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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.
| 26051. |
Vyguv |
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Answer» What is this ???????? |
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| 26052. |
P of x is x power 4 - 3x square +4x +5, g of x is x square +1 - x |
| Answer» X square +x - 3 | |
| 26053. |
In tringle abc write tan a+b/2 in term of angle c |
| Answer» | |
| 26054. |
Find the least positive integer which is exactly divisible by first five natural number |
| Answer» 1 | |
| 26055. |
Find the greatest five digit number which is exactly divisible by 12,18,24 |
| Answer» 6 | |
| 26056. |
Draw an equilateral triangle ABC of side 6cm each. |
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Answer» But the way is that ... We already know that an attitude bisects the side of triangle.. So use that Not possible here.. |
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| 26057. |
Important questions of of maths exercise number 4 |
| Answer» Every question is important | |
| 26058. |
Important important questions of off of maths exercise number 4 |
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Answer» Yes every question is important: ) All are important |
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| 26059. |
10 theta square equals to cos theta |
| Answer» | |
| 26060. |
Solve quadratic equation x2+(a/a+b+a+b/a)x+1=0 using factorisation |
| Answer» Aaaa... | |
| 26061. |
A dice is rolled twice. Find the probability that 3 will come exactly one time. |
| Answer» When a dice is rolled twice, the total outcomes = 62 = 36possible outcomes that 3\xa0will come one time is (1, 3), (2, 3) (3, 3) (4, 3) (5,3) (3, 1) (3, 2) (3, 4) (3, 5) (3, 6) = 10P(3\xa0will come up exactly one time) =\xa0{tex}\\frac { 10 } { 36 } = \\frac { 5 } { 18 }{/tex} | |
| 26062. |
Fcbvxvvxc |
| Answer» ?? | |
| 26063. |
What is zero ? |
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Answer» zero is the only integer that can neither be positive or neither negative 0 |
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| 26064. |
Why root 5-3root 2 |
| Answer» No idea.. | |
| 26065. |
Exam are coming and I have not prepared maths what should I do to score good marks |
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Answer» Oops I meant. Tricks As to me tignometry was the easiest chapter.remember do cover it up once concepts are very understanding n truck is almost common . And ncert Jo chptr zada mrks carry kr rhe h unhe prepare kr lijiye... Only sample papers u have to do |
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| 26066. |
If cos2a+cosa=1,prove thatSin12a+3sin10a+3sin8a+sin6a+2sin4a+2sin2a-2=1 |
| Answer» 1 | |
| 26067. |
Show that exactly one of the number n, n+2 or n+4 is divisible by 3 |
| Answer» Let n =3kthen n + 2 = 3k + 2and n + 4 = 3k + 4Case 1: When n=3k ,n is divisible by 3 ............(1)n + 2 = 3k + 2or, n + 2 is not divisible by 3n + 4 = 3k + 4= 3(k + 1) + 1or, n + 4 is not divisible by 3Case 2:When n=3k+1, n is not divisible by 3\xa0n + 2 = (3k + 1) + 2=3k + 3 = 3(k + 1){tex} \\Rightarrow{/tex}\xa0n+ 2 is clearly divisible by 3..........................(2)n + 4 = (3k + 1) + 4= 3k + 5= 3(k + 1) + 2{tex}\\Rightarrow{/tex}\xa0n + 4 is not divisible by 3Case 3:When n=3k+2,n is not divisible by 3\xa0n + 2 = (3k + 2) + 2= 3k + 4(n + 2) is not divisible by 3x + 4 = 3k + 6 = 3(k + 2){tex}\\Rightarrow{/tex}\xa0n + 4 is divisible by 3........................(3)Hence, from (1),(2) and (3) it is clear that\xa0exactly one of the numbers n, n + 2, n + 4, is divisible by 3. | |
| 26068. |
Which is the best sample paper for maths |
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Answer» RD sharma On internet you would get so many sample paper Oswaal question Bank Do ncert exampler Exam idea |
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| 26069. |
How many natural numbers are there between 200 and 500, which are divisible by 7? |
| Answer» Series are : 203 , 210, ......., 497 So now we know a, d and an By an =a + ( n -1 ) d We get ans is 43. | |
| 26070. |
,show that x3+4x+5 has no zeroes |
| Answer» | |
| 26071. |
Any positive square integer is in the form of 3m,3m+1,3m+2 |
| Answer» Let n be an arbitrary positive integer.On dividing n by 3, let m be the quotient and r be the remainder.The positive integer is in the form of 3m or (3m + 1) or (3m + 2)Then, by Euclid\'s division lemma, we haven = 3m + r, where {tex}0 \\leq r < 3{/tex}.{tex}\\therefore{/tex}\xa0n = 3m or (3m +1) or (3m + 2), for some integer m.Thus, any positive integer is of the form 3m or (3m + 1) or (3m + 2) for some integer m. | |
| 26072. |
4(x-1/x)2-4(x+1/x)+1=0 |
| Answer» | |
| 26073. |
If two positive integers p and q are written as p=a²b³ and q=a³b; a , b are prime number verify |
| Answer» LCM(p,q) = a3b3and HCF(p,q) = a2bLCM(p,q)\xa0{tex}\\times{/tex}\xa0HCF(p,q) = a3b3\xa0{tex}\\times{/tex}\xa0a2b{tex}\\begin{array}{l}\\mathrm{LCM}(\\mathrm p,\\mathrm q)\\;\\times\\mathrm{HCF}(\\mathrm p,\\mathrm q)\\;=\\mathrm a^5\\mathrm b^4\\;-----(1)\\\\\\mathrm{and}\\;\\mathrm{pq}=\\mathrm a^2\\mathrm b^3\\;\\times\\mathrm a^3\\mathrm b=\\mathrm a^5\\mathrm b^4\\;------(2)\\\\\\mathrm{from}\\;\\mathrm{eq}^\\mathrm n\\;(1)\\;\\mathrm{and}\\;(2)\\\\\\mathrm{LCM}(\\mathrm p,\\mathrm q)\\;\\times\\mathrm{HCF}(\\mathrm p,\\mathrm q)\\;\\;=\\mathrm{pq}\\\\\\\\\\end{array}{/tex}\xa0 | |
| 26074. |
Find the HCF of 65 and 117 and express it in the form 65m+117n. |
| Answer» m = 2 and n = –1. | |
| 26075. |
What Is the cube of 27 |
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Answer» Cube of 27 is 729Cube root of 27 is 3 27*27*27 Oh sorry it\'s wrong 3 |
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| 26076. |
If one of the zero of quadratic polynomial p(y)=5y |
| Answer» | |
| 26077. |
Show that there are infinitely many positive primes. |
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Answer» Thanks...☺☺ Let there be finite number of positive prime number\xa0a1,\xa0a2,\xa0a3\xa0.......,\xa0an\xa0.\xa0Such that\xa0a1<\xa0a2<\xa0a3\xa0< .......an+\xa0<\xa0an.Let\xa0m\xa0= 1 + a1,\xa0a2,\xa0a3\xa0.......,\xa0an\xa0. It is near that\xa0a1,\xa0a2,\xa0a3\xa0.......,\xa0an\xa0is divisible by\xa0a1,\xa0a2,\xa0a3\xa0.......,\xa0an\xa0.\xa0m\xa0is a prime number or it has factors other than\xa0a1,\xa0a2,\xa0a3\xa0.......,\xa0an\xa0. There exits a positive prime number other than\xa0a1,\xa0a2,\xa0a3\xa0.......,\xa0an\xa0.This contradicts the fact that there are finite number of positive primes.Hence, there infinite number of positive primes. |
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| 26078. |
5x²-6x-2=0,by complying the square method |
| Answer» 5x2 - 6x - 2 = 0Multiplying the above equation by 1/5{tex} \\Rightarrow {x^2} - \\frac{6}{5}x - \\frac{2}{5} = 0{/tex}{tex}\\Rightarrow x ^ { 2 } - \\frac { 6 } { 5 } x + \\left( \\frac { 3 } { 5 } \\right) ^ { 2 } - \\left( \\frac { 3 } { 5 } \\right) ^ { 2 } - \\frac { 2 } { 5 } = 0{/tex}{tex}\\Rightarrow \\left( x - \\frac { 3 } { 5 } \\right) ^ { 2 } = \\frac { 9 } { 25 } + \\frac { 2 } { 5 }{/tex}{tex}\\Rightarrow \\left( x - \\frac { 3 } { 5 } \\right) ^ { 2 } = \\frac { 9 + 10 } { 25 }{/tex}{tex}\\Rightarrow \\left( x - \\frac { 3 } { 5 } \\right) ^ { 2 } = \\frac { 19 } { 25 }{/tex}{tex}\\Rightarrow x - \\frac { 3 } { 5 } = \\pm \\frac { \\sqrt { 19 } } { 5 }{/tex}{tex}\\Rightarrow x = \\frac { 3 } { 5 } \\pm \\frac { \\sqrt { 19 } } { 5 }{/tex}{tex}\\Rightarrow x = \\frac { 3 + \\sqrt { 19 } } { 5 } \\text { or } x = \\frac { 3 - \\sqrt { 19 } } { 5 }{/tex} | |
| 26079. |
Find the l.c.m and h.c.f of 8/9,10/27,16/81 |
| Answer» Hcf=2/3 and lcm=640/6561 | |
| 26080. |
Related to math question |
| Answer» ?? | |
| 26081. |
(X-2)(X+1)=(X-1)(X+3) Check whether the following are quadratic equation |
| Answer» No | |
| 26082. |
In an equilateral triangle of side 3 root 3 cm find the length of the altitude |
| Answer» 4.5cm | |
| 26083. |
In an equilateral triangle ABC the side BC is trisected at D prove that 9AD×AD= 7AB×AB |
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Answer» It is also available in RS Aggarwal \'s book page no.249. The solution is in Rd sharma or in sample paper released by cbse in your cbse guide |
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| 26084. |
Given that HCF (306‚657)= 9‚find LCM (306‚657) |
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Answer» 306×657÷9=22338 as HCF×LCM=Product of numbers. 22338 |
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| 26085. |
How many prime numbers are there from 1 to ? |
| Answer» 25 prime numbers | |
| 26086. |
3(ax - by) + (a + 4b) = 0, 2(bx + ay) + (b - 4a) = 0 |
| Answer» The given equations may be written as2(ax - by) + (a + 4b) = 02ax - 2by + (a + 4b) = 02ax - 2by = - a - 4 b ... (i)and 2(bx + ay) + (b - 4a) = 0(2bx + 2ay) + (b - 4a) = 02bx + 2ay = 4a - b. ... (ii)Multiplying (i) by a and (ii) by b and adding, we get(2a2+ 2b2) x = (-a2\xa0- b2){tex} \\Rightarrow 2 \\left( a ^ { 2 } + b ^ { 2 } \\right) x = - \\left( a ^ { 2 } + b ^ { 2 } \\right) {/tex}{tex}\\Rightarrow x = \\frac { - 1 } { 2 }{/tex}Putting x = -{tex}\\frac 12{/tex}\xa0in (i), we get{tex}2 a \\times \\left( \\frac { - 1 } { 2 } \\right) - 2 b y = - a - 4 b{/tex}{tex}\\Rightarrow \\quad - a - 2 b y = - a - 4 b{/tex}{tex}\\Rightarrow \\quad 2 b y = 4 b {/tex}{tex}\\Rightarrow y = \\frac { 4 b } { 2 b } = 2{/tex}Hence, x =\xa0{tex}\\frac{-1}{2}{/tex} and y = 2 | |
| 26087. |
What is the difference between an and n in arithmetic progression? |
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Answer» An is the last term ofAP and n is number of terms in AP an\' is last term of A.P and \'n\' is no. Of terms in an AP an is the last term of ap and n is the number of the in ap |
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| 26088. |
The area of three adjacent faces of a cuboid are x,y,and z .If the volume is Vthen prove that V2 xyz |
| Answer» | |
| 26089. |
what is your opinion about new cbse exam system |
| Answer» According to me... As a coin has two faces..... It also have advantages as well as disadvantages | |
| 26090. |
Find the eleventh term from the last term of the ap 27,23,19,......-65 |
| Answer» | |
| 26091. |
A right triangle have sides 3,4,5 find it\'s allangles |
| Answer» It will be right angled triangle | |
| 26092. |
Carbon and its compound koi samja do |
| Answer» See PuStack to understand quick. | |
| 26093. |
1_1334_+;:)(455667= |
| Answer» | |
| 26094. |
shall I get 10th maths blue print? |
| Answer» You can check paper pattern here :\xa0https://mycbseguide.com/cbse-syllabus.html | |
| 26095. |
3+3 |
| Answer» 9 | |
| 26096. |
DIVIDE 12 INTO TWO PARTS SUCH THAT THEIR PRODUCT IS 32 |
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Answer» thanks for explaining briefly 8+4=128*4=32 4 nd 8 |
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| 26097. |
DIVIDE 12 INTO PART SUCH THAT THEIR PRODUCT IS 32 |
| Answer» 8+4=12 8*4=32 | |
| 26098. |
Pre board paper |
| Answer» Check last year papers here :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 26099. |
Give an AP |
| Answer» | |
| 26100. |
ST//RQ , PS=3 cm and SR=4 cm. Find the ratio of the area of triangle PST to the area OFTRIANGLE PRQ |
| Answer» PS = 3 cm, SR = 4 cm and ST || RQ.PR = PS + SR= 3 + 4 = 7 cmIn {tex}\\triangle{/tex}PST and {tex}\\triangle{/tex}PRQ{tex}\\angle{/tex}SPT\xa0{tex}\\cong{/tex}{tex}\\angle{/tex}RPQ (common angle){tex}\\angle{/tex}PST\xa0{tex}\\cong{/tex}{tex}\\angle{/tex}PRQ (Alternate angle){tex}\\triangle{/tex}PST\xa0{tex}\\sim{/tex}{tex}\\triangle{/tex}PRQ (AA configuration){tex}\\frac { \\text { ar } \\Delta P S T } { \\text { ar } \\Delta P Q R } = \\frac { P S ^ { 2 } } { P R ^ { 2 } } {/tex}{tex}= \\frac { 3 ^ { 2 } } { 7 ^ { 2 } } = \\frac { 9 } { 49 }{/tex}Hence required ratio = 9 : 49. | |