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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.
| 26451. |
Formula of minor segment |
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Answer» Area of circle-area of major secror sorry Cirle-major sector |
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| 26452. |
Two dairy owner Aand B |
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| 26453. |
Find the circumference of circle whose diameter is 55cm |
| Answer» 172.8 | |
| 26454. |
SinQ+cosQ=2 |
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| 26455. |
State basic proportional theorem |
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| 26456. |
3/cos^2-3tan^2 |
| Answer» 3 | |
| 26457. |
Theoram 10.1 |
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| 26458. |
Find the no. To be added to the polynomial x2 -5x +4 |
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| 26459. |
Find the quadratic polynomial while zeroes are √2+3 and √2-3 |
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| 26460. |
√64 |
| Answer» 8 | |
| 26461. |
x^2+2x-143 solve by factorisation |
| Answer» x²+2x-143 | |
| 26462. |
If cotø = 15/8 ,then ealuvate ( 2+2sinø)(1-sinø) /(1+cosø)(2-2cosø) |
| Answer» 8/15 | |
| 26463. |
Find the First term and next term AP is 9,15,21,27.... |
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Answer» A1=9 and next term is 33 a1=9 , a2=15 |
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| 26464. |
If cosA_sinA equal to root 2sinA, prove that cosA+sinA equal to root2 sinA |
| Answer» CosA - SinA = √2sinA => CosA = sinA + √2sinA => sinA ( 1+ √2 )= CosA => SinA = CosA / (1 + √2 ) rationalise it and get the answer | |
| 26465. |
Prove, a³+b³=(a+b)(a²+b²-ab) |
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| 26466. |
Mid point thoeom |
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| 26467. |
What was the population of Muslim in India in 2015 census |
| Answer» 138,188,240 | |
| 26468. |
Is -1 a term of an AP27,21,10.....-65?justify |
| Answer» Ans is no because we r getting n=(-43/3)so n can never be in - term | |
| 26469. |
What is distance formulae |
| Answer» Under root (x2-x1)^2 + (y2-y1)^2 | |
| 26470. |
Evaluate: sinAcosA-sinAcos(90-A)cosA/sec(90-A)-cosAsin(90-A)sinA/cosec(90-A) |
| Answer» Answer is 0SinAcosA - sinAcos^2A/secA - cosAsin^2A/cosecA= SinAcosA - sinAcos^3A - cosAsin^3A=SinAcosA(1- cos^2A - sin^2A)=SinAcosA(0)=0 | |
| 26471. |
What is the full form of ADP ?? |
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Answer» ***adenosine Adenosene di phosphate Adenosine diphosphate Anedosine di phosphate |
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| 26472. |
Value of cos 90 |
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Answer» Hey it\'s 0.. 0 |
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| 26473. |
N/2 =2a +n-1*d proof the statement |
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| 26474. |
In fig find perimeter of DEFG |
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| 26475. |
Mx+ny=m+n find the value of x and y m(1/m-n -1/m+n)x +N(1/n-m - 1/m+n)y=2m/m+n |
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| 26476. |
Which is best way for revision?? |
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| 26477. |
Find value of k for which one root for quadratic equation kx2-14x+8=0 is six times the other |
| Answer» Take one root =¥ and other 6¥ and u can use ¥+6¥ = -b/a formukae | |
| 26478. |
What is the probability that a leap year has 53 Sundays? |
| Answer» 2/7 | |
| 26479. |
tan 30 sec 45 + tan 60 sec 30 |
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Answer» Root 6+6/3 √3+9/3 1/underroot3*underroot 2+underroot 3 *2/underroot 3=2+underroot 2/3 |
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| 26480. |
How to prove trigonometry identity |
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| 26481. |
Exam date Aa gai kya? |
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| 26482. |
What is the meaning of Trigonometry |
| Answer» Trikonmiti | |
| 26483. |
Is everything in mathematics come from NCERT. |
| Answer» No...u have to use extrabooks | |
| 26484. |
Solve √(2x+9) +x=13 |
| Answer» (2x+9)=(13-x)^2 - --->(2x+9)=169+ x^2-26x - - - - >0=160+x^2-28x - - - >0=x^2-28x+160 - - - >0=x^2-20x-8x+160 - - \'>0=x(x-20) - 8(x-20) - -\' >(x-8)(x-20) - - - >x=8,20. | |
| 26485. |
If alpha and bita are zeroes of x*x-6x+k.What is the value of k, if 3alpha-2bita =20? |
| Answer» sum of zeroes=a+b =-b/a=6 a=6-b {given -> 3a-2b=20} 3*(6-b)-2b=20 - - - >18-3b-2b=20 - - - > -5b=2 - - - >b=-2/5 - - - > a=32/5 - - ->product of zeroes =a*b=k {32/5*(-2/5)=k) - - \'>k= - 64/25. | |
| 26486. |
What is pythagorous thearem |
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Answer» H^2=B^2+P^2 Square of hypotenuse is equal to sum of square of perpendicular and base In right angled triangle, H^2= P^2+ B^2 Study it from ncert Hypotenuse square=base square+perpetual square |
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| 26487. |
In the figure given find ‹QSR |
| Answer» Given: PQ and PR are tangents to a circle with centre O and {tex}\\angle{/tex}QPR = 50°.To find: {tex}\\angle{/tex}QSR{tex}\\angle{/tex}QOR + {tex}\\angle{/tex}{tex}QPR = 180°{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\angle{/tex}{tex}QOR + 50° = 180°{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\angle{/tex}{tex}QOR = 130°{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\angle{/tex}QSR = {tex}\\frac {1} {2}{/tex}\xa0{tex}\\angle{/tex}QOR{tex}\\Rightarrow{/tex}\xa0{tex}\\angle{/tex}QSR = {tex}\\frac {1} {2}{/tex}\xa0{tex}\\times{/tex}\xa0{tex}130° = 65°{/tex} | |
| 26488. |
solution of course study material of KV Pinjore maths |
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| 26489. |
Is 7×5×3×2+3 a composite number ? Justify your answer |
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| 26490. |
Find the value of k for which the quadratic equation (k-2)x + 2(2k-3) + (5k-6) =0 have equal roots |
| Answer» In quadratic equation, (k - 2)x2 + 2(2k - 3)x +\xa04(5k - 6) = 0a = (k-2) , b = 2(2k-3) , c = 4(5k-6)Since equation has equal roots,{tex}\\therefore {/tex}\xa0D = 0{tex}b ^ { 2 } - 4 a c = 0{/tex}{tex}\\{ 2 ( 2 k - 3 ) \\} ^ { 2 } - 4 ( k - 2 ) ( 5 k - 6 ) = 0{/tex}\xa0{tex}4 \\left( 4 k ^ { 2 } - 12 k + 9 \\right) - 4 ( k - 2 ) ( 5 k - 6 ) = 0{/tex}\xa0{tex}4 k ^ { 2 } - 12 k + 9 - 5 k ^ { 2 } + 6 k + 10 k - 12 = 0{/tex}\xa0{tex}k ^ { 2 } - 4 k + 3 = 0{/tex}\xa0{tex}k ^ { 2 } - 3 k - k + 3 = 0{/tex}{tex}k(k-3)-1(k-3)=0{/tex}\xa0{tex}(k-1)(k-3) =0{/tex}{tex}\\therefore k=1 \\ or \\ k=3{/tex} | |
| 26491. |
X+a/x-a + x+b/x-b + x+c/x-c=3 solve for x |
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| 26492. |
Board exams kab se hai? |
| Answer» Nhi nikla | |
| 26493. |
Example of sabadrup |
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| 26494. |
Define sabadrup....with example |
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| 26495. |
Which term is the first negetive term in the AP= 22,19,16,... |
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Answer» 9th 9th ...that is -1 9th term 9the term |
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| 26496. |
What formule applied in surface area and volume |
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Answer» Cuboid :- vol = lbh , surface area = 2[ lb + bh + lh ] Cube :- vol = a³ |
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| 26497. |
Sino |
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| 26498. |
Pragya??? |
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| 26499. |
How to form equations for speed related questions??? |
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| 26500. |
If one of the zeros of quadratic polynomial (k-1)x² +Kx+1 is -3 then the value of k is ? |
| Answer» Given that the sum of the zeros of the polynomial kx2\xa0- 3x + 5 is 1, we have to find the value of k.We have, f(x) =\xa0kx2\xa0- 3x + 5Sum of zeros =\xa0{tex}\\frac { - \\left( \\text { coefficient of } x ^ { 2 } \\right) } { \\text { coefficient of } x } = \\frac { - ( - 3 ) } { k } = \\frac { 3 } { k }{/tex}Given, sum of zeros = 1{tex}\\therefore \\frac{3}{k}{/tex}\xa0= 1{tex}\\Rightarrow{/tex} k = 3 | |