The sum of three numbers in G.P. is 14. If the first two terms are each increased by 1 and the third term decreased by 1,

Question: The sum of three numbers in G.P. is 14. If the first two terms are each increased by 1 and the third term decreased by 1, the resulting numbers are in A.P. Find the numbers. Solution: Let the numbers be $a$, ar and $a r^{2}$. Sum $=14$ $\Rightarrow a+a r+a r^{2}=14$ $\Rightarrow a\left(1+r+r^{2}\right)=14$ ...(1) According to the question,a + 1, ar + 1andar2 1 are in A.P. $\therefore 2(a r+1)=a+1+a r^{2}-1$ $\Rightarrow 2 a r+2=a+a r^{2}$ $\Rightarrow 2 a r+2=14-a r \quad[$ From (i) $]...

A box contains 90 discs which are numbered from 1 to 90.

Question: A box contains 90 discs which are numbered from 1 to 90. If one discs is drawn at random from the box. find the probability that it bears (i) a two digit number (ii) a perfect square number (iii) a number divisible by 5. Solution: GIVEN: A box contains 90 discs which are numbered from 1 to 90.If one disc is drawn at random from the box TO FIND: Probability that it bears (i) a two digit number (ii) a perfect square (iii) a number divisible by 5 Total number of discs numbered is 90 (i) d...

∆ABC and ∆BDE are two equilateral triangles such that D is the midpoint of BC.

Question: ∆ABCand ∆BDEare two equilateral triangles such thatDis the midpoint ofBC. Then, ar(∆BDE) : ar(∆ABC) = ? (a) $1: 2$ (b) $1: 4$ (c) $\sqrt{3}: 2$ (d) $3: 4$ Solution: (b) 1:4 ∆ABCand ∆BDEare two equilateral triangles​ andDis the midpoint ofBC.LetAB=BC=AC= a Then $\mathrm{BD}=B E=E D=\frac{a}{2}$ $\therefore \frac{\operatorname{ar}(\Delta B D E)}{\operatorname{ar}(\Delta A B C)}=\frac{\frac{\sqrt{3}}{4} A B^{2}}{\frac{\sqrt{3}}{4} B E^{2}}=\frac{\left(\frac{a}{2}\right)^{2}}{a^{2}}=\frac{...

Question: If $A=\left[\begin{array}{ll}1 0 \\ 0 1\end{array}\right], B=\left[\begin{array}{rr}1 0 \\ 0 -1\end{array}\right]$ and $C=\left[\begin{array}{ll}0 1 \\ 1 0\end{array}\right]$ then show that $A^{2}=B^{2}=C^{2}=l_{2}$. Solution: Here, $A^{2}=A A$ $\Rightarrow A^{2}=\left[\begin{array}{ll}1 0 \\ 0 1\end{array}\right]\left[\begin{array}{ll}1 0 \\ 0 1\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{ll}1+0 0+0 \\ 0+0 0+1\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{ll}...

(i) A lot of 20 bulbs contain 4 defective ones.

Question: (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? Solution: GIVEN: A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot TO FIND: Probability that it is defective Total number of bulbs is 20 (i) Total numbers of bulbs which are defective is 4 We know that PROBABILITY = Hence probabilities of bulbs which are defective is (ii) Suppose the bulb drawn in case (i) is n...

The sum of first three terms of a G.P.

Question: The sum of first three terms of a G.P. is $\frac{39}{10}$ and their product is 1 . Find the common ratio and the terms. Solution: Let the terms of the G.P be $\frac{a}{r}, a$ and $a r$. $\therefore$ Product of the G.P. $=1$ $\Rightarrow a^{3}=1$ $\Rightarrow a=1$ Now, sum of the G.P. $=\frac{39}{10}$ $\Rightarrow \frac{a}{r}+a+a r=\frac{39}{10}$ $\Rightarrow a\left(\frac{1}{r}+1+r\right)=\frac{39}{10}$ $\Rightarrow 1\left(\frac{1}{r}+1+r\right)=\frac{39}{10}$ $\Rightarrow 10 r^{2}+10 r...

A box contains 5 red marbels, 8 white marbles and 4 green marbles,

Question: A box contains 5 red marbels, 8 white marbles and 4 green marbles, One marble is taken out of the box at ramdom. What is the probability that the marble taken out will be (i) red? (ii) white? (iii) not green? Solution: GIVEN: A box contains 5 red, 4 green and 8 white marbles and a marble is drawn at random TO FIND: Probability of getting a marble (i) red (ii) white (iii) not green Total number of marble: (i) Total number red marble are 5 We know that PROBABILITY = Hence probability of ...

The product of three numbers in G.P.

Question: The product of three numbers in G.P. is 125 and the sum of their products taken in pairs is $871 / 2$. Find them. Solution: Let the required numbers be $\frac{a}{r}, a$ and $a r$. Product of the G.P. = 125 $\Rightarrow a^{3}=125$ $\Rightarrow a=5$ Sum of the products in pairs $=87 \frac{1}{2}=\frac{175}{2}$ $\Rightarrow \frac{a}{r} \times a+a \times a r+a r \times \frac{a}{r}=\frac{175}{2}$ $\Rightarrow \frac{a^{2}}{r}+a^{2} r+a^{2}=\frac{175}{2}$ Subs tituting the value of $a$ $\Right...

Evaluate the following

Question: Evaluate the following (i) $\left(\left[\begin{array}{rr}1 3 \\ -1 -4\end{array}\right]+\left[\begin{array}{rr}3 -2 \\ -1 1\end{array}\right]\right)\left[\begin{array}{lll}1 3 5 \\ 2 4 6\end{array}\right]$ (ii) $\left[\begin{array}{lll}1 2 3\end{array}\right]\left[\begin{array}{lll}1 0 2 \\ 2 0 1 \\ 0 1 2\end{array}\right]\left[\begin{array}{l}2 \\ 4 \\ 6\end{array}\right]$ (iii) $\left[\begin{array}{rr}1 -1 \\ 0 2 \\ 2 3\end{array}\right]\left(\left[\begin{array}{lll}1 0 2 \\ 2 0 1\en...

In the given figure ABCD and ABFE are parallelograms such

Question: In the given figureABCDandABFEare parallelograms such that ar(quad.EABC) = 17 cm2and ar(||gmABCD) = 25 cm2. Then, ar(∆BCF) = ? (a) $4 \mathrm{~cm}^{2}$ (b) $4.8 \mathrm{~cm}^{2}$ (c) $6 \mathrm{~cm}^{2}$ (d) $8 \mathrm{~cm}^{2}$ Solution: (d) $8 \mathrm{~cm}^{2}$ Since ||gmABCDand||gmABFEare on the same base and between the same parallel lines, we have: $\operatorname{ar}(\| g m A B F E)=\operatorname{ar}(\| g m A B C D)=25 \mathrm{~cm}^{2}$ $\Rightarrow \operatorname{ar}(\Delta B C F)...

A bag contains 3 red balls and 5 black balls. A ball is draw at random from the bag.

Question: A bag contains 3 red balls and 5 black balls. A ball is draw at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red? Solution: GIVEN: A bag contains 3 red and 5 black balls and a ball is drawn at random from the bag TO FIND: Probability of getting a (i) red ball (ii) not red ball Total number of balls (i) Total number red balls are 3 We know that PROBABILITY = Hence probability of getting red ball is (ii) Probability of getting red ball We know tha...

ABCD is a rhombus in which ∠C = 60°.

Question: ABCDis a rhombus in whichC= 60. Then,AC:BD= ? (a) $\sqrt{3}: 1$ (b) $\sqrt{3}: \sqrt{2}$ (c) $3: 1$ (d) $3: 2$ Solution: (a) $\sqrt{3}: 1$ ABCDis a rhombus. So all of its sides are equal.Now,BC=DC⇒ BDC=DBC=xo (Angles opposite to equal sides are equal)Also,BCD= 60oxo+xo+ 60o= 180o⇒​ 2xo= 120o⇒​xo=60oi.e., BCD =BDC=DBC=60oSo,​∆BCDis an equilateral triangle.BD=BC=aAlso,OB=a/2Now, in ∆OAB, we have: $A B^{2}=O A^{2}+O B^{2}$ $\Rightarrow O A^{2}=A B^{2}-O B^{2}=a^{2}-\left(\frac{a}{2}\right...

The sum of first three terms of a G.P. is 13/12 and their product is − 1.

Question: The sum of first three terms of a G.P. is 13/12 and their product is 1. Find the G.P. Solution: Let the first three numbers of the given G.P. be $\frac{a}{r}, a$ and $a r$. Product of the G.P. = 1 = a3= 1 = a = 1 Similarly, Sum of the G.P. $=\frac{13}{12}$ $\Rightarrow \frac{a}{r}+a+a r=\frac{13}{12}$ Substituting the value ofa= 1 $\frac{-1}{r}-1-r=\frac{13}{12}$ $\Rightarrow 12 r^{2}+25 r+12=0$ $\Rightarrow 12 r^{2}+16 r+9 r+12=0$ $\Rightarrow 4 r(3 r+4)+3(3 r+4)=0$ $\Rightarrow(4 r+3...

It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992

Question: It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday? Solution: Given: Probability that 2 students should not have the same birthday TO FIND: Probability that 2 students should have the same birthday CALCULATION:We know that sum of probability of occurrence of an event and probability of non occurrence of an event is 1. $P(E)+P(\bar{E})=1$ $0.992+P(\bar{E})=1$ $...

Show that AB ≠ BA in each of the following cases:

Question: Show thatABBAin each of the following cases: (i) $A=\left[\begin{array}{rrr}1 3 -1 \\ 2 -1 -1 \\ 3 0 -1\end{array}\right]$ and $B=\left[\begin{array}{lll}-2 3 -1 \\ -1 2 -1 \\ -6 9 -4\end{array}\right]$ (ii) $A=\left[\begin{array}{rrr}10 -4 -1 \\ -11 5 0 \\ 9 -5 1\end{array}\right]$ and $B=\left[\begin{array}{lll}1 2 1 \\ 3 4 2 \\ 1 3 2\end{array}\right]$ Solution: (i) $A B=\left[\begin{array}{ccc}1 3 -1 \\ 2 -1 -1 \\ 3 0 -1\end{array}\right]\left[\begin{array}{ccc}-2 3 -1 \\ -1 2 -1 \...

Find three numbers in G.P. whose sum is 38 and their product is 1728.

Question: Find three numbers in G.P. whose sum is 38 and their product is 1728. Solution: Let the terms of the the given G.P. be $\frac{a}{r}, a$ and $a r$. Then, product of the G.P. = 1728 = a3= 1728 = a = 12 Similarly, sum of the G.P. = 38 $\Rightarrow \frac{a}{r}+a+a r=38$ Substituting the value ofa $\frac{12}{r}+12+12 r=38$ $\Rightarrow 12 r^{2}+12 r+12=38 r$ $\Rightarrow 12 r^{2}-26 r+12=0$ $\Rightarrow 2\left(6 r^{2}-13 r+6\right)=0$ $\Rightarrow 6 r^{2}-13 r+6=0$ $\Rightarrow(3 r-2)(2 r-3...

A bag contains lemon flavoured candles only. Malini takes out one

Question: A bag contains lemon flavoured candles only. Malini takes out one candy without looking into the bag. what is the probability that she takes out(i) an orange flavoured candy?(ii) a lemon flavoured candy? Solution: GIVEN: A bag contains lemon flavored candies only. Malini takes out one candy without looking into the bag, TO FIND: Probability that she takes out (i) An orange flavored candy (ii) A lemon flavored candy (i) Probability of taking out orange flavored candy isas it is an impos...

From a pack of 52 playing cards Jacks, queens, Kings and aces of red colour are removed.

Question: From a pack of 52 playing cards Jacks, queens, Kings and aces of red colour are removed. From the remaining, a card is drawn at random. Find the probability that the card drawn is(i) a black queen(ii) a red card(iii) a black jack(iv) a picture card (Jacks, queens and Kings are picture cards). Solution: Given: The Kings, Queens, Aces and Jacks of red color are removed from a deck of 52 playing cards and the remaining cards are shuffled and a card is drawn at random from the remaining ca...

Find three numbers in G.P. whose sum is 65 and whose product is 3375.

Question: Find three numbers in G.P. whose sum is 65 and whose product is 3375. Solution: Let the terms of the the given G.P. be $\frac{a}{r}, a$ and $a r$. Then, product of the G.P. = 3375 = a3= 3375 = a = 15 Similarly, sum of the G.P. = 65 $\Rightarrow \frac{a}{r}+a+a r=65$ Substituting the value ofa $\frac{15}{r}+15+15 r=65$ $\Rightarrow 15 r^{2}+15 r+15=65 r$ $\Rightarrow 15 r^{2}-50 r+15=0$ $\Rightarrow 5\left(3 r^{2}-10 r+3\right)=0$ $\Rightarrow 3 r^{2}-10 r+3=0$ $\Rightarrow(3 r-1)(r-3)=...

Find the probability that a number selected at random from the numbers 1, 2, 3, ..., 35 is a

Question: Find the probability that a number selected at random from the numbers 1, 2, 3, ..., 35 is a(i) prime number(ii) multiple of 7(iii) a multiple of 3 or 5 Solution: GIVEN: A number is selected from numbers 1 to 35 TO FIND: Probability of getting a number (i) which is a prime number (ii) multiple of 7 (iii) multiple of 3 or 5 Total number of cards is 35 (i) Numbers that are primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31 Total prime numbers from 1 to 35 are11 We know that PROBABILITY...

Compute the products $A B$ and $B A$ whichever exists in each of the following cases:

Question: Compute the products $A B$ and $B A$ whichever exists in each of the following cases: (i) $A=\left[\begin{array}{rr}1 -2 \\ 2 3\end{array}\right]$ and $B=\left[\begin{array}{lll}1 2 3 \\ 2 3 1\end{array}\right]$ (ii) $A=\left[\begin{array}{rr}3 2 \\ -1 0 \\ -1 1\end{array}\right]$ and $B=\left[\begin{array}{lll}4 5 6 \\ 0 1 2\end{array}\right]$ (iii) $A=\left[\begin{array}{lll}1 -1 2 3\end{array}\right]$ and $B=\left[\begin{array}{l}0 \\ 1 \\ 3 \\ 2\end{array}\right]$ (iv) $[a, b]\left...

A bag contains 8 red, 6 white and 4 black balls. A ball is drawn at random from the bag.

Question: A bag contains 8 red, 6 white and 4 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is(i) red or white(ii) not black(iii) neither white nor black. Solution: GIVEN: A bag contains 8 red, 4 black and 6 white balls and a ball is drawn at random TO FIND: Probability of getting a (i) red or white ball (ii) not black ball (iii) neither white nor black Total number of balls (i) Total number red and white balls are We know that PROBABILITY = Hence ...

Find the probability that a number selected from the number

Question: Find the probability that a number selected from the number 1 to 25 is not a prime number when each of the given numbers is equally likely to be selected. Solution: GIVEN: A number is selected from numbers 1 to 25 TO FIND: Probability of getting a number which is not a prime. Total number of cards is 25.Total number of elementary events = 25 Cards bearing non prime numbers are 1,4,6,8,9,10,12,14,15,16,18,20,21,22,24, 25 Total number of cards bearing non-prime numbers = 16Number of favo...

In a quadrilateral ABCD, it is given that BD = 16 cm.

Question: In a quadrilateralABCD, it is given thatBD= 16 cm. IfALBDandCMBDsuch thatAL= 9 cm andCM= 7 cm, then ar(quad.ABCD)= ? (a) $256 \mathrm{~cm}^{2}$ (b) $128 \mathrm{~cm}^{2}$ (c) $64 \mathrm{~cm}^{2}$ (d) $96 \mathrm{~cm}^{2}$ Solution: (b) $128 \mathrm{~cm}^{2}$ ar(quadABCD) = ar (∆ABD) +​ar(∆DBC)​We have the following: $\operatorname{ar}(\Delta A B D)=\frac{1}{2} \times$ base $\times$ height $=\frac{1}{2} \times 16 \times 9=72 \mathrm{~cm}^{2}$ $\operatorname{ar}(\Delta D B C)=\frac{1}{2...

If the pth and qth terms of a G.P. are q and p, respectively,

Question: If the $p$ th and qth terms of a G.P. are $q$ and $p$, respectively, then show that $(p+q)$ th term is $\left(\frac{q^{p}}{p^{q}}\right)^{\frac{1}{p-q}}$. Solution: As, $a_{p}=q$ $\Rightarrow a r^{(p-1)}=q \quad \ldots$ (i) Also, $a_{q}=p$ $\Rightarrow a r^{(q-1)}=p \quad \ldots \ldots$ (ii) Dividing (i) by (ii), we get $\frac{a r^{(p-1)}}{a r^{(q-1)}}=\frac{q}{p}$ $\Rightarrow r^{(p-1-q+1)}=\frac{q}{p}$ $\Rightarrow r^{(p-q)}=\frac{q}{p}$ $\Rightarrow r=\left(\frac{q}{p}\right)^{\frac...