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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.
| 22001. |
Find sum of first 10 even number in an AP |
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Answer» n(n+1) its a formula to get sum of even number n 10(10+1)=110 110 d=4-2=2, d=6-4=2 yes AP=2,4,6...Sn=n÷2 (2a+(n_1)d) 110 |
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| 22002. |
State & prove Thale\'s Theorem. |
| Answer» Its too long soo u can take hlp from ncert txt book | |
| 22003. |
Formula of triangle |
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Answer» Kon sa formula 2 hai ik general formula 1/ 2 . B× h or ik ch. 7 ka formula hai 1/2 ( x1(y2 - y3)+x2(y3- y1 ) +x3 ( y2-y1) ) Its ar. of triangle......... U mean area of triangle 1/2 × base × height..... |
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| 22004. |
Use Euclid\'s Division Algorithm to find the HCF of 726 and 275 |
| Answer» 11 is the hcf of 726 and 275 | |
| 22005. |
From optional exercise question may come |
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Answer» From A. P like chapter even Yaa it can be asked Ya, it may come |
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| 22006. |
If the points A (k+1,2k),B (3k, 2k+3)and C (5k-1,5k) are collinear, then find the value of k. |
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Answer» Yur, welcome K=2, or k=1/2 |
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| 22007. |
Find 20th term from the last term of the AP 3,8,13 ,253 |
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Answer» 158 a=3 ,d=(8-3)=5 ,l=253Tn = l -(n-1)dT20= 253 -(20-1)5 =253-19*5 =253-95 = 158 a=253,d=5 but for the last term d=-5 a+(n-1)d. =253+(20-1)(-5). =253+(19)(-5)253-95 =158 158 Ap=3,8,13...............253a=3d=8-3=5a20=a+19da20=3+19×5a20=3+95a20=98_______________ |
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| 22008. |
Prove that 0/0=1 |
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| 22009. |
The hcf of 24456 by lemma |
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Answer» Divisor to hona chahiye na....tabhi division possible hh But kisse divide krna h...vo to btaeye |
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| 22010. |
Find the sum of the first 25 terms of an Ap whose nth term is given by an an=7-3n... |
| Answer» -800a=7-3*1a2=7-3*2d=1-4So therefore by putting the values in the formula we get the answer minus 800 | |
| 22011. |
If m,n,o,p and q,are integers,then m(n+o) (p-q) must be even when of the following is even |
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| 22012. |
Peove that √2 is irrational |
| Answer» Given √2 is irrational number. Let √2 = a / b wher a,b are integers b ≠ 0 we also suppose that a / b is written in the simplest form Now √2 = a / b ⇒ 2 = a2 / b2 ⇒ 2b2 = a2 ∴ 2b2 is divisible by 2 ⇒ a2 is divisible by 2 ⇒ a is divisible by 2 ∴ let a = 2c a2 = 4c2 ⇒ 2b2 = 4c2 ⇒ b2 = 2c2 ∴ 2c2 is divisible by 2 ∴ b2 is divisible by 2 ∴ b is divisible by 2 ∴a are b are divisible by 2 . this contradicts our supposition that a/b is written in the simplest form Hence our supposition is wrong ∴ √2 is irrational number. | |
| 22013. |
If A, B and C are interior angles of triangle ABC then show that cos(B+C/2)=sinA/2 |
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Answer» Hope u ll get it In a triangle, sum of all the interior anglesA + B\xa0+ C = 180°⇒ B\xa0+ C = 180° - A⇒ (B+C)/2 = (180°-A)/2⇒ (B+C)/2 = (90°-A/2)⇒ sin (B+C)/2 = sin (90°-A/2)⇒ sin (B+C)/2 = cos A/2 In a triangle abc we know that all three angles is equal to 180° so angle a +b+c=180. Now shift a on the other side i.e. b+c=180-a next divide both sides by 2 so it gives (b+c) /2= 90-a/2. further on put cos on both sides ---. Cos (b+c)/2= cos 90-a/2 and cos 90-a/2 = sin a/2. So substitute it and we get cos (b+c)/2 = sin a/2 Cos(B+C/2) CAN BE WRITTEN AS Sin(90-B+C/2)...by taking the L. C. M, we will get Sin(180-B+C/2) and we know that in a triangle 180-B+C/2 is equal to A . So, our answer will be SinA/2. |
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| 22014. |
Find AP whose nth term is 3n-5. |
| Answer» nth term = 3n - 5First term = a1 = 3 x 1 - 5 = -2Second term = a2 = 3 x 2 - 5 = 1Third term = a3 = 3 x 3 - 5 = 4Thus, the AP is -2, 1, 4, ...... | |
| 22015. |
Please anybody send me the date sheet for final board exam of 10th |
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Answer» You can download from :http://cbse.nic.in/newsite/circulars/2018/Class-X_datesheet.pdf It is given in google.. March 7 -maths 13- science19- hindi23- english29- ss |
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| 22016. |
Evaluate..cot12°cot38°cot52°cot60°cot78° |
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Answer» 1/root3 cot12cot38cot52cot60cot78=cot(90-78)cot(90-52)cot52cot60cot78=tan78cot78 x tan52cot52 x cot60=1x1x1 x cot60={tex}\\frac{1}{\\sqrt3}{/tex} 1/√3 1/√3 |
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| 22017. |
If the point (2,3) ,(4,4)and (6,_3) are co- linear then find the value of k |
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Answer» This question is Example 14 from chapter 7 page 169Example 14 : Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) arecollinear.Solution : Since the given points are collinear, the area of the triangle formed by themmust be 0.So {tex}\\begin{array}{l}\\frac12\\lbrack2(k+3)+4(-3-3)+6(3-k)\\rbrack=0\\\\\\frac12(2k+6-24+18-6k)=0\\\\\\frac12\\times-4k=0\\\\or\\;k=0\\end{array}{/tex}\xa0 Correct Question -\xa0Find the value of k if the points A(2,3) ,B(4,k) and C(6,-3) are collinear. Solution:Given that A, B and C are collinear. Hence ar(DABC) = 0\xa0 Where is k |
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| 22018. |
On dividing 3xcube -2xsquare+5x+5 by px then qx is x2-x+2 and rx is -7 |
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Answer» It is not complete Please,complete the question........?? |
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| 22019. |
Show that the no 7n cannot end with digit 0 |
| Answer» If 7ⁿ ends with a 0 the it must be divisible by 10 i.e., its prime factors should have the factors of both 2 and 5 since 2×5=10.But, 7ⁿ=(7×1)ⁿTherefore by the fundamental theorem of arithmetic (there is no prime factors of 7 other than 7 and 1) we can conclude that 7ⁿ can not end with zero. | |
| 22020. |
Sectheta =x+1x/4 prove that sectheta +tantheta =2x, 1/2x |
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Answer» Q. If sec theta = x+1/4x ,prove that :sec theta +tan theta=2x or 1/2x Solution: Please, check yur question, the value of sec thita given, is wrong according to me........... |
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| 22021. |
What is happening |
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Answer» Development refers to the progress, of something........ the process of creating something more advanced; a more advanced product...is known as development What is development |
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| 22022. |
Vqlue of sin90 |
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Answer» 1 Its....1....? 1.................. 1 1 |
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| 22023. |
Can i ask one question |
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Answer» Ya sure........? Yup |
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| 22024. |
168+208= |
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Answer» Do u have any toughest question from chapter light? 376 |
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| 22025. |
(5×7×13+7)is a composite number |
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Answer» Yes itis a composite No.7 (5*1*13+1)7(65+1)7×66 Is divided by 2 . 5...10.. 7. Yes 700 |
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| 22026. |
Find the distance between two parallel tangent of circle of radius 7cm |
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Answer» Radius = 07 cm so, diameter = 7×2 = 14cmthe distance between the parallel tangents of circle = diameter of circle = 14 cm Thank you?? 14cm |
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| 22027. |
Derive the formula for curved surface area and total surface area of the frustrum of a cone??? |
| Answer» CSA: pie L(R1+R2)TSA:pieL(R1+R2)+pie R1square+R2square | |
| 22028. |
For what value of n, are the nth term of two APs 63,65,67,....... And 3,10,17,...... equal? |
| Answer» Let the 1st AP be A =63,65,68.....Let the 2nd AP be B = 3,7,17......For A , common difference is : 65-63 = 2\xa0For B ,common difference is : 10-3 = 7\xa0Now Let the nth term be An and Bn respectively.\xa0According to question\xa0An = Bn\xa0=> 63+(n-1)2 = 3+(n-1)7=>60 + 2n-2 = 7n-7=>65 = 5n\xa0=> n = 13Thus 13th term of both AP\'s will be same\xa0 | |
| 22029. |
What is circumcentre of circle |
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Answer» My question circumcentre 2pie r |
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| 22030. |
Prove that 5+2√3is irrantionl |
| Answer» Let it is rational and =a/b2√3=a/b-5 | |
| 22031. |
Prove that cube any odd posivite is in form 4m+1 |
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| 22032. |
ax2+bx+c=0. roots are differ by 1. Prove,b2=a2+ac. |
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| 22033. |
How easy recall |
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| 22034. |
If sum of first ten terms are given in ap find the next ten terms in ap |
| Answer» what is AP | |
| 22035. |
More than ogive graph |
| Answer» It\'s too simple always remember just opposite more and less means lower limit and cf | |
| 22036. |
If the nth term of arithmetic progression is, -1,4,9,14...............129. Find the value of n. |
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Answer» Use formula an =a+(n-1)×d129=-1+(n-1)×5130=5n-5 135÷5=n27=n Its 27 Ram here l is not 149 its 129 N=31 Here a=-1, d=4-(-1)=4+1=5 , l=149so 149=-1+(n-1)x5149=-1+5n-55n=149+6=155so\xa0{tex}\\begin{array}{l}n=\\frac{155}5=31\\\\\\end{array}{/tex} |
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| 22037. |
A circle tonche the |
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| 22038. |
Application of bpt |
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| 22039. |
If alpha and beta are the zreoes of polynomial x²-5x +k and alpha - beta =-1.Find the value of k |
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Answer» How u solve this 6 |
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| 22040. |
Show that 3√3 is irrational |
| Answer» Assumed 3rout 3 is rational But we know rout 3 is irrational So, We assumption is wrong 3rout3 is irrational | |
| 22041. |
(C2 ab)x2-(a2 -bc)x+b2 ac=0 has equal roots prove that a=0 |
| Answer» We have,\xa0{tex}(c^2-ab)x^2-2(a^2-bc)x+b^2-ac=0{/tex}Here,\xa0{tex}A=(c^2-ab), B=-2(a^2-bc),C=b^2-ac{/tex}\xa0Now,\xa0{tex}B^2-4AC =0{/tex}{tex}4(a^2-bc)^2-4(c^2-ab)(b^2-ac)=0{/tex}{tex}4(a^2-bc)^2=4(c^2-ab)(b^2-ac){/tex}{tex}a^4-2a^2bc+b^2c^2=b^2c^2-ac^3-ab^3+a^2bc{/tex}{tex}a^4-2a^2bc+ac^3+ab^3-a^2bc=0{/tex}{tex}a^4-3a^2bc+ac^3+ab^3=0{/tex}{tex}a(a^3-3abc+c^3+b^3)=0{/tex}{tex}a=0 \\ or \\ a^3-3abc+c^3+b^3=0{/tex}{tex}a=0\\ or\\ a^3+b^3+c^3=3abc{/tex}Hence proved | |
| 22042. |
Each chapter weithage |
| Answer» Yu can check it in google.......? | |
| 22043. |
Underoot3/4r^2 konse triangle ka formula he please tell me fast |
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Answer» Its for equilateral triangle ..given in rs agarwall Its ok i got equilateral |
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| 22044. |
How we can solve the median related question |
| Answer» It is simple bro... first make cf table and then choose median class (class whose cf is nearly greater than n/2) and find f i.e. frequency of class preciding median class and put it on formula | |
| 22045. |
what is meaning of F1 , F0 & F2 when we find mode |
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Answer» He\'s right...... F1 is the frequency of modal class. F0 is the frequency of class preciding modal class.F2 is the frequency of class after modal class. |
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| 22046. |
Root 3 is irrational no. Prove |
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Answer» Let us assume that √3 is a rational numberThat is, we can find integers a and b (≠ 0) such that √3 = (a/b)Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.√3b = a⇒\xa03b2=a2 (Squaring on both sides) → (1)Therefore, a2 is divisible by 3Hence \x91a\x92 is also divisible by 3.So, we can write a = 3c for some integer c.Equation (1) becomes,3b2 =(3c)2⇒\xa03b2 = 9c2∴ b2 = 3c2This means that b2 is divisible by 3, and so b is also divisible by 3.Therefore, a and b have at least 3 as a common factor.But this contradicts the fact that a and b are coprime.This contradiction has arisen because of our incorrect assumption that √3 is rational.So, we conclude that √3 is irrational. Root3=irrational no. Hence proved. |
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| 22047. |
Find all zeroes of 2x^4-3x^3-3x^2+6x-2 if two of its zeroes are 1 and 1/2 |
| Answer» (x-1)(x-1/2) solve this and divide the polynomial by the equation solved u will get the other zeroes | |
| 22048. |
Show that the sum of all odd integers between1to 1000 divisible by 3is 83667 |
| Answer» A.P : 3,9,15,------------999. Here a= 3, d=6, An = 999. An= a+ (n-1)d999=3 + ( n-1)6 999=3+ 6n - 6999 = 6n- 3999+3=6n1002=6n1002/6=n167=n.Sn = n/2 [2a+(n-1) d ]S167 = 167/2 [2×3+(167-1)×6] = 167/2 [6+166×6] = 167/2 [1002] = 167/2× 1002 =167×501 = 83667. Hence proved | |
| 22049. |
How to capare the area |
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| 22050. |
What are dacimal |
| Answer» A decimal is a fraction that is written in the form of a dot followed by one or more numbers which represent tenths, hundredths, and so on | |