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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.
| 24951. |
Find the sum of first 5 multiple of 2 |
| Answer» 2,4,6,8,10 sn = n/2 [a+l] 5/2 [2+10] 2.5*12 30Ans = 30????????????????????? | |
| 24952. |
What is the sum of five positive integer divisible by6 |
| Answer» Five positive integers divisible by 6 are 6,12,18,24,30. a=6,l=30,d=6,n=5.Sum of five positive integer divisible by 6 = n/2(a+l) = 5/2(6+30) = 5/2(36) = 2.5×36 = 90 | |
| 24953. |
How to find if one side of triangle is propotional to the other |
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| 24954. |
Convers phythogerus |
| Answer» Abe, NCERT namak pustak KO khol me dhekh le.Stupiddddddd. | |
| 24955. |
Be hush-hush why |
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| 24956. |
Are two triangles |
| Answer» 3 bhi 4 5 6 7 8.. | |
| 24957. |
For what value of K will K+7, 2K-7 are in AP |
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| 24958. |
Can the numbet 6,n bring a natural number end with the digitb 5 give reason |
| Answer» If 6n\xa0ends with 0 or 5 then it must have 5 as a factor.Now 6n\xa0=\xa0{tex}( 2 \\times 3 ) ^ { n } = 2 ^ { n } \\times 3 ^ { n }{/tex}The prime factors of 6n are only 2 and 3And from the fundamental theorem of arithmetic, the prime factorization of every composite number is unique.{tex}\\therefore{/tex}\xa06n\xa0can never end with 0 or 5. | |
| 24959. |
How to find mode if v have 2 modal class |
| Answer» Consider only one of the.m | |
| 24960. |
1-cosA=sinA |
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| 24961. |
In a class interval 135-140 will 140 also be counted? |
| Answer» No 135 will be | |
| 24962. |
Find the greatest number by which 2053 and 967 are divided ,remainders left are 5 and 7 respectively |
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Answer» let us find the HCF of (2053 - 5 = 2048, 967 - 7 = 960)2048 = 2x960 + 128 960\xa0= 128x7 + 64 128 = 64x2 + 0therefore HCF = 64 That is 64 The largest no. Is HCF of (2053-5) and (967-7) |
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| 24963. |
Find the term of ap? what is formula of that? |
| Answer» a+d,a+2d,a+3d,a+4d,a+5d........ | |
| 24964. |
Divide (2x^+x+20)by (x+3) and verify the result by division algorithm |
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| 24965. |
2 can be divided by 6 |
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Answer» 0.3333333333......... No.. |
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| 24966. |
Square root of 200 |
| Answer» 40000 | |
| 24967. |
Let 3+2√5 is irrational |
| Answer» Let 3+2√5 is rational 3+2√5 = p\\qSubtract 3 both side2√5= p\\q-3 2√5= p-3q\\2Now, divide by 2 √5= p- 3q\\2q Hence, p and q are integers so p-3q \\ 2q are irrational numbers so √5 should rational number but √5 is irrational number so it is contradict a jact so 3+2√5 is irrational | |
| 24968. |
Where should we check for 2017-18 sample papers |
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| 24969. |
Prove that Cot A+cosecA-1/cotA-cosecA+1=1+cosA/sinA |
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| 24970. |
Give an example of two irrational whose product is rational |
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Answer» √2×2√2 √2 and 2√2 |
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| 24971. |
Evaluate:Sin70°/cos30° +codec30°/sec54° - 2cos43° cosec47° |
| Answer» 0 | |
| 24972. |
Formula of 8_15 chapters |
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| 24973. |
The surface area of cube is 625 . Find the side of cube. |
| Answer» Surface area of a cube=6a^2=6256a^2=625a^2=625/6a=√625/√6a=25/√6Rationalize it and you will get the answer | |
| 24974. |
1+3+5+6+-4-8 |
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Answer» It\'s 3 3 |
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| 24975. |
x^2-(√2+1)x+√2=0. Solve it completing square method |
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| 24976. |
Tan (A-B)=√3, and sin A =1, then find A and B. |
| Answer» tan (A-B)= root 3 = 60 (we know that tan 60=root 3)A-B = 60 ...... isin A = 1 = 90 (we know that sin 90=1)A=90 ........... iifrom i and ii90 - B=60B=30 , A=90 | |
| 24977. |
How many two digits number are divisible by 3 |
| Answer» The two -digit numbers divisible by 3 start from 12,15,18,21,...,99Here,\xa0{tex}a=12{/tex}{tex}d=3{/tex}{tex}a_n=a+(n-1)d{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}99=12+(n-1)(3){/tex}{tex}\\Rightarrow{/tex}{tex}\xa099=12+3n-3{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}90=3n{/tex}{tex}\\Rightarrow{/tex}\xa0n=30Thus, 30 two-digit numbers are divisible by 3. | |
| 24978. |
If cot A = 12/5 then find the value of (sin A + cos A ) cosec A |
| Answer» Given that cotA=12/5(sinA+cosA)cosecAsinA.cosecA+ cosA.cosecA1+cosA/sinA (cosecA=1/sinA ;cosA/sinA=cotA)1+cotA1+12/5=17/5 | |
| 24979. |
If ratioof corresponding sides of two triangle in5:6 then find the ratio of their areas |
| Answer» we know that in two similar triangle ABC and DEFratio of corrresponding sides =AB/DE = 5/6ratio f their area = AB2/DE2 = 52\xa0/ 62\xa0= 25/36 | |
| 24980. |
What is euclids divison lemma |
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Answer» A equal to bq+r where r less than bor equal to zero Is ka answer ncert ki book mein hai |
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| 24981. |
In a pythagoroes |
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| 24982. |
All triangles are similar or not |
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Answer» No Only all equilateral triangles r similar n isosceles triangles in some case No....all equilateral triangles r similar.......also isosceles in some cases |
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| 24983. |
Theorem 10.1 and 2 |
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| 24984. |
If sec A = x + 1/ x then prove that sec A. tan A = 2x or 1/x. |
| Answer» By the given condition of question\xa0{tex}\\sec \\theta = x + \\frac { 1 } { 4 x }{/tex}{tex}\\therefore \\quad \\tan ^ { 2 } \\theta = \\sec ^ { 2 } \\theta - 1{/tex}{tex}\\Rightarrow \\quad \\tan ^ { 2 } \\theta = \\left( x + \\frac { 1 } { 4 x } \\right) ^ { 2 } - 1 = x ^ { 2 } + \\frac { 1 } { 16 x ^ { 2 } } + \\frac { 1 } { 2 } - 1 = x ^ { 2 } + \\frac { 1 } { 16 x ^ { 2 } } - \\frac { 1 } { 2 } = \\left( x - \\frac { 1 } { 4 x } \\right) ^ { 2 }{/tex}{tex}\\Rightarrow \\quad \\tan \\theta = \\pm \\left( x - \\frac { 1 } { 4 x } \\right){/tex}{tex}\\Rightarrow \\quad \\tan \\theta = \\left( x - \\frac { 1 } { 4 x } \\right) \\text { or, } \\tan \\theta = - \\left( x - \\frac { 1 } { 4 x } \\right){/tex}CASE 1: When\xa0{tex}\\tan \\theta = - \\left( x - \\frac { 1 } { 4 x } \\right) :{/tex}\xa0In this case,{tex}\\sec \\theta + \\tan \\theta = x + \\frac { 1 } { 4 x } + x - \\frac { 1 } { 4 x } = 2 x{/tex}CASE 2: When\xa0{tex}\\theta = - \\left( x - \\frac { 1 } { 4 x } \\right) :{/tex}\xa0In this case,{tex}\\sec \\theta + \\tan \\theta = \\left( x + \\frac { 1 } { 4 x } \\right) - \\left( x - \\frac { 1 } { 4 x } \\right) = \\frac { 2 } { 4 x } = \\frac { 1 } { 2 x }{/tex}Hence,\xa0{tex}\\sec \\theta + \\tan \\theta = 2 x \\text { or } , \\frac { 1 } { 2 x }{/tex} | |
| 24985. |
If sec²ø(1+sin ø)(1- sinø)= k ‚then find the value of k |
| Answer» sec2\xa0{tex}\\theta{/tex}\xa0(1 + sin\xa0{tex}\\theta{/tex}) (1 - sin\xa0{tex}\\theta{/tex}) = k{tex}\\Rightarrow{/tex}\xa0sec2\xa0{tex}\\theta{/tex}\xa0(1 - sin2\xa0{tex}\\theta{/tex}) = k{tex}\\Rightarrow{/tex}\xa0sec2\xa0{tex}\\theta{/tex}\xa0. cos2\xa0{tex}\\theta{/tex}\xa0= k{tex}\\Rightarrow{/tex}\xa0k = 1 | |
| 24986. |
Find the value of k,if the quadratic equation 3xsq - k√3x + 4=0 has real roots |
| Answer» \xa0If Discriminant of quadratic equation is equal to zero, or more than zero then roots are real.3x2 -k{tex}\\sqrt3{/tex}x + 4 = 0Compare with ax2 + bx + c = 0then a = 3, -k{tex}\\sqrt3{/tex}\xa0and c = 4D = b2- 4acFor real roots, b2- 4ac > 0{tex}( - k \\sqrt { 3 } ) ^ { 2 } - 4 \\times 3 \\times 4 \\geq 0{/tex}3k2- 48 {tex}\\geq{/tex}0k2\xa0- 16\xa0{tex}\\geq{/tex}0(k - 4)(k + 4) {tex}\\geq{/tex}\xa00{tex}\\therefore{/tex}\xa0k {tex}\\leq{/tex}\xa0- 4 and k{tex}\\geq{/tex}\xa04 | |
| 24987. |
Xffg |
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| 24988. |
4x |
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| 24989. |
Area of sector |
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| 24990. |
Solve:6x + 3y - 6xy = 0 and 2x + 4y - 5xy =0 |
| Answer» The given pair of equation is6x + 3y = 6xy{tex}\\Rightarrow \\quad \\frac { 6 x } { x y } + \\frac { 3 y } { x y } = \\frac { 6 x y } { x y }{/tex}..............Dividing throughout by xy{tex}\\Rightarrow \\quad \\frac { 6 } { y } + \\frac { 3 } { x } = 6{/tex}.........................(1)2x + 4y = 5xy{tex}\\Rightarrow \\quad \\frac { 2 x } { x y } + \\frac { 4 y } { x y } = \\frac { 5 x y } { x y }{/tex} ..................Dividing throughout by xy{tex}\\Rightarrow \\quad \\frac { 2 } { y } + \\frac { 4 } { x } = 5{/tex}..........................(2)Put\xa0{tex}\\frac { 1 } { x } = u{/tex} .....................(3)And\xa0{tex}\\frac { 1 } { y } = v{/tex}..........................(4)Then, the equation (1) and (2) can be written as:{tex}6 v + 3 u = 6{/tex}..................(5)2 v + 4 u = 5 ..................(6)Multiplying equation (6) by 3, we get6 v + 12 u = 15 ..........................(7)Subtracting equation (5) from equation(7), we get 9u = 9\xa0{tex}\\Rightarrow \\quad u = \\frac { 9 } { 9 } = 1{/tex}{tex}\\Rightarrow \\quad \\frac { 1 } { x } = 1{/tex}.......................using (3){tex}\\Rightarrow \\quad x = 1{/tex}Substituting this value of u in equation (5), we get 6v + 3 X 1 = 6{tex}\\Rightarrow \\quad 6 v + 3 = 6{/tex}{tex}\\Rightarrow 6 v = 6 - 3 = 3{/tex}{tex}\\Rightarrow \\quad v = \\frac { 3 } { 6 } = \\frac { 1 } { 2 }{/tex}{tex}\\Rightarrow \\quad \\frac { 1 } { y } = \\frac { 1 } { 2 }{/tex}......................using (4){tex}\\Rightarrow y =2{/tex}Hence, the solution of the given pair of the equation is x =1, y =2Verification: Substituting x =1, y =2We find that both the equation (1) and (2) are satisfied as shown below:{tex}\\frac { 6 } { y } + \\frac { 3 } { x } - \\frac { 6 } { 2 } + \\frac { 3 } { 1 } - 3 + 3 - 6{/tex}{tex}\\frac { 2 } { y } + \\frac { 4 } { x } - \\frac { 2 } { 2 } + \\frac { 4 } { 1 } = 1 + 4 = 5{/tex}Hence, the solution is correct. | |
| 24991. |
Mark distribution for chapters for 2018 board |
| Answer» Check marks distribution here :\xa0https://mycbseguide.com/cbse-syllabus.html | |
| 24992. |
When class 10 board datesheet will be released for 2017-18 |
| Answer» Feb 2nd week is going to start | |
| 24993. |
If one zero of the polynomial 3x squar -5x+7 is negative of other than what is the zero |
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| 24994. |
Find first term of AP -168, -156 -144 |
| Answer» The first ofAPis -168 | |
| 24995. |
Solve for x by factorization: 1/x - 1/x-2 =3 |
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| 24996. |
Ex 10.2 Q.12 answer with clear diagram. |
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| 24997. |
If sn denotes ,the sum of the first n term of arthmetic progression then prove that s12=3(s8-s4) |
| Answer» S12=3(s8) | |
| 24998. |
Cos a_ sin a |
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| 24999. |
This is a question different from trigonometry brabch ,may be a stupid one-sinx/n |
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| 25000. |
2+8542+9588742555+9578955558-585355255•65785555+658955 |
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