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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 3451. |
EValuate sin^2 5 + sin^2 10 +................ sin^2 85 +cos^2 45 |
| Answer» 8 +√2 | |
| 3452. |
I love ?❤️❤️❤️❤️ |
| Answer» Bakwas band kar nahi to CBSE ko complain doonga | |
| 3453. |
I have one question |
| Answer» ??? | |
| 3454. |
if 9+17+25+....+x=450,then find value of x |
| Answer» 90 | |
| 3455. |
If p (y)=y6-3y4×2y2+6 and q (y)=y5-y3+2y2+y-6 then find p (y)+q(y) and p (y)-q (y) |
| Answer» | |
| 3456. |
What is x+x |
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Answer» 2x 2x |
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| 3457. |
Find the sum of first twelve terms of AP: 1, 6, 11, 16, 21,.........................101. |
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Answer» 342 162 |
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| 3458. |
Find middle term of AP 6,13,20............ 216 |
| Answer» 111 | |
| 3459. |
Prove that 5+√2 is irrational number |
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Answer» But I take a test of cbse they gives a answer or not I know bro its too easy Muje bhi and ata me dhek ra hu KO cbse wale bhejte he Kya nai So easy Bheje answer fir |
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| 3460. |
If the mode of data 3,5,8,9,8,12,7,12 and x is 8,find the value of x. |
| Answer» | |
| 3461. |
Show that points(a,b+c),(b,C+a) and (c,a+b) are collinear |
| Answer» Use the formula of area of triangle. It will come out 0 so it is collinear | |
| 3462. |
Evluate⅔cosec²58°-⅔cot²58°tan32°-5/3tan13° tan37°tan45° tan53°tan77° |
| Answer» 1 | |
| 3463. |
In figure angleBAC=90degree show DE square=BD intoEC |
| Answer» Given: {tex}\\triangle {/tex}ABC is right-angled at A and DEFG is a squareTo Prove: DE2 = BD {tex} \\times{/tex} ECProof: Let {tex}\\angle{/tex} C = x ........(i)Then, {tex}\\angle{/tex} ABC = 90o - x[ {tex}\\because {/tex} ABC is right angled]Also {tex}\\triangle {/tex} BDG is right-angled at D.{tex}\\angle{/tex}BGD = 90o - (90o - x) = x ..............(ii)From (i) and (ii), we get{tex}\\angle{/tex}BGD = {tex}\\angle{/tex}C.............(iii)Consider {tex}\\triangle {/tex} BGD and {tex}\\triangle {/tex} CEF{tex}\\angle{/tex}CEF = {tex}\\angle{/tex}BDG = 90o [ {tex}\\because {/tex}DEFG is square]{tex}\\angle{/tex}BGD = {tex}\\angle{/tex}C [From (iii)]{tex}\\therefore {/tex}{tex}\\triangle {/tex}BGD {tex}\\cong{/tex}{tex}\\triangle {/tex}FEC [By AA similarity]{tex}\\therefore {/tex}{tex}\\frac{{BD}}{{EF}} = \\frac{{DG}}{{EC}}{/tex}{tex}\\Rightarrow {/tex} EF{tex} \\times{/tex} DG = BD {tex} \\times{/tex} ECBut EF = DG {tex} \\times{/tex} DE [ {tex}\\because {/tex} side of a square]{tex}\\Rightarrow {/tex} DE {tex} \\times{/tex} DE = BD {tex} \\times{/tex} EC{tex}\\Rightarrow {/tex} DE2 = BD {tex} \\times{/tex} EC | |
| 3464. |
Find the value of sin(45°+ x)- cos(45°- x) |
| Answer» {tex}sin (45^o +{/tex}\xa0x{tex}) - cos (45^o -{/tex}\xa0x){tex}= sin 45^o cos{/tex}\xa0x\xa0{tex}+ cos 45^o sin{/tex}\xa0x\xa0{tex}- [cos 45^o cos{/tex}\xa0x\xa0{tex}+ sin 45^o sin{/tex}\xa0x][{tex}\\because{/tex}\xa0sin ( A + B ) = sin A cos B + cos A sin B and cos ( A - B ) = cos A cos B + sin A sin B ]=\xa0{tex}\\frac { 1 } { \\sqrt { 2 } } \\cos \\theta + \\frac { 1 } { \\sqrt { 2 } } \\sin \\theta - \\left[ \\frac { 1 } { \\sqrt { 2 } } \\cos \\theta + \\frac { 1 } { \\sqrt { 2 } } \\sin \\theta \\right]{/tex}= 0 | |
| 3465. |
why the value of pi is 22/7? |
| Answer» Yes3.14 and 22/7 | |
| 3466. |
Q: If the mean of the following data is 20.6. Find the value of p.x: 10 15 p 25 35 f: 3 10 25 7 5 |
| Answer» p= 20 | |
| 3467. |
CotA-CosA/CotA+CosA=CosecA-1/CosecA+1 |
| Answer» | |
| 3468. |
Can we get some discount on solved test paper of math |
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Answer» You are buying it from amazon? Yes 70 percent I think no but you can check I don\'t know much about it. |
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| 3469. |
If ∆ABC is similar to ∆DEF, if AB= 4cm.... |
| Answer» | |
| 3470. |
How many diagonals in triangle |
| Answer» | |
| 3471. |
Hlo..Everyone.. |
| Answer» | |
| 3472. |
TSA of cuboid |
| Answer» The total surface area of a cuboid is given by the formula SA = 2lh + 2wh + 2lw where l = length, w = width, and h = height. | |
| 3473. |
Hii friends kaise ho sab |
| Answer» Badeya | |
| 3474. |
How to find a width of rectangle whose length is 2m and the area is 4m |
| Answer» Area= lb4=2bB=2 | |
| 3475. |
Euclid Formula |
| Answer» Dividend =(divisor × quotient) + remainder | |
| 3476. |
RIMSHA PLZZ COME YRTR |
| Answer» | |
| 3477. |
Without using t tables find the value ofCos70/sin20+ cos57#cosec33_2cos60 |
| Answer» | |
| 3478. |
Solve the equation -4+(-1)+2+....+x = 437 |
| Answer» (-4) + (-1) + 2 + 5 + ---- + x = 437.Now,-1 - (-4) = -1 + 4 = 32 - (-1) = 2 +\xa01 = 35 - 2 = 3Thus, this forms an A.P. with a = -4, d = 3,l = xLet their be n terms in this A.P.Then,Sn\xa0=\xa0{tex}\\frac { n } { 2 } [ 2 a + ( n - 1 ) d ] {/tex}{tex}\\Rightarrow 437 = \\frac { n } { 2 } [ 2 \\times ( - 4 ) + ( n - 1 ) \\times 3 ]{/tex}{tex}\\Rightarrow{/tex}\xa0874 = n[-8 + 3n - 3]{tex}\\Rightarrow{/tex}874 = n[3n - 11]{tex}\\Rightarrow{/tex}874 = 3n2\xa0- 11n{tex}\\Rightarrow{/tex}3n2\xa0- 11n - 874 = 0{tex}\\Rightarrow{/tex}3n2\xa0- 57n + 46n - 874 = 0{tex}\\Rightarrow{/tex}3n(n - 19) + 46(n - 19) = 0{tex}\\Rightarrow{/tex}3n + 46 = 0 or n = 19{tex}\\Rightarrow n = - \\frac { 46 } { 3 }{/tex}\xa0or n\xa0= 19Numbers of terms cannot be negative or fraction.{tex}\\Rightarrow{/tex}\xa0n = 19Now, Sn\xa0=\xa0{tex}\\frac { n } { 2 } [ a + l ]{/tex}{tex}\\Rightarrow 437 = \\frac { 19 } { 2 } [ - 4 + x ]{/tex}{tex}\\Rightarrow - 4 + x = \\frac { 437 \\times 2 } { 19 }{/tex}{tex}\\Rightarrow - 4 + x = 46{/tex}{tex}\\Rightarrow x = 50{/tex} | |
| 3479. |
RIMSHA. ........... |
| Answer» | |
| 3480. |
3 sinA + 5 cosA = 5 prove that 5 sinA - 3 cosA = 3 |
| Answer» | |
| 3481. |
Find the roots of quadratic equation 9xsquare - 9(a+b)x +(2asquare +5ab +2bsquare) |
| Answer» D = b2 - 4ac{tex}= ( - 9 ( a + b ) ) ^ { 2 } - 4 \\times 9 \\times \\left( 2 a ^ { 2 } + 5 a b + 2 a b ^ { 2 } \\right){/tex}= 81(a + b)2 - 36(2a2 + 5ab + 2b2)= 9[9(a2 + b2 + 2ab - 8a2 - 20ab - 8b2)]= 9[a2 + b2 - 2ab]= 9(a - b)2{tex}x = \\frac { - b \\pm \\sqrt { D } } { 2 a } = \\frac { 9 ( a + b ) \\pm \\sqrt { 9 ( a - b ) ^ { 2 } } } { 2 \\times 9 }{/tex}{tex}{ \\Rightarrow x = 3 \\frac { [ 3 ( a + b ) \\pm ( a - b ) ] } { 2 \\times 9 } }{/tex}{tex}{ \\Rightarrow x = \\frac { ( 3 a + 3 b ) \\pm ( a - b ) } { 6 } }{/tex}{tex}\\Rightarrow x = \\frac { 3 a + 3 b + a - b } { 6 } \\text { or } x = \\frac { 3 a + 3 b - a + b } { 6 }{/tex}{tex}\\Rightarrow x = \\frac { 4 a + 2 b } { 6 } \\text { or } x = \\frac { 2 a + 4 b } { 6 }{/tex}{tex}\\Rightarrow x = \\frac { 2 a + b } { 3 } \\text { or } x = \\frac { a + 2 b } { 3 }{/tex} | |
| 3482. |
Angle of 170\' |
| Answer» | |
| 3483. |
In a Deck of cards find the probability of a spade or a red card |
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Answer» Spade 13/52 , red card 26/52 13 / 52 |
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| 3484. |
Show that the reciprocal of an irrational no. Is irrational ? |
| Answer» let x be irrational and 1/x be rational then 1/x=a/b where a and b are integer where (a)not = 0 and (B) not= 0 Now x= 1/1/x =1/a/b = b/a where a and b are integer not = 0 , hence x is rational but this is contradiction becoz x is irrational so, 1/x is irrational ☺ hope it\'s helpful. .. | |
| 3485. |
Formula of frustum |
| Answer» Frustum ka kya ?? | |
| 3486. |
Find the coordinates of the point on y -axis which is nearest to the point (-2,5). |
| Answer» The point on y-axis that is nearest to the point(-2,5) is (0,5). | |
| 3487. |
When the date sheet will be announced |
| Answer» | |
| 3488. |
pythogorus thearome |
| Answer» (H)^2=(P)^2 + (B)^2 | |
| 3489. |
2x +3y =8 |
| Answer» On a graph paper, draw a horizontal line X\'OX and a vertical line YOY\' as the x-axis and the y-axis respectively.Graph of {tex}2x + 3y = 8{/tex}{tex}2x + 3y = 8{/tex}{tex}\\Rightarrow{/tex}\xa0y = {tex}\\frac{8 - 2x}{3}{/tex}. ...(i)Thus, we have the following table for {tex}2x + 3y = 8{/tex}\tx{tex}1{/tex}{tex}-5{/tex}{tex}7{/tex}y{tex}2{/tex}{tex}6{/tex}{tex}-2{/tex}\tNow, plot the points{tex} A (1, 2), B(-5, 6)\\ and\\ C(7, -2){/tex} on the graph paper.Join AB and AC to get the graph line BC.Thus, the line AC is the graph of the equation {tex}2x + 3y = 8{/tex}.Graph of {tex}x - 2y + 3 = 0{/tex}{tex}x - 2y + 3 = 0{/tex}{tex}\\Rightarrow y = \\frac { x +3 } {2 }{/tex}........(ii)Thus, we have the following table for {tex}x - 2y + 3 = 0\xa0{/tex}\tx{tex}1{/tex}{tex}3{/tex}{tex}-3{/tex}y{tex}2{/tex}{tex}3{/tex}{tex}0{/tex}\tNow, plot the points {tex}P(3,3)\\ and\\ Q (-3, 0){/tex} on the same graph paper.The point {tex}A(1,2){/tex} has already been plotted.Join PA\xa0and QA\xa0to obtain the line PQ.Thus, the line PQ is the graph of the equation {tex}x - 2y + 3 = 0{/tex}Two graphs lines intersect at the point {tex}A(1,2){/tex}{tex}\\therefore{/tex}\xa0{tex}x = 1, y = 2{/tex}\xa0is the solution of the given system of equations. | |
| 3490. |
The Nth term of an AP is 6m+2 find the common different |
| Answer» t1 = 6(1)+2 = 6+2=8 t2 = 6(2)+2=12+2 =14 !! d = t2-t1 = 14-8=6 | |
| 3491. |
Prove that a parallelogram circumscribing a circle is a rhombus |
| Answer» Given ABCD is a parallelogram in which all the sides touch a given circleTo prove:- ABCD is a rhombusProof:-{tex}\\because{/tex}\xa0ABCD is a parallelogram{tex}\\therefore{/tex}\xa0AB = DC and AD = BCAgain AP, AQ are tangents to the circle from the point A{tex}\\therefore{/tex}\xa0AP = AQSimilarly, BR = BQCR = CSDP = DS{tex}\\therefore{/tex}(AP + DP) + (BR + CR) = AQ + DS + BQ + CS = (AQ + BQ) + (CS + DS){tex}\\Rightarrow{/tex}\xa0AD + BC = AB + DC{tex}\\Rightarrow{/tex}\xa0BC + BC = AB + AB [{tex}\\because{/tex} AB = DC, AD = BC]{tex}\\Rightarrow{/tex}\xa02BC = 2AB{tex}\\Rightarrow{/tex}\xa0BC = ABHence, parallelogram ABCD is a rhombus | |
| 3492. |
No one is speaking to me |
| Answer» Whatever | |
| 3493. |
Draw a circle of radius 4cm.draw a diameter POQ.through p or q draw tangent to the circle |
| Answer» Steps of construction:\tDraw a circle of radius 4 cm.\tDraw diameter POQ.\tConstruct.\xa0{tex}\\angle PQT = 90^\\circ {/tex}\tProduce PQ to T\', then TQT\' is the required tangent at the point Q. | |
| 3494. |
What is the exams date? |
| Answer» | |
| 3495. |
Area of square |
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Answer» Side ×side Side ×side Side*side |
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| 3496. |
If the numbers n - 2, 4n - 1 and 5n + 2 are in AP, find the value of n. |
| Answer» | |
| 3497. |
Sin2A+Cos2A=1.Proved that. |
| Answer» | |
| 3498. |
If the m term of AP is 1/n and n term 1/m than show that it\'s mn term=1 |
| Answer» Let a and d be the first term and common difference respectively of the given A.P. Thenan = a + (n - 1)d{tex}\\frac { 1 } { n } ={/tex}\xa0mth term\xa0{tex}\\Rightarrow \\frac { 1 } { n } {/tex}= a + ( m - 1 ) d\xa0...(i){tex}\\frac { 1 } { m }{/tex}= nth term{tex}\\Rightarrow \\frac { 1 } { m } {/tex}= a + ( n - 1 ) d\xa0...(ii)On subtracting equation (ii) from equation (i), we get{tex}\\frac { 1 } { n } - \\frac { 1 } { m } = {/tex} [a+ (m-1) d] -[ a+ (n -1)d]= a + md - d - a - nd + d{tex}= ( m - n ) d{/tex}{tex} \\Rightarrow \\frac { m - n } { m n } = ( m - n ) d {/tex}{tex}\\Rightarrow d = \\frac { 1 } { m n }{/tex}Putting d = {tex}\\frac { 1 } { m n }{/tex}\xa0in equation (i), we get{tex}\\frac { 1 } { n } = a + \\frac { ( m - 1 ) } { m n } {/tex}{tex}\\Rightarrow \\frac { 1 } { n } = a + \\frac { 1 } { n } - \\frac { 1 } { m n } {/tex}{tex}\\Rightarrow a = \\frac { 1 } { m n }{/tex}{tex}\\therefore{/tex}\xa0(mn)th term = a + (mn - 1) d=\xa0{tex}\\frac { 1 } { m n } + ( m n - 1 ) \\frac { 1 } { m n } {/tex}{tex}\\left[ \\because a = \\frac { 1 } { m n } = d \\right]{/tex}= {tex}\\frac { 1 } { m n } + \\frac { mn } { m n } - \\frac { 1 } { m n }{/tex}= 1 | |
| 3499. |
Explain why 1323334356715 is a composite number |
| Answer» {tex}13233343563715=5\\times11\\times73\\times10139\\times325079{/tex}.This number is having factors other than 1 and the number itself,therefore it is a composite number. | |
| 3500. |
What is alternative angle in a segment of a circle? |
| Answer» | |