InterviewSolution
Saved Bookmarks
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.
| 41251. |
(a-b)(a+b)=? |
|
Answer» a2-b2 a2-b2 a square - b square |
|
| 41252. |
If the 2nd term of an A.P is 8 and 5th term is 17 find its 1th term |
|
Answer» Second term (a2) = 8fifth term (a5) = 17we know that general formula of a term in ap is a+(n-1)d\xa0therefore, a2=a+d 8=a+da5=a+4d 17=a+4dsolving the two equations simultaneously....we get....d=3{tex}\\therefore{/tex}\xa0a=5 (a= First term) your answer is 5 |
|
| 41253. |
Find greatest no. That will divide 445,572&699 leaving remainder 4,5&6 respectively |
| Answer» We have to find the greatest number that divides 445, 572 and 699 and leaves remainders of 4, 5 and 6 respectively. This means when the number divides 445, 572 and 699 leaves remainders 4, 5 and 6 is that445 - 4 = 441, 572 - 5 = 567 and 699 - 6 = 693 are completely divisible by the required number.For the highest number which divides the above numbers can be calculated by HCF .Therefore, the required number is the H.C.F. of 441, 567 and 693 Respectively.First, consider 441 and 567.By applying Euclid’s division lemma, we get567 = 441 {tex}\\times{/tex}\xa01 + 126441 = 126 {tex}\\times{/tex}\xa03 + 63126 = 63 {tex}\\times{/tex}\xa02 + 0.Therefore, H.C.F. of 441 and 567 = 63Now, consider 63 and 693again we have to apply Euclid’s division lemma, we get693 = 63 {tex}\\times{/tex}\xa011 + 0.Therefore, H.C.F. of 441, 567 and 693 is 63Hence, the required number is 63. 63 is the highest number which divides 445,572 and 699 will leave\xa04,5 and 6 as remainder respectively. | |
| 41254. |
plz explain the chapter 7example 4 in detail |
| Answer» plz ans.dis 1 also | |
| 41255. |
If the distance of p(x,y) from A (6,2) and B(-2,6)are equal then prove that y=2x |
| Answer» If PA=PB therefore PAsq=PBsq (solve this by distance formula you will get the answers) | |
| 41256. |
ABC is a triangle in which |
|
Answer» Sorry Complete ur question |
|
| 41257. |
What to do in 5his question Find probability of 1st multiple of 3or5 |
| Answer» | |
| 41258. |
(a³–b³) |
|
Answer» ???? (a-b)(a^2+ab+b^2) Heyguys (a - b) { a² + b² + ab} a³ - b³ = (a - b) { a² + b² + ab}. |
|
| 41259. |
If tanA=n tanB and sinA=m sinB, prove that cos(square) =m square - 1 upon n square - 1 |
| Answer» Given,\xa0tan A = n tan B{tex} \\Rightarrow{/tex} tanB = {tex}\\frac{1}{n}{/tex}tan A{tex}\\Rightarrow{/tex}\xa0cotB =\xa0{tex}\\frac { n } { \\tan A }{/tex}..........(1)Also given,\xa0sin A = m sin B{tex}\\Rightarrow{/tex}\xa0sin B =\xa0{tex}\\frac{1}{m}{/tex}sin A{tex}\\Rightarrow{/tex}\xa0cosec B =\xa0{tex}\\frac { m } { \\sin A }{/tex}.....(2)We know that, cosec2B - cot2B = 1, hence from (1) & (2) :-{tex} \\quad \\frac { m ^ { 2 } } { \\sin ^ { 2 } A } - \\frac { n ^ { 2 } } { \\tan ^ { 2 } A } = 1{/tex}{tex}\\Rightarrow \\quad \\frac { m ^ { 2 } } { \\sin ^ { 2 } A } - \\frac { n ^ { 2 } \\cos ^ { 2 } A } { \\sin ^ { 2 } A } = 1{/tex}{tex}\\Rightarrow \\quad \\frac { m ^ { 2 } - n ^ { 2 } \\cos ^ { 2 } A } { \\sin ^ { 2 } A } = 1{/tex}{tex}\\Rightarrow{/tex}\xa0m2 - n2cos2A = sin2A{tex}\\Rightarrow{/tex}\xa0m2 - n2cos2A = 1 - cos2A{tex}\\Rightarrow{/tex}\xa0m2 - 1 = n2cos2A - cos2A{tex}\\Rightarrow{/tex}\xa0m2 - 1 = (n2 - 1) cos2A{tex}\\Rightarrow \\quad \\frac { m ^ { 2 } - 1 } { n ^ { 2 } - 1 } ={/tex}\xa0cos2A | |
| 41260. |
What is the value of m is mx(x+7)49 |
| Answer» {tex}mx(x - 7) +49 = 0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}mx^2 - 7mx + 49 = 0{/tex}Here a = m, b = -7m, c = 49{tex}\\Rightarrow{/tex}\xa0{tex}D = b^2 - 4ac{/tex}For equal roots, D = 0{tex}\\Rightarrow{/tex}\xa00 = (-7m)2 - 4 {tex}\\times{/tex}\xa0m {tex}\\times{/tex}\xa049{tex}\\Rightarrow{/tex}\xa0{tex}0 = 49m^2 - 196m{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}49m^2 - 196m = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}7m(7m - 28) = 0{/tex}{tex}\\Rightarrow{/tex}\xa07m = 0 or 7m - 28 = 0{tex}\\Rightarrow{/tex}\xa0m = 0 or m =\xa0{tex}\\frac{28}{7}{/tex}\xa0= 4but m\xa0{tex}\\ne{/tex}\xa00 [{tex}\\because{/tex}In quadratic a\xa0{tex}\\ne{/tex}\xa00]{tex}\\therefore{/tex}\xa0m = 4 | |
| 41261. |
Explain converse theorem |
| Answer» Kiska converse bro | |
| 41262. |
√sec-1/sec+1 + √sec+1/sec-1= 2 cosec |
| Answer» | |
| 41263. |
Find all the zeros of the polynomial 3x3_3x2+6x_2 if two zeros are |
| Answer» | |
| 41264. |
Find K if the sum of the zeroes of the polynomial x^2-(K+6)x+2(2K-1) is half of their product |
|
Answer» Given quadratic polynomial, x² - (k+6)x + 2(2k+1) = 0By comparing it with ax²+bx+c = 0,we geta = 1 , b = -(k+6) , c = 2(2k+1)Sum of zeroes = -b/a= -[-(k+6)]/1= k+6Product of zeroes = c/a= 2(2k+1)/1= 2(2k+1)Now..... Sum of zeroes = ½ × product of zeroes k+6 = ½ × 2(2k+1)k+6 = 2k+12k-k = 6-1k = 5Hope it helps Plz plz..... Answer fast guys it\'s urgent |
|
| 41265. |
If sinx=√cosx and cosx=√sinx find te value of sinsq.x -cossq. X |
| Answer» | |
| 41266. |
If nth term of an A. P is(2n+1) then what is the sum of its three termd |
| Answer» 15 | |
| 41267. |
Is the HCF of a and b is 15 and a into B is 4500 then find the LCM of a and b |
|
Answer» lcm ×15 = 4500=> lcm = 4500/15 => lcm = 300 in this ques appyl (hcf × lcm = product of hcf and lcm ) this formula this ques is right 300 The Question is incorrectBecause there given only b and to find LCM ,first we want to find a which is not possible in this question lcm = 500 |
|
| 41268. |
Find one rational and irarional no. Between √3 and √5. |
|
Answer» Thanks √3 = 1.732......√5 = 2.236.....The numbers which can be expressed in the form of p/q, where q ≠ 0 are called rational numbers. A terminating decimal and a non-terminating non-repeating decimal can be represented as a rational number.Therefore, a rational number between √3 and\xa0√5 is 2.So, three irrational numbers between √3 and √5 are: 1.79877985647984123564........ , 2.0100100010000100001110001.......... , 2.1212121212123121234....... |
|
| 41269. |
what are the important questions from chapter construction?? |
|
Answer» All are important question in that chapter????? There is no important questuons at all , you will trt to understand 1)how to construct a similar triangle in given ratio .2)how to cunstruct tangents of a circle. |
|
| 41270. |
Sn=4n-n×n.Find AP |
|
Answer» Put the value on Sn that n=1 as a first term. Then put n=2 as a second term. putting value n=1 This is a first term Putting value n=2This is a second term Show ap 3 4 .... n=1 , 4(1)-(1)*(1) = 3. n=2. , 4(2)-(2)*(2) = 4. n=3. , 4(3)-(3)*(3) = 3 |
|
| 41271. |
Q. For any positive integer n, prove that n3 – n is divisible by 6.????????? |
| Answer» | |
| 41272. |
How to get centum in board xam ? |
| Answer» | |
| 41273. |
finding hcf of69 and4 using Euclid, s division algorithms and represent in form of m+n. |
| Answer» | |
| 41274. |
The sum of the first nth term of an AP is 4n2 + 2n. Find the nth terms of an AP |
| Answer» It\'s a difficult question?? | |
| 41275. |
SinA+tanA=p then find value of cosecA |
| Answer» 1+sin | |
| 41276. |
1 upon 16 ka haff kitna hoga |
|
Answer» 1 upon 32 1/32 1/32 2/16 1 upon 32 |
|
| 41277. |
Formule\'s of surface Areas and Volumes |
|
Answer» HemisphereP.S.A.πr(square)C.S.A.- 2πr( square)T.S.A.- 3πr (square)Volume- 2/3πr (cuba) Kiska formula of surface area batao mai .....object toh batao Formule\'s of surface areas and volumes |
|
| 41278. |
Formules of chapter 13 |
| Answer» | |
| 41279. |
Prove that-sinA-cosA+1/sinA+cosA-1=1/secA-tanA |
|
Answer» ??????? gud Q.Prove that- sinA-cosA+1/sinA+cosA-1=1/secA-tanAAnswer:We have, |
|
| 41280. |
1/ sec x - tan x -1/cos x= 1/ cos x- 1/ sec x+ tan x |
| Answer» LHS =\xa0{tex}\\frac{1}{sec x - tan x}{/tex}\xa0-\xa0{tex}\\frac{1}{cos x}{/tex}={tex}\\frac{sec x + tan x}{(sec x - tan x)(sec x + tan x)}{/tex}\xa0-\xa0{tex}\\frac{1}{cos x}{/tex}=\xa0{tex}\\frac{secx + tan x}{sec^2 x - tan^2 x}{/tex}\xa0-\xa0{tex}\\frac{1}{cos x}{/tex}{tex}=sec x + tan x - sec x{/tex}{tex}[\\because sec^2\\theta-tan^2\\theta=1]{/tex}= tan xRHS =\xa0{tex}\\frac{1}{cosx - 1}{/tex}\xa0-\xa0{tex}\\frac{1}{sec x + tan x}{/tex}={tex}\\frac{1}{cos x}{/tex}\xa0-\xa0{tex}\\frac{sec x - tan x}{(sec x + tan x)(sec x - tan x)}{/tex}={tex}\\frac{1}{cos x}{/tex}\xa0-\xa0{tex}\\frac{sec x - tan x}{sec^2 x - tan^2 x}{/tex}= sec x -\xa0{tex}\\frac{sec x - tan x}{1}{/tex}{tex}=sec x - sec x + tan x{/tex}{tex}=tan x{/tex}Hence, LHS = RHS | |
| 41281. |
two chords ab and cd intersect at p outside the circle. if ab=6 cm, bp=2 cm and pd= 2.5 cm, find cd |
| Answer» CD=3.9 cm | |
| 41282. |
Sin^4x/a+cos^4x/b=1/a+1/b to prove that sin^8x /a^3+cosx^8/b^3=1/[a+b]^3 |
| Answer» | |
| 41283. |
Consider the following distribution, find the frequency of class 30-40 |
|
Answer» Oyeee bakwas Which distribution?? Complete your ques |
|
| 41284. |
If n is an odd integer then show that n2-1 is divisible by 8 |
| Answer» Let n = 4q + 1 (an odd integer){tex}\\therefore \\quad n ^ { 2 } - 1 = ( 4 q + 1 ) ^ { 2 } - 1{/tex}{tex}= 16 q ^ { 2 } + 1 + 8 q - 1 \\quad \\text { Using Identity } ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }{/tex}{tex}= 16{q^2} + 8q{/tex}{tex}= 8 \\left( 2 q ^ { 2 } + q \\right){/tex}= 8m, which is divisible by 8. | |
| 41285. |
To prove the sum of three sides of a triangle are equal |
| Answer» By sss rule | |
| 41286. |
1/2a+b+2x=1/2a+1/b+1/2xValue of x=? |
| Answer» Pl ease answer koi bolo | |
| 41287. |
Write the rational no between root2 and root3 |
| Answer» 3/2 | |
| 41288. |
Prove that tanA /100-cotA+cotA/1-tanA=1+secA.cosecA |
|
Answer» can you give me a solution of this question plzzz.? It is a sample papet question Lab activity I don\'t find |
|
| 41289. |
If the nth term of an AP is (3x +2) then find sum of n terms of an AP |
| Answer» | |
| 41290. |
For what value of p x=b is azero ofthe polynomial x2-a |
| Answer» | |
| 41291. |
Prove that cotthita _ tan thita=2cos square thita _1÷sin thita cos tjita |
|
Answer» Very good puja you are hardworking and helpful also cot theta - tan thetacos theta/sin theta-sin theta/cos thetacos²theta-sin²theta/sin theta * cos thetacos²theta-1(1-sin²theta) /sin theta * cos thetacos²theta-1+cos²theta/sin theta*cos theta2cos²theta-1/sin theta*cos thetaHence Proved. |
|
| 41292. |
If ∆FEC≅∆GDB, ∠1=∠2. Prove that ∆ADE~∆ABC |
| Answer» Given : ∆ FEC ≅ ∆ GBD ,\xa0SO From CPCTWE GET,BD = CE ----------------- ( 1 )ALSO GIVEN : ∠ 1 = ∠ 2 ,SO FROM BASE\xa0ANGLE THEOREM IN ∆ ADEWE GETAD =AE ------------------------ ( 2 )\xa0From equation 1 and 2 we getADBD = AECE , So from converse of B.P.T. we getDE | | BCTHEN ,∠ 1 = ∠ 3 ( CORRESPONDING ANGLES AS DE | | BC AND AB IS TRANSVERSAL LINE )AND∠ 2 = ∠ 4 ( CORRESPONDING ANGLES AS DE | | BC AND AC IS TRANSVERSAL LINE )FROM ABOVE TWO EQUATIONS WE CAN SAY THAT :∆ ADE ~ ∆ ABC ( ByAArule )( Hence proved ) | |
| 41293. |
What are the important theorems of unit 6. |
|
Answer» Converse* Its bpt and converse and area theorem and Pythagoras and its conclverse There are four most important theorem that i want to knoe Ya, lavanya is ryt All theorams are important |
|
| 41294. |
Use euclid\'s algorithm find HCF of 858 and 325 Express it in the form of 858 X + 325 y |
| Answer» | |
| 41295. |
Thé ratio of the length of the shadow is √3:1.find thé angle of elevation of the sun |
|
Answer» Yes answer is 60°. 60 degree |
|
| 41296. |
Real numbers example |
| Answer» | |
| 41297. |
Hii friends,can you evaluate this equation ? (x + 1)(\\/2x + 3). |
|
Answer» No akarsh , its root 2.? (x+1)(1/2x+3) hai equation ?Agr haan tox=. -1and -6. hllw Honey , I solve it but my answer is not correct, according to me its follow the quadratic equation.? Aashu (x+1)(\\/2x+1) is not equal to zero dude.? It is not an equation. Hii The value of x are. -1 & -3 / root 2 |
|
| 41298. |
If I scored 15/80 in board then will I got internal marks |
| Answer» Tell me | |
| 41299. |
What is different between sector and segment |
|
Answer» A sector is the part of a circle enclosed by two radii of a circle and their intercepted arc. The segment of a circle is the region bounded by a chord and the arc subtended by the chord. Sector is a part of circle made of a triangle and a segment eg-piece of pizza and segment is a part formed after triangle is removed from a sector. |
|
| 41300. |
In A.P if 1st S10=-80 & sum of nxt ten ter=-280...find A.P |
| Answer» Let A.P be a , a + d, a + 2d, ...a is the first term, d is the common differenceThe formula of sum of n terms is\xa0{tex}S_{n}=\\frac{n}{2}[2 a+(n-1) d]{/tex}Where n is the number of termsThe sum of its first 10 term is - 80. Therefore,{tex}-80=\\frac{10}{2}[2 a+(10-1) d]{/tex}-80 = 5[2a + 9d]-16 = 2a + 9d ...(1)Sum of its next 10 terms is -280.{tex}S_{20}=\\frac{20}{2}[2 a+(20-1) d]{/tex}{tex}S_{20}{/tex}\xa0= 10[2a + 19d]Since sum of the next 10 terms of AP is -280. Therefore,{tex}S_{20}-S_{10}{/tex}\xa0=-28010[2a + 19d] - (-80) = -28010[2a + 19d] = 3602a + 19d = -36 ...(2)Subtract (1) from (2)2a + 19d - 2a - 9d = -36 + 1610d = -20d = -2Put value of d in (1)-16 = 2a + 9d-16 = 2a + 9(-2)-16 = 2a - 182 = 2a\xa0a = 1Therefore, The AP series is 1, -1, -3, -5 ... | |