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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.
| 40601. |
Can anybody give me answer in detain . What is ogive |
| Answer» It\'s ozeev also | |
| 40602. |
How to prove theorem |
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Answer» Hence proved Which theorem |
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| 40603. |
How can we find class size when class intervals are different |
| Answer» adjust it by subtracting .5 from lower limit and add .5 to the upper limit | |
| 40604. |
If the point p (k-1,2)is equidistant from the point A (3, k) and B(k, 5),find the value of K. |
| Answer» We know that\xa0{tex}P A = \\sqrt { ( k - 1 - 3 ) ^ { 2 } + ( 2 - k ) ^ { 2 } }{/tex}and{tex}P B = \\sqrt { ( k - 1 - k ) ^ { 2 } + ( 2 - 5 ) ^ { 2 } }{/tex}Given that PA = PBSquaring both the sides,PA2 = PB2{tex}\\Rightarrow{/tex}\xa0(k - 1 - 3)2 + (2 - k)2 = (k - 1 - k)2 + (2 - 5 )2{tex}\\Rightarrow{/tex}\xa0(k - 4)2 + (2 - k)2 = (-1)2 + (3)2{tex}\\Rightarrow{/tex}\xa0k2 + 16 - 8k + 4 + k2 - 4k = 1 + 9{tex}\\Rightarrow{/tex}\xa02k2 - 12k + 10 = 0{tex}\\Rightarrow{/tex}\xa0k2 - 6k + 5 = 0{tex}\\Rightarrow{/tex}\xa0k2 - 5k - k + 5 = 0{tex}\\Rightarrow{/tex}\xa0k(k - 5) -1(k - 5)= 0{tex}\\Rightarrow{/tex}\xa0(k - 5)(k - 1) = 0{tex}\\Rightarrow{/tex}\xa0k = 5 or k = 1 | |
| 40605. |
Best samples papers for 2019 |
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Answer» CBSE issued Sample Papers for current session. Check that . Changes have been made in papers. Oswaal , u like |
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| 40606. |
which constant should be added and subtracted to solve quadratic equation 5xsq.-root2x+3 |
| Answer» | |
| 40607. |
If the point p(_1,2)is equidistant from the point A(3,k)and B(k, 5),find the value of K. |
| Answer» K=-1 | |
| 40608. |
Can 2 number have 16 as their HCF and 380 as their LCM? give reason |
| Answer» Here HCF=16Let the numbers be a and b.We know that a and b both are divisible by HCF\xa0and LCM is divisible by a and b bothHence, LCM must be divisible by HCF i.e; 16Now 380=23{tex}\\times{/tex}16 + 12We get that 12 is the remainder while dividing 380 by 16, It means that 380 is not divisible by HCF 16Therefore\xa0two numbers cannot have 16 as their HCF and 380 as their LCM | |
| 40609. |
What is the area of equilateral traiangle |
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Answer» √3/4 a² Root 3by 4 side square Sum of all sides Root 3 upon 4 side square |
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| 40610. |
If sin theta +cos theta=√2 . Then find the value of----- tan theta+ cot theta=? |
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Answer» here given ...sin θ + cos θ = √2we have to findtanθ + cot θ =?solution:-sin θ + cos θ = √2now square on both side= ( sin θ + cos θ )² = √2²= (sin² θ + cos² θ )+ 2 sin θ cos θ = 2= 1+ 2sin θ cos θ = 2=> sin θ cos θ = 1/2nowtanθ + cot θ =sin θ/cos θ + cos θ/sin θ=( sin² θ +cos² θ) / sin θ cos θ= 1 / (1/2) = 2 answer 2 2 |
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| 40611. |
When cbse will realease datesheet for class 10 2019? |
| Answer» | |
| 40612. |
21st |
| Answer» | |
| 40613. |
3root7is irrational numbeer prove |
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Answer» Product of ratinal and irrational is always irrational As a rational no. is wrong and 2√7is a irrational no. HENCE PROVED This is the complete answer . Let 2√7 as a rational no. So, it can be written in form p\\q where q is not equal to 0 .So, 2√7= p\\q Now, we will solve it further so it will be,√7=P\\2q But we already know that √7 is an irrational no, so our contraprediction of 2√7 is a |
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| 40614. |
-2+3 |
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Answer» 1 +1 1 +1 |
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| 40615. |
4π25-5π55 |
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Answer» -550 -550 -78.5398163397 15 |
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| 40616. |
A farmer connect a pipe of |
| Answer» | |
| 40617. |
All formulas of probability |
| Answer» | |
| 40618. |
If cot(x) = 2 then find (2+2sinx)(1−sinx)(1+cosx)(2−2cosx) |
| Answer» | |
| 40619. |
Let d be the HCF of 24 and 36 find the two numbers a and b such that d = 24a+36b |
| Answer» Let d be the HCF of 24 and 36.24 = 23{tex}\\times{/tex}336 = 22{tex}\\times{/tex}32HCF = 22{tex}\\times{/tex}3 = 12{tex}\\Rightarrow{/tex}{tex}d = 12{/tex}Now {tex}d = 24a + 36b{/tex}{tex}12 = 24a + 36b{/tex}When\xa0{tex} a = -1{/tex} and {tex}b = 1{/tex}, we get12 = 24{tex}\\times{/tex}(-1) + 36{tex}\\times{/tex}(1)= -24 + 3612 = 12So, a = -1 , b = 1 satisfy the equation d = 24a + 36b{tex}\\therefore{/tex} One possible value of a and b is -1 and 1. | |
| 40620. |
Prove cos1°.cos2°.cos3°.........cos90°=0 |
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Answer» As cos90°=0 so if cos90° is multiplyed by any other number the answer will be 0 •.• cos90°=0•.• cos1°.cos2°.cos3°..........(0)=0 |
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| 40621. |
What is geometric AP |
| Answer» | |
| 40622. |
Find the value of K for which the points (3K-1,K-2),(K,K-7)and (K-1,-K-2) are collinear. |
| Answer» Given points are A(3k -1, k - 2), B(k, k - 7) and C(k-1, -k - 2)We know that points A, B, C will be collinear, if the area of the ΔABC =0Area of ΔABC={tex}\\frac{1}{2}{/tex}|x1(y2 -\xa0y3)+x2(y3\xa0- y1)+x3(y1\xa0- y2)|here, x1\xa0=3k-1, x2\xa0= k, x3= k-1, y1= k-2, y2= k-7, y3= -k-2Area of ΔABC = 0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{1}{2}{/tex}|(3k−1)[(k−7)−(−k−2)]+k[(−k−2)−(k−2)]+(k−1)[(k−2)−(k−7)]|=0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{1}{2}{/tex}|(3k−1)(k−7+k+2)+k(−k−2−k+2)+(k−1)(k−2−k+7)|=0{tex}\\Rightarrow{/tex}\xa0|(3k−1)(2k−5)+k(−2k)+(k−1)(5)|=0{tex}\\Rightarrow{/tex}\xa0|3k(2k−5)−1(2k−5)−2k2+5k−5|=0{tex}\\Rightarrow{/tex}\xa0|6k2−15 k−2k+5−2k2+5k−5|=0{tex}\\Rightarrow{/tex}\xa0|4k2 - 10k - 2k|=0{tex}\\Rightarrow{/tex}\xa04k2 - 12k = 0{tex}\\Rightarrow{/tex}\xa04k(k-3) = 0{tex}\\Rightarrow{/tex}\xa0k = 0 or k - 3 =0{tex}\\Rightarrow{/tex}\xa0k = 0 or k = 3 | |
| 40623. |
Chapter 15 probability. Ka 14 ka answer |
| Answer» No. Of cards in a well shuffed deck=521= a king of red color=2/52=1/262= a face card =12/52=3/133=a red color card=6/52=3/264=the Jack of hearts=1/525=a spade=13/52=1/46= the queen of diamonds=1/52 | |
| 40624. |
If the mean of 1,2,3,...,x is 6x/11,find the value of x. |
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Answer» Answer is 11 9 |
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| 40625. |
Find the sum of all natural numbers from 1 to 100 which are divisible by 2 or 5 |
| Answer» 2 will divide the numbers:2, 4, 6, 8, 10, 12…98, 100it will cover the multiples of 10 which will also be divisible by 5.so for 5 the remaining divisible numbers are:5, 15, 25, …85, 95both the sequences are A.P whose sum are = 50/2(2 + 100) = 2550 and 10/2(5 + 95) = 500 so total sum is 3050 | |
| 40626. |
Prove the area ratio theorem? |
| Answer» Ntg | |
| 40627. |
If sin A =3/4, calculate cos A and tan A. |
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Answer» 3/root 5?????? =3/√5 We haveSinA= 3/4Then p/h =3/4Let p=3k and h=4kThen,by Pythagoras\' theoremb=root5Hence, cosA=root5/4 tanA =3/root5. cos a =5/4 and tan a=3/5 |
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| 40628. |
Any blue print by cbse |
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Answer» You can check the current latest\xa0sample papers uploaded by CBSE on its official website. No |
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| 40629. |
P(a Sin a, a cos a) q(a cos a, - a sin a) |
| Answer» Given that p=2sinA / 1+cosA+sinA and q=cosA / 1+sinA We have to prove p+q = 1 Proof: LHS | |
| 40630. |
If sec theta + tan theta =p then find csc theta |
| Answer» | |
| 40631. |
Area of triangle=what |
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Answer» For right triangle root 3/4×side×side root is only at 3 1/2×base×height |
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| 40632. |
Proove that (secA-tanA)^2 (1+sinA)=1-sinA |
| Answer» Plz send this fast | |
| 40633. |
If x=2 and x=3 are the solution of 3x sq - 2kx + 2m = 0 then what is final value of k and m..... |
| Answer» M=-2.1 K=-7.5 | |
| 40634. |
Using prime factorization method find the hcf and LCM=product of 3 numbers??? |
| Answer» The prime factors of:378 = 2 {tex} \\times{/tex}\xa033\xa0{tex}\\times{/tex} 7180 = 23\xa0{tex}\\times{/tex} 32\xa0{tex}\\times{/tex} 5420 = 22\xa0{tex}\\times{/tex} 3 {tex}\\times{/tex} 7{tex}\\times{/tex} 5HCF = 2 {tex}\\times{/tex} 3=6and LCM(378, 180, 420) = 22\xa0{tex}\\times{/tex} 33\xa0{tex}\\times{/tex} 5 {tex}\\times{/tex} 7 = 3780HCF {tex}\\times{/tex} LCM = 6 {tex}\\times{/tex} 3780 = 22680Product of given numbers is = 378 {tex}\\times{/tex} 180 {tex}\\times{/tex} 420 = 28576800Hence, HCF\xa0{tex}\\times{/tex}\xa0LCM {tex}\\ne{/tex}\xa0Product of three numbers. | |
| 40635. |
x^2+(a/a+b+a+b/a)x+1=0 by quadratic formula |
| Answer» We have, {tex}x^2 +({a \\over a+b}+ {a+b \\over a})x +1 = 0{/tex}{tex}\\implies x^2 + {a \\over a+b}x + {a+b \\over a}x + 1=0{/tex}{tex}\\implies x (x + {a \\over a+b}) +{a+b \\over a}(x + {a \\over a+b}) = 0 {/tex}{tex}\\implies (x + {a \\over a+b}) (x +{a+b \\over a}) = 0{/tex}{tex}Either \\, x + {a \\over a+b} = 0 \\,or\\, x + {a+b \\over a}=0{/tex}{tex}\\implies x = -{a \\over a+b} ,\\, - {a+b \\over a}{/tex}{tex}\\therefore x = -{a \\over a+b} ,\\, - {a+b \\over a}{/tex} are the required roots. | |
| 40636. |
If G be the centroid of a triangle ABC then prove that AB^2 + BC^2+ AC^2=3(AG^2+GB^2+GC^2) |
| Answer» Here,G is\xa0the centroid of a triangle ABC.Let A(b, c), B(0, 0) and C(a, 0) be the coordinates of {tex}\\Delta{/tex}ABC then coordinates of centroid are {tex}G\\left[ {\\frac{{a + b+0}}{3},\\frac{c+0+0}{3}} \\right]{/tex}To prove:-(AB)2 + (BC)2 + (CA)2 = 3(GA2 + GB2 + GC2)Consider : L.H.S.=(AB)2 +( BC)2 + (CA)2= b2 + c2 + a2 + (a - b)2 + c2= b2 + c2 + a2 + a2 + b2 - 2ab + c2= 2a2 + 2b2 + 2c2 - 2abConsider :\xa0R.H.S.=3(GA2 + GB2 + GC2){tex}=3\\left[ {{{\\left( {\\frac{{a + b}}{3} - b} \\right)}^2} + {{\\left( {c - \\frac{c}{3}} \\right)}^2} + {{\\left( {\\frac{{a + b}}{3}} \\right)}^2}} \\right.{/tex}{tex}\\left. { + {{\\left( {\\frac{c}{3}} \\right)}^2} + \\left( {\\frac{{a + b}}{3} - a} \\right) + {{\\left( {\\frac{c}{3}} \\right)}^2}} \\right]{/tex}{tex} = 3{\\left[ {{{\\left( {\\frac{{a - 2b}}{3}} \\right)}^2} + {{\\left( {\\frac{{2c}}{3}} \\right)}^2} + \\left( {\\frac{{a + b}}{3}} \\right)} \\right.^2}{/tex}{tex}\\left. { + {{\\left( {\\frac{c}{3}} \\right)}^2} + {{\\left( {\\frac{{b - 2a}}{3}} \\right)}^2} + {{\\left( {\\frac{c}{3}} \\right)}^2}} \\right]{/tex}{tex} = 3\\left[ {\\frac{{{a^2} + 4{b^2} - 4ab}}{9} + \\frac{{4{c^2}}}{9} + \\frac{{{a^2} + {b^2} + 2ab}}{9}} \\right.{/tex}{tex}\\left. { + {{\\frac{c^2}{9}}} + \\frac{{{b^2} + 4{a^2} - 4ab}}{9} + {{\\frac{c^2}{9}}}} \\right]{/tex}{tex}= 3 \\left[ {\\frac{{{a^2} + 4{b^2} - 4ab + 4{c^2} + {a^2} + {b^2} + 2ab + {c^2} + {b^2} + 4{a^2} - 4ab + {c^2}}}{9}} \\right]{/tex}{tex} = 3\\left[ {\\frac{{6{a^2} + 6{b^2} + 6{c^2} - 6ab}}{9}} \\right]{/tex}{tex} = 3 \\times 3\\left[ {\\frac{{2{a^2} + 2{b^2} + 2{c^2} - 2ab}}{9}} \\right]{/tex}= 2a2 + 2b2 + 2c2 - 2abL.H.S. = R.H.S.Therefore,\xa0(AB)2 + (BC)2 + (CA)2 = 3(GA2 + GB2 + GC2) | |
| 40637. |
Prove that:(1+cotA+tanA)(sinA-cosA)=sinAtanA-cotAcosA |
| Answer» (1 + cotA + tanA) (sinA - cosA)= sinA - cosA + cotA sinA - cotA cosA + sinA tanA - tanA cosA{tex}= \\sin A - \\cos A + \\frac{{\\cos A}}{{\\sin A}} \\times \\sin A - \\cot A\\cos A + \\sin A\\;\\tan A - \\frac{{\\sin A}}{{\\cos A}} \\times \\cos A{/tex}= sinA - cosA + cosA - cotA cosA + sinA tanA - sinA= sinA tanA - cotA cosA | |
| 40638. |
Ncert exemplar |
| Answer» What\'s the question give me your number WhatsApp | |
| 40639. |
Sin90+cos45+sec60 |
| Answer» Sin90-a+cos45+sec90-60=cos90+cos45+sec30=0+1÷√2+2÷√3 | |
| 40640. |
Prove that √1-sin theta /1+sin theta=sec theta-tan theta |
| Answer» | |
| 40641. |
divide P(x) by g(x)p(x)=x^4-3x^2+4x+5g(x)=x^2+1-x |
| Answer» {tex}p ( x ) = x ^ { 4 } - 3 x ^ { 2 } + 4 x + 5 , \\quad g ( x ) = x ^ { 2 } + 1 - x{/tex}p(x) is in standard form.g(x), in standard form, is x2 - x + 1.Now, we apply the division algorithm to the given polynomial p(x) and g(x).We stop here since degree (8) = 0 < degree (x2 - x + 1)So, quotient = x2 + x - 3, remainder = 8Therefore,{tex}\\text { Quotient } \\times \\text { Divisor } + \\text { Remainder }{/tex}{tex}= \\left( x ^ { 2 } + x - 3 \\right) \\left( x ^ { 2 } - x + 1 \\right) + 8{/tex}{tex}= x ^ { 4 } - x ^ { 3 } + x ^ { 2 } + x ^ { 3 } - x ^ { 2 } + x - 3 x ^ { 2 } + 3 x - 3 + 8{/tex}{tex}= x ^ { 4 } - 3 x ^ { 2 } + 4 x + 5{/tex}= DividendTherefore, the division algorithm is verified. | |
| 40642. |
Solve by formula method x^2+4x+5 |
| Answer» x^2+4x+5x^2+5x-x+5x(x+5)-1(x+5)(x+5)(x-1)x=-5,1 | |
| 40643. |
Largest positive integer that will divide 398 ,436and 542 leaving remainder 7, 11 and 15 |
| Answer» We need to find the largest number that will divide 398, 436 and 542 leaving remainders 7, 11 and 15 respectively. Hence, we need to find the HCF of 3 numbers (398-7), (436-11) and (542-15). Hence, the largest number that will divide 398,436 and 542 leaving remainders 7, 11 and 15 respectively is 17. | |
| 40644. |
Value of 9! |
| Answer» The\xa0formula for the calculating of factorial of a positive integer n is illustrated below:n! = n . (n - 1) . (n -\xa02) ... 3 .\xa02 .\xa01\xa09! =\xa09 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 =\xa0362880 | |
| 40645. |
If sn denotes the sum of first n terms of an a.p.,prove that s30=3(s20-s10) |
| Answer» Hii | |
| 40646. |
How to Find a rational no b/w root 2 and root3 |
| Answer» Bhai root 2 ki value 1.41 and root 3 ki 1.732 ke beec ke leekh de jaise 1.68,1.49,1.63 etc | |
| 40647. |
Find greatest number that will divide 445,552 and 699 leaving remainder 4, 5 and 6 respectively. |
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Answer» sorry,at the place of 696 there will be 693 445-4=441552-5=547699-6=696now find the HCF of 441,547,696 |
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| 40648. |
What is the rules of congurency |
| Answer» Three rules of congurency 1.sss(side side side)2.ASA(angle side angle)3.SAS(side angle side) For prove two equal triangle | |
| 40649. |
What is the weightage of marks in all chapter of math |
| Answer» | |
| 40650. |
Cot2A-1/Sin2A |
| Answer» | |