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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.
| 28351. |
What is common difference between of an A. P. in which a21 - a7 =84 |
| Answer» an=a+(n-1)dFor a21a21=a+(21-1)da21=a+20d-------[1]Now for a7,a7=a+(7-1)da7=a+6d---------[2]a21-(a7)=a+20d-(a+6d) 84 =a+20d-a-6d 84=14d d=84/14 d=6Answer:The common difference of this arithmetic progression is 6. | |
| 28352. |
If the diameter of a semi circular protactor is 14cm, then find its perimeter |
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Answer» Diameter=2×radius14=2×rr=7cmCircumference of a semi circle=1/2×2πr =22/7×7 =22cm 22 222 7 36cm..... |
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| 28353. |
If p(E) and P(E bar) are zeroes of polynomial then find the polynomial and verify the relationship |
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| 28354. |
What is the area of triangle Formed by the line bx+ay=ab with the coordinate axes? |
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| 28355. |
Tri to trygonmetry |
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Answer» Im unable to study .....Bcoz of loveTrigono metry used to access concentration ,So..what i do.. What\'s your question ??? |
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| 28356. |
Determine the vertex which contains a right side in ∆ABC where A(4,-2) B(7,9) C(7,-2). |
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| 28357. |
nahi |
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Answer» Bus nahi Kya nhi.......??? |
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| 28358. |
Determine no problem |
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Answer» What????? ?? ????? |
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| 28359. |
Find the hcf of 448,1008,&168 using fundamental theorems of arithmetic |
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Answer» 7*2*2*2=56 56 |
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| 28360. |
three men and eight women can complete a peice of work together in 10 hour |
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Answer» Kaamchor.... Akele nahi kar sakte,????? Arre itne chote kam ke 11 bande ko employe karne ki kya jarurat??? Question incomplete hai..???????? Hmm, to kam pura ho gya......???, question to complete kr diya kro??? |
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| 28361. |
the HCF and LCM of two number are 9and 360 respectively . If one number is 45find the other number |
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Answer» The HCF and LCM of two numbers are 9 and 360.HCF = 9 and LCM 360if a and b are two numbers , a = 45 then find value of bHCF × LCM = Product of two numbers9 × 360 = 45 × bb = 9 × 360/45b = 360/5b = 72 Use the formaulaeLcm×hcf=prdouct of 2 numbers....to find the number Other no is 72 72 |
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| 28362. |
1/2a+b+2x=1/2a+1/b+1/2x solve for x |
| Answer» x= -a and x =-b/2 | |
| 28363. |
DEW \' AB parallel EW AD = 4 cm DE = 12 cmDW = 24 cm Find the value of DB |
| Answer» DB=8cm | |
| 28364. |
Ok bye friends |
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Answer» Byee? Bye bye |
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| 28365. |
Double cone form by right triangle |
| Answer» Samajh ni aya question.........??? | |
| 28366. |
Solve for x:{1/(a+b+x)}=1/a+1/b+1/x |
| Answer» Transfer 1/x to lhs and then solve it | |
| 28367. |
prove that [ 1/(sec^-2xcos^2x) +1/(cosec^2x-sin^2x)]sin^2xcos^2x= 1-sin^2xcos^2x/2+sin^2xcos^2x |
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| 28368. |
b/a x+a/by=a2+b2 x+y=2ab Solve and find x and y |
| Answer» b/a.x+a/b.y=a^2+b^2-------(1)x+y=2ab-------(2)Multiplying (2) by a/ba/b.x+a/b.y=2ab.a/ba/b.x+a/b.y=2a^2----------(3)(1)-(3)b/a.x-a/b.x=b^2-a^2Taking x commonx(b^2-a^2)/ab=b^2-a^2x=abSimilarly y=ab | |
| 28369. |
2[sin6A+cos6A]-3[sin4A+cos4A]+1=0 |
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Answer» Sorry, put these in LHS Sin^6A+cos^6A=(sin^2A+cos^2A)^3-3sin^2Acos^2A(sin^2A+cos^2A)=1-3sin^2Acos^2And. Sin^4A+cos^4A=(sin^2A+cos^2A)^2-2sin^2Acos^A=1-2sin^2Acos^2APut these values in Rhs 2[sinA +sin (90-A)] - 3[ sin4A + sin(90-4A)] +12[1] -3[1] +1=0 |
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| 28370. |
Find the roots of the quadratic equations x2-x/5+1/100=0 |
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Answer» X2+x/5-1/100 =0100x2-20x+1/100=0.[Both sides multiply by 100]100x2-20x+1=0100x2-10x-10x+1=010x(10x-1)-1(10x-1)=0(10x-1)(10x-1)=0(10x-1)=0 , 10x-1=010x=1, 10x=1X=1/10, x=1/10 Multiplying both sides by 100100x^2-20x+1=0100x^2-10x-10x+1=010x(10x-1)-(10x-1)=0(10x-1)(10x-1)=0X=1/10 |
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| 28371. |
Apne rd Sharma complete kar liya ky |
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Answer» Mistake ky instead of kya Hmm..... |
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| 28372. |
Soo gaye kya.... |
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Answer» Answer aa raha hai but explain me how... Bhai thoda sbr to rkho... insan hun computer thodi na hun??? |
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| 28373. |
Solve hua kya |
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Answer» Rs.5365.80 Are answer bata to diya |
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| 28374. |
Koi help me ... |
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Answer» Paii mere ta haal khud hi maade hai..Google pa dekh le ques!! Mera questions solve karo plzzz ... Oo paii Kya ho gya??Jd mrne lgunga ta mainu ds deyi..Mainu bda mn kr ra hai kisi di arthi chakne da!!! |
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| 28375. |
Please give me all formulae ch. 4 math class x |
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Answer» Mai aise hi pareshaan ho rha tha ale pehle Kyoo NHi btaya Bt kuch jyada hi bda hai... smjha kro... vrna isi app pe chapter 4 ke notes pe jao kuch help mil jayegi Book m se nikaalne hote to yha Par kiyo bolta Book....me check kro...... |
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| 28376. |
In a tri ABC , DE||BC .if AD is 2.4 AE is 3.2DE is 2 Bc is 5Find BD nd CE |
| Answer» We have,DE || BCNow, In {tex}\\triangle{/tex}ADE and {tex}\\triangle {/tex}ABC{tex}\\angle A = \\angle A{/tex}\xa0[common]{tex}\\angle A D E = \\angle A B C{/tex}\xa0[{tex}\\because{/tex}\xa0DE || BC {tex}\\Rightarrow{/tex}\xa0Corresponding angles are equal]{tex}\\Rightarrow \\triangle A D E= \\triangle A B C{/tex}\xa0[By AA criteria]{tex}\\Rightarrow \\frac { A B } { B C } = \\frac { A D } { D E }{/tex}\xa0[{tex}\\because{/tex}\xa0Corresponding sides of similar triangles are proportional]{tex}\\Rightarrow \\frac { A B } { 5 } = \\frac { 2.4 } { 2 }{/tex}{tex}\\Rightarrow A B = \\frac { 2.4 \\times 5 } { 2 }{/tex}{tex}\\Rightarrow{/tex}\xa0AB = 1.2 {tex}\\times{/tex}\xa05= 6.0 cm{tex}\\Rightarrow{/tex}\xa0AB = 6 cm{tex}\\therefore{/tex}\xa0BD = AB - AD= 6 - 2.4= 3.6 cm{tex}\\Rightarrow{/tex}\xa0DB = 3.6 cmNow,{tex}\\frac { A C } { B C } = \\frac { A E } { D E }{/tex}\xa0[{tex}\\because{/tex}\xa0Corresponding sides of similar triangles are equal]{tex}\\Rightarrow \\frac { A C } { 5 } = \\frac { 3.2 } { 2 }{/tex}{tex}\\Rightarrow A C = \\frac { 3.2 \\times 5 } { 2 }{/tex}= 1.6 {tex}\\times{/tex}\xa05= 8.0 cm{tex}\\Rightarrow{/tex}\xa0AC = 8 cm{tex}\\therefore{/tex}\xa0CE = AC - AE= 8 - 3.2= 4.8 cmHence, BD = 3.6 cm and CE = 4.8 cm | |
| 28377. |
Using Euclid division leema. Show that the qube of any odd integer is 9q+1 or 9q+8 |
| Answer» Let a be any positive integer and b = 3a = 3q + r, where q ≥ 0 and 0 ≤ r < 3Therefore, every number can be represented as these three forms. There are three cases.Case 1: When a = 3q,\xa0Where m is an integer such that m = Case 2: When a = 3q + 1,a\xa03\xa0= (3q +1)\xa03\xa0a\xa03\xa0= 27q\xa03\xa0+ 27q\xa02\xa0+ 9q + 1\xa0a\xa03\xa0= 9(3q\xa03\xa0+ 3q\xa02\xa0+ q) + 1a\xa03\xa0= 9m + 1\xa0Where m is an integer such that m = (3q\xa03\xa0+ 3q\xa02\xa0+ q)\xa0Case 3: When a = 3q + 2,a\xa03\xa0= (3q +2)\xa03\xa0a\xa03\xa0= 27q\xa03\xa0+ 54q\xa02\xa0+ 36q + 8\xa0a\xa03\xa0= 9(3q\xa03\xa0+ 6q\xa02\xa0+ 4q) + 8a\xa03\xa0= 9m + 8Where m is an integer such that m = (3q\xa03\xa0+ 6q\xa02\xa0+ 4q)\xa0Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8. | |
| 28378. |
(1+cotA+tanA)(sinA-cosA) =sinAtanA-cotAcosA |
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Answer» Taking l.h.s. (1+cosA/sinA+sinA/cosA)(sinA-cosA) ,=(sinA.cosA+sin sq A+cos sq A)/sinA.cosA-(sinA-cosA)=1+sinA.cosA/sinA.cosA (sinA-cosA)=(cosecA.secA+1)(sinA-cosA) similarly prove r.h.s into sinA&cosA proving bahut long hai. (1+cotA+tanA)(sinA-cosA)= (sinA+sinAcotA +tanAsinA-cosA -cosAcotA -tanAcosA)=(sinA+sinAcosA/sinA+tanAsinA-cosA-cosAcotA -cosAsinA/cosA)= (sinA+cosA+tanAsinA -cosA- cosAcotA -sinA)= tanAsinA -cosAcotA\xa0= sinAtanA-cotAcosA |
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| 28379. |
Sorry ha p v.. 3 hai |
| Answer» ?????? | |
| 28380. |
Duniya Mein Paanch Badshah Hai Char to Taash Ki Gaddi ke Har EK Jiska Tum status padho ho |
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Answer» Yaar to dildar h Koi n Bhai Re nawab tu Mujhe bhul GA mai Tera bhi badshah hu..... Vase Sahi kya tunee mujhe apne saath count na krke.... Tu kha mamuli sa badshah or Mai Tera badshah..... |
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| 28381. |
Star jee aap itna gussa kyon ho. |
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Answer» Bhai m ch dadri t su Bro aap garhara se bol rahe ho kya. Bhai mera naam atul h |
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| 28382. |
(X+3/x-2) - (1-x/x) = 17/4 find the value of x |
| Answer» (X+3/x-2)-(1-x/x)=17/4=>(x^2+3x-3x+x^2+2)/(x^2-2x)=17/4=>(17x^2-34x)=(8x^2+8)=>9x(x-4)+2(x-4)=0=>(9x+2)(x-4)=0Either x=4 or x=-2/9 | |
| 28383. |
Find the value of p fir which one root of the quadratic equation px²-14x+8 =0 is 6 times the other . |
| Answer» Puja ab mera answer do | |
| 28384. |
How to prove that the 2root3 /5 is irrational |
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| 28385. |
Find the roots of the equation x2 minus 3x minus m(m+3) |
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Answer» Wlcm thank u so much By splitting the middle term x2 -3x - m(m-3)=0x2+mx-(m+3)x-m(m+3)=0x(x+m)-(m+3)(x+m)=0(x+m)(x-m-3)=0Therefore x= -m and m+3 |
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| 28386. |
Find the perimeter in(cm) of a square circumscribing a circle of radius a cm |
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Answer» The radius of circle is a so the diameter will become 2a . Therefore 2a will be the side of the square The area of square is 4*length = 4*2a=8a |
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| 28387. |
If two zeros of tje polynomial x4_6x3_26x2+138x_38 are 2 +_root3 find other zeroes |
| Answer» Two zeros are {tex}2\\pm\\sqrt3{/tex}Sum of Zeroes\xa0{tex}2 + \\sqrt { 3 } + 2 - \\sqrt { 3 } = 4{/tex}and product of zeroes =\xa0{tex}( 2 + \\sqrt { 3 } ) ( 2 - \\sqrt { 3 } ) = 4 - 3 = 1{/tex}Hence quadratic polynomial formed out of this will be a factor of given polynomial,So, x2 - (sum of zeroes)x + product of zeroes= x2 - 4x + 1 will be a factor of given polynomial,Divide given polynomial with x2 - 4x + 1 to get other zeroes.Now,x2 -2x - 35= x2 - 7x + 5x - 35= x(x - 7) + 5(x - 7)= (x - 5) (x - 7){tex}\\therefore {/tex}\xa0Zeros arex = 7 and x = -5{tex}\\therefore {/tex} Other two zeros are 7 and -5\xa0 | |
| 28388. |
any number by o is infinity than why not o by o is infinity doubt |
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Answer» ????, no idea... apka question hi samajh ni aya most brilliant question in the world...........only 5 percent people told answer correctly |
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| 28389. |
SinA-CosA+1/SinA+CosA-1=1/SecA-TanA Proov it |
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| 28390. |
How to solve X=1/2-1/2-1/2-xFor getting the value of x |
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| 28391. |
Let the 3 no\'s x+1,x+2,x+7 are in AP find value of x |
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Answer» They are not in ap They re not in ap |
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| 28392. |
Circle theorm |
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| 28393. |
Factorization of x- 1/x |
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| 28394. |
Solve for x; 1/a+b+x=1/a+1/b+1/x |
| Answer» X=-a and-b | |
| 28395. |
Theorem 10.1 class 10th mathematics explanation |
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| 28396. |
how many marks in all chapter |
| Answer» Vo google karke dekh mil jayega teko?? | |
| 28397. |
Express the trigonmetric ratios sin A, Sec A, and tan A in terms of cot A |
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Answer» I cant understand what r u telling??? You should convert them by using firmukae |
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| 28398. |
Find the area of the largest ∆ that can be inscribed in a semicircle of radius r units |
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| 28399. |
Write the sum of first n even numbers..? |
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Answer» कबीर जी सैड सैड थे आज कुछ आप ? राम राम जी Hii kabir jii n( n+1 )..... |
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| 28400. |
Find the sum of 8 multiples of 3 |
| Answer» The AP will form 3,6,9,12,15,18,21,24 A=3 d=6-3=3 an=24S8=n÷2(a + an ) =8÷2(3+24) =4(27) =108 | |