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InterviewSolution
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.
| 41351. |
find the gratest number of 6 digit exactly divisible by 18,24and 36. |
| Answer» 999936 | |
| 41352. |
Formula of find a class mark |
| Answer» Lower limit + Upper limit/2 | |
| 41353. |
2y_12y_432 factories |
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Answer» Factories ? I think it should be written as factorize....?? y=18 or y= -12...... |
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| 41354. |
(100/100)-(100/100)=2 Prove that |
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Answer» Saras bhidu Mr. Vikas these are useless things still for you I give the explainationbut for (100-100)/(100-100){tex}\\begin{array}{l}=\\frac{\\displaystyle10^2-10^2}{10(10-10)}=\\frac{(10-10)(10+10)}{10(10-10)}\\\\=\\frac{10+10}{10}\\\\=\\frac{\\displaystyle20}{10}=2\\\\But\\;this\\;is\\;coming\\;because\\;we\\;used\\;\\frac{10-10}{10-10}=\\;\\frac00=1\\;which\\;is\\;not\\;possible,So\\;such\\;results\\;are\\;useless\\end{array}{/tex} |
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| 41355. |
If cos A +cos square A=1 then find the value of sin square A and sin4 A |
| Answer» Cos A = 1- sin²A = Cos²A Sin²A+sin⁴A = cosA + cos²A (given) = 1 | |
| 41356. |
Show me 2014 board exam paper |
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Answer» DOWNLOAD FROM:https://www.vedantu.com/previous-year-question-paper/cbse-maths-class-10-year-2014 Go |
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| 41357. |
Find the zeroes of polynomial 2x+1÷6x-2 |
| Answer» -1/2 | |
| 41358. |
In an ap of 50 terms the sum of 10 terms is 210 and the sum of its last 15 terms is 2565.find the ap |
| Answer» 3,7,11,15........ | |
| 41359. |
Hcf of two number is 20 find the other two number |
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Answer» Ques is incomplete...plz check once Complete ur question Plz check your question again....... |
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| 41360. |
Draw a line segment of length 7.6 cm and divide it in the ratio 5:8 . Measure the two parts. |
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Answer» Can\'t draw here How can i draw here....... U can see it on cbse guide app maths solution Ncert 11.1 question no 1 |
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| 41361. |
(x-2)=(x-2)(3+7) |
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Answer» (x-2)=(x-2)(10)X-2=10x-2020-2=10x-x18=9xX=2 I hope quadratic equation ????? ???? |
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| 41362. |
Find the condition that the zeroes of the polynomial f(x)=x3+3px2+3qx+r may be in AP |
| Answer» No | |
| 41363. |
Cos 45/ sec 30 + cosec 30 please tell me how to do this? |
| Answer» cos 45 = 1/ √2sec 30= 2/ √3Cosec 30 = 2Cos 45/sec 30 + cosec 30=1/ √2 / 2/ √3 + 2= 1/ √2 * √3/2 + 2= √3/2 √2 + 2\xa0 | |
| 41364. |
Can two number have 18 as there HCF and 380 as there LCM. Give reasons |
| Answer» No, two numbers can\'t have 18 as their HCF and 380 as LCM because HCF of the numbers must be a factor of the LCM.Here 380 is not divisible with 18. | |
| 41365. |
Sin2+cos2=? |
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Answer» 1... Sin+cos2gkxnlckvkg office l One 1..... |
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| 41366. |
Find the sum and product of polynimial 3x²+7x+8=0 |
| Answer» Sum=-7/3 and product=8/3 | |
| 41367. |
Solve the eqation 2x²-ax-1=0 |
| Answer» I think 2✓2 | |
| 41368. |
Find the common difference of the AP 1/p ,1-P/P ,1-2P / P ...…... |
| Answer» -1 | |
| 41369. |
√3/2=3/h |
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Answer» 2√3 If u will simplify than answer will be 2✓3 H=6✓3/3 |
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| 41370. |
Please tell me all the areas and perimeters of all the shapes please please please please please |
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Answer» So simple Yar my cbse guides me na math sub. Khol osme revision notes pade he os se padh le |
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| 41371. |
If A(a,2) , B (-3,-a) ,c (7,-5) are co linear .find n |
| Answer» Since the points are collinear, then,\xa0Area of triangle = 0{tex}\\frac { 1 } { 2 } \\left[ x _ { 1 } \\left( y _ { 2 } - y _ { 3 } \\right) + x _ { 2 } \\left( y _ { 3 } - y _ { 1 } \\right) + x _ { 3 } \\left( y _ { 1 } - y _ { 2 } \\right) \\right] = 0{/tex}{tex}\\frac { 1 } { 2 } [ x ( - 4 + 5 ) + ( - 3 ) ( - 5 - 2 ) + 7 ( 2 + 4 ) ] = 0{/tex}x + 21 + 42 = 0x = -63 | |
| 41372. |
Theorem of BPT |
| Answer» If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio | |
| 41373. |
Prove Pythagoras thoerm |
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Answer» Proved in ncert..... In a right ?, the sq. of hypotenuse is equal to the sum of the sq. of the other two sides |
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| 41374. |
if two vertical of Qudaterial ∆A(3,0) and (6,0) find third vertex |
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Answer» X=9/2 and Y=0 therefore (9/2,0) Chapter 7 coordinate geometry Sorry wrong ans 12 units |
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| 41375. |
If cosA=2/5, find the value of 4 +4tan²A |
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Answer» First of all u have to find ( Perpedicular) by pythagoras theorem. Value of P is root21. Then 4 + 4×(root21/2)^24+ 4×( 21/4)4+2125 Ans 25 |
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| 41376. |
Find perimeter of ∆ with verticals A(0,4) ,B(0,0) and c(3,0) |
| Answer» 12 units | |
| 41377. |
If I got 15/80 in maths of board then will i pass |
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Answer» No bro u will fail if u get 15 marks in maths.Passing marks=27out of 80.passing %=33. Out of 100 marks 20 will be given by school on oral,PT, etc. To pass only 33%out of 100 you will have to score. Be patient!! If I suggest then Just be positive .Don\'t think negative that you will get less marks in board exams.???? Practical means You may if you will get 15+ marks in Practical!! But I am too weak in maths No you need minimum 30 /80 |
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| 41378. |
If x=3 is one root of the quadratic eq x2-2kx-6=0 |
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Answer» x2-2kx-6=0(3)square-6k-69-6k-6=03-6k=0-6k=-3k=-3/-6k =1/2 (x)2 -2kx -6=0,(3)2-2k(3)-6=09-6k-6=03-6k=0So, k=1/2 k = 1/2 sorry wrong solution added.Correct solution isx2 - 2kx - 6 = 0x = 3(3)2 - 2k(3) - 6 = 09 - 6k - 6 = 0- 6k + 3 = 06k = 3k = 3/6k = 1/3 x2 - 2kx - 6 = 0x = 3(3)2 - 2k(3) - 6 = 09 - 6k - 6 = 02 - 6k = 06k = 2k = 2/6k = 1/3 Let p(x)=x^2-2kx-6Let X = 3 in p(x)p(3)=(3)^2 - 2k(3) -6 =9-6k-6 =3-6k -3=-6k 3=6k 3/6=k 1/2=k Value of k=6 |
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| 41379. |
1-sin A ÷1+sinA = (secA+tanA)² |
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Answer» Very nice explain gaurav seth 1-sinA/1+sinA=(secA-tanA)²L.H.S1-sinA/1+sinARationalising the denominator1-sinA (1-SinA) /1+sinA(1-SinA)(1-SinA)²/ 1² -(Sin²A)(1-SinA)²/ 1 -(Sin²A)(1-SinA)² /Cos² A[ 1 -Sin²A = cos²A](1-SinA/CosA)²(1/CosA-SinA/CosA)²(SecA-tanA)²L.H S = R H.S |
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| 41380. |
Find the value of .tan²30+3cos²30 |
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Answer» 1/3+ 9/4=4+27/12=31/12 Tan²30°=√1/3Cos²30°=√3/2So,(√1/3)²+3(√3/2)²1/9+3x3/41/9+9/44+81/3685/36 is the answer Sorry answer is 11/12......by mistAke wrong answer posted 1/3 + 3/2=5/6 |
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| 41381. |
FROM NCERT 8.4 ,5 question and part i |
| Answer» | |
| 41382. |
X² +100x-300000=0Solve for x |
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Answer» X is equal to - 600 or 500 We see that it is a quadratic polynomial.so, splitting the middle term,x²+100x-300000\xa0x²+600x - 500x - 300000 =0x(x+600) -500(x+600) = 0(x-500)(x+600) = 0X = 500 or X = -600 x2 + 100x - 300000 = 0x2 + 600x - 500x - 300000 = 0x(x + 600) - 500 (x + 600) = 0(x + 600) (x - 500) = 0x = -600 and x = 500 |
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| 41383. |
3/x+1_1/2 =2/3x_1=5/x+1 |
| Answer» | |
| 41384. |
If 1+cos theta /1-cos theta =16/9 ,find 1+cot theta /1-cot theta |
| Answer» | |
| 41385. |
Prove that CosA-SinA+1/CosA+SinA-1 =1/CosecA+CotA |
| Answer» | |
| 41386. |
to prove root 2is irrational |
| Answer» let us assume that root 2 is rational no. which is in the form of p/q , where q is not equal to 0.therefore, root 2 = p/q squaring both the sides 2 = p^2/q^2 2q^2 = p^2 ....(1) Here, 2 is divisible by p^2 therefore, 2 is also divisible by plet p = 2c [for some integer c]now, substituting p = 2c in equation (1).therefore, 2q^2 = 4c^2 that is q^2 = 2c^2 this means that 2 is divisible q^2and so 2 is also divisible by q since p and q have 2 as a common factor this contradict the fact that p & q have no common factor other than 1.But this contradiction has arisen due to our wrong assumption that is root 2 is rational no.so, we conclude that root 2 is irrational. | |
| 41387. |
How can we draw construction |
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Answer» Which pencil(company)is preferable while drawing By using paper , pencil , scale ,compass ,eraser ,and never forget to use sharped pencil as it would deduct your marksAlso never use pen in condtruction. Try to write the dimensions of every angle and line as it gives a good presentation.Also write the step of construction. |
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| 41388. |
How to draw the graph of zeros of a polynomial |
| Answer» First of find zeros and Make the table likeEx. X when put the value of X then the and. Ans. Is y | |
| 41389. |
Find lcm of 120 and140 by using fundamental theorm of arithmetic |
| Answer» 144=2*2*2*2*3*3120=2*2*2*3*5lcm=2*2*2*2*3*3*5=720hcf=2*2*2*3=24 | |
| 41390. |
(cos^2 20+cos^2 70 divide by sin^2 20+sin^2 70)+sin 64 +cos64×sin20 |
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Answer» Ceck yur question, i think it should be, cos64×sin26 at last.......... Plz give answer fast |
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| 41391. |
The tangents drawn are epual |
| Answer» Yes but whenThe tangents drawn from an external point are equal not any two tangents are equal | |
| 41392. |
find the distance between the points A(ax2,2ay) and B(ay2,2ax) |
| Answer» ✓x-y is the answer | |
| 41393. |
Solve for x(in term of a. And. b):a/x-b + b/x-a. = 2 |
| Answer» The given equation can be rewritten as{tex}\\frac { a ( x - a ) + b ( x - b ) } { ( x - b ) ( x - a ) } = 2{/tex}a(x -a) + b(x- b) = 2[x2 - (a + b)x + ab]ax- a2\xa0+ bx - b2 = 2x2- 2(a + b)x + 2ab2x2- 3(a + b)x + (a + b)2 = 02x2 - 2(a + b) x - (a + b)x + (a + b)2 = 0{tex}2x[x-(a+b)]-(a+b)[x-(a+b)]{/tex}[2x - (a + b)] [x - (a + b)] = 0{tex}x = a + b , \\frac { a + b } { 2 }{/tex} | |
| 41394. |
hcf and LCM of 9 nd360 one number is 45 ND find the another number |
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Answer» lcm=360 and hcf=9Let the numbers be a and ba=45(given)We know that LCM(a,b)×HCF(a,b)=a×bSo, 360×9=45×b 3240=45×b b=3240÷45 b=72Hence the other number is 72. 72 |
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| 41395. |
In the right triangle B is a point on AC such that AB+AD=BC+CD if AB= x.BC= h and CD= d then find x |
| Answer» Here we are given that{tex}\\therefore \\quad A B + A D = B C + C D{/tex}or,\xa0{tex}AD = BC+CD-AB{/tex}or,\xa0{tex}AD =h+d-x{/tex}In the right angled\xa0{tex}\\Delta A C D,{/tex}\xa0{tex}A D ^ { 2 } = A C ^ { 2 } + D C ^ { 2 }{/tex}or,\xa0{tex}( h + d - x ) ^ { 2 } = ( x + h ) ^ { 2 } + d ^ { 2 }{/tex}or,\xa0{tex}( h + d - x ) ^ { 2 } - ( x + h ) ^ { 2 } = d ^ { 2 }{/tex}\xa0{tex}( h + d - x - x - h ) ( h + d - x + x + h ) = d ^ { 2 }{/tex}\xa0{tex} Because \\ a^2-b^2=(a-b)(a+b){/tex}or,\xa0{tex}( d - 2 x ) ( 2 h + d ) = d ^ { 2 }{/tex}or,\xa0{tex}2 h d + d ^ { 2 } - 4 h x - 2 x d = d ^ { 2 }{/tex}or,\xa0{tex}2hd = 4hx+2xd{/tex}\xa0{tex}2hd= 2x(2h+d){/tex}Hence {tex}x = \\frac { h d } { 2 h + d }{/tex} | |
| 41396. |
(SinA/1+cosA + 1+cosA/sinA) *( sinA/1-cosA - 1- cosA) =4cosecA*cotA |
| Answer» | |
| 41397. |
System of card |
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Answer» 52 cards 26 RED and 26 BLACK KING . QUEEN. JACK.THESE THREE CARDS ARE CALLED FACE CARDS. 10. 9 .8 .7.6.5.4.3.2.1 (ACE)♤-SPADE.♡-HEART.♢-DIAMOND.♧-CLUB.♤&♧ are BLACK and ♡&♢ are RED 52 |
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| 41398. |
if x = sin A cos B , y = r sin A sin B and z = r cos A , show that x^2 + y^2 + z^2 = r^2 . |
| Answer» We have,x2\xa0+ y2 + z2 = r2sin2{tex}\\alpha{/tex}cos2{tex}\\beta{/tex}\xa0+ r2sin2{tex}\\alpha{/tex}\xa0sin2{tex}\\beta{/tex}\xa0+ r2cos2{tex}\\alpha{/tex}= r2sin2{tex}\\alpha{/tex}\xa0(cos2{tex}\\beta{/tex}\xa0+ sin2{tex}\\beta{/tex}) + r2cos2{tex}\\alpha{/tex}= r2sin2{tex}\\alpha{/tex}\xa0+ r2cos2{tex}\\alpha{/tex}\xa0[{tex}\\because{/tex}\xa0cos2{tex}\\beta{/tex}\xa0+ sin2{tex}\\beta{/tex}\xa0= 1]= r2(sin2{tex}\\alpha{/tex}\xa0+ cos2{tex}\\alpha{/tex}) = r2 [{tex}\\because{/tex}\xa0sin2{tex}\\alpha{/tex}\xa0+ cos2{tex}\\alpha{/tex}\xa0= 1].Hence, (x2 + y2 + z2) = r2. | |
| 41399. |
Find the area of cordilateral is vertexes taken in order or (-4,-2) (-3,-5) (3,-2) (2,3). |
| Answer» 28sq.units | |
| 41400. |
Cartoon |
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Answer» ??? What do you mean by cartoon Yes |
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