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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.
| 42251. |
please help me kon kon syllabus ka chapters delete Hua 2020 ma |
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Answer» See at cbse website...I hope it will help you!?? See at cbse website UNIT I-NUMBER SYSTEMS ChapterTopicsREAL NUMBERSEuclid’s division lemmaPOLYNOMIALSStatement and simple problems on division algorithm for polynomials with real coefficients.PAIR OF LINEAR EQUATIONS IN TWO VARIABLEScross multiplication methodQUADRATIC EQUATIONSSituational problems based on equations reducible to quadratic equationsARITHMETIC PROGRESSIONSApplication in solving daily life problems based on sum to n termsCOORDINATE GEOMETRYArea of a triangleTRIANGLESProof of the following theorems are deleted · The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. · In a triangle, if the square on one side is equal to sumof the squares on the other two sides, the angle opposite to the first side is a right angle.CIRCLESNo DeletionCONSTRUCTIONSConstruction of a triangle similar to a given triangle.UNIT V- TRIGONOMETRY ChapterTopicsINTRODUCTION TO TRIGONOMETRYMotivate the ratios whichever are defined at 0o\xa0and 90oTRIGONOMETRIC IDENTITIESTrigonometric ratios of complementary anglesHEIGHTS AND DISTANCESNo deletionAREAS RELATED TO CIRCLESProblems on central angle of 120°SURFACE AREAS AND VOLUMESFrustum of a coneUNIT VI-STATISTICS & PROBABILITY ChapterTopicsSTATISTICS· Step deviation Method for finding the mean Cumulative Frequency graphPROBABILITYNo deletion Google par dekh lo you can search it on google.......... |
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| 42252. |
If the point p (2,1)lies on the line segment joining points A (4,2)and B (8,4)then, |
| Answer» Triangle | |
| 42253. |
Which term of the AP 3 8 13 18 is 88 |
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Answer» I love you 18th term |
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| 42254. |
A candidate scored 40 marks in a test if |
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Answer» Bruhh atleast write the question complete Bro plz.. write full que |
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| 42255. |
Maths exercise -8. 4 question 5 solutin |
| Answer» To check CBSE guide in math ncert solutions | |
| 42256. |
Solve the elimination method X + y equal to 14 x minus y equal to 4 |
| Answer» Here\'s your answer....x+y=14Let this equation be equation 1x-y=4Let this equation be equation 2.Solve equation 1 and 2You can either add or subtract the equations.Let us subtract the equation...x+y=14x-y=4_______2y=10y=10/2Y=5Substitute y=5 in equation 1x+y=14x+5=14x=14-5x=9So, x=9 and y =5 | |
| 42257. |
? and ? are zeroes of the quadratic polynomial ?2 − 6? + ?. Find the value of ? if 3? + 2? = 20. |
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Answer» Ache se hi toh likha hai?? Question acche se likho |
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| 42258. |
Find the value of ? in the polynomial 2?2 + 2?? + 5? + 10 if (? + ?) is one of its factors. |
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Answer» Thanks for your answer? Let f(a) = 2a2\xa0+ 2xa + 5x + 10As (a + x) is a factor of 2a2\xa0+ 2xa + 5x + 10, f(-x) = 0So, f(-x) = 2x2\xa0– 2x2\xa0– 5x + 10 = 0Or, -5x + 10 = 0Thus, x = 2 |
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| 42259. |
The sum of two numbers is 16 and their difference is 18 |
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Answer» 510 = 92 x 5 + 5092 = 50 x 1 + 4250 = 42 x 1 + 842 = 8 x 5 + 28 = 2 x 4 + 0∴ HCF of 510 and 92 = 2Product of two numbers = Product of their LCM and HCF510 x 92 = 2 x LCMLCM = (510 x 92) / 2 = 23460∴ LCM of 510 and 92 = 23460. X+Y =16------1eq. X-Y=18-----2nd. BY eliminate method X-Y=18 X+Y=16 THEN,-Y AND +Y Is cut After that 2X=16+18THEREFORE , 2X=34 THEN,X= 17 Y= -1 Suppose one number is X,And another be Y.By first condition,X+Y=16………….(1)By second condition,X-Y =18………….(2)Adding both the equation,X+Y=16+X -Y=18…………….2x =34 (y,-y get cancelled)X=17.Substitute x=17 in (1),17+y=16Y=16–17Y= -1(x=17,y=-1). |
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| 42260. |
What is the probability that two friends have different birthday |
| Answer» It is a non leap year. That means the year consist of 365 days.NowThe birthday of first friend will be on any day of the total 365 days.So, P(1) =\xa0AgainThe birthday of the 2nd friend will be on any day of the total of 364 days as his birthday date will be on the different day than the first friend.So, P(2) =\xa0So, P =\xa0P =\xa0 | |
| 42261. |
=°^ |
| Answer» | |
| 42262. |
For an AP Tm-Tn =............. |
| Answer» Let \'a\' be the first term and \'d\' be common difference of the A.P⇒Tm\u200b=a+(m−1)d=n⇒a−d=n−md...(1)and\xa0Tn\u200b=a+(n−1)d=m⇒a−d=m−nd..(2)Using (1) and (2) we get,\xa0d=−1\xa0and\xa0a=n+m−1∴Tp\u200b=a+(p−1)=n+m−1+(p−1)(−1)=n+m−pHence, option \'A\' is correct. | |
| 42263. |
find a quadratic polynomial whose zeroes are (5+√2) and (5-√2) |
| Answer» Given zeroes,5+√2 and 5-√2Sum of the zeroes=5+√2+5-√2Sum of the zeroes=10Product of zeroes=(5+√2)(5-√2)Product of zeroes=25-2=23We know that quadratic polynomial is in the form of,=k{x²-(sum of zeroes)x+product of zeroes}By putting required values we get,=k{x²-10x+23}=x²-10x+23Hence x²-10x+23 is the required polynomial.\xa0OrSo, f(x) = k {x2\xa0- 10x + 23}, where, k is any non-zero real number. | |
| 42264. |
What is the division lemba |
| Answer» It is removed from shllybus | |
| 42265. |
(2)1/2*(12)1/2*(27)1/2*(5)1/2 divide by (8)1/2*(10)1/3*(18)1/2*(81)1/4 |
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| 42266. |
Find the first 51 term of AP whose second and first term are 14 and 18 |
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Answer» #ß$%&--JJ hu v C #ß$%&--JJ hu v C |
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| 42267. |
Tan=60o__ |
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Answer» Tan 60degree is equal to root3 1.73205080757 |
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| 42268. |
Trigonometric table |
| Answer» Steps to Create Trigonometric Table:Step 1: Draw a tabular column with the required angles such as 0, 30o, 45o, 60o, 90o, 180o, 270o, 360o\xa0in the top row and all 6 trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent in first column.Step 2:Step 3:Step 4:Step 5:Step 6:Step 7:\xa0 | |
| 42269. |
In an AP the sum of first n terms is 3n²/2+5n/2. Find its 25th term. |
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Answer» Yes it is very helpful ?✌ 76 is the 25th term... Hope it would be helpful to you?✌ |
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| 42270. |
What is digree of cubic polynomial |
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Answer» 3 3 3 3 3 |
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| 42271. |
If a(3,y) is equidistant from p (8,-3) q (7, 6) find the value of y |
| Answer» pa=qa pa= √ (8-3)²+(-3-y)²= qa =√(7-3)²+(6-y)² √25+y² +6y+9=√16+y²+12y+3634+y²+6y = 52 +y² +12y 6y = -18y = -18/6 = -3 Hence y=-3 | |
| 42272. |
Find the greatest number which leaves a reminder 10 in each case whendivided by 56,84 and 140 |
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| 42273. |
In a triangle ABC angle C = 3angle B = 2(angle A + angle B |
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Answer» Let it be...i hv solved it.. Ya... Mate I don\'t think it is a complete question.You may repost it. |
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| 42274. |
If P x,y is equidistant from the point A 7,-2 and B 3,1, express y in terms of x |
| Answer» Pls dont answer i hv got it... | |
| 42275. |
Find the length of median through 5,1 to triangle ABC where other vertices are 1,5 and -3,-1 |
| Answer» Pls dont answer i hv got it... | |
| 42276. |
The emperical relationship bet mean, mode and median ??? |
| Answer» ANSWEREmpirical relationship between mean, median and mode is: Mode = 3 Median - 2 Mean⇒ Mode - Mean = 3 Median - 2 Mean - Mean⇒ Mode - Mean = 3 Median - 3 Mean⇒ Mode - Mean = 3 [Median - Mean] | |
| 42277. |
After how many palces will the rational number 1251/125 terminate |
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Answer» It will end after 3 decimal places u should use the trick to do it instead of dividing Q u e s t i o n :After how many palces will the rational number 1251/125 terminateA n s w e r :1251 / 1250 = 1.0008After 4 decimal places the rational number end. |
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| 42278. |
if 0.3528 is expanded in the form of p/2m5n find the smallest value of m, n, p. |
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| 42279. |
-2x=14 |
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Answer» -7 -7 = -7 Khud hi question ask kiya aur khud hi answer de diya?????? =7 |
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| 42280. |
find k for real roots 2x²+kx+3=0 |
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Answer» Here, a=2, b=k, c=3D=b^2-4ac/2a =k^2-4×2×3/2×2=K^2-24/4=0 k^2=24K=√24.......It may helpful to you. We have to find the values of\xa0k\xa0for quadratic equations 2x² + kx + 3 = 0 so that they have two\xa0equal\xa0roots.we know, quadratic equation will be equal only whendiscriminant, D = b² - 4ac = 0on comparing 2x² + kx + 3 = 0 with general form of quadratic equation , ax² + bx + c = 0 we get, a = 2, b = k and c = 3so Discriminant , D = (k)² - 4(2)(3) = 0or, k² - 24 = 0or, k = ± √24 = ±2√6hence, the value of k =\xa02√6 or -2√6 |
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| 42281. |
Product of its prime factors of 7429 |
| Answer» 7429 = 17 x 19 x 23 =17 x 19 x 23 | |
| 42282. |
Hii guys kon kon ghr pr bor ho rha h meri trha???? |
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Answer» Mai buk Serious mt lena majak kr rha tha Hii guys to ghr ke kaam kr lo alsi logo Me... Hii mai bor ho raye hoo ghar per bhyte bhyte ???? |
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| 42283. |
The sum of first 5 multiple of 3 is ……………………….. |
| Answer» Sum of first 5 multiples of 3:Sum of an AP is:\xa0n/2 \u200b× (2a+(n−1)d)a=3,d=3,n=5Then,\xa0Sum =\xa05/2 ×(2×3+(5−1)×(3))=\xa05/2 \u200b×(6+12)= 5/2\xa0×18= 5\xa0×9=\xa045 | |
| 42284. |
10th term of the AP: 1, 7, 13 ………….. is ………………………..Your answer |
| Answer» a = 1 d = 4-1 = 3an\xa0= a + (n-1)da10\xa0= 1 + (10-1)3a10\xa0= 1 + (9)3a10= 1 + 27a10= 28 | |
| 42285. |
If Discriminant = 0, then the nature of the roots of a quadratic equation is ………………………..Your answer |
| Answer» When a, b, and c are real numbers, a ≠ 0 and the discriminant is zero, then the roots α and β of the quadratic equation ax2+ bx + c = 0 are real and equal. | |
| 42286. |
A quadratic polynomial has maximum three zeroes.\xa0*TRUEFALSE |
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Answer» False A quadratic polynomial has maximum three zeroes.False,\xa0A quadratic polynomial has two zeroes. |
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| 42287. |
The value of 1/2 cosec²45⁰ is 1/4.\xa0*TRUEFALSE |
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| 42288. |
A(4,3) is a trisection point of a line joining B(3,6) and C( 6,3). Justify\xa0*TRUEFALSE |
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| 42289. |
Which of the following is not a quadratic equation\xa0* |
| Answer» For example by-c=0 is not a quadratic equation because it is not have an degree of 2 | |
| 42290. |
The value of cosec 30 is\xa0* |
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Answer» =2 The value of cosec 30 iscosec 30 = 1/sin 30cosec 30 = 1/ 1/2= 2 |
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| 42291. |
5 tan² A – 5 sec² A + 1 is equal to\xa0* |
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Answer» 5(tan^2-sec^2) +15(-1)+1-5+1-4 -4 |
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| 42292. |
If a=10 and d=10, then first four terms will be:\xa0* |
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Answer» a=10 ,a+d=20,a+2d=30,a+3d=40 a4\xa0=40 a=10 and d=10\xa0a1 = 10a2 = a1 + d = 10 + 10 = 20a3 = a2 + d = 20 + 10 = 30a4 = a3 + d = 30 + 10 = 40If a=10 and d=10, then first four terms will be: 10, 20, 30, 40,......... |
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| 42293. |
The points (–4, 0), (0, 0), (0, 3) are the vertices of a\xa0* |
| Answer» Let the points A(-4,0),B(4,0)and C(0,3)are the vertices\xa0∴AB=√(4−(−4))2+(0−0)2\u200b=√(8)2\u200b=√64\u200b=8∴BC=√(0−4)2+(0−3)2\u200b= √16+9\u200b = √25 \u200b= 5∴CA=√(−4−0)2+(0−3)2\u200b=√16+9\u200b=√25\u200b=5As we know that two points are equal in the given vertices\xa0∴This is the vertices of a isosceles triangle. | |
| 42294. |
(sin30° + cos30°) – (sin 60° + cos60°) |
| Answer» (sin30+cos30) - (sin60+cos60)sin30 = 1/2sin60= √3/2cos30= √3/2cos60= 1/2= (1/2+√3/2) - (√3/2+1/2)=1+√3/2 - √3+1/2= 1+1+√3-√3 /2= 2/2= 1 | |
| 42295. |
If x*sin (90-theta) *cot(90-theta) = coc (90-theta) then x is equal to |
| Answer» x* cos thetha * tan thetha = sin the thax * sin thetha = sin thethax= 1 | |
| 42296. |
For any natural number "n |
| Answer» Please answer??? | |
| 42297. |
How to solve problems related to LCM and HCF |
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Answer» Okayy but how do we interpret when to use LCM and when HCF By prime factorisation methods |
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| 42298. |
using prime factorisation find the hcf \' and Lcm of 23\' 31 |
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Answer» solve ho gya HCF - 1 LCM- 713 |
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| 42299. |
Can anyone send me important question of maths which come in board exam 2020-21 |
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Answer» Up-to chapter 7 I told remaining I will send later Chapter 1: fundamental theorem of arithmetic 3marks Chapter 2:exam practice 5mark. Chapter 3: algebraic method, equation reducible , and exam practice problems : 6marks. Chapter 4: solution of a quadratic equations by quadratic formula in that topic any 3marks sum. Chapter 5: own you have to practice. No important questions. Chapter 6: proof of Thales and Pythagorean property and all exercise sums and exam practice sums. 8marks Chapter 7: distance and section formula and exam practice 6marks. |
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| 42300. |
ABCD is a trapezium with AB parallel DC . E and F are points on non - parallel of AD and BC |
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Answer» ABCD is a trapezium with AB || DC, E and F are points on non-parallel sides AD and\xa0BC respectively such that EF is parallel to AB show that\xa0 Given trap. ABCD in which AB || DC and EF || AB.To prove:\xa0Const: Join A-C.Proof: ∵ AB || DC and EF || AB⇒ AB || EF || DCNow, in ∆ADC, we have ...(i)[using Basic proportionality theorem]Now, in ∆ABC, we have ...(ii)Comparing (i) and (ii), we get |
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