one head. When two coins are tossed then the outcome will be TT, HT, TH, HH. Hence total number of outcome is 4. At least one head means 1H or 2H Hence total number of favorable outcome i.e. at least one head is 3 We kn...
Read More →In a || gm ABCD, if P and Q are midpoints of AB and CD respectively and ar(|| gm ABCD) = 16 cm2,
Question: In a $\| \mathrm{gm} A B C D$, if $P$ and $Q$ are midpoints of $A B$ and $C D$ respectively and $\operatorname{ar}(\| \operatorname{gm} A B C D)=16 \mathrm{~cm}^{2}$, then $\operatorname{ar}(\| \operatorname{gm} A P Q D)=?$ (a) $8 \mathrm{~cm}^{2}$ (b) $12 \mathrm{~cm}^{2}$ (c) $6 \mathrm{~cm}^{2}$ (d) $9 \mathrm{~cm}^{2}$ Solution: (a) $8 \mathrm{~cm}^{2}$ Let the distance betweenABandCDbehcm.Thenar(||gmAPQD) =APh $=\frac{1}{2} \times A B \times h \quad\left(\mathrm{AP}=\frac{1}{2} ...
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Question: If $A=\left[\begin{array}{rr}4 2 \\ -1 1\end{array}\right]$, prove that $(A-2 D)(A-3 l)=0$ Solution: Given: $(A-2 I)(A-3 I)$ $\Rightarrow(A-2 I)(A-3 I)=\left(\left[\begin{array}{cc}4 2 \\ -1 1\end{array}\right]-2\left[\begin{array}{ll}1 0 \\ 0 1\end{array}\right]\right)\left(\left[\begin{array}{cc}4 2 \\ -1 1\end{array}\right]-3\left[\begin{array}{ll}1 0 \\ 0 1\end{array}\right]\right)$ $\Rightarrow(A-2 I)(A-3 I)=\left(\left[\begin{array}{cc}4 2 \\ -1 1\end{array}\right]-\left[\begin{a...
Read More →The product of three numbers in G.P. is 216. If 2, 8, 6 be added to them,
Question: The product of three numbers in G.P. is 216. If 2, 8, 6 be added to them, the results are in A.P. Find the numbers. Solution: Let the terms of the given G.P. be $\frac{a}{r}, a$ and $a r$. Product =216 $\Rightarrow a^{3}=216$ $\Rightarrow a=6$ It is given that $\frac{a}{r}+2, a+8$ and $a r+6$ are in A.P. $\therefore 2(a+8)=\frac{a}{r}+2+a r+6$ Putting $a=6$, we get $\Rightarrow 28=\frac{6}{r}+2+6 r+6$ $\Rightarrow 28 r=6 r^{2}+8 r+6$ $\Rightarrow 6 r^{2}-20 r+6=0$ $\Rightarrow(6 r-2)(r...
Read More →Five cards − the ten, jack, queen, king and ace of diamonds,
Question: Five cards the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.(i) What is the probability that the card is the queen?(ii) If the queen is drawn and put a side, what is the probability that the second card picked up is (a) an ace? (b) a queen? Solution: GIVEN: Five cards-the ten, jack, queen, king and ace of diamond are well shuffled with their face downwards of cards. One card is drawn at random. TO FIND: Pr...
Read More →The product of three numbers in G.P. is 216. If 2, 8, 6 be added to them,
Question: The product of three numbers in G.P. is 216. If 2, 8, 6 be added to them, the results are in A.P. Find the numbers. Solution: Let the terms of the given G.P. be $\frac{a}{r}, a$ and $a r$. Product =216 $\Rightarrow a^{3}=216$ $\Rightarrow a=6$ It is given that $\frac{a}{r}+2, a+8$ and $a r+6$ are in A.P. $\therefore 2(a+8)=\frac{a}{r}+2+a r+6$ Putting $a=6$, we get $\Rightarrow 28=\frac{6}{r}+2+6 r+6$ $\Rightarrow 28 r=6 r^{2}+8 r+6$ $\Rightarrow 6 r^{2}-20 r+6=0$ $\Rightarrow(6 r-2)(r...
Read More →12 defective pens are accidently mixed with 132 good ones. It is not possible to just
Question: 12 defective pens are accidently mixed with 132 good ones. It is not possible to just look at pen and tell whether or not it is defective. one pen is taken out at random from this lot. Determine the probability that the pen taken out is good one. Solution: GIVEN: 12 defective pens are accidently mixed with 132 good ones. One pen is taken out at random from this lot. TO FIND: Probability that the pen taken out is good. Total number of bulbs is Total numbers of bulbs which are good are 1...
Read More →A lot consists of 144 ball pens of which 20 are defective and others good.
Question: A lot consists of 144 ball pens of which 20 are defective and others good. Nutri will buy a pen if it is good, but will not buy if it is defective. the shopkeeper draws one pen at random and gives it to her. What is the probability that(i) She will buy it?(ii) She will not buy it? Solution: GIVEN: A lot consists of 144 ball pens of which 20 are defective and others good Nuri will buy a pen if it is good but will not buy if it is defective. The shop keeper draws one pen at random and gi...
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Question: If $A=\left[\begin{array}{rr}2 -1 \\ 3 2\end{array}\right]$ and $B=\left[\begin{array}{rr}0 4 \\ -1 7\end{array}\right]$, find $3 A^{2}-2 B+1$ Solution: Given : $A=\left[\begin{array}{cc}2 -1 \\ 3 2\end{array}\right]$ Now, $A^{2}=A A$ $\Rightarrow A^{2}=\left[\begin{array}{cc}2 -1 \\ 3 2\end{array}\right]\left[\begin{array}{cc}2 -1 \\ 3 2\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{cc}4-3 -2-2 \\ 6+6 -3+4\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{cc}1 -4 \...
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