If 5th, 8th and 11th terms of a G.P. are p. q and s respectively,
Question: If 5th, 8thand 11thterms of a G.P. arep.qandsrespectively, prove thatq2=ps. Solution: Let $a$ be the first term and $r$ be the common ratio of the given G.P. $\therefore p=5^{\text {th }}$ term $\Rightarrow p=a r^{4}$ ...(1) $q=8^{\text {th }}$ term $\Rightarrow q=a r^{7}$ ...(2) $s=11^{\text {th }}$ $\Rightarrow s=a r^{10}$ ...(3) Now, $q^{2}=\left(a r^{7}\right)^{2}=a^{2} r^{14}$ $\Rightarrow\left(a r^{4}\right)\left(a r^{10}\right)=p s \quad[$ From $(1)$ and $(3)]$ $\therefore q^{2}...
Read More →In a lottery there are 10 prizes and 25 blanks.
Question: In a lottery there are 10 prizes and 25 blanks. What is the probability of getting a prize? Solution: GIVEN: In a lottery there are 10 prizes and 25 blanks. TO FIND: Probability of winning a prize Total number of tickets is Total number of prize carrying tickets is 10 We know that PROBABILITY = Hence probability of winning a prize is...
Read More →Tickets numbers from 1 to 20 are mixed up and a ticket is
Question: Tickets numbers from 1 to 20 are mixed up and a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 7? Solution: GIVEN: Tickets are marked from 1 to 20 are mixed up. One ticket is picked at random. TO FIND: Probability that the ticket bears a multiple of 3 or 7 Total number of cards is 20 Cards marked multiple of 3 or 7 are 3, 6, 7, 9, 12, 14, 15 and18 Total number of cards marked multiple of 3 or 7 are 8 We know that PROBAB...
Read More →If the G.P.'s 5, 10, 20, ... and 1280, 640, 320, ... have their nth terms equal,
Question: If the G.P.'s 5, 10, 20, ... and 1280, 640, 320, ... have theirnth terms equal, find the value ofn. Solution: Given : First term, $a=5$ Common ratio, $r=2$ $a_{n}=(5)(2)^{n-1}$ .. (1) Similarly, $a_{n}=(1280)\left(\frac{1}{2}\right)^{n-1} \ldots(2)$ From $(1)$ and $(2)$ $(5)(2)^{n-1}=(1280)\left(\frac{1}{2}\right)^{n-1}$ $\Rightarrow \frac{1}{256}=\left(\frac{1}{4}\right)^{n-1}$ $\Rightarrow\left(\frac{1}{4}\right)^{4}=\left(\frac{1}{4}\right)^{n-1}$ $\Rightarrow n-1=4$ $\Rightarrow n=...
Read More →In the given figure, ABCD and ABPQ are two parallelograms and M is a point on AQ and BMP is a triangle.
Question: In the given figure,ABCDandABPQare two parallelograms andMis a point onAQandBMPis a triangle. Then, $\operatorname{ar}(\Delta B M P)=\frac{1}{2} \operatorname{ar}(\| \operatorname{gm} A B C D)$ is (a) true(b) false Solution: We know parallelogram on the same base and between the same parallels are equal in area.Here, AB is the common base and AB || PDHence, ar(ABCD) = ar(ABPQ) .....(1)Also, when a triangle and a parallelogram are on the same base and between the same parallels then the...
Read More →A bag contains 5 white and 7 red balls.
Question: A bag contains 5 white and 7 red balls. One ball is drawn at random. What is the probability that ball drawn is not black? Solution: GIVEN: A bag contains 7 red, and 5 white balls and a ball is drawn at random TO FIND: Probability that the ball drawn is white Total number of balls Total number of white balls is 5 We know that PROBABILITY = Probability of getting a white ball...
Read More →A bag contains 5 white and 7 red balls.
Question: A bag contains 5 white and 7 red balls. One ball is drawn at random. What is the probability that ball drawn is not black? Solution: GIVEN: A bag contains 6 red, 8 black and 4 white balls and a ball is drawn at random TO FIND: Probability that the ball drawn is not black Total number of balls Total number of black balls is 8 We know that PROBABILITY = Probability of getting a black ball We know that sum of probability of occurrence of an event and probability of non occurrence of an ev...
Read More →Show that AB ≠ BA in each of the following cases:
Question: Show thatABBAin each of the following cases: (i) $A=\left[\begin{array}{rr}5 -1 \\ 6 7\end{array}\right]$ and $B=\left[\begin{array}{ll}2 1 \\ 3 4\end{array}\right]$ (ii) $A=\left[\begin{array}{rrr}-1 1 0 \\ 0 -1 1 \\ 2 3 4\end{array}\right]$ and $B=\left[\begin{array}{lll}1 2 3 \\ 0 1 0 \\ 1 1 0\end{array}\right]$ (iii) $A=\left[\begin{array}{lll}1 3 0 \\ 1 1 0 \\ 4 1 0\end{array}\right]$ and $B=\left[\begin{array}{lll}0 1 0 \\ 1 0 0 \\ 0 5 1\end{array}\right]$ Solution: (i) $A B=\lef...
Read More →The seventh term of a G.P. is 8 times the fourth term and 5th term is 48.
Question: The seventh term of a G.P. is 8 times the fourth term and 5th term is 48. Find the G.P. Solution: Let $a$ be the first term and $r$ be the common ratio. $\therefore a_{7}=8 a_{4}$ and $a_{5}=48$ $\Rightarrow a r^{6}=8 a r^{3}$ and $a r^{4}=48$ $\Rightarrow r^{3}=8$ $\Rightarrow r^{3}=2^{3}$ $\Rightarrow r=2$ Putting $r=2$ in $a r^{4}=48$ $a(2)^{4}=48$ $\Rightarrow a=3$ Thus, the given G.P. is $3,6,12, \ldots$...
Read More →What is the probability that a number selected
Question: What is the probability that a number selected from the numbers 1, 2, 3, ...,15 is a multiple of 4? Solution: GIVEN: Numbers are from 1 to 15. One number is selected TO FIND: Probability that the selected number is multiple of 4 Total number is 15 Numbers that are multiple of 4 are 4,8,12, Total number which is multiple of 4 is 3 We know that PROBABILITY = Hence probability of selecting a multiple of 4 is...
Read More →Two parallelograms are on equal bases and between the same parallels.
Question: Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is(a) 1 : 2(b) 1 : 1(c) 2 : 1(d) 3 : 1 Solution: Parallelograms on the same base and between the same parallels are equal in area.So, the ratio of their areas will be 1 : 1.Hence, the correct answer is option (b)....
Read More →A bag contains 3 red balls, 5 black balls and 4 white balls.
Question: A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is :(a) white?(b) red?(c) black?(d) not red? Solution: GIVEN: A bag contains 3 red, 5 black and 4 white balls TO FIND: Probability of getting a (i) White ball (ii) Red ball (iii) Black ball (iv) Not red ball Total number of balls (i) Total number white balls is 4 We know that PROBABILITY = Hence probability of getting white ball = (ii) Total n...
Read More →The area of ||gm ABCD is
Question: The area of ||gmABCDis(a)AB BM(b)BC BN(c)DC DL(d)AD DL Solution: Area of a parallelogram is base into height.Height = DL = NBBase = AB = CDSo, area of parallelogram ABCD = DCDLHence, the correct answer is option (c)....
Read More →The fourth term of a G.P. is 27 and the 7th term is 729,
Question: The fourth term of a G.P. is 27 and the 7th term is 729, find the G.P. Solution: Let $a$ be the first term and $r$ be the common ratio of the given G.P. $\therefore a_{4}=27$ and $a_{7}=729$ $\Rightarrow a r^{3}=27$ and $a r^{6}=729$ $\Rightarrow \frac{a r^{6}}{a r^{3}}=\frac{729}{27}$ $\Rightarrow r^{3}=3^{3}$ $\Rightarrow r=3$ Putting $r=3$ in $a r^{3}=27$ $a(3)^{3}=27$ $\Rightarrow a=1$ Thus, the given G.P. is $1,3,9, \ldots$...
Read More →An urn contains 10 red and 8 white balls. One ball is drawn at random.
Question: An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white. Solution: GIVEN: A bag contains10 red, and 8 white balls TO FIND: Probability that one ball is drawn at random and getting a white ball Total number of balls Total number of white balls is 8 We know that PROBABILITY = Hence probability of getting a white ball is...
Read More →In a lottery of 50 tickets numbered 1 to 50, one ticket is drawn.
Question: In a lottery of 50 tickets numbered 1 to 50, one ticket is drawn. Find the probability thatn the drawn ticket bears a prime number. Solution: GIVEN: Tickets are marked with one of the numbers 1 to 50. One ticket is drawn at random. TO FIND: Probability of getting a prime number on the drawn ticket Total number of tickets is 5. Tickets marked prime number are 1,3,5,7,11,13,17,19,23,29,31,37,43,47,49 Total number of tickets marked prime is 15 We know that PROBABILITY = Hence probability ...
Read More →Find the 4th term from the end of the G.P.
Question: Find the 4 th term from the end of the G.P. $\frac{1}{2}, \frac{1}{6}, \frac{1}{18}, \frac{1}{54}, \ldots, \frac{1}{4374}$ Solution: After reversing the given G.P., we get another G.P. whose first term, $l$ is $\frac{1}{4374}$ and common ratio is 3 . $\therefore 4^{\text {th }}$ term from the end $=l\left(\frac{1}{r}\right)^{4-1}$ $=\left(\frac{1}{4374}\right)(3)^{4-1}$ $=\left(\frac{27}{4374}\right)$ $=\frac{1}{162}$...
Read More →Two unbiased dice are thrown.
Question: Two unbiased dice are thrown. Find the probability that the total of the numbers on the dice is greater than 10. Solution: GIVEN: A pair of dice is thrown TO FIND: Probability that the total of numbers on the dice is greater than 10 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), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,...
Read More →In the given figure, ABCD is a || gm in which DL ⊥ AB. If AB = 10 cm and DL = 4 cm, then the ar(||gm ABCD) = ?
Question: In the given figure,ABCDis a || gm in whichDLAB. IfAB= 10 cm andDL= 4 cm, then the ar(||gmABCD) = ? (a) $40 \mathrm{~cm}^{2}$ (b) $80 \mathrm{~cm}^{2}$ (c) $20 \mathrm{~cm}^{2}$ (d) $196 \mathrm{~cm}^{2}$ Solution: (a) $40 \mathrm{~cm}^{2}$ $\operatorname{ar}(\| g m A B C D)=$ base $\times$ height $=10 \times 4=40 \mathrm{~cm}^{2}$...
Read More →A and B throw a pair of dice. If A throws 9, find B's chance of throwing a higher number.
Question: AandBthrow a pair of dice. IfAthrows 9, findB's chance of throwing a higher number. Solution: GIVEN: A pair of dice is thrown TO FIND: Probability that the total of numbers on the dice is greater than 9 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), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3)...
Read More →Which term of the progression 18, −12, 8, ... is
Question: Which term of the progression $18,-12,8, \ldots$ is $\frac{512}{729} ?$ Solution: Here, first term, $a=18$ and common ratio, $r=\frac{-2}{3}$ Let the $n^{\text {th }}$ term be $\frac{512}{729}$. $\therefore a r^{n-1}=\frac{512}{729}$ $\Rightarrow(18)\left(\frac{-2}{3}\right)^{n-1}=\frac{512}{729}$ $\Rightarrow\left(\frac{-2}{3}\right)^{n-1}=\frac{512}{729} \times \frac{1}{18}=\frac{256}{6561}$ $\Rightarrow\left(\frac{-2}{3}\right)^{n-1}=\left(\frac{-2}{3}\right)^{8}$ $\Rightarrow n-1=8...
Read More →What is the probability that a leap year has 53 Tuesdays and 53 Mondays?
Question: What is the probability that a leap year has 53 Tuesdays and 53 Mondays? Solution: GIVEN: A leap year TO FIND: Probability that a leap year has 53 Tuesdays and 53 Mondays Total number of days in a non leap year is 366days Hence number of weeks in a non leap year is In a non leap year we have 52 complete weeks and 2 day which can be any pair of the day of the week i.e. (SUNDAY, MONDAY) (MONDAY, TUESDAY) (TUESDAY, WEDNESDAY) (WEDNESDAY, THURSDAY) (THURSDAY FRIDAY) (FRIDAY, SATURDAY) SATU...
Read More →Compute the indicated products:
Question: Compute the indicated products: (i) $\left[\begin{array}{rr}a b \\ -b a\end{array}\right]\left[\begin{array}{rr}a -b \\ b a\end{array}\right]$ (ii) $\left[\begin{array}{rr}1 -2 \\ 2 3\end{array}\right]\left[\begin{array}{rrr}1 2 3 \\ -3 2 -1\end{array}\right]$ (iii) $\left[\begin{array}{rrr}2 3 4 \\ 3 4 5 \\ 4 5 6\end{array}\right]\left[\begin{array}{rrr}1 -3 5 \\ 0 2 4 \\ 3 0 5\end{array}\right]$ Solution: (i) $\left[\begin{array}{cc}a b \\ -b a\end{array}\right]\left[\begin{array}{cc...
Read More →In the given figure, ABCD is a || gm in which diagonals AC and BD intersect at O.
Question: In the given figure,ABCDis a || gm in which diagonalsACandBDintersect atO. If ar(||gmABCD) is 52 cm2, then the ar(∆OAB) = ? (a) $26 \mathrm{~cm}^{2}$ (b) $18.5 \mathrm{~cm}^{2}$ (c) $39 \mathrm{~cm}^{2}$ (d) $13 \mathrm{~cm}^{2}$ Solution: (d) $13 \mathrm{~cm}^{2}$ The diagonals of a parallelogram divides it into four triangles of equal areas. $\therefore$ Area of $\Delta O A B=\frac{1}{4} \times \operatorname{ar}(\| \mathrm{gm} A B C D)$ $\Rightarrow \operatorname{ar}(\Delta O A B)=\f...
Read More →What is the probability that an ordinary year has 53 Sundays?
Question: What is the probability that an ordinary year has 53 Sundays? Solution: GIVEN: An ordinary year TO FIND: Probability that a non leap year has 53 Sundays. Total number of days in an ordinary year is 365days Hence number of weeks in an ordinary year is In an ordinary year we have 52 complete weeks and 1 day which can be any day of the week i.e. SUNDAY, MONDAY, TUESDAY, WEDNESDAY, THURSDAY, FRIDAY and SATURDAY To make 53 Sundays the additional day should be Sunday Hence total number of da...
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