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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.
| 24601. |
Illiana you are what type of girl.are you good or beautiful girl |
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Answer» Heyy yash she is not a girl first of all and changes her name daily today her name is bajarang can u relly believe her I think u should not..??? Give me answer |
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| 24602. |
If the circumference of a circle exceeds it\'s diameter by 16.8 cm what is diameter of circle |
| Answer» Let, radius of the circle = r cmAccording to questionCircumference - Diameter = 16.8 cm{tex}\\Rightarrow 2\\pi r - 2r = 16.8{/tex}{tex}\\Rightarrow 2 \\times \\frac{{22}}{7} \\times r - 2r = 16.8{/tex}{tex}\\Rightarrow \\frac{{44r - 14r}}{7} = 16.8{/tex}{tex} \\Rightarrow \\frac{{30r}}{7} = 16.8{/tex}{tex} \\Rightarrow r = \\frac{{16.8 \\times 7}}{{30}} = \\frac{{117.6}}{{30}} = 3.92cm{/tex}{tex}\\therefore {/tex}\xa0Diameter of circle\xa0= 2r= 2 x 3.92 = 7.84 cm | |
| 24603. |
If sin^2A =2sinA find the value of A |
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| 24604. |
2 2. 2(a+b). = a. + 2ab+b. 2Then what is the answer of (√7+√9) |
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| 24605. |
How to construct a more than ogive graph |
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| 24606. |
Chapter no. 14 exercise 14.1 question no. 8 solved by the step divation method |
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| 24607. |
X3-6x2+11x-6 find the roots |
| Answer» The given polynomial f(x)={tex}\\text{x}^3-\\text{6x}^2+\\text{11x-6}{/tex}Since 3 is a zero of p(x), so (x - 3) is a factor of f(x).On dividing f(x) by (x - 3), we get{tex}\\therefore{/tex}\xa0f(x) = (x2\xa0- 3x + 2)(x - 3)= ( x2\xa0- 2x - x + 2)( x - 3)= [x(x - 2) -1(x - 2)](x - 3)= (x - 1)(x - 2)(x - 3)Now f(x)=0 if x - 1 = 0 or x - 2 = 0 or x - 3 = 0{tex}\\Rightarrow{/tex}\xa0x = 1 or x = 2 or x = 3{tex}\\mathrm{Hence}\\;\\mathrm{the}\\;\\mathrm{remainig}\\;\\mathrm{roots}\\;\\;\\mathrm{of}\\;\\mathrm f(\\mathrm x)\\;\\mathrm{are}\\;1\\;\\mathrm{and}\\;2\\;{/tex} | |
| 24608. |
If the number a,9,b,25form an so, a and b |
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| 24609. |
CosA -sinA + 1/cosA +sinA -1 = cosecA + cot A |
| Answer» I can solve it.....but cant type the solution on this becoz solution is large......so how can i help u miss... | |
| 24610. |
ANTHE ka result kya aaya |
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| 24611. |
Is there any AAA congruency criterion for triangles |
| Answer» Yes | |
| 24612. |
How to find perimeter of circle? |
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Answer» 2πr Answer is the circumference in equals to the perimeter in circle????? |
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| 24613. |
If sec0 + tan0 =7 then sec0-tan 0=? |
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| 24614. |
8 to the power x=5 to the power y=40 to the power 6. What is the value of x+y/xy |
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| 24615. |
Find roots 2x+7 |
| Answer» X= - 7/2 | |
| 24616. |
Chapter 10(Circles) of Xth class theorem 10.1 & 10.2 |
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| 24617. |
Show that 6 n cannot end with the digit 0 or 5 for any natural number N |
| Answer» If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5Prime factorisation of 6n = (2 ×3)nIt can be observed that 5 is not in the prime factorisation of 6n.Hence, for any value of n, 6n will not be divisible by 5.Therefore, 6n cannot end with the digit 0 or 5 for any natural number n. | |
| 24618. |
12+14-3 |
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Answer» 23 23 |
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| 24619. |
The sum of two no. Is 11and the sum of their reciprocal 1/2 find the no. |
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Answer» let two nos are x and yx+y=11x=11-y ...... i1/x + 1/y=1/2x+y/xy=1/22(x+y)=xyfrom i2(11-y+y)=xy22=xy.......... iix=22/y11-y=22/y11y -y2=22y2-11y+22=0\xa0 Its answer IS (11+ \\/33)÷2••••••(i)(11-\\/33)÷2••••••••(ii) OK TRY X+Y=111/X+1/y=1/2Equate |
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| 24620. |
Hi guys. What thing u find in girls. |
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| 24621. |
A-b= bIs it possible.give reason or example for ur answer |
| Answer» Yes it\'s possible.If A=2 & b=1Then, A-b= 2-1=1I.e A-b=b=1 | |
| 24622. |
If radius of cylinder is 25cm and height is 20cm .find surface area of cylinder. |
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| 24623. |
Blue print marks wise all 30 questions |
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| 24624. |
A-b =c |
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| 24625. |
How many terms of ap 18,16,14 taken so that there sum is 0. |
| Answer» As Sn=1/2(2a+(n-1)d) So, 0=1/2(2(18)+(n-1)(-2)) 0=36+(n-1)(-2) -36=(n-1)(-2) -36/-2=(n-1) 18=n-1 18+1=n n=19 | |
| 24626. |
What is the probability that a leap |
| Answer» We have to find the probability that a leap year has 53 Tuesdays and 53 Mondays.We know that, a leap year has 365 days which means 52 complete weeks and 2 days.Therefore,if 52 weeks end in Mon, then 2 days will be = Tue, WedIf 52 weeks end in Tue, then 2 days will be = Wed, ThuIf 52 weeks end in Wed, then 2 days will be = Thu, FriIf 52 weeks end in Thu, then 2 days will be = Fri, SatIf 52 weeks end in Fri, then 2 days will be = Sat, SunIf 52 weeks end in Sat, then 2 days will be = Sun, Monif 52 weeks end in Sun, then 2 days will be = Mon, TueTherefore,Total number of outcomes = 7Also,number of cases favourable to the event = 1Therefore,required probability that a leap year has 53 Sundays and 53 Mondays =\xa0{tex}\\frac{1}{7}{/tex} | |
| 24627. |
How much marks will be come from ncert book in maths paper of board exam ? |
| Answer» May be 30% | |
| 24628. |
To prove root 3 is a irrational number |
| Answer» \xa0let us assume that\xa0{tex}\\sqrt 3{/tex}\xa0be a rational number.{tex}\\sqrt { 3 } = \\frac { a } { b }{/tex}, where a and b are integers and co-primes and b{tex} \\neq{/tex}0Squaring both sides, we have{tex}\\frac { a ^ { 2 } } { b ^ { 2 } } = 3{/tex}or,\xa0{tex}a ^ { 2 } = 3 b ^ { 2 }{/tex}--------(i)a2\xa0is divisible by 3.Hence a is divisible by 3..........(ii)Let a = 3c ( where c is any integer)squaring on both sides we get(3c)2\xa0= 3b29c2\xa0= 3b2b2\xa0= 3c2so b2\xa0is divisible by 3hence, b is divisible by 3..........(iii)From equation(ii) and (iii), we have3 is a factor of a and b which is contradicting the fact that a and b are co-primes.Thus, our assumption that\xa0{tex}\\sqrt 3{/tex} is rational number is wrong.Hence,\xa0{tex}\\sqrt 3{/tex}\xa0is an irrational number. | |
| 24629. |
In a circle of centre o and r =5cm Ab is a third of length 5√3cm . Find the area of the sector AOB |
| Answer» It is given that AB = 5{tex}\\sqrt3{/tex}\xa0cm.{tex}\\Rightarrow \\quad A L = B L = \\frac { 5 \\sqrt { 3 } } { 2 } \\mathrm { cm }{/tex}Let\xa0{tex}\\angle {AOB}=2\\theta{/tex} . Then,\xa0{tex}\\angle A O L = \\angle B O L = \\theta{/tex}In\xa0{tex}\\triangle{/tex}OLA, we have{tex}\\sin \\theta = \\frac { A L } { O A } = \\frac { \\frac { 5 \\sqrt { 3 } } { 2 } } { 5 } = \\frac { \\sqrt { 3 } } { 2 }{/tex}{tex}\\Rightarrow \\quad \\theta = 60 ^ { \\circ }{/tex}{tex}\\Rightarrow \\quad \\angle A O B = 120 ^ { \\circ }{/tex}{tex}\\therefore{/tex}Area of sector AOB =\xa0{tex}\\frac { 120 } { 360 } \\times \\pi \\times 5 ^ { 2 } \\mathrm { cm } ^ { 2 } = \\frac { 25 \\pi } { 3 } \\mathrm { cm } ^ { 2 }{/tex} | |
| 24630. |
What is perimeter of circle |
| Answer» 2. Pie .r(radius) | |
| 24631. |
Show that the sequence defined by an=3n2- 5 is not an A.P. |
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| 24632. |
If a sin θ+ b cos θ = c then prove that a cos θ - b sin θ =Г a2 + b2-c2 |
| Answer» We have, {tex}asin\\theta+bcos\\theta=c{/tex}On squaring both sides, we get{tex}(asin\\theta+bcos\\theta)^2=c^2{/tex}(a sin θ)2\xa0+ (b cos θ)2\xa0+ 2(a sin θ) (b cos θ) = c2⇒ a2\xa0sin2\xa0θ + b2\xa0cos2\xa0θ + 2ab sin θ cos θ = c2⇒ a2(1 – cos2\xa0θ) + b2\xa0(1 – sin2\xa0θ) + 2 ab sin θ cos θ = c2 {tex}[\\because sin^2\\theta+cos^2\\theta=1]{/tex}⇒ a2\xa0– a2\xa0cos2\xa0θ + b2\xa0– b2\xa0sin2\xa0θ + 2ab sin θ cos θ = c2⇒ –a2\xa0cos2\xa0θ – b2\xa0sin2\xa0θ + 2ab sin θ cos θ = c2\xa0– a2\xa0– b2 Taking Negative common,⇒ a2\xa0cos2\xa0θ + b2\xa0sin2\xa0θ – 2ab sin θ cos θ = a2\xa0+ b2\xa0– c2⇒ (a cos θ)2\xa0+ (b sin θ)2\xa0– 2(a cos θ) (b sin θ) = a2\xa0+ b2\xa0– c2⇒ {tex}(acos\\theta-bsin\\theta)^2=a^2+b^2-c^2{/tex}⇒{tex}acos\\theta-bsin\\theta{/tex} =\xa0{tex}\\pm \\sqrt { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } }{/tex}\xa0Hence proved, {tex}acos\\theta-bsin\\theta{/tex} =\xa0{tex}\\sqrt{a^2+b^2-c^2}{/tex} | |
| 24633. |
Why astica called zero |
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| 24634. |
2+5_6+6/7 |
| Answer» And its a silly question | |
| 24635. |
if x=1 is a common root of quadratic equations ax2+ax+3=0and x2+x+b=0,then find ab. |
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| 24636. |
If x=y thenz=? |
| Answer» Answer is v | |
| 24637. |
2+4+6+8+10+12+14+16+18+20 |
| Answer» Great question? | |
| 24638. |
What is value of x0 |
| Answer» 0 | |
| 24639. |
How we can download the videos from my cbse app |
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| 24640. |
CosecQ + CotQ = x then CosecQ +CotQ =? |
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| 24641. |
Find k questions |
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| 24642. |
Root3 |
| Answer» 1.732 | |
| 24643. |
what is the sum of 2 3 4 5 6 if they are in AP |
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Answer» 20 20 |
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| 24644. |
Show that 5-_/3 is irrational where _/3 is an irrational number |
| Answer» We will solve this by contradiction method i.e.,\xa0Assume\xa0{tex}5 - \\sqrt { 3 } = \\frac { p } { q }{/tex}\xa0be a rational number.{tex}\\therefore 5 - \\frac { p } { q } = \\sqrt { 3 }{/tex}\xa0or\xa0{tex}\\frac { 5 q - p } { q } = \\sqrt { 3 }{/tex},Since p,q are integers, therefore\xa0{tex}\\frac { 5 q - p } { q }{/tex} is a rational number, which is a contradiction,since\xa0{tex}\\sqrt 3{/tex} is an irrational number.Therefore, our supposition is wrong and hence, 5 - {tex}\\sqrt 3{/tex}\xa0is irrational. | |
| 24645. |
4s square -4s+1 |
| Answer» g(s)=4s2 - 4s + 1Here, a = 4, b = -4 and c = 1We have, 4s2 - 4s + 1= 4s2 - 2s - 2s + 1= 2s\xa0(2s\xa0− 1) − 1 (2s\xa0− 1)= (2s\xa0− 1) (2s\xa0− 1)g(s) =0 if 2s-1=0\xa0Hence\xa0{tex}\\text{s=}\\frac12\\text{,}\\frac12{/tex}Sum of zeroes\xa0{tex}\\text{=}\\frac12+\\frac12=1\\operatorname{=-}\\frac{-4}4=-\\frac{\\mathrm b}{\\mathrm a}=-\\frac{\\mathrm{coefficient}\\;\\mathrm{of}\\;\\mathrm s}{\\mathrm{coefficient}\\;\\mathrm{of}\\;\\mathrm s^2}{/tex}{tex}{/tex}Product of Zeroes= {tex}\\frac12\\text{×}\\frac12=\\frac14=\\frac{\\mathrm c}{\\mathrm a}=\\frac{\\mathrm{constant}\\;\\mathrm{term}}{\\mathrm{coefficient}\\;\\mathrm{of}\\;\\mathrm s^2}{/tex} | |
| 24646. |
5x+5 |
| Answer» 25 | |
| 24647. |
X-3/5+x-4/7=6-2x-1/35 |
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| 24648. |
x+a/b+c + x+b/c+a +x+c/a+b +3 =0 a>0 b>a c>0 then x is equal to |
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| 24649. |
Cot sqaure theta -1 by sin square theta =-1 (prove) |
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| 24650. |
How √2 irrational & ranation |
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