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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.
| 39351. |
Factor of x(x+1)-306 |
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Answer» X=-18and 17 X= -17 and 18 |
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| 39352. |
a×b |
| Answer» ab | |
| 39353. |
Solve it -1.3(8+3) |
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Answer» 14.3 33 14.3 14.3 14.1 |
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| 39354. |
1+tan2= |
| Answer» Sec2 | |
| 39355. |
X+2y=3/2 2x+y=3/2 |
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Answer» X=1/6 and Y=2/3 X=2,y=2 |
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| 39356. |
Find the k value (2k-1)x+(k-1)y=2k+1 |
| Answer» here what is the value of x&y???? | |
| 39357. |
decimal expansion of 1717/2×5™3 |
| Answer» 1.717 | |
| 39358. |
In a a.p the sum of first n terms is 3n2+n find its 22nt term. |
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Answer» pls explain it write the question properly 130 |
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| 39359. |
Polynomial types |
| Answer» | |
| 39360. |
What is the HCF of smallest prime number and the composite number? |
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Answer» 2 Smallest prime no. is 2 and smallest composite no. is 4 thus HCF is 2 itself. 2 1 |
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| 39361. |
(3+√5)×(3-√5) |
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Answer» We use (a+b)(a-b)identity then Answer is 4 (3+√5)(3-√5)=(3)^2-(√5)^2=9-5=4As there is a identity (a+b)(a-b)=a^2-b^2 |
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| 39362. |
X²+7x +10 |
| Answer» X^2+5x+2x+10=0X(x+5)+2(x+5)=0(X+2)or(x+5)=0X=-2,x=-5 | |
| 39363. |
If the area of two similar triangle are equal, prove that they are congruent |
| Answer» Given:{tex}∆ABC\\sim∆PQR{/tex}and\xa0{tex}ar∆ABC=ar∆PQR{/tex}To prove:\xa0{tex}∆ABC\\cong∆PQR{/tex}Proof:\xa0{tex}∆ABC\\sim∆PQR{/tex}\xa0Also\xa0{tex}\\operatorname { ar } ( \\Delta A B C ) = \\operatorname { ar } ( \\Delta P Q R ){/tex}\xa0(given)or,\xa0{tex}\\frac { \\operatorname { ar } ( \\Delta A B C ) } { \\operatorname { ar } ( \\Delta P Q R ) } = 1{/tex}Or\xa0{tex}\\frac{AB^2}{PQ^2}=\\frac{BC^2}{QR^2}=\\frac{CA^2}{RP^2}=1{/tex}Or\xa0{tex}\\frac{AB}{PQ}=\\frac{BC}{QR}=\\frac{CA}{RP}=1{/tex}Hence we get thatAB=PQ,BC=QR and\xa0CA=RPHence\xa0{tex}∆ABC\\cong ∆PQR{/tex} | |
| 39364. |
Root x+y=7. X+ root y =11 |
| Answer» Que is wrong. √X+Y=11X+√Y=7 | |
| 39365. |
Prove that both the roots of equations(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 |
| Answer» The given equation is; (x - a)(x - b)+(x - b)(x - c)+(x - c)(x - a) =0\xa0{tex}\\Rightarrow{/tex}\xa0(x2\xa0- ax - bx + ab) + (x2- bx - cx + bc) +(x2\xa0- cx - ax + ac) = 0{tex}\\Rightarrow{/tex}3x2\xa0- 2x(a + b + c) + (ab + bc + ca) = 0.....(1).\xa0Discriminant \'D\' of quadratic equation (1) is given by;{tex}\\therefore{/tex}\xa0D = 4(a + b + c)2\xa0- 12(ab + bc\xa0+ ca)= 4[(a + b + c)2\xa0- 3(ab + bc + ca)]= 4(a2 + b2 + c2\xa0- ab - bc\xa0- ca)= 2(2a2\xa0+ 2b2 + 2c2 -\xa02ab - 2bc\xa0- 2ca)= 2 [(a - b)2 + (b - c)2 + (c - a)2] ≥ 0[{tex}\\because{/tex}\xa0(a - b)2\xa0≥ 0, (b - c)2\xa0{tex}{/tex}\xa0≥ 0 and (c - a)2\xa0{tex}{/tex}\xa0≥ 0].This shows that both the roots of the given equation are real.For equal roots, we must have D = 0.Now, D = 0 {tex}\\Rightarrow{/tex}\xa0(a - b)2\xa0+ (b - c)2\xa0+ (c - a)2= 0{tex}\\Rightarrow{/tex}\xa0(a - b) = 0, (b - c) = 0 and (c - a) = 0 (sum of squares can be zero only if they all are equal to 0){tex}\\Rightarrow{/tex}\xa0a = b = c.Hence, the roots are equal only when a = b = c | |
| 39366. |
1/2-0.5= |
| Answer» =0 | |
| 39367. |
X2 -6x+3=0 |
| Answer» D =b2 -4ac =36-12 =24x=-b +root D /2a x=6+2root 6/2x=2(3 +root 6)/2 x=3+root6x =-b -root D /2a x=6-2root 6/2x=3-root6 | |
| 39368. |
For what value of K , (-4) is a zero of P(X)=x square - x - (2k-2)? |
| Answer» P(X) =X²- X - (2K-2) P(-4) =(-4)² - (-4) - (2K-2 IF P(X) =0 16+4-2K+2=0 22-2K=0 22=2K 22/2=K 11=K | |
| 39369. |
Any thoghest question of maths chapter triangles for class 10th |
| Answer» example number 13 | |
| 39370. |
-4,_,_,_,_,6 |
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Answer» -2 -2 |
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| 39371. |
what is the syllabus for2019 |
| Answer» Ncert me zitna chapter hai bss utna hi padhna hai | |
| 39372. |
What is the formula of substitution method |
| Answer» No formula will be there only we substitute the value of x and y and put into other equation | |
| 39373. |
5x+2x =2 (3x+1 |
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Answer» X=1 X=2 X=2 X=1 |
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| 39374. |
ax square +7x +b =0 find value of a and b if x=2/3 and x=-3 |
| Answer» a=3 and b=6 | |
| 39375. |
How money will add to paytam |
| Answer» | |
| 39376. |
11term of the AP -3,-1/2, 2 ........is\xa0 |
| Answer» The given AP is\xa0{tex} - 3,\\frac{1}{2},2,......{/tex}Here, a = -3{tex}d = - \\frac{1}{2} - ( - 3) = - \\frac{1}{2} + 3 = \\frac{5}{2}{/tex}and n = 11a11 = ?We have, an = a + (n - 1)dSo,\xa0{tex}{a_{11}} = - 3 + (11 - 1)\\left( {\\frac{5}{2}} \\right){/tex}{tex} \\Rightarrow {a_{11}} = - 3 + 25{/tex}{tex} \\Rightarrow {a_{11}} = 22{/tex} | |
| 39377. |
What is rational and irrational .. |
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Answer» a no. of the form p/q or which could be express in the form of p/q where p and q are integers and q is not equal to 0 is known as rational no. a no. which is not of the form p/q or which could nog be expressed in the form p/q where p and q are integers and q is not equal to 0 is known as irrational no. p/q=0 rational number p/q=/o |
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| 39378. |
How many terms of the ap: 9,17,25...must be taken to give a sum of 636 |
| Answer» According to the question,we have,\xa0a=9. Therefore, common difference d =17-9=8let the required number of terms be n.Therefore, Sn=636{tex}\\Rightarrow{/tex}{tex}\\frac{n}{2}{/tex}[2a+(n-1)d]=636{tex}\\Rightarrow{/tex}{tex}\\frac{n}{2}{/tex}[2(9)+(n-1)8]=636{tex}\\Rightarrow{/tex}n[18+8n-8]=1272{tex}\\Rightarrow{/tex}8n2+10n-1272=0{tex}\\Rightarrow{/tex}4n2+5n-636=0{tex}\\Rightarrow{/tex}4n2+53n-48n-636=0{tex}\\Rightarrow{/tex}n(4n+53)-12(4n+53)=0{tex}\\Rightarrow{/tex}(4n+53)(n-12)=0{tex}\\Rightarrow{/tex}4n+53=0 or n-12=0{tex}\\Rightarrow{/tex}n={tex}\\frac{{ - 53}}{4}{/tex}\xa0or n=12Since number of terms cannot neither be negative nor fraction, n=12hence, the required number of terms is 12. | |
| 39379. |
Can give Solution of 14.1 by all three methods of mean |
| Answer» Fhvcnhg FL | |
| 39380. |
1/2-/2-1/2-x |
| Answer» | |
| 39381. |
How to verify the sums of intesers |
| Answer» | |
| 39382. |
Did cbse put question only ncert book |
| Answer» Questions are based on information given in ncert books but the questions are asked in different ways. | |
| 39383. |
How can we solve math and understand very well |
| Answer» By continuous practise | |
| 39384. |
If alpha and beta aer the zeroes of p(x)-x2-x-1 find alpha and beta |
| Answer» (1+√5)/2 and (1-√5)/2 | |
| 39385. |
Show that sin b+c/2 =cos a/2 |
| Answer» \xa0A, B, C, are interior angles of a\xa0{tex}\\Delta {/tex}{tex}\\because A + B + C = 180 ^ { 0 }{/tex}{tex}\\Rightarrow B + C = 180 ^ { 0 } - A \\Rightarrow \\frac { B + C } { 2 } = 90 ^ { 0 } - \\frac { A } { 2 }{/tex}{tex}\\Rightarrow \\sin \\frac { \\mathrm { B } + \\mathrm { C } } { 2 } = \\sin \\left( 90 ^ { \\circ } - \\frac { \\mathrm { A } } { 2 } \\right) \\left[ \\because \\sin \\left( 90 ^ { 0 } - \\theta \\right) = \\cos \\theta \\right]{/tex}{tex}\\Rightarrow \\sin \\frac { \\mathrm { B } + \\mathrm { C } } { 2 } = \\cos \\frac { \\mathrm { A } } { 2 } \\text { proved }{/tex}LHS = RHS | |
| 39386. |
Sin+cos=√2sin then find value of sin -cos |
| Answer» | |
| 39387. |
Find the area of the shaded region if PQ=24cm PR=7 cm and O is the centre of the circle |
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Answer» Give me solution of these question 84 sq.cm |
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| 39388. |
Is b/a solution of a^2x^2-3a^2b^2x+2b^2 |
| Answer» | |
| 39389. |
A one root of quadratic equation2x^2+px-15=0 is -5 find the value of p solution |
| Answer» Given that -5 is the root of\xa0{tex}2 x^{2}+p x-15=0{/tex}Put x = -5 in\xa0{tex}2 x^{2}+p x-15=0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}2(-5)^{2}+p(-5)-15=0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}50-5 p-15=0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}35 - 5p = 0\xa0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}5p = 35\xa0{/tex}{tex}\\therefore{/tex}\xa0{tex}p = 7{/tex}Hence the quadratic equation p {tex}(x^2 + x) + k = 0{/tex} becomes,\xa0{tex}7\\left(x^{2}+x\\right)+k=0{/tex}{tex}7 x^{2}+7 x+k=0{/tex}\xa0Here {tex}a = 7,\\ b = 7\\ and\\ c = k{/tex} Given that this quadratic equation has equal roots\xa0{tex}\\therefore{/tex}\xa0{tex}b^{2}-4 a c=0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}7^{2}-4(7)(\\mathrm{k})=0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}49 - 28k = 0\xa0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}49 = 28k\xa0{/tex}{tex}\\therefore{/tex}\xa0k = {tex}\\frac{49} {28}{/tex}\xa0=\xa0{tex}\\frac{7} {4}{/tex} | |
| 39390. |
2/√x+3/√y=2;4/√x-9/√y |
| Answer» Question incomplete | |
| 39391. |
If 3sinA + 5cosA =5 prove that 5sinA - 3cosA=+-3 |
| Answer» | |
| 39392. |
If x= y and x and y are some integers then prove that 1= 2. |
| Answer» | |
| 39393. |
What do you mean by Euclid division algorithm |
| Answer» If a and b are positive integers such that a=bq+r,then every common divisor of a and b is a common divisor of b and r, and vice-versa | |
| 39394. |
1-5-5+5442-*75-/55 |
| Answer» | |
| 39395. |
(sec theta -tan theta)^2 =1-sin theta/1+sin theta |
| Answer» | |
| 39396. |
Sin theta = cos theta prove |
| Answer» | |
| 39397. |
cosA-sinA+1/cosA+sinA-1=cosecA-cotA |
| Answer» LHSDIVIDE NUMERATOR AND DENOMINATOR BOTH BY sin A =(CosA÷sinA - sinA÷sinA +1÷sinA)/(cosA÷sinA + sinA÷sinA - 1÷sinA)=(CotA-1+cosecA)/(cotA+1-cosecA)At the place of 1 at the numerator write (cosec2A -cot2A)={CotA+cosecA-(cosec2A-cot2A)}/(cotA-cosecA+1)={CotA+cosecA-(cosecA+cotA)(cosecA-cotA)}/(cotA-cosecA+1)=(CotA+cosecA){1-(cosecA-cotA)}/(cotA-cosecA+1)={(cotA+cosecA)(1-cosecA+cotA)}/(cotA-cosecA+1)=CotA+cosecAHENCE PROVEDLHS=RHS | |
| 39398. |
cosA-sinA+1---------------------. = cosecA+cotAcosA+sinA-1 |
| Answer» | |
| 39399. |
ax-by=a2-b2,x+y=a+b |
| Answer» x=a ,y=b | |
| 39400. |
Find yX+y=14 |
| Answer» | |