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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.
| 21401. |
PE+Pnot E=1 |
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| 21402. |
Find the other zero of the quadratic polynomial y square+7y_60 if one zero is _12 |
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Answer» Sum of zeroes= -b/a= -12+X= -7/1= X = -7+12=5 5 |
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| 21403. |
The ratio of7th to the 3rd term of an AP is 12:5. Find the ratio of 13th to the 4th term |
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Answer» Take a7/a3 = 12/5 you will get valuez then put those values in second part ? 10:3 |
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| 21404. |
SinA-sin^2A=1 then find cos^2A-cosA=? |
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| 21405. |
What is the formula of circle |
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Answer» πr^2 PaiR(2) |
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| 21406. |
Prove that √3+√5 are irrational |
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Answer» We have to assume that it is rational no. Sum of irrational and irrational is always irrational |
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| 21407. |
If sec=x+1/4x then prove that sec+tan=2x |
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| 21408. |
2x+3y=112x-4y=-24Y=mx+3 find the value of m |
| Answer» M=-1 | |
| 21409. |
If for some angle ,cot 2¤=1/√3,then the value of sin 3¤,where 2¤is |
| Answer» sin 3 theta is 1 | |
| 21410. |
in the given figure if AD=AE= 2 find area DB EC |
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| 21411. |
Find the sim of all three digiy number which leaves the remainder 3 when divided by 7 |
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Answer» 21 Find the sum of the three digit number which leaves the remainder 3 when divided by 7 |
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| 21412. |
in a city 30% are females 40% are male and remaining are the children . What percent are children |
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Answer» 30% will be the answer because the maximum probability is of 1 or 100% and since male and female constitutes 70%.Therefore, the remaining 30% will be children. 30% I think it would be more than 30% 30% Children also come in males and females |
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| 21413. |
(Sec 60+tan 60)(1-sin 60) |
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Answer» than 60 = √3sec 60 = 2sin 60 = √3/2( sec 60 + tan 60)(1 - sin60)= (2+ √3)( 1 - √3/2)= (2 + √3) [(2 -√3)/2]= ](2)2 - (√3)2]/ 2= 4 - 3 /2= 1/2 1/2 1/2 |
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| 21414. |
Solve forx :-(X/x-1)-x-1/x=34/15,x not equal 0,x not equal -1 |
| Answer» Given,{tex} \\frac { x } { x + 1 } + \\frac { x + 1 } { x } = \\frac { 34 } { 15 }{/tex}Taking LCM, we get{tex} \\Rightarrow \\quad \\frac { x ^ { 2 } + ( x + 1 ) ^ { 2 } } { x ( x + 1 ) } = \\frac { 34 } { 15 }{/tex}{tex} \\Rightarrow \\quad \\frac { x ^ { 2 } + x ^ { 2 } + 2 x + 1 } { x ^ { 2 } + x } = \\frac { 34 } { 15 }{/tex}After cross multiplication,we get{tex} \\Rightarrow{/tex}\xa034x2 + 34x = 15x2 + 15x2 + 30x +15{tex} \\Rightarrow{/tex}\xa04x2 + 4x - 15 = 0{tex} \\Rightarrow{/tex}\xa04x2 + 10x - 6x - 15 = 0{tex} \\Rightarrow{/tex}\xa0(2x - 3)(2x + 5) = 0\xa0{tex} \\Rightarrow{/tex}\xa02x - 3 = 0 or, 2x + 5 = 0\xa0{tex} \\Rightarrow{/tex}\xa0{tex}x = \\frac { 3 } { 2 } \\text { or } x = - \\frac { 5 } { 2 }{/tex} | |
| 21415. |
Give me questions I will answer |
| Answer» . Two metallic wires A and B are connected in second wire A has length l and radius r,while wire B has length 2l and radius 2r. Find the ratio of total resistance of seriescombination and the resistance of wire A, if both the wires are of same material? | |
| 21416. |
Question 3 of exercise 8.2 |
| Answer» tan (A+B)=60°....1=> tan (A-B)=30°.......2From equation 1&2=>2A=90°=>A=45°=> B=15° | |
| 21417. |
Prove that cosA-sinA+1÷cosA+sinA-1=cosecA+cotA |
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Answer» Just open 1 in "cosA-sinA+1" then take common cosA+sinA divide both numerator and denominator by sinQand then put the value of 1 from1+cosscQ =cotQ. |
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| 21418. |
Class interval 0-1010-2020-3030-4040-5050-6060-70Frequency 34715107 4 find mode |
| Answer» Modal class is 30-40. Formula is l+(f1-f0/2f1-f0-f2)*h here l is 30 f1 is 15 ; f2 is 10 and f0 is 7 ; h is 10 put all the values in the formula u get ur ans...? | |
| 21419. |
Prove that (sin A + sec A)^2 + (cos A + cosec A)^2 = (1+ sec A.cosec A)^2. |
| Answer» LHS = (sin A + sec A)2 + (cos A + cosec A)2{tex}= \\left( \\sin A + \\frac { 1 } { \\cos A } \\right) ^ { 2 } + \\left( \\cos A + \\frac { 1 } { \\sin A } \\right) ^ { 2 }{/tex}{tex}= \\sin ^ { 2 } A + \\frac { 1 } { \\cos ^ { 2 } A } + 2 \\frac { \\sin A } { \\cos A } + \\cos ^ { 2 } A + \\frac{1}{sin^2A} + 2\\frac{cosA}{sinA}{/tex}=\xa0{tex}sin^2A+cos^2A +\\frac { 1 } { \\sin ^ { 2 } A } + \\frac { 1 } { \\cos ^ { 2 } A } +2 \\left( \\frac { \\sin A } { \\cos A } + \\frac { \\cos A } { \\sin A } \\right){/tex}= 1 +\xa0{tex}\\frac { \\sin ^ { 2 } A + \\cos ^ { 2 } A } { \\sin ^ { 2 } A \\cos ^ { 2 } A } + 2 \\left( \\frac { \\sin ^ { 2 } A + \\cos ^ { 2 } A } { \\sin A \\cos A } \\right){/tex}= 1 +\xa0{tex}\\frac { 1 } { \\sin ^ { 2 } A \\cos ^ { 2 } A } + \\frac { 2 } { \\sin A \\cos A }{/tex}{tex}= \\left( 1 + \\frac { 1 } { \\sin A \\cos A } \\right) ^ { 2 }{/tex}=\xa0{tex}(1+secAcosecA)^2{/tex}\xa0= RHS | |
| 21420. |
16 x minus 10 upon X is equals to 27 find the roots |
| Answer» Roots | |
| 21421. |
1/a+x+b= 1/b+1/x+1/aSolve for x |
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Answer» X=-a or x=-b X is -a,-b x= -a, -b |
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| 21422. |
Try thus if you dare:What is square root of 14+root132? |
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Answer» Not write answer 160.847754 approx |
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| 21423. |
In a triangle ABC ad=3.6 ab=10 and ae =4.5 find EC and AC |
| Answer» Incomplete queation | |
| 21424. |
2+2÷2=? |
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Answer» ????3 3 3 3 |
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| 21425. |
What is Square root of 14+root132? |
| Answer» 25.48912529.. and in extract form 14+2√33 | |
| 21426. |
Sin 4 Alpha/ sin 2 beeta + cos 4 Alpha/cos 2 beeta =1 prove that Alpha=beeta |
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| 21427. |
(1-tanA÷1-cotA)^2=tan^2A |
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| 21428. |
Find the value of k so that the zeros of the quadratic polynomial 3x^2-kx+14are the ratio 7:6 |
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Answer» let the roots are 7x and 6x. So, sum of roots are 13x , but sum of roots is also -b/a and -b/a = -(-k)/3 So, 13x=k/3 39x=k and product of roots are c/a=42x2 14/3 =42x2 So,x2=1/9 So, x=1/3 So, k=39/3 So, k=13............ Give full solution K=13...... |
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| 21429. |
Of Sn denotes the sum of first n terms of an AP, prove that,S30=3(S20-S10) |
| Answer» S30=3(S20-S10),S30=S60-S30S30=S30L. H. S=R.H.S | |
| 21430. |
Find the distance between the point p(-6,7) and q(-1,-5) |
| Answer» 13 unitssssssssssss | |
| 21431. |
CosecA+cotA=pSecA=p+1 ------- p-1 |
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| 21432. |
3x-3y=57x-2y=2 |
| Answer» 3x-3y=5.....eq 17x-2y=2....eq 2 X=3y+5/3 Put in eq 1 7(3y+5/2)-2y=2 | |
| 21433. |
If mean and median of an observation are 10and8 respectively then find the mode the observation |
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Answer» 4 Mode = 3Median - 2 Mean= 3 × 8 - 2 × 10= 24 - 20= 4 4 4....... |
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| 21434. |
the common difference (d) = -4, and the seventh term (a7) is 4, then find the first term |
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Answer» 28 a+6×-4=4a-24=4a=24+4=28 a=28....... |
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| 21435. |
Find the zeros of 4 x^2 - 4 x + 1 |
| Answer» For zeroes -4x² - 4x +1 = 04x² - 2x - 2x + 1 = 02x(2x - 1) - 1(2x - 1) = 0(2x - 1)(2x - 1) = 02x - 1 = 0 or 2x - 1 = 0x = 1/2 or x = 1/2Hence, the zeroes are 1/2,1/2. | |
| 21436. |
find area if triangle ABC with A(1,-4) and midpoint of sides through A being (2,-1) and (0,-1)? |
| Answer» Let E be the midpoint of AB.{tex}\\therefore \\quad \\frac { x + 1 } { 2 } = 2{/tex}\xa0or x = 3and\xa0{tex}\\frac { y + ( - 4 ) } { 2 } = - 1{/tex} or, y = 2or, B(3, 2)Let F be the mid-point of AC.Then,{tex}0=\\frac{x_1+1}{2}{/tex}\xa0or\xa0{tex}x_1=-1{/tex}and {tex}\\frac { y _ { 1 } + ( - 4 ) } { 2 }{/tex}\xa0= -1 or, y1\xa0= 2or, C= (-1, 2)Now the co-ordinates are A(1, - 4), B(3,2), C (-1,2)Area of triangle{tex}= \\frac { 1 } { 2 } \\left[ x _ { 1 } \\left( y _ { 2 } - y _ { 3 } \\right) + x _ { 2 } \\left( y _ { 3 } - y _ { 1 } \\right) + x _ { 3 } \\left( y _ { 1 } - y _ { 2 } \\right) \\right]{/tex}{tex}= \\frac { 1 } { 2 } [ 1 ( 2 - 2 ) + 3 ( 2 + 4 ) - 1 ( - 4 - 2 ) ]{/tex}{tex}= \\frac { 1 } { 2 } [ 0 + 18 + 6 ]{/tex}= 12 sq units. | |
| 21437. |
If mth term of an AP is 1/n and nth term is 1/m then show that (mn)th term is 1 |
| Answer» Let a and d be the first term and common difference respectively of the given A.P. Thenan = a + (n - 1)d{tex}\\frac { 1 } { n } ={/tex}\xa0mth term\xa0{tex}\\Rightarrow \\frac { 1 } { n } {/tex}= a + ( m - 1 ) d\xa0...(i){tex}\\frac { 1 } { m }{/tex}= nth term{tex}\\Rightarrow \\frac { 1 } { m } {/tex}= a + ( n - 1 ) d\xa0...(ii)On subtracting equation (ii) from equation (i), we get{tex}\\frac { 1 } { n } - \\frac { 1 } { m } = {/tex} [a+ (m-1) d] -[ a+ (n -1)d]= a + md - d - a - nd + d{tex}= ( m - n ) d{/tex}{tex} \\Rightarrow \\frac { m - n } { m n } = ( m - n ) d {/tex}{tex}\\Rightarrow d = \\frac { 1 } { m n }{/tex}Putting d = {tex}\\frac { 1 } { m n }{/tex}\xa0in equation (i), we get{tex}\\frac { 1 } { n } = a + \\frac { ( m - 1 ) } { m n } {/tex}{tex}\\Rightarrow \\frac { 1 } { n } = a + \\frac { 1 } { n } - \\frac { 1 } { m n } {/tex}{tex}\\Rightarrow a = \\frac { 1 } { m n }{/tex}{tex}\\therefore{/tex}\xa0(mn)th term = a + (mn - 1) d=\xa0{tex}\\frac { 1 } { m n } + ( m n - 1 ) \\frac { 1 } { m n } {/tex}{tex}\\left[ \\because a = \\frac { 1 } { m n } = d \\right]{/tex}= {tex}\\frac { 1 } { m n } + \\frac { mn } { m n } - \\frac { 1 } { m n }{/tex}= 1 | |
| 21438. |
13.4 Q no. 5 |
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| 21439. |
How to find out the value of √3 |
| Answer» By using long division method that we had started learning from 8th class..........? | |
| 21440. |
Please give me a blueprint for math |
| Answer» Check it from Google.......? | |
| 21441. |
The sum of 2 no. is 9 . the sum of their reciprocal is 1/2 . Find the numbers. |
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Answer» thanks a lot dear Given that the sum of two numbers is 9 Let the two numbers be x and 9-xBy the given hypothesis, we have1/x + 1/9-x = 1/2(9 - x + x)/x(9-x) = 1/218 = 9x - x2x 2 - 9x + 18 = 0(x - 6)(x - 3) = 0x = 6 or x = 3Therefore the numbers are 3 and 6. |
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| 21442. |
If the points p(-3,9),q(a,b)and R(4,-5)are collinear and a+b=2 find the value of A and B |
| Answer» I\xa0guess the correct question is :Q:\xa0If the points P(-3, 9), Q(a, b) and R(4, -5) are collinear and a+b=1, find the value of a and b.\xa0 | |
| 21443. |
Find HCF of 65 and 117 and find a pair of integral values of m&n such that HCF =65m+117n. |
| Answer» By Euclid\'s division algorithm\xa0117 = 65x1 + 52.65 = 52x1 + 1352 = 13x4 + 0Therefore 13 is the HCF (65, 117).Now work backwards:13 = 65 + 52x(-1)13\xa0= 65 + [117 + 65x(-1)]x(-1)13\xa0= 65x(2) + 117x(-1).∴ m = 2 and n = -1. | |
| 21444. |
Find HCF of 378,180, 420. Is HCF *LCM is equal to the product of three numbers? |
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Answer» Factor all three numbers common number in all Three number then hcf and power mutiply then lcmLcm×hcf=a×b×c 378 = 2x3x3x3x7180 = 2x2x3x5x7420 = 2x2x3x5x7.L.C.M ( 378,180,420) = 2x2x3x3x3x5x7 = 3780H.C.F ( 378,180,420) = 2x3 = 6⇒ L.C.M ( 378,180,420) x H.C.F ( 378,180,420) ≠ 378 x 180 x 420.⇒ 3780 x 6 ≠ 378 x 180 x 420.\xa0∴ 22680 ≠ 28576800.∴HCF and LCM of three number is not equal to product of three number. |
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| 21445. |
Determine the values of y and q so that the prime factorization of 2520 is expressible as 23*y*q*7. |
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Answer» But 23 is not its factor Since, 2520 = 2 x 2 x 2 x 3 x 3 x 5 x 7\xa0= 23 x 32 x 5 x 7Therefore, p = 2 and q = 5 |
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| 21446. |
If secθ + tanθ = p, then find the value of cosecθ ?? |
| Answer» Given, sec θ + tan θ = p ...........1We know that sec2\xa0θ\xa0- tan2\xa0θ =\xa01=> (sec θ\xa0- tan θ)(sec θ + tan θ) = 1=>\xa0(sec θ\xa0- tan θ) * p = 1=>\xa0sec θ\xa0- tan θ = 1/p ............2Add equation 1 and 2, we get 2 * sec θ = p + 1/p=>\xa02 * sec θ = (p2\xa0+ 1)/p=>\xa0sec θ = (p2\xa0+ 1)/2pand\xa0cos θ = 1/sec θ=>\xa0cos θ = 2p/(p2\xa0+ 1)Subtract equation 1 and 2, we get 2 *\xa0tan θ = p\xa0- 1/p=>\xa02 *\xa0tan θ = (p2\xa0- 1)/p=>\xa0tan θ = (p2\xa0- 1)/2pand\xa0cot θ = 1/tan θ=>\xa0cot θ = 2p/(p2\xa0- 1)We know that tan θ =\xa0sin θ/ cos θ=>\xa0sin θ =\xa0tan θ *\xa0cos θ=>\xa0sin θ =\xa0{(p2\xa0- 1)/2p} *\xa0{2p/(p2\xa0+ 1)}=>\xa0sin θ = (p2\xa0- 1)/(p2\xa0+ 1)and\xa0cosec θ = 1/sin\xa0θ\xa0=> cosec θ = (p2\xa0+ 1)/(p2\xa0- 1)\xa0 | |
| 21447. |
Prove that the product of any 3 consecutive positive integer is divisible by 6 |
| Answer» Let three consecutive positive integers be, n, n + 1 and n + 2.When a number is divided by 3, the remainder obtained is either 0 or 1 or 2.\xa0∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.If n = 3p, then n is divisible by 3.If n = 3p + 1, ⇒\xa0n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.If n = 3p + 2, ⇒\xa0n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.\xa0⇒ n (n + 1) (n + 2) is divisible by 3.Similarly, when a number is divided 2, the remainder obtained is 0 or 1.∴ n = 2q or 2q + 1, where q is some integer.If n = 2q\xa0⇒\xa0n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2.If n = 2q + 1 ⇒\xa0n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2.⇒ n (n + 1) (n + 2) is divisible by 2.Hence n (n + 1) (n + 2) is divisible by 2 and 3.∴ n (n + 1) (n + 2) is divisible by 6. | |
| 21448. |
Find the value of : (-1) + (-1)2n +(-1)2n+1 +(-1)4n+1 where n is positive odd integer. |
| Answer» Isme 2n, 2n+1,4n+1 inke exponents hai | |
| 21449. |
If sintheta equalto cos theta ,then find the valueof 2tantheta +cossquare theta |
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Answer» 5/2 will b d ryt answer 5/2 is correct answer 5/2 in place of theetha put 45 because cos = sin at angle 45° |
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| 21450. |
Find all zeroes of of the polynomial (2x4-9x3+5x2+3x-1) if two zeroes are (2+ |
| Answer» Given:f(x) = (2x4\xa0– 9x3\xa0+ 5x2\xa0+ 3x – 1)Zeroes = (2 + √3) and (2 – √3)Given the zeroes, we can write the factors = (x – 2 + √3) and (x – 2 – √3){Since, If x = a is zero of a polynomial f(x), we can say that x - a is a factor of f(x)}Multiplying these two factors, we can get another factor which is:((x – 2) + √3)((x – 2) – √3) = (x – 2)2 –\xa0(√3)2⇒x2\xa0+ 4 – 4x – 3 = x2\xa0– 4x + 1So, dividing f(x) with (x2\xa0– 4x + 1)f(x) = (x2\xa0– 4x + 1) (2x2\xa0– x – 1)Solving (2x2\xa0– x – 1), we get the two remaining roots as{tex}x = {-b \\pm \\sqrt{b^2-4ac} \\over 2a}{/tex}where f(x) = ax2\xa0+ bx + c = 0(using Quadratic Formula){tex}\\mathrm{x}=\\frac{-(-1) \\pm \\sqrt{(-1)^{2}-4(2)(-1)}}{2(2)}{/tex}{tex}\\mathrm{x}=\\frac{-1 \\pm 3}{4}{/tex}{tex}\\Rightarrow \\mathrm{x}=1,-\\frac{1}{2}{/tex}Zeros of the polynomial =\xa0{tex}1,-\\frac{1}{2}, 2+\\sqrt{3}, 2-\\sqrt{3}{/tex} | |