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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.
| 7051. |
What is trigonometry ? |
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Answer» Three side measurement Tri -three gon -side metry measurement |
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| 7052. |
use euclid\'s divisiono algorithm to find the HCF of the following numbers 1. 65 and 117 |
| Answer» 117 = 13 {tex}\\times{/tex}\xa03 {tex}\\times{/tex}\xa0365 = 13 {tex}\\times{/tex}\xa05HCF (117, 65) = 13LCM(117,65) = 13 {tex}\\times{/tex}\xa05 {tex}\\times{/tex}\xa03 {tex}\\times{/tex}\xa03 = 585Here is given that:{tex}HCF =65m-117{/tex}{tex}13=65m-117{/tex}{tex}65m=130{/tex}m =\xa0{tex}\\frac { 130 } { 6 5 } ={/tex}2 | |
| 7053. |
The sum of the first nth odd term of an ap |
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Answer» Thanks bhi nahi kaha mehnat bekar kerdi me thanks ke deta ho thanks kunki mene ye padhe bahut accha N^2is right and {tex}s_n ={n\\over2}[{2a+(n-1)d}]{/tex}in case of odd number series{tex}s_n ={n\\over2}[{2(1)+(n-1)1}]{/tex}{tex}s_n ={n\\over2}[{2+n-1}]{/tex}{tex}s_n ={n\\over2}[{n+1}]{/tex}is the required answer, |
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| 7054. |
An observer 1.5 m tall is 20.5 m away f |
| Answer» let ht of observer=CD\xa0=1.5mht of tower = AB=22mdistance between observer and tower = AC=20.5mht of remaining tower from his eye D\'B=22-1.5=20.5mangle of elevation tan D=20.5/20.5=1 thus Angle=45o | |
| 7055. |
sum of all 2-digit multiples of 3 |
| Answer» Sum of 2digit multiples of 3 are:12,15,18,21..........................99a=12d=15-12 =3an=99an=a+(n-1)d99=12+(n-1)399-12 =3n-387=3n-33n=87+33n=90N=30Sn=n÷2(a+an)Sn=30÷2(12+99)Sn=15(111)Sn=1665 | |
| 7056. |
In a right angle triangle ABC ,BC=36015 ,AB=48020 ,find hypotenuse AC |
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| 7057. |
What I\'d area of triangle |
| Answer» 1/2*b*h or root s(s-a)(S-b)(s-c) | |
| 7058. |
The value of √6+√6+√6+.......is |
| Answer» Let\xa0{tex}x = \\sqrt { 6 + \\sqrt { 6 + \\sqrt { 6 + \\dots } } }{/tex}\xa0......(i){tex}\\Rightarrow{/tex}\xa0x2 =\xa0{tex}( \\sqrt { 6 + \\sqrt { 6 + \\sqrt { 6 . . } } } ) ^ { 2 }{/tex}\xa0[Squaring both sides]{tex}\\Rightarrow{/tex}\xa0x2 =\xa0{tex}6 + \\sqrt { 6 + \\sqrt { 6 + \\sqrt { 6 + \\ldots } } }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}x^2 = 6 + x{/tex} [From (i)]{tex}\\Rightarrow{/tex}\xa0{tex}x^2 - x - 6 = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}(x - 3) (x + 2) = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}x = 3, x = -2{/tex}{tex}\\therefore{/tex}\xa0x = 3 [{tex}\\because{/tex}\xa0x > 0] | |
| 7059. |
tan 48° tan 23°tan 42°tan 67° = 1 |
| Answer» tan48 tan42 tan23 tan67tan48 tan(90-48) tan23 tan(90-67) tan48 cot48 tan23 cot231×1=1 | |
| 7060. |
I can\'t understand triangles and solve the q can u help me plzz |
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Answer» I can help u Yeah give your no |
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| 7061. |
if ABC is right angle triangle at C if tan A =1/rout3 .find the value of sinA cosB +codA sinB |
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| 7062. |
How to do mid term factorization......how to find mid term |
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| 7063. |
Can 15 n ends zero for any natural number |
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Answer» No No 15 n can never end with zero no because every natural no should have 2 and 5 as a prime factors to end with digit 0. Yes |
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| 7064. |
SinA+cosA = 1 |
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Answer» Not possible No.it is not a identity. True will- sin2+ cos2=1 |
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| 7065. |
What is series |
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| 7066. |
Prove root3 is irrational |
| Answer» Given in ncert | |
| 7067. |
(a+b)(a-b) |
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Answer» a2-b2 (a+b)(a-b)=a2-b2(a+b)(a-b)=a(a-b)+b(a-b)=a2-ab+ab-b2 =a2-b2\xa0\xa0\xa0\xa0 |
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| 7068. |
how to find the value of sin35 ? Give answer please |
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| 7069. |
An a. P consists 50 terms and its last term is 20th and the d=2.find the first term of an ap |
| Answer» n=50,no. of termsl=20, last termd=2, common differencea=?, first termFormula l=a+(n-1)d20=a+(50-1)2a=-78\xa0\xa0 | |
| 7070. |
A cube + B cube=10A sq. + B sq.=7A + B= ????? |
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| 7071. |
(a+b)³ = |
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Answer» a^3 + b^3+ 3a^2b + 3 ab^2 a^3 +b^3+3ab(a+b) |
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| 7072. |
class 10 ch.6 ex 6.2 |
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| 7073. |
If sec^ - tan^ =4 prove that cos^= 8/17????????????..... Let ^ be theta...???☺☺?? |
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| 7074. |
X^2-6x=0 |
| Answer» x2\xa0- 6x = 0x2\xa0= 6xx = 6 | |
| 7075. |
if P (-5,7) Q (0,y) PQ = 13 find y |
| Answer» (pq)2=(-5-0)2 + (7-y)213x13=25+49+y2-14y169=y2-14y+74y2-14y-95=0y2-19y+5y-95=0y(y-19) + 5(y-19)=0y=19 or y=-5 | |
| 7076. |
Find the discrimination of the equation 3xsqaure -2x+8=0 |
| Answer» -92=D | |
| 7077. |
What it AP |
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Answer» A sequence in which each term except the first is obtained from the previous by adding a constant value, known as the common difference of the arithmetic progression Arithmetic progression |
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| 7078. |
Is the maths easy or not ??? Give reason |
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Answer» It is very easy but you need BRAIN to solve. If you don\'t have brain you will hate it and if you have brain you will like it.. I have brain so I love it ❤❤❤❤❤. Easy hai but itna bhi nahi |
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| 7079. |
What is fundamental theorem |
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| 7080. |
How many methods are ther in chapter 3 |
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Answer» Graphical method also Elimination,Subsitation and Cross multiplication |
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| 7081. |
Represent 9.3 on number line |
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| 7082. |
Represent 0n number line |
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| 7083. |
SinA+CosA is equal to root2.cosa then prove that sinA-cosA is equal to root2.sinA |
| Answer» sinA+cosA=root2.cosAS.B.S.(sinA+cosA)^2=root 2 sq.cos^2Asin^2A+cos^2A+2sinA.cosA=2.cos^2A1+2sinA.cosA=2.cos^2A---(sin^2A+cos^2A=1)2sinA.cosA=2cos^2A-1*L.H.S.*sinA-cosAS.B.S.(sinA-cosA)^2=sin^2A+cos^2A-2sinA.cosA=1-2sinA.cosA (sin^2A+cos^2A=1)=1-(2cos^2A-1) (2sinA.cosA=2cos^2A-1)=1-2cos^2A+1=2-2cos^2A=2(1-cos^2A)=2.sin^2A(SinA-cosA)^2=2.sin^2AsinA-cosA=root2.sin^2A (whole root)sinA-cosA=root2.sinA | |
| 7084. |
Sectetha=x+1/x (given) prove that sectetha×tantetha=2X or 1/2X |
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| 7085. |
23-23 |
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Answer» 0.. 0 0 0 |
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| 7086. |
tanA+secA-(sec square A - tan square A)÷tan A - secA |
| Answer» 0 | |
| 7087. |
Diffine polynomial |
| Answer» Polynomial:- It is an expression consisting of variables (or indeterminates) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.An example of a polynomial of a single indeterminate x is x2 − 4x + 7. | |
| 7088. |
What is theorm 6.1 proved it? |
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| 7089. |
2.÷9 |
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Answer» Ans. Is 0.2 4.5 ans of this question ? |
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| 7090. |
If sin3A=cos(A-26°)where 3A is an acute angle,find value of A |
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Answer» Sin3A=cos(A-26)Sin3A=sin263A=26A=26÷3 Thanks Bolne ki jarurat bahu hai SinA= cos(90-A) sin 3A = sin(90-A+26) Sin3A = sin(116-A)3A=116-A4A=116A=116/4A= 29 29 26 |
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| 7091. |
What is Ap? |
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Answer» Arithmetic progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. For instance, the sequence 5, 7, 9, 11, 13, 15, is an arithmetic progression with common difference of 2. Arithmetic progression |
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| 7092. |
sin 10 + sin 20 + sin 50 + sin 40 = sin 70 + sin 80 prove |
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| 7093. |
What you mean by secant???? |
| Answer» Sec | |
| 7094. |
Kx(x-2)+6 find the value of k |
| Answer» We have, kx(x - 2) + 6 = 0{tex}\\Rightarrow k x^{2}-2 k x+6=0{/tex}........(1)We know that quadratic equation {tex}ax^2+bx+c=0{/tex},a≠0, has equal roots if its discriminant D is 0.D = 0\xa0{tex}\\Rightarrow{/tex}{tex}b^2-4ac=0{/tex}{tex}\\Rightarrow{/tex}4k2 - 4(k)6 = 0 [from given quad. equ.(1); a=k≠0, b= -2k, c= 6]{tex}\\Rightarrow{/tex}4k2 - 24k = 0\xa0{tex}\\Rightarrow{/tex}4k(k - 6) = 0 {tex}\\Rightarrow{/tex}k - 6 = 0 [ since, k≠0]{tex}\\Rightarrow{/tex}k = 6\xa0Hence the value of K is 6. | |
| 7095. |
Bpt theorem |
| Answer» IN A TRIANGLE THE LINE DRAWN PARLLEL TO ANYSIDE DIVIDES REMAINING SIDES PROPORYIONALLY | |
| 7096. |
Find HCF and LCM of 8/9,10/27,16/81 |
| Answer» HCF of fractions = HCF\xa0of numerators/LCM of denominatorsHCF= HCF\xa0of 8,10,16 / LCM\xa0of 9,27,81 = 2 / 81LCM of fractionss =LCM of numerators / HCF\xa0of denominatorsLCM = LCM\xa0of 8,10,16 / HCF\xa0of 9,27,81 = 80 / 9REMEMBER formula for finding HCF of fractions = hcf of numerators / lcm of denominators LCM ,,,,,,,,,,,,,,,,, = lcm of numerators / hcf of denominators | |
| 7097. |
how to find common difference and write the first term |
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Answer» \'a\'is first term, \'d\' is common difference Use the formula -last term(An)=a+(n-1)d |
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| 7098. |
If sinA+2cosA=1prove that 2sinA-cosA=2 |
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| 7099. |
HOW TO PROVE THAT A QUADRILATERAL IS TRAPEZIUM |
| Answer» Given in the quadrilateral ABCD{tex}\\frac{AO}{BO}=\\frac{CO}{DO}{/tex}or,\xa0{tex}\\frac { A O } { C O } = \\frac { B O } { D O }{/tex} ...(i)Draw\xa0EO {tex}\\parallel{/tex} AB onIn\xa0{tex}\\triangle A B D,{/tex}\xa0EO {tex}\\parallel{/tex} AB (By construction){tex}\\therefore {/tex}\xa0{tex}\\frac { A E } { E D } = \\frac { B O } { D O }{/tex}\xa0(By BPT)...(ii)From (i) and (ii) we get\xa0{tex}\\frac{AO}{CO}=\\frac{AE}{ED}{/tex}Hence by converse of BPT in {tex}\\triangle{/tex}ADC\xa0{tex}EO\\|CD{/tex}But\xa0{tex}EO \\|AB {/tex}So {tex}AB\\|CD {/tex}Therefore ABCD is a trapezium | |
| 7100. |
Find the number of natural numbers between 101 and 999 which are devisible by both 2 and 5 |
| Answer» we know that 1st number div by 2 and 5 = 110thus a = 110d = 10 (2 x 5 = 10)last number div by 2 and 5 = an = 990now we know that an = a + (n-1)d990 = 110 + (n-1)10880 = 10n - 10890 = 10nn = 89thus number of numbers between 101 and 999 are 89 | |