eSaral - JEE | NEET | Class 8-10 Exam Prep App

Three coins are tossed together. Find the probability of getting:

Question: Three coins are tossed together. Find the probability of getting:(a) exactly two heads(b) at most two heads(c) at least one head and one tail.(d) no tails Solution: GIVEN: Three coins are tossed simultaneously. TO FIND: We have to find the following probability When three coins are tossed then the outcome will be TTT, THT, TTH, THH. HTT, HHT, HTH, HHH. Hence total number of outcome is 8. (i) For exactly two head we get favorable outcome as THH, HHT ,HTH Hence total number of favorable ...

Which term of the G.P. :

Question: Which term of the G.P. : (i) $\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2 \sqrt{2}}, \frac{1}{4 \sqrt{2}}, \ldots$ is $\frac{1}{512 \sqrt{2}}$ ? (ii) $2,2 \sqrt{2}, 4, \ldots$ is 128 ? (iii) $\sqrt{3}, 3,3 \sqrt{3}, \ldots$ is 729 ? (iv) $\frac{1}{3}, \frac{1}{9}, \frac{1}{27} \ldots$ is $\frac{1}{19683}$ ? Solution: (i) Here, first term, $a=\sqrt{2}$ and common ratio, $r=\frac{1}{2}$ Let the $n^{\text {th }}$ term be $\frac{1}{512 \sqrt{2}}$. $\therefore a_{n}=\frac{1}{512 \sqrt{2}}$ $\R...

A die is thrown, Find the probability of getting:

Question: A die is thrown, Find the probability of getting:(a) a prime number(b) 2 or 4(c) a multiple of 2 or 3(d) an even prime number(e) a number greater than 5(f) a number lying between 2 and 6 Solution: GIVEN: A dice is thrown once TO FIND (i) Probability of getting a prime number (ii) Probability of getting 2 or 4 (iii) Probability of getting a multiple of 2 or 3. (iv) Probability of getting an even number (v) Probability of getting a number greater than five. (vi) Probability of lying betw...

In the given figure, ABCD is a || gm in which AB = CD = 5 cm and BD ⊥ DC such that BD = 6.8 cm.

Question: In the given figure,ABCDis a || gm in whichAB=CD= 5 cm andBDDCsuch thatBD= 6.8 cm. Then, the area of || gmABCD= ? (a) $17 \mathrm{~cm}^{2}$ (b) $25 \mathrm{~cm}^{2}$ (c) $34 \mathrm{~cm}^{2}$ (d) $68 \mathrm{~cm}^{2}$ Solution: (c) 34 cm2ar(parallelogramABCD) = base height = 5 6.8 = 34 cm2...

The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7.

Question: The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves Rs 15,000 per month, find their monthly incomes using matrix method. This problem reflects which value? Solution: Let the monthly incomes of Aryan and Babban be 3xand 4x, respectively.Suppose their monthly expenditures are 5yand 7y, respectively.Since each saves Rs 15,000 per month, Monthly saving of Aryan : $3 x-5 y=15,000$ Monthly saving of Babban : $4 x...

The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow?

Question: The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow? Solution: Given: Probability that it will rain TO FIND: Probability that it will not rain CALCULATION:We know that sum of probability of occurrence of an event and probability of non occurrence of an event is 1. $P(E)+P(\bar{E})=1$ $0.85+P(\bar{E})=1$ $P(\bar{E})=1-0.85$ $P(\vec{E})=0.15$ Hence the probability that it will not rain is $=0.15$...

A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is

Question: A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is(i) a black king(ii) either a black card or a king(iii) black and a king(iv) a jack, queen or a king(v) neither a heart nor a king(vi) spade or an ace(vii) neither an ace nor a king(viii) neither a red card nor a queen.(ix) other than an ace(x) a ten(xi) a spade(xii) a black card(xiii) the seven of clubs(xiv) jack(xv) the ace of spades(xvi) a queen(xvii) a heart(xviii) a red card(xix) neither ...

The area of trapezium ABCD in the given figure is

Question: The area of trapeziumABCDin the given figure is (a) $62 \mathrm{~cm}^{2}$ (b) $93 \mathrm{~cm}^{2}$ (c) $124 \mathrm{~cm}^{2}$ (d) $155 \mathrm{~cm}^{2}$ Solution: (c) $124 \mathrm{~cm}^{2}$ In the right angle triangleBEC,we have: $E C=\sqrt{17^{2}-15^{2}}=\sqrt{289-225}=\sqrt{64}=8 \mathrm{~cm}$ $\operatorname{ar}($ trapez $. A B C D)=\frac{1}{2} \times($ sum of parallel sides $) \times$ distance between them $=\frac{1}{2} \times 31 \times 8=124 \mathrm{~cm}^{2}$...

In a simultaneous throw of pair of dice, find the probability of getting:

Question: In a simultaneous throw of pair of dice, find the probability of getting:(i) 8 as the sum(ii) a doublet(iii) a doublet of prime numbers(iv) a doublet of odd numbers(v) a sum greater than 9(vi) an even number on first(vii) an even number on one and a multiple of 3 on the other(viii) neither 9 nor 11 as the sum of the numbers on the faces(ix) a sum less than 6(x) a sum less than 7(xi) a sum more than 7(xii) at least once(xiii) a number other than 5 on any dice.(xiv) even number on each d...

The area of quadrilateral ABCD in the given figure is

Question: The area of quadrilateralABCDin the given figure is(a) 57 cm2(b) 108 cm2(c) 114 cm2(d) 195 cm2 Solution: (c) $114 \mathrm{~cm}^{2}$ ar (quad.ABCD) = ar(∆ABC) + ar(∆ACD)In right angle triangleACD,we have: $A C=\sqrt{\left(17^{2}-8^{2}\right)}=\sqrt{225}=15 \mathrm{~cm}$ In right angle triangleABC, we have: $B C=\sqrt{\left(15^{2}-9^{2}\right)}=\sqrt{144}=12 \mathrm{~cm}$ Now, we have the following: $\operatorname{ar}(\Delta A B C)=\frac{1}{2} \times 12 \times 9=54 \mathrm{~cm}^{2}$ $\op...

Which term of the progression 0.004, 0.02, 0.1,

Question: Which term of the progression 0.004, 0.02, 0.1, ... is 12.5? Solution: We have; $\frac{a_{2}}{a_{1}}=\frac{0.02}{0.004}=5, \frac{a_{3}}{a_{2}}=\frac{0.1}{0.02}=5$ $\Rightarrow \frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=5$ The given progression is a G.P. whose first term, $a$ is $0.004$ and common ratio, $r$ is 5 . Let the $n$th term be $12.5$. $\therefore a_{n}=12.5$ $\Rightarrow a r^{n-1}=12.5$ $\Rightarrow(0.004)(5)^{n-1}=12.5$ $\Rightarrow(5)^{n-1}=\frac{12.5}{0.004}$ $\Rightarrow(5)^{...

In a certain city there are 30 colleges.

Question: In a certain city there are 30 colleges. Each college has 15 peons, 6 clerks, 1 typist and 1 section officer. Express the given information as a column matrix. Using scalar multiplication, find the total number of posts of each kind in all the colleges. Solution: Number of different types of posts in any college is given by $x=\left[\begin{array}{c}15 \\ 6 \\ 1 \\ 1\end{array}\right]$ Total number of posts of each kind in all the colleges = 30X $=30\left[\begin{array}{c}15 \\ 6 \\ 1 \\...

Find the 4th term from the end of the G.P.

Question: Find the 4th term from the end of the G.P. $\frac{2}{27}, \frac{2}{9}, \frac{2}{3}, \ldots, 162$ Solution: Here, first term, $a=\frac{2}{27}$ $C$ ommon ratio, $r=\frac{a_{2}}{a_{1}}=\frac{\frac{2}{9}}{\frac{2}{27}}=3$ Last term, $l=162$ After reversing the given G.P., we get another G.P. whose first term is $l$ and common ratio is $\frac{1}{r}$. $\therefore 4$ th term from the end $=l\left(\frac{1}{r}\right)^{4-1}=(162)\left(\frac{1}{3}\right)^{3}=6$...

The median of a triangle divides it into two

Question: The median of a triangle divides it into two(a) triangles of equal areas(b) congruent triangles(c) isosceles triangles(d) right triangles Solution: (a)triangles of equal areas...

If X and Y are 2 × 2 matrices, then solve the following matrix equations for X and Y.

Question: IfXandYare 2 2 matrices, then solve the following matrix equations forXandY $2 X+3 Y=\left[\begin{array}{ll}2 3 \\ 4 0\end{array}\right], 3 X+2 Y=\left[\begin{array}{cc}-2 2 \\ 1 -5\end{array}\right]$ Solution: We have, $3(2 X+3 Y)-2(3 X+2 Y)=3\left[\begin{array}{ll}2 3 \\ 4 0\end{array}\right]-2\left[\begin{array}{cc}-2 2 \\ 1 -5\end{array}\right]$ $\Rightarrow 6 X+9 Y-6 X-4 Y=\left[\begin{array}{cc}6 9 \\ 12 0\end{array}\right]+\left[\begin{array}{cc}4 -4 \\ -2 10\end{array}\right]$ ...

In which of the following figures, you find polynomials on the same base and between the same parallels?

Question: In which of the following figures, you find polynomials on the same base and between the same parallels? Solution: In this figure, the following polygons lie on the same base and between the same parallel lines:a) ParallelogramABCDb) ParallelogramABPQ...

Find:

Question: Find: (i) the ninth term of the G.P. 1, 4, 16, 64, ... (ii) the 10 th term of the G.P. $-\frac{3}{4}, \frac{1}{2},-\frac{1}{3}, \frac{2}{9}, \ldots$ (iii) the 8th term of the G.P. 0.3, 0.06, 0.012, ... (iv) the 12 th term of the G.P. $\frac{1}{a^{3} x^{3}}, a x, a^{5} x^{5}, \ldots$ (v) $n$th term of the G.P. $\sqrt{3}, \frac{1}{\sqrt{3}}, \frac{1}{3 \sqrt{3}}, \ldots$ (vi) the 10th term of the G.P. $\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2 \sqrt{2}}, \ldots$ Solution: (i) Here, First ...

Out of the following given figures which are on the same base but not between the same parallels?

Question: Out of the following given figures which are on the same base but not between the same parallels? Solution: In this figure, both the triangles are on the same base (QR) but not on the same parallels....

If a 1.5 m tall girl stands at a distance of 3 m from a lamp-post

Question: If a 1.5 m tall girl stands at a distance of 3 m from a lamp-post and casts a shadow of length 4.5 m on the ground, then the height of the lamp-post is(a) 1.5 m(b) 2 m(c) 2.5 m(d) 2.8 m Solution: LetABbe the lamp post andCD=1.5 m be the girl. The given information can be represented as Here, shadow of girl isDE= 4.5 m andBD= 3 m. In $\triangle C D E, \tan \theta=\frac{C D}{D E}=\frac{1.5}{4.5}$....(1) In $\triangle A B E, \tan \theta=\frac{A B}{B E}=\frac{A B}{4.5+3}=\frac{A B}{7.5}$.....

In the adjoining figure, ABCD and BQSC are two parallelograms.

Question: In the adjoining figure,ABCDandBQSCare two parallelograms. Prove that ar(∆RSC) = ar(∆PQB). Solution: In $\triangle \mathrm{RSC}$ and $\triangle \mathrm{PQB}$ $\angle \mathrm{CRS}=\angle \mathrm{BPQ}$ (CD || AB) so, corresponding angles are equal) $\angle \mathrm{CSR}=\angle \mathrm{BQP}$ ( SC || QB so, corresponding angles are equal) SC = QB (BQSC is a parallelogram) So, $\Delta \mathrm{RSC} \cong \Delta \mathrm{PQB}$ (AAS congruency) Thus, $\operatorname{ar}(\Delta \mathrm{RSC})=\oper...

The tops of two poles of height 16 m and 10 m are connected by a

Question: The tops of two poles of height 16 m and 10 m are connected by a wire of lengthlmetres. If the wire makes an angle of 30 with the horizontal, thenl=(a) 26(b) 16(c) 12(d) 10 Solution: LetABandCDbe the poles such thatAB= 16 m andCD= 10 m. The given information can be represented as Here,ACis the length of wire which is. Also,AE=ABBE= 16 m 10 m = 6 m We have to find the length of wire. So we use trigonometric ratios. In triangleACE, $\sin C=\frac{A E}{E C}$ $\Rightarrow \sin 30^{\circ}=\f...

Two poles are 'a' metres apart and the height of one is double of the other.

Question: Two poles are 'a' metres apart and the height of one is double of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the smaller is (a) $\sqrt{2} a$ metres (b) $\frac{a}{2 \sqrt{2}}$ metres (c) $\frac{a}{\sqrt{2}}$ metres (d) $2 \mathrm{a}$ metres Solution: LetABandCDbe the two posts such thatABCD. Then, letAB=hso thatCD= 2h Now, the given information can be represented as, He...

Find x, y, z and t, if

Question: Findx,y,zandt, if (i) $3\left[\begin{array}{ll}x y \\ z t\end{array}\right]=\left[\begin{array}{rc}x 6 \\ -1 2 t\end{array}\right]+\left[\begin{array}{cc}4 x+y \\ z+t 3\end{array}\right]$ (ii) $2\left[\begin{array}{cc}x 5 \\ 7 y-3\end{array}\right]+\left[\begin{array}{ll}3 4 \\ 1 2\end{array}\right]=\left[\begin{array}{rc}7 14 \\ 15 14\end{array}\right]$ Solution: (i) $3\left[\begin{array}{ll}x y \\ z t\end{array}\right]=\left[\begin{array}{cc}x 6 \\ -1 2 t\end{array}\right]+\left[\beg...

In the adjoining figure, D and E are respectively the midpoints of sides AB and AC of ∆ABC.

Question: In the adjoining figure,DandEare respectively the midpoints of sidesABandACof ∆ABC.IfPQ||BCandCDPandBEQare straight lines then prove that ar(∆ABQ) = ar(∆ACP). Solution: In $\triangle \mathrm{PAC}$ PA || DE and E is the midpoint of ACSo, D is the midpoint of PC by converse of midpoint theorem. Also, $\mathrm{DE}=\frac{1}{2} \mathrm{PA}$ ........(i) Similarly, $\mathrm{DE}=\frac{1}{2} \mathrm{AQ}$ ...........(ii) From (1) and (2) we havePA = AQ∆ABQand ∆ACP areon same base PQ and between ...

Show that the sequence

Question: Show that the sequence $a_{n}$, defined by $a_{n}=\frac{2}{3^{n}}, n \in N$ is a G.P. Solution: We have: $a_{n}=\frac{2}{3^{n}}, n \in N$ Putting $n=1,2,3, \ldots$ $a_{1}=\frac{2}{3^{1}}=\frac{2}{3}, a_{2}=\frac{2}{3^{2}}=\frac{2}{9}, a_{3}=\frac{2}{3^{3}}=\frac{2}{27}$ and so on. Now, $\frac{a_{2}}{a_{1}}=\frac{\frac{2}{9}}{\frac{9}{3}}=\frac{1}{3}, \frac{a_{3}}{a_{2}}=\frac{\frac{2}{27}}{\frac{2}{9}}=\frac{1}{3}$ and so on. $\therefore \frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=\ldots=\...

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