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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.
| 38701. |
Find 4 numbers in AP such that their sum is 32 and the sum of their squares is 336 |
| Answer» Let the required number be (a - 3d), (a - d), (a + d) and (a + 3d)Sum of these numbers = (a - 3d) + (a - d)+ (a + d) + (a + 3d)According to the question, sum of the numbers=32{tex}\\therefore{/tex}4a = 32\xa0{tex}\\Rightarrow{/tex}\xa0a = 8Sum of the squares of these numbers=(a-3d)2+(a-d)2+(a+d)2+(a+3d)2=4(a2+5d{tex}^2{/tex})Now, sum of the squares of numbers=336{tex}\\therefore{/tex}4(a2+5d2)=336{tex}\\Rightarrow{/tex}a2+5d2=84 [{tex}\\because {/tex}a=8]{tex}\\Rightarrow{/tex}5d2= 84-64{tex}\\Rightarrow{/tex}5d2=20{tex}\\Rightarrow{/tex}d2=4{tex}\\Rightarrow{/tex}d={tex} \\pm {/tex}2Hence, the required numbers (2, 6, 10, 14). | |
| 38702. |
How we solve (X+2)3=x3-4 |
| Answer» 3x+6 =3x-4;=6+4=10 | |
| 38703. |
Find the value of k for which the equations 2x+5y=7,3x-ky=5 has a unique solution |
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Answer» 2/3 not= 5/K2K not= 15 ,K not = 15/2. Sorry for previous answer of this question For unique solution A1/A2 not equal to b1/b2Then, 2/5 not= 5/K ~2K not = 25. ~K not= 25/2. Accept 25/2 other value can be written as K. |
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| 38704. |
All formula of ch-1 |
| Answer» Which subject | |
| 38705. |
5-[(1×2)+4÷2+1] |
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Answer» 0 0 0 6 5-(1x 2) +4/2+15-(2)+2+15-2+33+36 1 |
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| 38706. |
Solve the linear equation:-4/x+3y=143/x-4y=23 |
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Answer» Ok thanx i\'ll check it a=5 ,y=-2. Sorry for previous answer I done silly mistake of sign there so check it Let 1/x= a then , 4/x= 4a,3/x=3a Then our equations are. 4a+3y=14 and 3a-4y= 23 by substitution method we get ,a=(14-3y)/4....(1) by putting the value of \'a\' on another eq. We get 3a-4y=23. ,3{(14-3y)/4}-4y=23 ,(42-9y-16y)/4=23. 42-25y=92 .25y=-92+42. Therefore, -y=-50/25. Y=2,. Again in eq. (1). a=(14-3y)/4 =(14-6)/4, a=8/4 =2So a =2 ,y= 2. Guyz plz solve this ques |
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| 38707. |
,If tanA=√3, find other trignometric ratiosvof A |
| Answer» TanA=√3/1. We know TanA=p/b Then,p=√3,b=1,h=√(√3)^2+(1)^2Therefore,h=2. SinA=p/h=√3/2,cosA=b/h=1/2,cotA=b/p=1/√3,secA=h/b=2/1,cosecA=h/p=2/√3. | |
| 38708. |
Factorise-100-9x^2 |
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Answer» 100-9x²) (10²-9x²)(10x²-3x²) (a²-b²=(a+b)(a-b) a=10&b=3x. (10+3x) (10-3x) multiply 3 by -1 to get -3 (10+3x) (10-3x)...? 100-9x^2=10^2-(3x)^2=(10+3x)(10-3x) |
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| 38709. |
If sinA+coda=√3 then prove that tanA+cotA=1 |
| Answer» | |
| 38710. |
√3+√5 is not a prime number |
| Answer» root3 + root5 is not a rational number. (Prove this yourself)Now since root3 + root5 is not a rational number therefore it\'s not a prime number | |
| 38711. |
Compartment |
| Answer» | |
| 38712. |
how much chapterwill come in board of2018_2019 |
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Answer» All chapters in all the books of ncert All chapters All the chapters will come |
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| 38713. |
Solve the x /alpha +y /beta =2, ax square - by =ax-by=a square - b square by all the methods? |
| Answer» | |
| 38714. |
Solve the x/alpha + y/beta =2, ax square - by=a square - b square by all the methods? |
| Answer» Please don\'t answer this question is wrong | |
| 38715. |
If 8 is zero of polynomial x square - 10x +k=0, then find the value of k? |
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Answer» Put x=8 in f(x) we get 8square - 10×8 + k=0We get 64-80+k=0Or k=80-64=16So k=16 16 is the value of k. f (x) = x square - 10x + kf (8) = (8×8) - (10×8) + k 64 - 80 + k = 0 -16 + k = 0 k = 16 Putting the value of = 8 so, 10X 8 +k= 0 80+ k= 0K= - 80 |
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| 38716. |
If roots of equation 2x²+7x+4=0are in ratio p:q then find the value of√p/q+√q/p |
| Answer» Yaar plz answer my both questions.??? | |
| 38717. |
Find the zero : h(t) =t² - 15 |
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Answer» Sorry put t=root15 or -root15. Then see Put x=root15 or -root15. Then see There is no root on 15 Root15 or -root15 |
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| 38718. |
®2-1/3 |
| Answer» 2/1 - 1/36/3-1/35/3 | |
| 38719. |
Ifx+1/x=3, find the value of x2+1/4x |
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Answer» Solve it 5/4 |
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| 38720. |
Irrational +irrational =? |
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Answer» "The sum of two irrational numbers is SOMETIMES irrational." The sum of two irrational numbers, in some cases, will be irrational. However, if the irrational parts of the numbers have a zero sum (cancel each other out), it will rational irrational Maybe rational or irrational |
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| 38721. |
Send me Facebook req I will make a group on fb Fb name karan oberoi) |
| Answer» | |
| 38722. |
1st word is 38, 16th word 73 find out 31 th word |
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Answer» Use APA=38D =d73 =38+(16-1)d73-38=15dD=7/3Use thisN=a+(n-1)d=38+30×7/338+70108???? 108 |
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| 38723. |
Find the HCF of 1305,1365 by euclid division algorithim |
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Answer» Their HCF is 15 Nhi ata to solve kr k du Divide kr do yr simple |
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| 38724. |
Show that any positive odd integer is of the from 4q+1 or 4q+3 were q is some integer |
| Answer» suppose that a is any positive integer then it is divided by 4 then we get q as a quotient and r as a remainder then by euclid\'s division lemma we get that a=4q + r where 0<= 4 | |
| 38725. |
12+1235488 |
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Answer» 1235500 1235500 |
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| 38726. |
If roots are equal then find k, 2x2-(k-1)x+8 |
| Answer» | |
| 38727. |
If 7sin ka power2+ cos ka power2 = 4, show that tan=1/3 |
| Answer» Wrong question | |
| 38728. |
8.1 ke formulae |
| Answer» Check formulae in notes\xa0:\xa0https://mycbseguide.com/cbse-revision-notes.html | |
| 38729. |
10 application promblems related to edl |
| Answer» | |
| 38730. |
Square of 10 |
| Answer» 100 | |
| 38731. |
Sin A=a÷bFind the value of cot A |
| Answer» CotA=√b^2-a^2÷a | |
| 38732. |
Use euclid division algorithm to find the hcf of1= 135 and 225 |
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Answer» Not 45 it is actually (3+3)*5 45 45 |
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| 38733. |
Explain full ap |
| Answer» Arithmetic progressions | |
| 38734. |
Trignomery |
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Answer» The most boring samajh na aay to sir ke upar Trignometery hota hai |
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| 38735. |
a³-12a(a-4)-64 |
| Answer» a³- 12a² + 48a - 64 | |
| 38736. |
Find the sume of 16terms of an ap -1,-5,-9,..... |
| Answer» Put the formula of sum i.e. n/2 (2a + n-1) d .now by putting the value , u will get the ans. -496 | |
| 38737. |
if n is any prime number and a square is divisible by n . then n also divide a , justify |
| Answer» | |
| 38738. |
If 9×5×2=529and 4×7×2=724 then find value of 3×9×8 |
| Answer» 983 | |
| 38739. |
I wamt to learn substituting metod of ch 3 |
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Answer» Substitution is the method of substituting a value of let\'s say "x" from an equation to another equation. Ch-3 example7 pdhlo vha s smz aa jayega Obviously in math substitution method math m hii h From which subject?? |
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| 38740. |
9x^2 -9(a+b)x+(2a^2+5ab+2b^2)=0 |
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Answer» What to do What to di |
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| 38741. |
Difference between oxidation and reduction? |
| Answer» OXIDATION->it means1.(adding of O2)2.(removal of H2)3.(loss of electrons).Ex-->. C+O2----->CO2,here C is getting oxidised. REDUCTION->it means 1. (removal of O2)2.(adding of H2)3.(adding/gain of electrons).Ex->CO2-->C+O2 ,here CO2 us getting reduced. | |
| 38742. |
Is it necessary to write x in between the formula of rust? |
| Answer» Yes,because we don\'t know that how many atoms are there. | |
| 38743. |
What is UPWARD PERA BOLA AND DOWNWARD PERABOLA |
| Answer» Upward perabola means related to value of x and downward means related to y | |
| 38744. |
How can we factorise X square + 400 x - 960000 |
| Answer» x²+400x-960000 ......................................... here,product=-960000 and sum=400.Factors=-1200 and800 .................. x²-1200x+800x-960000........... ..................... x(x-1200)+800(x-1200)............. .............. (x+800)(x-1200)=0....................... ............... x=-800or+1200 | |
| 38745. |
Solve: 1/x-3-1/x+5=1/6 by quadratic method |
| Answer» 258555 | |
| 38746. |
Prove that underroot 5 |
| Answer» | |
| 38747. |
if a and b are positive integers root to always lies between a/b and a-2b/a+b |
| Answer» We do not know whether\xa0{tex} \\frac { a } { b } < \\frac { a + 2 b } { a + b } \\text { or, } \\frac { a } { b } > \\frac { a + 2 b } { a + b }{/tex}.Therefore, to compare these two numbers, let us compute\xa0{tex} \\frac { a } { b } - \\frac { a + 2 b } { a + b }{/tex}We have,{tex} \\frac { a } { b } - \\frac { a + 2 b } { a + b } = \\frac { a ( a + b ) - b ( a + 2 b ) } { b ( a + b ) }{/tex}\xa0{tex} = \\frac { a ^ { 2 } + a b - a b - 2 b ^ { 2 } } { b ( a + b ) } = \\frac { a ^ { 2 } - 2 b ^ { 2 } } { b ( a + b ) }{/tex}{tex} \\therefore \\quad \\frac { a } { b } - \\frac { a + 2 b } { a + b } > 0{/tex}{tex} \\Rightarrow \\quad \\frac { a ^ { 2 } - 2 b ^ { 2 } } { b ( a + b ) } > 0{/tex}{tex} \\Rightarrow{/tex} a2 - 2b2 > 0{tex} \\Rightarrow{/tex} a2> 2b2{tex} \\Rightarrow \\quad a > \\sqrt { 2 } b{/tex}and,\xa0{tex} \\frac { a } { b } - \\frac { a + 2 b } { a + b } < 0{/tex}{tex} \\Rightarrow \\quad \\frac { a ^ { 2 } - 2 b ^ { 2 } } { b ( a + b ) } < 0{/tex}{tex} \\Rightarrow{/tex} a2 - 2b2 < 0{tex} \\Rightarrow{/tex}a2 <2b2{tex} \\Rightarrow \\quad a < \\sqrt { 2 } b{/tex}Thus,\xa0{tex} \\frac { a } { b } > \\frac { a + 2 b } { a + b }{/tex}, if\xa0{tex}a > \\sqrt { 2 b }{/tex}\xa0and\xa0{tex} \\frac { a } { b } < \\frac { a + 2 b } { a + b }{/tex},\xa0if\xa0{tex} a < \\sqrt { 2 } b{/tex}.So, we have the following cases:CASE I When\xa0{tex} a > \\sqrt { 2 } b{/tex}In this case, we have{tex} \\frac { a } { b } > \\frac { a + 2 b } { a + b } \\text { i.e., } \\frac { a + 2 b } { a + b } < \\frac { a } { b }{/tex}We have to prove that{tex} \\frac { a + 2 b } { a + b } < \\sqrt { 2 } < \\frac { a } { b }{/tex}We have,{tex} a > \\sqrt { 2 } b{/tex}{tex} \\Rightarrow{/tex} a2> 2b2 [Adding a2 on both sides]{tex} \\Rightarrow \\quad 2 a ^ { 2 } + 2 b ^ { 2 } > \\left( a ^ { 2 } + 2 b ^ { 2 } \\right) + 2 b ^ { 2 }{/tex}\xa0[Adding\xa02b2 on both sides]{tex} \\Rightarrow \\quad 2 \\left( a ^ { 2 } + b ^ { 2 } \\right) + 4 a b > a ^ { 2 } + 4 b ^ { 2 } + 4 a b{/tex}\xa0[Adding 4ab on both sides]{tex} \\Rightarrow \\quad 2 \\left( a ^ { 2 } + 2 a b + b ^ { 2 } \\right) > a ^ { 2 } + 4 a b + 4 b ^ { 2 }{/tex}{tex} \\Rightarrow \\quad 2 ( a + b ) ^ { 2 } > ( a + 2 b ) ^ { 2 }{/tex}{tex} \\Rightarrow \\quad \\sqrt { 2 } ( a + b ) > a + 2 b{/tex}{tex} \\Rightarrow \\quad \\sqrt { 2 } > \\frac { a + 2 b } { a + b }{/tex} ........(i)Again,{tex} a > \\sqrt { 2 } b {/tex}{tex}\\Rightarrow \\frac { a } { b } > \\sqrt { 2 }{/tex} .......(ii)From (i) and (ii), we get{tex} \\frac { a + 2 b } { a + b } < \\sqrt { 2 } < \\frac { a } { b }{/tex}CASE II When\xa0{tex} a < \\sqrt { 2 } b{/tex}In this case, we have{tex} \\frac { a } { b } < \\frac { a + 2 b } { a + b }{/tex}We have to show that\xa0{tex} \\frac { a } { b } < \\sqrt { 2 } < \\frac { a + 2 b } { a + b }{/tex}We have,{tex} a < \\sqrt { 2 } b{/tex}{tex} \\Rightarrow \\quad a ^ { 2 } < 2 b ^ { 2 }{/tex}{tex} \\Rightarrow \\quad a ^ { 2 } + a ^ { 2 } < a ^ { 2 } + 2 b ^ { 2 }{/tex}\xa0[Adding a2 on both sides]{tex} \\Rightarrow \\quad 2 a ^ { 2 } + 2 b ^ { 2 } < a ^ { 2 } + 2 b ^ { 2 }+ 2 b ^ { 2 }{/tex}\xa0[Adding 2b2 on both sides]{tex}\\Rightarrow \\quad 2 a ^ { 2 } + 2 b ^ { 2 } < a ^ { 2 } + 4 b ^ { 2 }{/tex}{tex} \\Rightarrow \\quad 2 a ^ { 2 } + 4 a b + 2 b ^ { 2 } < a ^ { 2 } + 4 a b + 4 b ^ { 2 }{/tex}\xa0[Adding 4ab on both sides]{tex} \\Rightarrow \\quad 2 ( a + b ) ^ { 2 } < ( a + 2 b ) ^ { 2 }{/tex}{tex} \\Rightarrow \\sqrt { 2 } ( a + b ) < a + 2 b{/tex}{tex} \\Rightarrow \\quad \\sqrt { 2 } < \\frac { a + 2 b } { a + b }{/tex}\xa0. ...(iii){tex} \\Rightarrow \\quad a < \\sqrt { 2 } b \\Rightarrow \\frac { a } { b } < \\sqrt { 2 }{/tex} ....(iv)From (iii) and (iv), we get{tex} \\frac { a } { b } < \\sqrt { 2 } < \\frac { a + 2 b } { a + b }{/tex}Hence,\xa0{tex} \\sqrt { 2 }{/tex}\xa0lies between\xa0{tex} \\frac { a } { b }{/tex}\xa0and\xa0{tex} \\frac { a + 2 b } { a + b }{/tex}. | |
| 38748. |
Kis chapter se adhik number ata hai papers mei |
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Answer» Mensuration ,Trigonometary, geomertyAre importantBUT YOU SHOULD TAKE WHOLE CHAPTERS AS PAPER PATTERN.... Trigonometry Maths |
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| 38749. |
If an=9-Sn write 10th term of an AP |
| Answer» | |
| 38750. |
Traingle kai liya kuch tips doo |
| Answer» Triangle | |