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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.
| 22451. |
Sec+tan=P than prove that p2_1 upon p2+1=sin |
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| 22452. |
General formula of polynomial |
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| 22453. |
1+tan thita+sec thita |
| Answer» Cos thita +sin thita+1------------------------------------- Cos thita. | |
| 22454. |
Cbse class 10 last ten year question paper |
| Answer» Check last year papers here :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 22455. |
If √x+y=11And x+√y=7Then find x and y WITHOUT HIT AND TRIAL METHOD |
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| 22456. |
Is maths 2 papers in cbse board exam confirm |
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Answer» Kyaaaaaa Next yr sai h .. |
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| 22457. |
CosA/1+sinA+1+sinA/cosA=2secA |
| Answer» L.C.M. se solve kar le bhai | |
| 22458. |
Prove that sin(A+B)=sinACosB+sinBcosA |
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| 22459. |
If nth term of an AP is (2n+1) what is the sum of its three terms |
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Answer» 15 which three times Wkt AP=2n+1Let n=1,2,3Then, 2.1 |
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| 22460. |
Triangle ABC similar to triangle DEF , area of ABC =162 sqcm , if BC =18 cm . Find EF |
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| 22461. |
3 upon 8 -1 upon 12 |
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Answer» Mera dhyaan kahi aur tha us time pe 1/3 is wrong okk but 7/24 is 3.428 its correct 1/3 is wrong sry guys {tex}\\frac38-\\;\\frac1{12}\\;\\;=\\;\\frac{3(3)\\;-1(2)}{24}=\\;\\frac{9\\;-\\;2}{24}=\\;\\frac7{24}{/tex} , 24 is not coming in table of 7 Its correct answer is 1/3 because 7/24 can be divided more so after dividing 1/3 will be the answer 7/24 7 upon 24 |
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| 22462. |
find the quadratic polynomial whose sum and product of the zeros are 21_8 AND 5_16 respectively |
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Answer» The quadratic polynomials are - xsqaure -21x+8 and xsquare-5x+16 Wait 2mins |
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| 22463. |
Find the perpendicular distance of a 5 and 12 from the y axis |
| Answer» 5 | |
| 22464. |
Easy method to learn cube cuboid cylinder all formulas |
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Answer» Solve problems more and more. Co relate kr kai yaad kr skte ho |
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| 22465. |
What stream u r thinking to take in 11th guys ? |
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Answer» PCB Pata nhi Medical Arts PCMB Maths Commers |
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| 22466. |
Which questions are important from circles chapter?? |
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Answer» Only q 9,12,and13 are most important and the tangents theoram is also importamt 8,9,11,12,13 of ex.10.2 and one is ready to come in board exam |
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| 22467. |
If radius of a circle is increased by 10% then area is increased by how many % |
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Answer» 21℅ not 1.21 times because question ask by what ℅ not times the earlier New radius =1.1rArea=π(1.1r)2=1.21πr2Hence area becomes 1.21 times |
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| 22468. |
Find cosec30 and cos 60 geometrically |
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Answer» Answer is 2 Aise question bhi aaenge kya exam me?? Good ques. Ans.mile to batana |
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| 22469. |
a squre of diagonal 8cm is inscribed in a circle . find the area of shaded part |
| Answer» Since it is a square, all 4 sides are equal.. the diagonal is also given..2AB^2=AC^2AC=4AB^2=8AB=2root2 cmSide of the square is equal to diametre of the circle..Therefore radius of the circle is root2 cm..Area of the circle= 22/7×2root2×2root2 = 22×8/7Area of square = 2root2×2root2=8cmTherefore area of shaded region = 8-(22×8/7) = 8-25/-18/ .. negative areas within bar brackets | |
| 22470. |
(Cos^2alpha-1)(cot^2alpha+1)+1=0 |
| Answer» So value of alpha is 90°because by solving cos alpha=0 | |
| 22471. |
Find the HCF of 4052 and 420 and their linear combination |
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Answer» Thanks for answer 4052=420×9+272420=272×1+148272=148×1+124148=124×1+24124=24×5+424=4×6+0hence hcf is 4 2 |
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| 22472. |
5-3.5/2 |
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Answer» Apply BODMAS rule,Then, 5-3.5/2=5-1.75=3.25. 3.25 is the answer Answer=3.25 |
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| 22473. |
A cylinder a cone and a hemisphere have same base and same height find the ratio of their volume |
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Answer» Their ratio is1 : 2 : 3 H : h/3 : 2/3 1:1/3:2/3.?? |
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| 22474. |
If the sum of first p terms of an A.P is ap2 +bp,find its common difference |
| Answer» We are given that sum\xa0of first p terms of this APSp= ap2 + bpLet the first term= x and common difference = d{tex}{\\mathrm S}_1=\\mathrm a(1)^2+\\mathrm b(1)=\\mathrm a+\\mathrm b{/tex}So first term x = a+bNow S2 = a(2)2 + b(2)S2 = 4a + 2b,{tex}\\mathrm{So}\\;{\\mathrm s}_2={\\mathrm a}_1+{\\mathrm a}_2=\\mathrm x+\\mathrm x+\\mathrm d=4\\mathrm a+2\\mathrm b{/tex}2x + d = 4a+2b2(a+b)+d = 4a+2b2a+2b+d=4a+2bd=4a+2b-2a-2bd = 2aTherefore, the common difference of the given AP is 2a. | |
| 22475. |
If the median of the distribution given below is 28.5find the values of x and y |
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| Answer» | Monthly Consumption | \t\t\t0-105510-20x5 + x20-302025 + x30-401540 + x40-50y40 + x + y50-60545 + x + yTotal\xa0{tex}\\sum f _ { i } = n = 60{/tex}\xa0\tHere, {tex}\\sum f _ { i } = n = 60{/tex}, then {tex}\\frac { n } { 2 } = \\frac { 60 } { 2 } = 30{/tex}, also, median of the distribution is 28.5, which lies in interval 20 – 30.{tex}\\therefore{/tex}\xa0Median class = 20 – 30So, l = 20, n = 60, f = 20, cf = 5 + x\xa0and h = 10{tex}\\because 45 + x + y = 60{/tex}{tex}\\Rightarrow x + y = 15{/tex}\xa0………...........(i)Now, Median = {tex}l + \\left[ \\frac { \\frac { n } { 2 } - c f } { f } \\right] \\times h{/tex}{tex}\\Rightarrow { 28.5 = 20 + \\left[ \\frac { 30 - ( 5 + x ) } { 20 } \\right] \\times 10 }{/tex}{tex}\\Rightarrow 28.5 = 20 + \\frac { 30 - 5 - x } { 2 }{/tex}{tex}\\Rightarrow { 28.5 } = \\frac { 40 + 25 - x } { 2 }{/tex}{tex}\\Rightarrow 2 ( 28.5 ) = 65 - x{/tex}{tex}\\Rightarrow 57.0 = 65 - x{/tex}{tex}\\Rightarrow x = 65 - 57 = 8{/tex}{tex}\\Rightarrow{/tex}\xa0x = 8Putting the value of x\xa0in eq. (i), we get,8 + y = 15{tex}\\Rightarrow{/tex}\xa0y\xa0= 7Hence the value of x\xa0and y\xa0are\xa08 and 7 respectively. | |
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| 22476. |
Which term of the AP :121,117,113,109....in first negative term? |
| Answer» 14 | |
| 22477. |
Evaluate 2tan^2 30 degree +cos^2 19 degree +cos^2 71 degree |
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Answer» 2*(1/√3)^2+sin^2(90-19)+cos^2 71=2/3+sin^2 71+cos^2 71=2/3+1=5/3 I think 5/3 |
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| 22478. |
What is the LCM of smallest prime number and smallest composite number? |
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Answer» 4 4 4 Smallest prime number is 2 and smallest composite number is 4.LCM of 2 and 4 is 4 |
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| 22479. |
Tsa and csa of sphere........... .. |
| Answer» Sphere has only TSA that is 4πrsq. It does not have any side as it is spherical round. Therefore CSA of sphere is not applicable | |
| 22480. |
If cot theta =7/8 then find the value of (1+sin theta)/(1+cos theta) |
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Answer» Say theta =xCotx=7/8Cosec x=(1+49/64)1\\2/=√113/8{tex}\\frac{1+sinx}{1+cosx}=\\frac{cosecx+1}{cosecx+cotx}{/tex}=(√113/8+1)/(√113/8+7/8)= (√113+8) / (√113+7)\xa0 √113+8/√113+7 |
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| 22481. |
Find the area of a square inscribed in a circle of radius x cm |
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Answer» 2x=diagonal of square,if a is side then4x2=2a2a\u200b\u200b\u200b\u200b\u200b2=2x2 is the area of square 2x^2 |
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| 22482. |
Prove that sinx /Cotx+cosexx=2+sinx/cotx-cosecx |
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| 22483. |
AAA similarity criteria |
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Answer» Hii guys U can see in ncert So what |
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| 22484. |
The length of mint hand clock is 14cm . Find the area swept by the mint hand in 20 mint |
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Answer» Length will be assumed as radiusSo r=14cmAngle swept by minute hand in 1 minute =6 degreeAngle swept in 20 minutes 120degreeArea of sector=theta/360*πr²120/360*22/7*14*14=1/3*22*2*14=205.33cm²I hope that my answer will help you 154 616/3 |
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| 22485. |
Cot square theta -1/sinsquare theeta |
| Answer» _1 | |
| 22486. |
Where hindi site mathematics |
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| 22487. |
If two vertices of an equilateral triangle are (0,0), (3,3) then find the third vertex |
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| 22488. |
QR is a tangent at Q, PR parallel PQ where AQ is a chord prove BR is tangent at B |
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| 22489. |
PlZ yaar this app is use to homework help no personal chat plzzzzzzzzzzzzz |
| Answer» U r right Aryan. ? | |
| 22490. |
Kx+3y=2k+1 & 2(k+1)x+9=7k+1 find k |
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Answer» Infinity many solution Mention hi ni hai ki .....infinitely many hai ya exactly one...ya no solutions....?? Solution fast dena jaldi h |
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| 22491. |
find the middle term of the AP 6 13 20 ................216 |
| Answer» a=6, d=7an= a+(n-1)×d216=6+(n-1)x7216=6+7n-7216=-1+7n217 = 7nn= 217/7n= 31total no. of terms = 31middle term=1/2 × (31+1) = 1/2 × 32 = 16tha16 = 6+(16-1)×7 = 6+15×7 = 6+105 111 | |
| 22492. |
What is the value under root3 |
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Answer» 1.732142857.... 1.732142857 |
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| 22493. |
1/a+1/b+1/c=1/a+b+c |
| Answer» Krna kya hai | |
| 22494. |
Is there any option in mathematics chapter? |
| Answer» Pta ni | |
| 22495. |
Make a time table for class 10 student |
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Answer» Haha timetable No Yes please Should i share my timetable ?? |
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| 22496. |
The sum of 3 numbers i n Ap is 12 and sum of their cubes is 288.find Ap |
| Answer» Let Three numbers be a-d ,a ,a+d a-d+a+a+d=123a=12a=4Now ,(a-d)3+a3+(a+d)3=288 Substitute the value of a(4-d)3 +64 +(4+d)3=288Substitute the formula (a +b)3You will get d Then you will get the required A.P | |
| 22497. |
If the area of two similar triangles drawn from an external point to a circle are equal |
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| 22498. |
If Sn denotes the sum of first n terms of an A.P , prove that S30=3(S20 -S10) |
| Answer» Let a be the first term and d be the common difference of the given AP.Then sn = n/2(2a + (n-1)d).\xa0LHS :S30 = (30/2)(2a + (30 - 1)d) = 15(2a + 29d) = 30a + 435d. ---------------- (1).RHS:(S20 - S10) = (20/2)(2a + (20 - 1) * d) - (10/2)(2a + (10 - 1) * d) = 10(2a + 19d) - 5(2a + 9d) = 20a + 190d - 10a - 45d = 10a + 145d 3(S20 - S10) = 3(10a + 145d) = 30a + 435d ------------- (2)Therefore From (1) and (2), It is proved that S30 = 3(S20 - S10).LHS = RHS. | |
| 22499. |
For thita i use 0If cot0+tan0=xSec0-cos0=yProve that (X.square×Y)ki whole power2/3 |
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| 22500. |
The sum of 5th and 9th terms of an A.P is 30. If its 25th term is thrice its 8th term.find A.p |
| Answer» The sum of 5th and 9th term is 30.⇒ a + 4d + a + 8d = 30⇒ 2a + 12d = 30 ..................(1)And,Its 25th term is 3 times its 8th term.⇒ a + 24d = 3(a + 7d)⇒ a + 24d = 3a + 21d⇒ a + 24d - 3a - 21d = 0⇒ - 2a + 3d = 0 ...................(2)Adding equation (1) and (2).⇒ 2a +12d = 30\xa0 - 2a + 3d = 0________________ 15d = 30________________⇒ 15d = 30⇒ d = 30/15⇒ d = 2So, common difference, \'d\' is 2.Putting the value of d = 2 in the equation (1).2a + 12d = 30⇒ 2a + (12*2) = 30⇒ 2a = 24 = 30⇒ 2a = 30 - 24⇒ 2a = 6⇒ a = 6/2⇒ a = 3So, the first term of the required AP is 3So, the required AP is 3, 5, 7, 9, 11, 13, 15, 19,............. | |