InterviewSolution
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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.
| 41151. |
How many 3 digit nos. Are divisible by 7? |
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Answer» a=105d=7 l=994An=a+(n-1)d994=105 + ( n-1)7 (-105)889=(n-1)7 (÷7) 127=n-1 ( +1)128=nn=128There are 128 there digit numbers which are divisible by 7. a=105d=7An=994An=a+(n-1)d994=105+(n-1)7n=896 |
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| 41152. |
Find all the zeros Of The polynomial p(x)=2xrais4 -7xcube-19x sq +14x+30 if two zeroes are √2&-√2 |
| Answer» x-√2and X+√2 are factorsSo x^2-2 is factorOn dividing 2x^4-7x^3-19x^2+14x+30 byx^2-2 we get2x^2-7x-15=0(x-5)(2x+3)=0other two zeros are 5,-3/2 | |
| 41153. |
X= secA + sinA and Y= secA-sinA prove that 2uponX+ykawholesqare+x-yupon2ka whole sqare =y |
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| 41154. |
show that the points (1 ,7) (4, 2) (-1, -1) and(-4 ,4 )are the vertices of a square |
| Answer» Let the given vertices be A (1,7), B (4,2), C (-1,-1), D (-4,4)AB = √ (4-1)2 + (2 -7)2 = √ 9 + 25 = √ 34BC = √ (-1-4)2 + (-1-2)2 = √ 25 + 9 = √ 34CD = √ (-4+1)2 + (4 +1)2 = √ 9 + 25 = √ 34AD = √ (1 +4)2 + (7 -4)2 = √ 25 + 9 = √ 34AC = √ (-1-1)2 + (-1-7)2 = √ 4 + 64 = √ 68BD = √ (-4-4)2 + (4 - 2)2 = √ 64 + 4 = √ 68Clearly, AB = BC = CD = DA and AC = BDHence, given points are the vertices of a square | |
| 41155. |
Convert the decimals into ascendig order |
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| 41156. |
Prove that 7 - √5 is an irrational number. |
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Answer» This is so easy yar Let if possible it is assumed that 7-√5 is rational no. 7- √5 = p/q. where p nd q are integers and q. qis not equal to 0 √5 = p/q - 7 Since p and q are integers so, p/q - 7 is rational and so √5 is rationalBut this contradicts the fact that √5 is irrational This is bcoz of our wrong assumptionSo we conclude that 7 - √5 is irrational |
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| 41157. |
Prove that : cosA-sinA+1=cosecA+cotA _____________ cosA+sinA-1 |
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Answer» Hope yu can understand.... Divide numerator and denominator by sin A. Then , put value of 1 as cosec square A - cot square A=1, in the numerator uske bad ap common lijiye or numerator and denominator se cancel kijiye uske bad prove ho jayega. Firstly divide L. H. S by sinA Mai yahan pe answer kaise dun apko By mistake question is written wrong,Prove that:cosA-sinA+1/cosA+sinA-1=cosecA+cotA |
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| 41158. |
tan2alpha =1+ 2 tan2 beta prove that 2sin2 alpha = 1+ sin2 beta |
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Answer» Thanks but. I cant dear Hope, yu can understand.......... I cant type the answer here sorry,.......bt ap pehele tan alpha or tan beta dono side me sin or cos ki terms me convert kijiye and name it equation (1) now 1 ko l.h.s pe le ayiye or ise phir solve kijiye yu will get (cos square bita =2 cos square alpha) now put this value in equation 1 and yu will get the answer Please anyone |
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| 41159. |
chapter 13.3 last question |
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| 41160. |
Find the lcm of 16/9 and 32/27 |
| Answer» LCM is (48/27.32/27).... | |
| 41161. |
divide 3k^2 - k^3 - 3k + 5 ÷ k - 1- k^2 |
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| 41162. |
tanA + secA -1/tanA-secA +1 |
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| 41163. |
Sin squared +1by 1 tan squared=? |
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| 41164. |
Find the value of p so that the quadratic equation px(x-3)+9=0has two equal roots |
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Answer» 0,4 After solving px(x-3)+9=0 u will get px square - 3px +9=0 now u can take discriminant equal to zero for equal roots |
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| 41165. |
Vol. Of hemisphere is 2425.5cm^3. Find its CSA.(use π=22/7) |
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Answer» Given : Volume of hemisphere = 2425.5\xa0cm3Volume of hemisphere = 2/3πr³2425.5 = 2/3*22/7*r³r³ = (2425.5*21)/44r³ = 50935.5/44r³ = 1157.625r = 10.5 cmNow,\xa0Curved Surface Area of hemisphere = 2πr²= 2*22/7*10.5*10.5= 693\xa0cm2CSA of hemisphere is 693 cm2 U can first find the radius of hemisphere as the volume is already given and after finding the radius, u can find its csa |
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| 41166. |
Which are given below by monika |
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| 41167. |
How you can say this give me complete solutions of this 2 questions |
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Answer» Which are given below by monika mudgal Which |
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| 41168. |
2. Write the number of solution of the following pair of linear equation: x+2y-8=0 , 2y+4y = 16 |
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Answer» Is 2y+4y=16 correct check Infinite solutions. |
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| 41169. |
The zeros of the quadratic polynomial x2+99x+127 |
| Answer» X=(-99+-√99×99-4×1×127)/2=(-99+-√9293)/2 | |
| 41170. |
Divide 15000by 2:3:2 |
| Answer» R1+R2+R3=2+3+2=71st part=2/7×15000=2142.85×2=4285.72nd part=3/7×15000=6428.63rd part=4285.7 | |
| 41171. |
K wala ncert ka |
| Answer» ?? | |
| 41172. |
What are the zeroes of a polynomial? |
| Answer» Solutions of a poly. OR roots of a poly. | |
| 41173. |
What is adfected quadratic equation |
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| 41174. |
Solve quadratic equation. 1/x+3+2/2x+1=3/x+4 |
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Answer» Anshika are u frm gujart Bolo 14xsquare -11x -4=0ye equation he kya boli Shi hain, pata thodi hain Juthe honey tum to just paas the aur sare questions ke answer de rhe ho waah topper ban gye kya (1 +-. root 39 )/2 |
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| 41175. |
If Mth term of an A.P. is 1÷n and the nth term is 1÷m then find the sum of its first mn term |
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Answer» Let a and d be the first term and common difference respectively of the given AP,then\xa0mth\xa0term = a+(m-1)d ==> a+(m-1)d = 1/n -----(1)\xa0and nth\xa0term=a+(n-1)d ==> a+(n-1)d = 1/m -----(2)\xa0subtracting (2) from (1), we have Let the first term of the A.P be \'a \' and its common difference be \'d \'Now, mth term of the A.P is 1/nwhich means, a + (m - 1)d = 1/n --- (i)Again,its nth term is 1/mwhich means, a + (n - 1)d = 1/m --- (ii)on substracting eqn. (ii) from (i) we get,a + (m - 1)d - a - (n - 1)d = 1/n - 1/m=> d (m - 1 - n + 1) = (m - n)/mn=> d (m - n) = (m - n) / mn=> d = 1/mnso, a = 1/m - (n - 1)(1/mn) = n - n + 1 / mn =1/mnie., a = d = 1/mnNow, Smn = mn/2 [2a + (mn - 1)d]= mn/2 [2/mn + (mn - 1)1/mn]= mn/2 [2 + mn - 1]/mn= (mn + 1)/2 mn+1by2 |
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| 41176. |
If cosec thita + cot thita = p prove that cos thita = p^2-1/p^2+1 |
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| 41177. |
In an AP:[an] if common difference is 3 find a37-a15 |
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Answer» Most most most most most wlcm Most wlcm neha Thanks to all 66 Yes 66 is correct answer ....... Just make equation like a37= a+36d and then solve accordingly but I solved it the answer is 66 66 |
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| 41178. |
22+79+76+75-72+65+34 |
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Answer» Right answer is 423 Neha is right No no no no 276 Is 579 423 galat ans 279 |
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| 41179. |
cross multiplication |
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| 41180. |
Any one help me to prepare chapter A. P....... |
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Answer» Don\'t worry jitna jyada questions solve karoge utne hi expert hote jaoge You can tell if u have problem in any question or any querie Ya tell the question Just try to understand the questions first... Read it atleast two times and then answer it. Also learn all the formulae of this chapter |
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| 41181. |
Find the zeros of the polynomial x square - 4 x + 4 |
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Answer» DHAMA bhi gujjar hote hai ....Kabir bro?? zeroes are 2 and 2. Zeroes r 2,2 ☺☺ 2,2 |
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| 41182. |
How can learn squares in easy way...? |
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Answer» Number ending 5.........for example (25)^2 use the trick 1st term ×(1st term +2×(2+1 By learning By writing |
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| 41183. |
The point of a ( x, y ) ,b (x2,y2), c ( x3,y3) are the vertics of triangle abc |
| Answer» I think, question is incomplete... | |
| 41184. |
Root6+wholeroot6+whole root 6 |
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| 41185. |
If costheta + sintheta =root 2 costheta then prove that costheta - sintheta =root2 sin theta |
| Answer» I\'m taking A instead of theta.... cosA+sinA=root2 cosA.......... =>sinA=root2cosA-cosA...... =>sinA=cosA(root2-1).......... =>cosA=sinA/root2-1............ =>cosA=sinA(root2+1)/2-1.....(rationalise the denominator)......... =>cosA=root2sinA+sinA....... =>cosA-sinA=root2 sinA....... proved..? | |
| 41186. |
How many terms are there in AP 7,10,13,.......43 |
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Answer» Thanks 43=7+(n-1)336+3=3n39=3nn=13 There are 13 terms...... |
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| 41187. |
Find the ratio in which p(4,m) divides the line segment joining the points a(2,3) and b(6,-3). |
| Answer» Req. ratio is 1:1 and the value of m is 0 | |
| 41188. |
Plz tell me the blue print |
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Answer» Blue print means kisi bhi chiz ki copy I also want Exact cpoy |
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| 41189. |
Sir Hindi medium math |
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Answer» What is ur question ?? Where is question???? What?? What is the question? |
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| 41190. |
TanA+secA=p find cosecA |
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| 41191. |
ln the given figure, if PQ=QR then prove that PE=ER |
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Answer» Where is the figure ??? You forget to draw the figure Figure?? But, where is the figure??? |
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| 41192. |
How can we find zeroes of -1 2 |
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| 41193. |
Ratinal number |
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Answer» In mathematics, a rational number is any number that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non-zero denominator q.Since q may be equal to 1, every integer is a rational number. Plz post the complete question!! Complete yur question..... Whts ur ques......,,,..?_? It is .............m |
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| 41194. |
Explain why 13233343563715 is a composite no. |
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Answer» Because it has more than two factors As the number is ending with 5 it has a factor of 5 as well as 1Therfore the given number is a composite number |
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| 41195. |
Dosto 50 50 offer main wallet ka use kese karte hain |
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Answer» aap hi use kar lo Don\' t know? |
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| 41196. |
Express the hcf of number 72 and 124 as a linear combination of 72 and 124 |
| Answer» HCF of 72,124 is 44=72×2-124×1-20 | |
| 41197. |
A+b ka hole squre |
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Answer» a sq + bsq +2ab (a+b)²====> a²+b²+2ab... A square + b square + 2Ab. A^2+B^2+2AB Hi |
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| 41198. |
Sin (50-Ø)-cos (40-Ø) |
| Answer» 0 | |
| 41199. |
SecA+TanA=pFind the value of CosecA. |
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Answer» P sq.+1/p sq.-1 and -1 (P²+1)/(P²-1)And -1 Ji hoga ..... Yeh ques. bahut important hain kya ?sab yhi poochte hain P square +1/p square-1 6th times i am solving this ques. (p sq.+1)/(p sq.-1) |
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| 41200. |
What is the formula of frustum(tsa) how we get it |
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Answer» Piel(r1+r2) + pie(r1square + r2square) Csa of frustum=πl(R1+R2)In tsa area of two more circles is added im csaFirst radius is taken as r1 and second as r2Area of circle=πr²So,TSA=πl(r1+r2)+πr1²+πr2²Hope u will understand |
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