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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.
| 27001. |
3+root 5 and 3-root 5 are the sum and product |
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Answer» Why Sum=6,product=4 |
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| 27002. |
Hiii...hlo koi h ... |
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Answer» Hlo...hii mai hu... Ji bestie..........hloo |
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| 27003. |
Find the zeros of the polynomials 8×€(÷£÷¥292 (×999 |
| Answer» | |
| 27004. |
Sol. Pair of linear equations 6x+3y=6xy : 2x+4y=5xy |
| Answer» Not good | |
| 27005. |
IN AN AP. GIVEN d=5,s9=75 find a and a9 |
| Answer» Here, d = 5S9 = 75We know that{tex}{S_n} = \\frac{n}{2}\\left[ {2a + (n - 1)d} \\right]{/tex}{tex} \\Rightarrow {S_9} = \\frac{9}{2}\\left[ {2a + (9 - 1)d} \\right]{/tex}{tex} \\Rightarrow {S_9} = \\frac{9}{2}\\left[ {2a + 8d} \\right]{/tex}{tex} \\Rightarrow {S_9} = 9\\left[ {a + 4d} \\right]{/tex}{tex} \\Rightarrow {S_9} = 9\\left[ {a + 4 \\times 5} \\right]{/tex}{tex} \\Rightarrow {/tex}\xa0S9 = 9[a + 20]{tex} \\Rightarrow {/tex}\xa075 = 9a + 180{tex} \\Rightarrow {/tex}\xa09a = 75 - 180{tex} \\Rightarrow {/tex}\xa09a = -105{tex} \\Rightarrow a = - \\frac{{105}}{9}{/tex}{tex} \\Rightarrow a = - \\frac{{35}}{3}{/tex}Again, we know thatan = a + (n - 1)d{tex} \\Rightarrow {/tex}\xa0a9 = a + (9 - 1)d{tex} \\Rightarrow {/tex}\xa0a9 = a + 8d{tex} \\Rightarrow {a_9} = - \\frac{{35}}{3} + 8(5){/tex}{tex} \\Rightarrow {a_9} = - \\frac{{35}}{3} + 40{/tex}{tex} \\Rightarrow {a_9} = \\frac{{ - 35 + 120}}{3}{/tex}{tex} \\Rightarrow {a_9} = \\frac{{85}}{3}{/tex} | |
| 27006. |
ABCD is a cyclic quadrilateral .find tha sngle if tha cyclic quadrilatrtal. |
| Answer» We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°.{tex}\\therefore{/tex}{tex}\\angle{/tex}A + {tex}\\angle{/tex}C = 180°{tex}\\Rightarrow{/tex} 4y + 20 - 4x = 180°{tex}\\Rightarrow{/tex} -4x + 4y = 160°{tex}\\Rightarrow{/tex} x - y = -40° ....(1)\xa0Also {tex}\\angle{/tex}B + {tex}\\angle{/tex}D = 180°{tex}\\Rightarrow{/tex} 3y - 5 - 7x + 5 = 180°{tex}\\Rightarrow{/tex} -7x + 3y = 180°............... (2)Multiplying equation (1) by 3, we obtain:3x - 3y = -120° ..... (3)Adding equations (2) and (3), we obtain:-4x = 60°{tex}\\Rightarrow{/tex} x = -15°Substituting the value of x in equation (1), we obtain:-15 - y = 40°{tex}\\Rightarrow{/tex} y = -15 + 40 = 25°{tex}\\therefore{/tex}{tex}\\angle{/tex}A = 4y + 20 = 4 {tex}\\times{/tex}\xa025 + 20 = 120°{tex}\\angle{/tex}B = 3y - 5 = 3 {tex}\\times{/tex}\xa025 - 5 = 70°{tex}\\angle{/tex}C = -4x = -4 {tex}\\times{/tex}\xa0(-15) = 60°{tex}\\angle{/tex}D = -7x + 5 = -7(-15) + 5 = 110° | |
| 27007. |
Hlo...koi h.. |
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Answer» Hii... Yaa ...hello |
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| 27008. |
Prime factories,5005 |
| Answer» So, 5005 = 5 {tex}\\times{/tex}\xa07{tex}\\times{/tex}\xa011 {tex}\\times{/tex}\xa013. | |
| 27009. |
(m*2+n*2)x*2 - 2(mp+nq)x + p*2 + q*2 = 0 |
| Answer» We have the following equation,{tex}(m^2+n^2)x^2-2(mp+nq)x+p^2+q^2=0{/tex}where, a =\xa0(m2 + n2), b =\xa0-2(mp + nq), and c = p2 + q2The given equation will have equal roots,if D = 0{tex}\\Rightarrow{/tex}\xa0{tex}b^2-4ac=0{/tex}{tex}\\Rightarrow{/tex}{tex}[-2(mp+nq)]^2-4(m^2+n^2)(p^2+q^2)=0{/tex}{tex}\\Rightarrow{/tex}{tex}4m^2p^2+4n^2q^2+8mnpq-4m^2p^2-4m^2q^2-4n^2q^2-4n^2p^2=0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}-4m^2q^2-4p^2n^2+8mnpq=0{/tex}\xa0{tex}\\Rightarrow{/tex}{tex}-4(m^2q^2+p^2n^2-2mnpq)=0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}m^2q^2+p^2n^2-2mnpq=0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}(mq-pn)^2=0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}mq-pn=0{/tex}\xa0{tex}\\Rightarrow{/tex}{tex}mq=pn{/tex} | |
| 27010. |
6+√2 |
| Answer» It is irrational as √2 is an irrational number and sum of rational and irrational number is irrational. First show that √2 is irrational. | |
| 27011. |
2x - 5y +8=0X - 4y+7=0Solve Elimination method |
| Answer» X=1. And Y=2 | |
| 27012. |
(x-1) (2x-1) find discrimination |
| Answer» (x - 1)(2x - 1) = 0{tex}\\Rightarrow{/tex} 2x2 - x - 2x + 1 = 0{tex}\\Rightarrow{/tex} 2x2 - 3x + 1 = 0Here, a = 2, b = -3, c = 1D = b2 - 4ac{tex}= ( - 3 ) ^ { 2 } - 4 \\times 2 \\times 1{/tex}= 9 - 8 = 1\xa0 | |
| 27013. |
Root 7 root 8 root 18 root 32 |
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| 27014. |
3x +7y = 10y - 2x - 3 |
| Answer» The given equations are2x - y + 3 = 0{tex}\\Rightarrow {/tex} 2 x - y = - 3.............(i)3x - 7y + 10= 0{tex}\\Rightarrow {/tex}3 x - 7 y = - 10...........(ii)Multiplying (i) by -7 ,we get-14x + 7y = 21 ............ (iii)Add (ii) and (iii) , we get3x - 14x = -10 + 21{tex}\\Rightarrow - 11 x = 11{/tex}{tex}\\Rightarrow x = - 1{/tex}Substituting x = -1\xa0in (i),we get2(-1) - y = -3-2 - y = -3y = -2 + 3y = 1So, the solution of given system of equations is x = -1 and y = 1. | |
| 27015. |
The zeros of a quadratic polynomial p(x) = 2xsquare - 9 ?? |
| Answer» | |
| 27016. |
2/3cosec²58°-2/3cot58°Tan32°-5/3Tan13°Tan37°Tan45°Tan53°Tan77° |
| Answer» According to the question,\xa0{tex}\\frac { 2 } { 3 } \\operatorname { cosec } ^ { 2 } 58 ^ { \\circ } - \\frac { 2 } { 3 } \\cot 58 ^ { \\circ } \\tan 32 ^ { \\circ } - \\frac { 5 } { 3 }{/tex}{tex}\xa0tan13° tan37° tan45° tan53° tan77°{/tex}=\xa0{tex}\\frac { 2 } { 3 } \\operatorname { cosec } ^ { 2 } 58 ^ { \\circ } - \\frac { 2 } { 3 } \\cot 58 ^ { \\circ } \\tan \\left( 90 ^ { \\circ } - 58 ^ { \\circ } \\right) - \\frac { 5 } { 3 }{/tex}tan 13° tan 37° tan 45°tan (90° - 37°) tan (90° -13°)=\xa0{tex}\\frac { 2 } { 3 } \\operatorname { cosec } ^ { 2 } 58^\\circ - \\frac { 2 } { 3 } \\cot ^ { 2 } 58^\\circ - \\frac { 5 } { 3 }{/tex}{tex}tan 13° tan37° tan 45° cot 37° cot 13°{/tex}=\xa0{tex}\\frac { 2 } { 3 } \\left( \\operatorname { cosec } ^ { 2 } 58 ^ { \\circ } - \\cot ^ { 2 } 58 ^ { \\circ } \\right) - \\frac { 5 } { 3 } \\tan 13 ^ { \\circ } \\tan 37 ^ { \\circ } \\times 1 \\times \\frac { 1 } { \\tan 37 ^ { \\circ } } \\times \\frac { 1 } { \\tan 13 ^ { \\circ } }{/tex}=\xa0{tex}\\frac { 2 } { 3 } \\times 1 - \\frac { 5 } { 3 } = \\frac { 2 } { 3 } - \\frac { 5 } { 3 } = -1{/tex} | |
| 27017. |
Where the trigonometry works and used??¿¿ tell me more than 2 uses |
| Answer» \tTrigonometry\xa0is used in oceanography in calculating the height of tides in oceans.\tThe sine and cosine functions are fundamental to the theory of periodic functions, those that describe the sound and light waves.\tCalculus is made up of Trigonometry and Algebra. | |
| 27018. |
Find the zeroes of 6x |
| Answer» Solving the equation g(x) = 0, we get3 - 6x = 0, which gives us {tex}x = \\frac{1}{2}{/tex}So,\xa0{tex}\\frac{1}{2}{/tex} is a zero of the polynomial 3 - 6x. | |
| 27019. |
How many irrational number lie between 2and 3write any 2 |
| Answer» InfiniteAny two- 2.121221222... 2.151551555... | |
| 27020. |
If p is a prime number then p is _______number |
| Answer» | |
| 27021. |
The volume of two spheres are in the ratio 64:27. The ratio of there surface area is |
| Answer» Suppose {tex}r_1\\\xa0and\\ r_2{/tex} be the radii of two spheres.{tex}\\therefore{/tex} the ratio of their volumes =\xa0{tex}\\frac{\\frac{4}{3} \\pi r_{1}^{3}}{\\frac{4}{3} \\pi r_{2}^{3}}=\\frac{64}{27}{/tex}\xa0{tex}\\left(\\frac{r_{1}}{r_{2}}\\right)^{3}=\\left(\\frac{4}{3}\\right)^{3}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{r_{1}}{r_{2}}{/tex}\xa0=\xa0{tex}\\frac{4}{3}{/tex}Ratios of surface areas of two spheres =\xa0{tex}\\frac{4 \\pi r_{1}^{2}}{4 \\pi r_{2}^{2}}{/tex}\xa0{tex}=\\left(\\frac{r_{1}}{r_{2}}\\right)^{2}{/tex}\xa0= ({tex}\\frac{4}{3}{/tex})2 =\xa0{tex}\\frac{16}{9}{/tex}{tex}\\therefore{/tex}Required ratio {tex}= 16: 9.{/tex} | |
| 27022. |
syllabus of quadratic equationFor2020 |
| Answer» | |
| 27023. |
Solution for sin 30theeta+ tan45 theeta - cosec60 / sec30 theeta + cos 60theeta + cot45 |
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| 27024. |
What is the value of theta if 2cos3theta = 1 |
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| 27025. |
5y+6y |
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Answer» Factorise of 5y +6y =y(5+6) Itna saa b nhi aata.... 5y+6y=11y ...hota h .. |
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| 27026. |
589999+88888 |
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Answer» 678887 678887 |
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| 27027. |
The distance of the point p (2,3) from the x axis is |
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Answer» 3 unit 3 |
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| 27028. |
Prove that the square of any positive integer is of the form of 5q, 5q+1, 5q+4 for some integer q. |
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Answer» First 5qLet the value of q be 55(5)255mValue of m is 5 5q +1 Let the value of q be 105(10)+1by this take the value in bracket be m (because 10 is integer and we can take the value of 10 as m) 5m+15q+4Let the value of q be 105(10)+45m+4Solved Very easy |
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| 27029. |
xsquare -2x-underroot x-5=0 |
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| 27030. |
What is prodhyodiki |
| Answer» | |
| 27031. |
Find the area of quadrilateral ABCD in which AB=3cm BC=4cm. CD=4cm. DA =5cm and AC =5cm |
| Answer» For ΔABCa = 4 cmb = 5 cmc = 3 cm∵ a2\xa0+ c2\xa0= b2\xa0∴ ΔABC is right angled with ∠B = 90°.∴ Area of right triangle ABC\xa0For ΔACDa = 4 cm b = 5 cmc = 5 cm\xa0∴ Area of the ΔACD= 2 x 4.6 cm2\xa0(approx.)= 9.2 cm2\xa0(approx.)∴ Area of the quadrilateral ABCD= Area of ΔABC + Area of ΔACD= 6 cm2\xa0+ 9.2 cm2= 15.2 cm2, (approx.) | |
| 27032. |
What is balanced reaction |
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Answer» A balanced chemical equation has an eqal no. Of atoms of different elements in reactants and products. Same no. Of atom in both side product and reactant called blance chemical reaction A balanced chemical reaction occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side....... |
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| 27033. |
For what value of n, 2n X 5n ends ends with |
| Answer» {tex}\\begin{array}{l}(2\\times5)^{\\mathrm n}=2^{\\mathrm n}\\times5^{\\mathrm n}\\end{array}{/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. | |
| 27034. |
7 + 7 x 7 - 7 = ? Please solve this |
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Answer» 49 7+7*7-7=7+49-7=56-7=49 49 7+49-756-7=49 |
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| 27035. |
X/2+2y/3 =-1 and x-y/3=3 |
| Answer» {tex}\\frac{x}{2} + \\frac{{2y}}{3} = - 1{/tex} ....(1){tex}x - \\frac{y}{3} = 3{/tex} ...(2)\tElimination method: Multiplying equation (2) by 2, we get (3)\t{tex}2x - \\frac{2}{3}y = 6{/tex} ....(3)\t{tex}\\frac{x}{2} + \\frac{{2y}}{3} = - 1{/tex} ....(1)\tAdding (3) and (1), we get\t{tex}\\frac{5}{2}x = 5{/tex}\t\xa0⇒ x =2\tPutting value of x in (2), we get\t2− {tex}\\frac{y}{3}{/tex}= 3\t⇒ y =−3\tTherefore, x =2 and y =−3\tSubstitution method:{tex}\\frac{x}{2} + \\frac{{2y}}{3} = - 1{/tex} ....(1)\t{tex}x - \\frac{y}{3} = 3{/tex} ....(2)\tFrom equation (2), we can say that {tex}x = 3 + \\frac{y}{3} = \\frac{{9 + y}}{3}{/tex}\tPutting this in equation (1), we get\t{tex}\\frac{{9 + y}}{6} + \\frac{2}{3}y = - 1{/tex}\t{tex} \\Rightarrow \\;\\frac{{9 + y + 4y}}{6} = - 1{/tex}\t⇒ 5y +9=−6\t⇒ 5y =−15\t⇒ y =−3\tPutting value of y in (1), we get\t{tex}\\frac{x}{2} + \\frac{2}{3}( - 3) = - 1{/tex}\t⇒ x = 2\tTherefore, x =2 and y =−3. | |
| 27036. |
7 + 7 x 7 - 7 = ? |
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Answer» 49 wrong 49 |
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| 27037. |
If the common difference of an A.P is, find a6_a12 |
| Answer» Let the first term of an A.P be {tex} a{/tex}, common difference {tex}d{/tex},{tex}d = - 6{/tex}an = a + (n -1)da16 = a + (16 - 1)(- 6)= a + (15)(- 6)= a - 90a12\xa0= a + (12 - 1)(-6 )= a + 11(-6)= a - 66a16\xa0-a12 = (a- 90) - ( a - 66)= a - 90 - a + 66= - 24 | |
| 27038. |
(a+b) (a-b)= ? |
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Answer» Its an identity which is (a+b)(a-b)=a^2-b^2 A^2 - B^2 |
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| 27039. |
4√6x² - 13x - 2√6 =0 |
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| 27040. |
Show that 1089 is a perfect ,also find the number whose square is 1089 |
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| 27041. |
If n is an natural number then 9^2n-4^2n will be divisible by... |
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| 27042. |
2+3=4 kaisa ? |
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Answer» Correct 2+3=4 kaise....glti se...??? Galti se..?? Sorry sorry bhai ne type Kiya????? |
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| 27043. |
How can i find out square root easily? |
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Answer» By multipying it...for example 6²=>6×6=36!!!!! By dividing |
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| 27044. |
If polynomial y^4+4y^2 + 5 have zeros or not |
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| 27045. |
Is their are two papers of math I mean basic and standard...???? |
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Answer» Yes... Yess Yes get more refreshers |
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| 27046. |
How can we make more easier mathmatics for all |
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Answer» More and more practice can help to improve maths... When you cannot solve any questions then help with side book is reliable For better mind and concentrate as well as focus |
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| 27047. |
25337+0000000445784377(0) |
| Answer» 445809714 | |
| 27048. |
Given that hcf (306,657)=9 find lcm (306,657) |
| Answer» HCF×LCM=PRODUCT OF 2 NO.SLCM=PRODUCT OF TWO NO.S/HCFLCM=306×657/9LCM=22338 | |
| 27049. |
√5 is irrational nunber |
| Answer» Yes | |
| 27050. |
If the zeroes of a qudratic polynomial ax square+bx+c are both negative a,b amd c have same sign |
| Answer» Answer is yes but how | |