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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.
| 34551. |
4-6>5-6 |
| Answer» Complete yur question? | |
| 34552. |
Prove thatvany positive odd integer is of the form 4q + 1 or 4q + 3 |
| Answer» Let\xa0\xa0be any positive integerWe know by Euclid\'s algorithm, if a and b are two positive integers, there exist unique integers q and r satisfying,\xa0where.Take\xa0Since 0 ≤\xa0r\xa0< 4, the possible remainders are 0, 1, 2 and 3.That is,\xa0\xa0can be\xa0, where\xa0q\xa0is the quotient.Since\xa0\xa0is odd,\xa0\xa0cannot be 4q\xa0or 4q\xa0+ 2 as they are both divisible by 2.Therefore, any odd integer is of the form 4q\xa0+ 1 or 4q\xa0+ 3. | |
| 34553. |
Example 9 in chapter 2 pg no 35 |
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Answer» They gave u two zeros na.. Take it as (x-√2)(x+√2)=x*2-2And then do the long division.. Simple Write the ques |
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| 34554. |
If ( x+a) is a factor of 2xsq +2ax + 5x + 10 find a |
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Answer» 2(-a)sq+2a.(-a) +5(-a) +10=02asq-2asq-5a+10=0-5a =-10a=2 2xsq +2ax +5x + 10X=-a2(-a)sq+ 2a(-a)+5(-a)+102asq-2asq-5a+10-5a+10=0a=2 |
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| 34555. |
if bita and alpa are the zeroes of the quadratic polynomial x²-7x+10 |
| Answer» = x2-5x-2x+10. =x(x-5)-2(x-5) =(x-2) (x-5) =x-2=0=x=2 and x-5 =0=x=5. =a+b= 2+5=7===-(-7/1)==-b/a ==== ab =2*5= 10/1 c/a | |
| 34556. |
The sum of a number and it\'s square is 63/4, find the numbers. |
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Answer» let \'x\' be the numberThen,according to the questionx + x² = (63/4)4x²+ 4x -63 = 0Using splitting the middle term,4x² + 18x - 14x -63 = 02x(2x-9) - 7(2x-9) = 0(2x - 7)(2x - 9) = 0⇒ x = (7/2) or (9/2)x = (7/2) satisfy the condition given in the questionso, x =(7/2) is the number\xa0 Solve it like x +x^2=63/4 where x is the no. |
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| 34557. |
Solution of a quadratic equation by completing the square |
| Answer» {tex}x^2 +\xa06x - 16 = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}x^2 + 6x = 16{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}x^2 + 6x + 9 = 16 + 9{/tex} [Adding on both sides square of coefficient of x, i.e. ({tex}\\frac{6}{2}{/tex})2]{tex}\\Rightarrow{/tex}\xa0{tex}(x + 3)^2 = 25{/tex}{tex}\\Rightarrow{/tex}\xa0x + 3 =\xa0{tex}\\pm{/tex}{tex}\\sqrt{25}{/tex}{tex}\\Rightarrow{/tex}\xa0x + 3 =5 or x + 3 = -5{tex}\\Rightarrow{/tex}\xa0x = 2 or x = -8 | |
| 34558. |
Ch-8trignometery |
| Answer» Get NCERT solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 34559. |
Find the cir cum frence of the triangle whose vertics are (-2,-3) (-1,0) (7,-6) |
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Answer» Use dist. Formula and find all 3 distances then add all the dist. To get perimeter Add all the distances by using the formula √(x2 -x1 )^2 +(y2-y1)^2 |
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| 34560. |
sin0+cos0=p,sec0+cosec0=q.prove that q(p square-1)=2p (0=theta) |
| Answer» Given : cos{tex}\\theta{/tex} + sin{tex}\\theta{/tex}= p ....(i)\xa0sec{tex}\\theta{/tex} + cosec{tex}\\theta{/tex} = q....(ii)L.H.S = q(p2- 1)= (sec{tex}\\theta{/tex} + cosec{tex}\\theta{/tex}) [(cos{tex}\\theta{/tex} + sin{tex}\\theta{/tex})2 - 1]= (sec{tex}\\theta{/tex} + cosec{tex}\\theta{/tex}) (cos2{tex}\\theta{/tex} + sin2{tex}\\theta{/tex} + 2 sin{tex}\\theta{/tex} cos{tex}\\theta{/tex} - 1)\xa0{tex}[\\because (a+b)^2=a^2+b^2+2ab]{/tex}{tex}= \\left( \\frac { 1 } { \\cos \\theta } + \\frac { 1 } { \\sin \\theta } \\right) ( 2 \\sin \\theta \\cos \\theta ){/tex}\xa0{tex}[\\because sin^2\\theta+cos^2\\theta=1]{/tex}{tex}= 2sin\\theta cos\\theta. \\frac{1}{cos\\theta}+2sin\\theta cos\\theta.\\frac{1}{sin\\theta}{/tex}=\xa0{tex}2sin\\theta+2cos\\theta{/tex}= {tex}2(sin\\theta+cos\\theta){/tex}= {tex}2p{/tex} ......( using eq.i )= R.H.S.Hence Proved. | |
| 34561. |
for what value of k ,(-4) is the zero of polynomial x2-x-(2k+2) |
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Answer» Thanks Zero of the polynomial is -4p(x) =x² - x - (2k + 2)p(x) = x²-x-2k-2p(-4)=00=-4²-(-4)-2k-20=16+4-2k-20=18-2k2k=18k=18/2=9 k=9 |
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| 34562. |
Prove that 2+3√5 is irrational |
| Answer» Firt assume 2+3√5 is rationalSo 2+3√5=p/q (where p andq are co-prime)2+3√5=p/q 3√5=p/q-23√5=5/q-2q/q√5=15/q-6q/qBut this contradicts the fact √5 is irrational.2+3√5 is rational Hence proved. | |
| 34563. |
Write the coordinates of a point p on x axis which is equidistant form the point A(-2,0)and B (6,0) |
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Answer» Pagal Let A(-2, 0) and B(6, 0) be the given points.Let P(x, 0) be the point on x -axis\xa0such that PA = PBPA = PBPA2 = PB2(x +\xa02)2 + (0 - 0)2 = (x -\xa06)2 + (0 - 0)2{tex}\\Rightarrow{/tex}\xa0x2 + 4 + 4x = x2 + 36\xa0-\xa012x{tex}\\Rightarrow{/tex}\xa016x = 32{tex}\\Rightarrow{/tex}\xa0x = 2{tex}\\therefore{/tex}\xa0The point\xa0on x-axis is (2, 0). |
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| 34564. |
What are polinomials |
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Answer» An algebraic expression in which power of each term is a whole no. is called polynomial. Ex- x²+5x+2, x³+3 etc. An expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s). |
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| 34565. |
3x²+5x-8 |
| Answer» 3x^-3x+8x-83x(x-1)+8(x-1)(3x+8)(x-1)X=-8/3 or x=1 | |
| 34566. |
What is the probability of getting a number 3 in a dice of 6faces |
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Answer» 1/6 1/6 1/6 1/6 1/6 4 1/6 |
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| 34567. |
Frustum of pymarid |
| Answer» The frustum of a pyramid or truncated pyramid is the result of cutting a pyramid by a plane parallel to the base and separating the part containing the apex. The height of the pyramidal frustum is the perpendicular distance between the bases. The apothem is the height of any of its sides. | |
| 34568. |
Find the nature of the roots of the quadratic equation |
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Answer» If D>0 then roots are real and equal Which Quadratic Equation? D>0 real no |
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| 34569. |
Solve the substitution 2x + y = 15 and 8x - 5y = -11 |
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Answer» X=32/9 and y=71/9 Here x=32/9 and y=70/9 |
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| 34570. |
1/0=? |
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Answer» 0 0 Not defined.... oo Infinity Not defined |
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| 34571. |
a,a2,a3,a4----- is AP? If it is AP,find the common difference and write three more terms |
| Answer» Common difference is 1 | |
| 34572. |
If H is the HCF of 4052 and 12376 and H=4052×A+12576×B then find the value of (H+A+B) |
| Answer» {tex}12576 = 4052 \\times 3 + 420{/tex}{tex}4052 = 420 \\times 9 + 272{/tex}{tex}420 = 272 \\times 1 + 148{/tex}{tex}272 = 148 \\times 1 + 124{/tex}{tex}148 = 124 \\times 1 + 24{/tex}{tex}124 = 24 \\times 5 + 4{/tex}{tex}24 = 4 \\times 6 + 0{/tex}HCF of 12576 and 4052 is \'4\'. | |
| 34573. |
P(x)=x3 -2x2-4x-1,g(x)=2x+1 |
| Answer» 3/8 | |
| 34574. |
Did Anyone know the blue print of class 10th maths/sst/science? |
| Answer» You can check the marks distribution here :\xa0https://mycbseguide.com/cbse-syllabus.html | |
| 34575. |
1/a+1/b+1/x = 1/a+b+c |
| Answer» Given,{tex}\\frac { 1 } { ( a + b + c ) } = \\frac { 1 } { a } + \\frac { 1 } { b } + \\frac { 1 } { c }{/tex}{tex}\\Rightarrow \\quad \\frac { 1 } { ( a + b + c ) } - \\frac { 1 } { c } = \\frac { 1 } { a } + \\frac { 1 } { b } \\Rightarrow \\frac { c - ( a + b + c ) } { c ( a + b + c ) } = \\frac { b + a } { a b }{/tex}{tex}\\Rightarrow \\quad \\frac { - ( a + b ) } { c ( a + b + c ) } = \\frac { ( a + b ) } { a b }{/tex}On dividing both sides by (a+b){tex}\\Rightarrow \\quad \\frac { - 1 } { c ( a + b + c ) } = \\frac { 1 } { a b }{/tex}Now cross multiply{tex}\\Rightarrow{/tex}\xa0c(a + b + c) = -ab\xa0{tex}\\Rightarrow{/tex}\xa0c2 + ac + bc + ab = 0{tex}\\Rightarrow{/tex}\xa0c(c +a) + b(c +a) = 0{tex}\\Rightarrow{/tex}\xa0(c\xa0+ a) (c + b) = 0{tex}\\Rightarrow{/tex}\xa0c + a = 0 or c + b = 0{tex}\\Rightarrow{/tex}\xa0c = -a or c = -b.Therefore, -a and -b\xa0are the roots of the equation. | |
| 34576. |
Express the hcf of 234 and 111 as 234x+111y, where x and y are integers |
| Answer» HCF of 234 and 111234 = 2 x 3 x 3 x 13111 = 3 x 37Therefore, HCF = 3So we can write it3 = 111 - (12 x 9)3 = 111 - {(234 – 111 x 2) x 9}[where 12 = 234 - 111 x 2]3 = 111 - {234 x 9 – 222 x 2 x 9}3 = 111 - (234 x 9) + (111 x 18)3 = 111 + (111 x 18) - (234 x 9)3 = 111[1 + 18] – 234 x 93 = 111 x 19 – 234 x 93 = 234 x (-9) + 222 x 19Hence, HCF of 234 and 111 in the form of234x + 111y is 234 x (-9) + 111 x 19\xa0 | |
| 34577. |
ax+by=(a-b)bx-ay=(a+b)This question of pair of linear equation in two variable...plz solve it |
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Answer» Thnku.. Given,\xa0ax + by = a - b .......1 bx - ay = a + b .......2Multiply by a in equation 1 and b in equation 2, we geta2 x + aby =\xa0a2 - ab .......3 b2 x - aby = ab + b2 .......4Add equation 3 and 4, we get a2\xa0x + aby +\xa0b2\xa0x - aby =\xa0a2\xa0- ab +\xa0ab + b2=>\xa0a2\xa0x +\xa0b2\xa0x =\xa0a2\xa0+ b2\xa0=> x(a2\xa0+\xa0b2\xa0) =\xa0a2\xa0+ b2=> x =\xa0(a2\xa0+\xa0b2\xa0)/(a2\xa0+\xa0b2\xa0)\xa0=> x = 1Put value of x in equation 1, we get=>\xa0a + by = a - b=>\xa0by = a - b - a=> by = -b=> y = -b/b=> y = -1So, x = 1, y = -1 |
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| 34578. |
Tan5° tan25° tan30° tan65°tan85° =1/ root 3 |
| Answer» We have,tan 5° tan 25° tan 30° tan 65° tan 85°= (tan 5° tan 85°) (tan 25° tan 65°) tan 30° {tex}\\left[ \\begin{array} { c } { \\because \\tan 85 ^ { \\circ } = \\tan \\left( 90 ^ { \\circ } - 5 ^ { \\circ } \\right) = \\cot 5 ^ { \\circ } } \\\\ { \\tan 65 ^ { \\circ } = \\tan \\left( 90 ^ { \\circ } - 25 ^ { \\circ } \\right) = \\cot 25 ^ { \\circ } } \\end{array} \\right] {/tex}= (tan 5° cot 5°) (tan 25° cot 25°) tan 30°={tex}\\left(tan5^\\circ\\times\\frac1{\\tan5^\\circ}\\right)\\left(tan25^\\circ\\times\\frac1{\\tan25^\\circ}\\right){/tex}{tex}\\left[cot\\theta=\\frac1{\\tan\\theta}\\right]{/tex}{tex}= 1 \\times 1 \\times \\frac { 1 } { \\sqrt { 3 } } = \\frac { 1 } { \\sqrt { 3 } } {/tex} | |
| 34579. |
If x=1,then find the value of y in the equation 4/x+3/y=5. |
| Answer» 3 | |
| 34580. |
Why only used x,y,z in variable and a,b,c in costants |
| Answer» It is conventional. | |
| 34581. |
For what value of n, 2^4 * 5^n, ends with 5 |
| Answer» {tex}2^{\\mathrm n}\\times5^{\\mathrm n}{/tex}{tex}\\text{=10}^n{/tex}{tex}\\text{If n=0 then 10}^0\\text{=1}{/tex}{tex}\\text{If n>0 then 10}^n\\text{ will end with 0 }{/tex}{tex}\\mathrm{If}\\;\\mathrm n<0\\;\\mathrm{then}\\;10^{\\mathrm n}\\;\\mathrm{ends}\\;\\mathrm{with}1\\;(\\mathrm e.\\mathrm g.\\;0.1,0.01,0.001){/tex}Hence for all values of n, {tex}2^n\\times 5^n{/tex} can never end with 5. | |
| 34582. |
Prove that root 7/5 +4 is an irrational |
| Answer» Let a=√7/4+5 is a rational number.a×4+5=√7Since a×4+5 is a rational number.But we know that,√7 is anirrational number.This contradicts due to our wrong supposition.Therefore,a=√7/4+5 is an irrational number.Hence prooved. | |
| 34583. |
Find the greatest common factor of 2730 and 9350 |
| Answer» Now,9350 = 2730{tex}\\times {/tex}3 + 11602730 = 1160\xa0{tex}\\times {/tex}2 + 4101160 = 410{tex}\\times {/tex}2 + 340410 = 340\xa0{tex}\\times {/tex}1 + 70340 = 70\xa0{tex}\\times {/tex}4 + 6070 = 60\xa0{tex}\\times {/tex}1 + 1060 = 10\xa0{tex}\\times {/tex}6 + 0Therefore, HCF = 10 | |
| 34584. |
7*11*13*15+15is a which type of no. |
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Answer» Composite number composite number |
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| 34585. |
If a is rational and b is irrational and ab is rational find a |
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Answer» A is 0 I don\'t know |
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| 34586. |
Rational numbers... |
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Answer» A rational no. Can be written in the foam of p/q and q is not equal to zero A rational number is any number that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non-zero denominator q. Since q may be equal to 1, every integer is a rational number. |
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| 34587. |
Hiii..... |
| Answer» Hello | |
| 34588. |
Is there anyone who is not suspended |
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Answer» Yes Yup |
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| 34589. |
Reducing method |
| Answer» Reduction of order is a technique in mathematics for solving second-order linear ordinary differential equations. It is employed when one solution is known and a second linearly independent solution is desired. The method also applies to n-th order equations. | |
| 34590. |
Find hcf 344 |
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Answer» HCF of 344=2 HCF of 344 and??? HCF OF 344 = 2 |
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| 34591. |
Triangle formulas |
| Answer» {tex}A = \\frac { h _ { b } b } { 2 }{/tex} | |
| 34592. |
ax-by=a²+b²,x+y=2a |
| Answer» X = 2a-ya(2a-y)-by = a^2+b^2a^2-ay-b^2=0 | |
| 34593. |
What is the arithmetic progression |
| Answer» \xa0An\xa0arithmetic progression\xa0is a sequence of numbers such that the difference of any two successive members is a constantLet us consider following series(1)1,5,9,13,17....(2)1,2,3,4,5,...(3)7,7,7,7.....All these sets follow certain rules. In first set 5−1=9−5=13−9=17−13=4In second set 2−1=3−2=4−3=1and so on.Here the difference between any successive members is a constantSuch series are called Arithmetic Progression | |
| 34594. |
Which of the following are polynomial in x? |
| Answer» Linear polynomial | |
| 34595. |
if cosec a -cot a =1/3 the value of cosec a +cot a is |
| Answer» \xa0cosec a -cot a =1/3\xa0cosec2 a -cot2a =1\xa0cosec a +cot a =cosec2 a -cot2a /cosec a -cot a=1/1/3=3\xa0 | |
| 34596. |
If alpha and beata are zeros of kx^2+5x+2such that 1/alpha ^2+1/beata^2=17/4.find the value of k |
| Answer» let f(x)=kx2+5x+2{tex}\\begin{array}{l}\\mathrm\\alpha\\;\\;\\mathrm{are}\\;\\mathrm\\beta\\;\\mathrm{are}\\;\\mathrm{zer}o\\;\\mathrm{of}\\;\\mathrm f(\\mathrm x)\\\\\\mathrm{so}\\;\\alpha+\\mathrm\\beta=-\\frac5{\\mathrm k},\\mathrm{αβ}=\\frac2{\\mathrm k}\\end{array}{/tex}{tex}\\begin{array}{l}\\frac1{\\alpha^2}+\\frac1{\\beta^2}\\\\=\\frac{\\alpha^2+\\beta^2}{\\alpha^2\\beta^2}=\\frac{{\\displaystyle\\left(\\alpha+\\beta\\right)^2}{\\displaystyle-}{\\displaystyle2}{\\displaystyle\\alpha}{\\displaystyle\\beta}}{\\alpha^2\\beta^2}\\\\=\\frac{\\displaystyle25/k^2-4/k}{\\displaystyle\\left({\\displaystyle\\frac2k}\\right)^2}=\\frac{\\displaystyle\\frac{25-4k}{k^2}}{\\displaystyle\\frac4{k^2}}=\\frac{25-4k}4=\\frac{17}4\\\\\\\\\\end{array}{/tex}{tex}\\begin{array}{l}4\\ast(25-4k)=17\\ast4\\\\100-16k=68\\\\16k=100-68\\\\16k=32\\\\k=2\\\\\\\\\\\\\\end{array}{/tex} | |
| 34597. |
If cos ∅=7/25, find the values of all T-ratios of ∅ |
| Answer» All T-ratio of thita be...Sin thita = 24/25 , tan thita = 24/7,cosec thita = 25/24,sec thita =25/7,cot thita = 7/24 nd cos thita is given.... | |
| 34598. |
The sum of two numbers is 8 and the difference of their squares is 16, find the numbers |
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Answer» 5 and 3 x+y=8 _________ 1x\'2-y\'2=16(X+y)(x-y)=168(x-y)=16x-y=16/8x-y=2 _________ 2Add eqn 1 and 2x+y+x-y=8+22x=10x=10/2x=5Put value of x in eqn 2x-y=25-y=25-2=y3=y 5 & 3Because square of 5 is 25 and square of 3 is 9 |
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| 34599. |
Using euclid division on lemma find hcf of 1260 and 280 |
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Answer» Now divide 1260 by 280 write in the form of above equation where A is 1260 and B is 280 and Q is quotient and R Is remainder A=BQ+R |
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| 34600. |
Find hcf of 125 and 75 and express it in linear combination |
| Answer» 125 = 75 × 1 + 5075 = 50 × 1 + 2550 = 25 × 2 + 0In linear combination 50 = 125 - 75 ×1 equation....125 = 75 - 50 × 1 equation....2From equation 2 we get 25 = 75 - 50 × 1 = 75 - (125 - 75 × 1) × 1 = 75 -125 × 1 - 75 × 1×1 =75(1-1) -125×1 = 75- 125 | |