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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.
| 21651. |
Is 4x2 +8√x + 6 is a quadratic polynomial |
| Answer» No | |
| 21652. |
A unbiased dice is thrown, then what is the probability of getting even number? |
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Answer» 1/2 The probabilty of getting even no. Is 1/2 1/2 |
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| 21653. |
The 19th term of an AP is equal to three times its sixth term.if its 9th term is 19,find the AP |
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Answer» a19=3(a6)a9=19a+8d=19a=19-8d .......(1) a19=3(a6)a+18d=3(a+5d)a+18d=3a+15d18d-15d=3a-a3d=2a3d=2(19-8d) ........(from 1)3d=38-16d3d+16d=3819d=38d=38/2d=2. ......(2)Put d=2 in eq.1a=19-8×2a=3now put values of a &d in above equationa19=3(a6)a+18d=3(a+5d)3+18×2=3(3+5×2)3+36=3(3+10)39=9+3039=39 .•. AP= 3,5,7......H.P.Hope this helps.?Regards |
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| 21654. |
The diametre of frustum of cone is 10 and 30 and hieght is 24 find area if metal sheet and volume? |
| Answer» | |
| 21655. |
Find the value of sin25/cos65+tan23/cot67 |
| Answer» 2 | |
| 21656. |
7,13,19,....205. Ap |
| Answer» Plz continue ur ques | |
| 21657. |
Sum of first n` numbers |
| Answer» Sum of first n numbers = n/2(2a+(n-1)d) | |
| 21658. |
In figure if AD =4cm DB =6cm AE = 6cm and EC =9 and angle ADE = 52 find angle ABC |
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Answer» Angle ABC =52° because through the theorem BPT the ratio of triangle are equal and hence the DE||BC therfore as they form corresponding angles ABC is 52...Hope you will understand angle ABC = 52° |
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| 21659. |
If the points P(x,y) is equidistant from two points A(-3,2) and B(4,-5),,prove that y=x-2 |
| Answer» X=m1x2 +m2x1/m1+m2=4+(-3)/1+1=4-3/2=1/2Y=m1y2+m2y1/m1+m2=-5+2/2=-3Y=x-2-3/2=1/2-2(cross multiply)-3/2=1-4/2-3/2=-3/2 | |
| 21660. |
How to justify the construction? |
| Answer» Prove similar triangles included in your construction. | |
| 21661. |
solve for x and y:3x-y=39x-3y=9 |
| Answer» 3x - y = 3........ (i)9x - 3y = 93x - y = 3 .......... (ii)(i) and (ii) are equal Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by y = 3x - 3 | |
| 21662. |
Find 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 | |
| 21663. |
Find the mode Classes140-145145-150150-155155-160160-165165-170170-175Frequency3,7,10,7,6,2,3 |
| Answer» 155 | |
| 21664. |
Nth tearm of an ap is 2n+1 and find the sum of first three terms |
| Answer» If nth term of an A.P = 2n+1then, first term of A.P = 2(1)+1= 2+1= 3then, second term = 2 (2) + 1 = 4 +1 = 5third term = 2 (3) +1 = 6+1 = 7sum of terms = 3+5+7sum of terms = 15 | |
| 21665. |
Solve for x and y:x+2y-4=02x+4y-12=0 |
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Answer» x + 2y - 4 = 0x + 2y = 4.......... (i)2x + 4y - 12 = 02x + 4y = 12x + 2y = 4 ......... (ii)Here, both the equation (i) and (ii) are same and we know that when both equations are exactly same there are infinitely many solutions and the value of x and y can\'t be predicted. There is no solution ,lines are parallel ?????? |
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| 21666. |
Find the area of minor and major sector with radius 14cm and angle 120° |
| Answer» Minor sector=205/33,. Major sector =41•66 | |
| 21667. |
Find all the zeroes of polynomial(2x⁴_9x³+5x²+3x_1).if two of its zeroes are (2+√3)and(2_√3) |
| Answer» Given:f(x) = (2x4\xa0– 9x3\xa0+ 5x2\xa0+ 3x – 1)Zeroes = (2 + √3) and (2 – √3)Given the zeroes, we can write the factors = (x – 2 + √3) and (x – 2 – √3){Since, If x = a is zero of a polynomial f(x), we can say that x - a is a factor of f(x)}Multiplying these two factors, we can get another factor which is:((x – 2) + √3)((x – 2) – √3) = (x – 2)2 –\xa0(√3)2⇒x2\xa0+ 4 – 4x – 3 = x2\xa0– 4x + 1So, dividing f(x) with (x2\xa0– 4x + 1)f(x) = (x2\xa0– 4x + 1) (2x2\xa0– x – 1)Solving (2x2\xa0– x – 1), we get the two remaining roots as{tex}x = {-b \\pm \\sqrt{b^2-4ac} \\over 2a}{/tex}where f(x) = ax2\xa0+ bx + c = 0(using Quadratic Formula){tex}\\mathrm{x}=\\frac{-(-1) \\pm \\sqrt{(-1)^{2}-4(2)(-1)}}{2(2)}{/tex}{tex}\\mathrm{x}=\\frac{-1 \\pm 3}{4}{/tex}{tex}\\Rightarrow \\mathrm{x}=1,-\\frac{1}{2}{/tex}Zeros of the polynomial =\xa0{tex}1,-\\frac{1}{2}, 2+\\sqrt{3}, 2-\\sqrt{3}{/tex} | |
| 21668. |
Draw aline segment AB of lenght 7cm using ruler compass find apoint P onAB such that AP/AB =3/5 |
| Answer» Refer to ncert or RSA | |
| 21669. |
In chapter 14 how do we come to know that which method of mean we have to use if it is not given |
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Answer» when the frequency and the class intervals are small then the direct mean method is used.......if the class intervals are large and the height of class interval is same then the step deviation is used......and if the class intervals are large but height of the class interval is different then the assumed mean method is used where no height is required.......and if we have to find the value of \'x or y\' then always use direct method......HOPE IT WILL HELP YOU!! U cn use any..if it is not given |
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| 21670. |
If the total marks is 400 and marks secured is 359 then calculate the percentage? |
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Answer» %age will be 89.75% 89.75 The percentage will be 89.75%. |
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| 21671. |
Find the lent of the longest roof that can be fit in a room of 10m.10m.5m dimensions |
| Answer» | |
| 21672. |
7 time 7th term is equal to11 time 11th term find 18th term |
| Answer» 7(a7)=11(a11).....7(a+6d)=11(a+10d).....7a+42d=11a+110d.......a-17d=0.......a18=0 | |
| 21673. |
Introduction of trignometry all formulae |
| Answer» SinA = P / HCosA = B / HTanA = P / BCotA = B / PCosecA = H / PSecA = H/BSin2A + Cos2A = 1Tan2A + 1 = Sec2ACot2A + 1 = Cosec2AIn order to find a relationship between various trigonometric identities, there are some important formulas:1. TanA = SinA / CosA2. CotA = CosA / SinA3. CosecA = 1 / SinA4. SecA = 1 / CosA Thank you?? | |
| 21674. |
2 2 2 Sol: x- 4ax +4a - b = 0 |
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| 21675. |
Which book is best for prActice |
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Answer» NCERT Oswaal books |
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| 21676. |
If the (m)th term of ap be 1/n and (n)th term be 1/m .then show that its (mn)th term is 1 |
| Answer» a+(m-1)d=1/n (i) & a+(n-1)d=1/m (ii) sub eq (ii) from (i) we get (m-n)d=1/n-1/m therefore ,d=1/mn put d value in (i) we get , a=1/mn ,see how,a+(m-1)1/mn=1/n , a+1/n-1/mn=1/n , a=1/mn , now, 1/mn+(mn-1)1/mn=1 , proved,? | |
| 21677. |
Sina+cosa= coseca+seca prove |
| Answer» When we divide both sides by cosA We get cotA+1=CotA +1. | |
| 21678. |
If 5 sin theta is equal to 3 then find the value of cos square theta |
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Answer» 16/25 Value of cos square theta is 16/25 |
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| 21679. |
(x+2)(x+4)(x+6)(x+8) =945 |
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| 21680. |
If a b a 3, a, b, - 3 then find a and b |
| Answer» | |
| 21681. |
If (2m-1)x + 3y=5, 3x +(n-1)y=2 has infinitely many solutions. Find the value of m and n? |
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Answer» Solution is very lengthy so I can\'t post full solution. m=4.25n=2.2 |
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| 21682. |
12 or 21 kya hoga. |
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Answer» 12 Mean |
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| 21683. |
What is the LCM of P and Q where P=a3b2 and Q=b3a2? |
| Answer» a^3b^3 | |
| 21684. |
1+1+1+1+1+11+1+1+1+1+1 |
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Answer» 21 12 / 21 |
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| 21685. |
By selling a motor cycle for $23,000 a dealer gains 15percent it\'s cost price |
| Answer» | |
| 21686. |
Can numbers have 15 as their Hcf and 175 as their lcm....give reason |
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Answer» No AB\'S oxford international school....... No, Because LCM>HCF and HCF of 2 numbers should divide the LCM completely. Here, 175 is not completely divisible by 15. |
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| 21687. |
12cos+9cos=15Find cot |
| Answer» 5/2√6 | |
| 21688. |
12cos+9sin#15Find cot |
| Answer» | |
| 21689. |
Find x in terms of a, b, c:a/x-a+b/x-b=2c/x-c, x not equal to a, b, c |
| Answer» | |
| 21690. |
If secA+tanA=P then find the value of SinA? |
| Answer» SecA=1/cosATanA=sinA/cosA1/cosA+sinA/cosA=P1+sinA/cosA=P1+sinA=PcosASinA=Pcos-1 | |
| 21691. |
EvaluationSinA (1+tanA)+cosA(1+cotA)=secA+cosecA |
| Answer» | |
| 21692. |
Tangent pq |
| Answer» | |
| 21693. |
A,b,c are interior angle of triangle abc. Prove that cosec(A+B)÷2 =sec c÷2 |
| Answer» We know that, A\xa0+ B + C = 180o\xa0{tex}\\Rightarrow{/tex}\xa0A + B =\xa0180o - CDividing both sides with 2, we get{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{A + B}{2}{/tex}\xa0= 90o -\xa0{tex}\\frac{C}{2}{/tex}Applying cosec on both sides, we get{tex}\\Rightarrow{/tex}\xa0cosec({tex}\\frac{A + B}{2}{/tex}) = cosec(90o -\xa0{tex}\\frac{C}{2}{/tex}){tex}\\Rightarrow{/tex}\xa0cosec({tex}\\frac{A + B}{2}{/tex}) = sec\xa0{tex}\\frac{C}{2}{/tex} | |
| 21694. |
2x+7y=19,3x-5y=5 |
| Answer» | |
| 21695. |
How many hours are you spending in maths? |
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Answer» 3 hrs daily....doing a sample paper 2 days...for complete revision 1hr 30 min or 2 hrs daily.......... Three hours. |
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| 21696. |
If sec theta + tan theta =p ,then find the value of cosec theta |
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Answer» SecA+tanA=p ----------------------------(1)∵, sec²A-tan²A=1.......(secA+tanA)(secA-tanA)=1......... secA-tanA=1/p -----------------------(2)Subtracting (2) from (1) we get,2tanA=p-1/p......, tanA=(p²-1)/2p.....cotA=2p/(p²-1)Now, cosec²A-cot²A=1.......cosec²A=1+cot²A...... cosec²A=1+{2p/(p²-1)}²....... cosec²A=1+4p²/(p²-1)²..,..cosec²A=(p⁴-2p²+1+4p²)/(p²-1)²......cosec²A=(p⁴+2p²+1)/(p²-1)².............., cosec²A=(p²+1)²/(p²-1)²...... cosecA=(p²+1)/(p²-1) (P²+1)/(p²-1) (p²+1)/(p²-1) Ans. |
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| 21697. |
Calculate the hcf of 3 |
| Answer» Their have no h.c.f of single digit | |
| 21698. |
EC AP with that is what he is APK trick madhe kay 3 Pado ka Yog 225 Hai A.P gyat karo |
| Answer» | |
| 21699. |
Which chapter take more marks in maths? |
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Answer» Thanks Each ch is equally distribution in marks. Read cbse guidelines about it in maths column. Co-ordinate geometry ch 7 AlgebraWhich includes linear equations,polynomials,quadratic equation |
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| 21700. |
Prove that bpt theoram |
| Answer» See in ur ncert textbook | |