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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.
| 3501. |
If two vertices of an equilateral triangle are (3,0) and (6,0) find the third vertices |
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Answer» Find distance b/w them and put in next eq. Form othee 2 eq by using x,y put the distance value Take out distance from these two Easy hai |
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| 3502. |
What is the formula of area of minor and major segment??? |
| Answer» Minor =pie r2thita /360 - 1/2 r2 sin thita. Major =pie r2- area of the minor segment | |
| 3503. |
To prove: 3+2=7 |
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Answer» No the ans.is by mistake as we do in our childhood Both the side whole and ×0 Easy hai |
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| 3504. |
What is the difference between mesn and averave |
| Answer» Mean is computed by adding the values and dividing by the number of values. Average is the value obtained by dividing the sum of a set of quantities by the number of quantities in the set. | |
| 3505. |
In a certain AP the 24term term is twice the10term .prove that 72term is twice the 34 term |
| Answer» Given,a24\xa0= 2{tex}\\times{/tex}\xa0a10{tex}\\Rightarrow{/tex}\xa0a + (24\xa0- 1)d = 2{tex}\\times{/tex}[a + (10 - 1)d]{tex}\\Rightarrow{/tex}a + 23d = 2[a + 9d]{tex}\\Rightarrow{/tex}a + 23d = 2a + 18d{tex}\\Rightarrow{/tex}a - 2a = 18d - 23d{tex}\\Rightarrow{/tex}-a = -5d{tex}\\Rightarrow{/tex}a = 5d...........(i)To prove: a72\xa0= 2a34Proof:LHS = a72= a + (72\xa0- 1)d= 5d\xa0+ 71d [From (i)]= 76dRHS = 2a34= 2[a + (34\xa0- 1)d]= 2[5d + 33d]= 2\xa0{tex}\\times{/tex}38d= 76d{tex}\\therefore{/tex} LHS = RHS | |
| 3506. |
Is maths paper is simple |
| Answer» Very simple | |
| 3507. |
Define math |
| Answer» Mathematics is the study of topics such as quantity (numbers), structure, space and change. | |
| 3508. |
Find the value of k , x=a is a solution of equation x×x (a+b)x+k=0 |
| Answer» We have, x2- x (a+b) + k = 0Since x = a is the solution of the given equation.{tex}\\therefore{/tex}\xa0x = a satisfies the given equation.a2- a (a+b) + k = 0{tex}\\implies{/tex}a2 - a2 -ab +k = 0{tex}\\implies{/tex}k - ab = 0 or k =ab | |
| 3509. |
If diagonals of a rhombus are 12 cm and 16 cm then the perimeter of rhombus is |
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Answer» 40 So easy ?? |
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| 3510. |
Find the no. Of three digit natural no. Which are divisible by 11 |
| Answer» Three digit natural numbers are 100,101,102.....,998,999. From which,\xa0multiples of 11 are 110, 121, 132, ...., 990.so,this forms an AP 110,121....,990.\xa0where, first term ({tex}a {/tex})=110 and common difference({tex}d{/tex})=121-110=11, last term({tex}l{/tex}) = 990..Let,\xa0{tex}l = {a_n} = a + (n - 1)d{/tex}{tex}\\therefore 990 = 110 + (n - 1) \\times 11{/tex}{tex} \\Rightarrow n = 81{/tex}There are 81 three - digit numbers which are divisible by 11. | |
| 3511. |
Who is the CEO of my CBSE guide |
| Answer» | |
| 3512. |
Who invente zero |
| Answer» Arya Bhatt | |
| 3513. |
What we do in night |
| Answer» Ur wish | |
| 3514. |
acute angle = |
| Answer» The angle which is less than 90 degree is called acute angle | |
| 3515. |
Find the value of tan 15* .... Solve it on app |
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Answer» Tan45-tan30 So easy ?? |
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| 3516. |
In triangle ABC,right angle at C if tanA = 1,then verify that 2 sinA cosA=1 |
| Answer» In triangle ABC , tan A = 1 so A=45*2sinAcosA = 2x1/√2x1/√2= 1 | |
| 3517. |
Trigonometry shortcut |
| Answer» | |
| 3518. |
If the distance P(x,y) from the points A(3,6) B(-3,4) are equal, prove that 3x+y-5=0. |
| Answer» Ok I had proved it | |
| 3519. |
Define prime number |
| Answer» Mazak Hai Kya ???? | |
| 3520. |
tan2a-1/tan a-1=sec2a+tan a |
| Answer» | |
| 3521. |
Construction Mai steps aur justification likhna kyo necessary hai??? |
| Answer» So easy ?? | |
| 3522. |
Cos 90 minus theta is equal to |
| Answer» Sin theta | |
| 3523. |
How\'s going preparation for board exam?? |
| Answer» | |
| 3524. |
Divide x^3-6x^2+11x-6 by x^2+x+1 |
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| 3525. |
In which city we can go but never come |
| Answer» Electricity | |
| 3526. |
Please tell me something about coordinate geometry |
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Answer» It is a branch of mathematics Only 3 formulae r to b learned, n u prepare ur full chapter! Based on Midpoint formula , section formula, distance formula |
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| 3527. |
Hiii everyone!! |
| Answer» | |
| 3528. |
If the sum of n terms of an AP is given by Sn=4n^2-3n, then find the nth term of an AP |
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| 3529. |
A very good evening To all of my dear friend |
| Answer» | |
| 3530. |
Anyone do practice from r.d Sharma nowadays?? |
| Answer» | |
| 3531. |
3 cubes each side 5cm are joined end to end. Find the surface area of the resulting cuboid. |
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Answer» uuyy 350 |
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| 3532. |
the length of tbe minute hand is 14 cm find the area swept by the minute hand in 5 min |
| Answer» Here, r= 14 cm and {tex}\\theta = \\frac { 90 ^ { \\circ } } { 3 } = 30 ^ { \\circ }{/tex}{tex}\\therefore{/tex}Area swept = {tex}\\frac { \\theta } { 360 ^ { \\circ } } \\times \\pi r ^ { 2 }{/tex}={tex}\\frac { 30 ^ { \\circ } } { 360 ^ { \\circ } } \\times \\frac { 22 } { 7 } \\times 14 \\times 14{/tex}= {tex}\\frac { 154 } { 3 }{/tex}\xa0cm2 | |
| 3533. |
what is the value of k if 3x2 |
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| 3534. |
Find k so that 13, k, - 3, are in AP |
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Answer» 5 K=5 5 5 K=5 |
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| 3535. |
pin triangle abc de paralel to bc find value of x |
| Answer» Given in triangle, {tex}{/tex}ABC DE\xa0{tex}\\|{/tex}\xa0BC\xa0{tex}\\frac { \\mathrm { AD } } { \\mathrm { DB } } = \\frac { \\mathrm { AE } } { \\mathrm { EC } }{/tex}\xa0(by BPT){tex}\\Rightarrow \\quad \\frac { x } { x - 2 } = \\frac { x + 2 } { x - 1 }{/tex}{tex}\\Rightarrow{/tex}\xa0x(x - 1) = (x + 2)(x - 2){tex}\\Rightarrow{/tex}\xa0x2\xa0- x = x2\xa0- 22\xa0{tex}\\Rightarrow{/tex}\xa0x2\xa0- x = x2\xa0- 4{tex}\\Rightarrow{/tex} x = 4 | |
| 3536. |
Find k so that 13, k, - 3 are in AP |
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Answer» 6 Maybe 5 |
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| 3537. |
4x^2+2x+4 please sum |
| Answer» | |
| 3538. |
2x+3k=19 |
| Answer» | |
| 3539. |
If the hcf of 85 and 153 is expressible in the form of 85m-153then find the value of m |
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Answer» May be m=1 Hcf of 85 and 153 is 17Now,85m-153=17So, 85m=170 m=170÷85 m=2 |
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| 3540. |
In adjoining figure PQ=6cm, QR=26cm angle PAR=90,PA=6CMAND AR=8cm find angle QPR |
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| 3541. |
Prove that thales theorm |
| Answer» So easy theoram | |
| 3542. |
A chord of a circle of a radius 15 subtends of 60° at the centre. Find the areas |
| Answer» r = 15 cm,\xa0θ = 60oArea of the minor sector = {tex}\\frac\\theta{360^\\circ}\\mathrm{πr}^2\\;=\\;\\frac{\\displaystyle60^\\circ}{\\displaystyle360^\\circ}\\times3.14\\;\\times15\\times15{/tex} = 117.75 cm2In {tex}\\triangle{/tex}AOB, draw OM\xa0{tex}\\perp{/tex} ABIn right triangle OMA and OMB,OA = OB .........Radii of the same circleOM = OM .........Common side{tex}\\therefore{/tex}\xa0{tex}\\triangle{/tex}OMA\xa0{tex}\\cong{/tex}{tex}\\triangle{/tex}OMB .........RHS congruence criterion{tex}\\therefore{/tex}\xa0AM = BM .......CPCT{tex}\\Rightarrow{/tex}\xa0AM = BM = {tex}\\frac 12{/tex}AB{tex}\\angle{/tex}AOM = {tex}\\angle{/tex}BOM .......CPCT{tex}\\Rightarrow{/tex}\xa0{tex}\\angle{/tex}AOM = {tex}\\angle{/tex}BOM = {tex}\\frac 12{/tex}{tex}\\angle{/tex}AOB =\xa0{tex}\\frac 12{/tex}\xa0{tex}\\times{/tex} 60o = 30o{tex}\\therefore{/tex}\xa0In right triangle OMA, cos30o = {tex}\\frac {OM}{OA}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{\\sqrt3}2{/tex}= {tex}\\frac {OM}{15}{/tex}{tex}\\Rightarrow{/tex}\xa0OM = {tex}\\frac{15\\sqrt3}2{/tex}cmsin30o = {tex}\\frac {AM}{OA}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac 12{/tex}= {tex}\\frac {AM}{15}{/tex}{tex}\\Rightarrow{/tex}\xa0AM = {tex}\\frac{15}2{/tex}cm{tex}\\Rightarrow{/tex}\xa0AB = 15 cm{tex}\\therefore{/tex}\xa0Area of {tex}\\triangle{/tex}AOB =\xa0{tex}\\frac 12{/tex}\xa0{tex}\\times{/tex} AB\xa0{tex}\\times{/tex} OM= {tex}\\frac 12{/tex}\xa0{tex}\\times{/tex} 15\xa0{tex}\\times{/tex}{tex}\\frac{15\\sqrt3}2{/tex} = {tex}\\frac{225\\sqrt3}4{/tex}= {tex}\\frac {225 × 1.73}4{/tex} = 97.3125 cm2{tex}\\therefore{/tex}\xa0Area of the corresponding minor segment of the circle = Area of minor sector - Area of {tex}\\triangle{/tex}AOB= 117.75 - 97.3125 = 20.4375 cm2and, area of the corresponding major segment of the circle = {tex}\\pi{/tex}r2 - area of the corresponding minor segment of the circle= 3.14\xa0{tex}\\times{/tex} 15\xa0{tex}\\times{/tex} 15 - 20.4375= 706.5 - 20.4375 = 686.0625 cm2 | |
| 3543. |
Who is Elli in novel the story of young girls |
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| 3544. |
Find the value of sin 60degree geometrically . |
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| 3545. |
2 under root 45 + 3 under root 20÷2under root 5 |
| Answer» | |
| 3546. |
Yaha par AKSHITA THAKUR kon h |
| Answer» | |
| 3547. |
Explain bpt tehoram |
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Answer» Thales theorem ......... It\'s easy ?? So easy ?? |
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| 3548. |
Why math subject was going on |
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| 3549. |
Hii... Eva |
| Answer» | |
| 3550. |
Determine the 10th term the end of the ap 4,9,14......254 |
| Answer» 164 | |