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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.
| 37001. |
Show that the square of any odd integer is of the form 8m+1 for some integer m. |
| Answer» Since, any odd positive integer n is of the form 4m\xa0+ 1 or 4m + 3.if n = 4m + 1n2 = (4m + 1)2= 16m2 + 8m + 1= 8(2m2 + m) + 1So n2 = 8q + 1 ........... (i)) (where q = 2m2 + m is a positive integer)If n = (4m + 3)n2 = (4m + 3)2= 16m2 + 24m + 9= 8(2m2 + 3m + 1) + 1So n2 = 8q + 1 ...... (ii) (where q = 2m2 + 3m + 1 is a positive integer)From (i) and (ii) we conclude that the square of an odd positive integer is of the form 8q + 1, for some integer q. | |
| 37002. |
Give me 1st chapter |
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| 37003. |
Why is maths so tough |
| Answer» Maths is not tough if you practice it daily | |
| 37004. |
If the diagonal of a cube is 3 cm . Then find its length |
| Answer» 3into3^1/2/3 | |
| 37005. |
Term 1 ka syllabus kya haiWhat is the syllabus of term 1 2019 2020 |
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| 37006. |
What is the lenth of a cylinder |
| Answer» Hey Ayush here is ur answer!!The diameter, or the distance across a cylinder that passes through the center of the cylinder is 2R (twice the radius). The surface area of an open ended cylinder.....? | |
| 37007. |
Given that LCM(26,169)=338 write hcf (26,169) |
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Answer» Hcf=product of given two no.s÷lcm 26×169÷338=13 The answer will be LCM=338HCF=?One number=26Other number=169LCM×HCF=Product of two numbers338×HCF=26×169HCF=26×169/338HCF=13 |
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| 37008. |
4 x + 3 Y is equals to 2 and 7 x + 2 Y is equals to 1 solve the equation by substitution method |
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| 37009. |
After how many place 37/1250 is terminated |
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| 37010. |
Middle term split of x^2+x+1 |
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| 37011. |
Polynomial 2.3 exercise question number 1 ka 2 |
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| 37012. |
Ex.1.1 Questions no.5 |
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| 37013. |
X°2+2x+1 |
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| 37014. |
(Cos^2 20°+cos^2 70°/sec^2 50°-cot^2 40°)+2cosec^2 58°-2cot58° tan32° |
| Answer» We have,{tex}\\frac { \\cos ^ { 2 } 20 ^ { \\circ } + \\cos ^ { 2 } 70 ^ { \\circ } } { \\sec ^ { 2 } 50 ^ { \\circ } - \\cot ^ { 2 } 40 ^ { \\circ } }{/tex}\xa0+2 cosec258° - 2cot58° tan 32° - 4 tan13° tan37° tan45° tan53° tan77°=\xa0{tex}\\frac { \\cos ^ { 2 } 20 ^ { \\circ } + \\cos ^ { 2 } \\left( 90 ^ { \\circ } - 20 ^ { \\circ } \\right) } { \\sec ^ { 2 } 50 ^ { \\circ } - \\cot ^ { 2 } \\left( 90 ^ { \\circ } - 50 ^ { \\circ } \\right) }{/tex}\xa0+ 2cosec258° - 2cot58° tan (90° - 58°)- 4tan13° tan37° tan45° tan(90° - 37°) tan(90° -13°)=\xa0{tex}\\frac { \\cos ^ { 2 } 20 ^ { \\circ } + \\sin ^ { 2 } 20 ^ { \\circ } } { \\sec ^ { 2 } 50 ^ { \\circ } - \\tan ^ { 2 } 50 ^ { \\circ } }{/tex}\xa0+ 2cosec258° - 2cot258° - 4 tan13° tan37° tan45° cot37° cot13°=\xa0{tex}\\frac { 1 } { 1 }{/tex}\xa0+ 2(cosec258° - cot258°) - 4(tan13° cot13°) (tan37° cot37°) tan45°= 1 + 2 - 4 {tex}\\times{/tex}\xa01\xa0{tex}\\times{/tex}\xa01\xa0{tex}\\times{/tex}\xa01\xa0= 3 -\xa04 = -1 | |
| 37015. |
SinΦ+cosΦ=sinΦα=cosβ=? |
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| 37016. |
If 8 is the root of equation x square -10 x + p =0 then find the value of p . |
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Answer» If 8 is the root of x²then x= 8 , put the value -10x + p =0 -10 *8 + p = 0 P= 80 Ans. 16 16 |
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| 37017. |
solve the following equation 3x-2y+3=0,4x+3y-47=0 by. cross multiplication method |
| Answer» {tex}3x - 2y + 3 = 0{/tex}........(i){tex}4x + 3y - 47 = 0{/tex}......(ii)By cross multiplication, we have{tex}\\therefore \\frac { x } { [ ( - 2 ) \\times ( - 47 ) - ( 3 \\times 3 ) ] }{/tex}{tex}= \\frac { y } { [ ( 3 \\times 4 ) - ( - 47 ) \\times 3 ] }{/tex}{tex}= \\frac { 1 } { [ 3 \\times 3 - ( - 2 ) \\times 4 ] }{/tex}{tex}\\Rightarrow \\quad \\frac { x } { ( 94 - 9 ) } = \\frac { y } { ( 12 + 141 ) } = \\frac { 1 } { ( 9 + 8 ) }{/tex}{tex}\\Rightarrow \\quad \\frac { x } { 85 } = \\frac { 1 } { 17 } , \\frac { y } { 153 } = \\frac { 1 } { 17 }{/tex}{tex}17x = 85, \\ 17y = 153{/tex}{tex}\\Rightarrow \\quad x = \\frac { 85 } { 17 } , y = \\frac { 153 } { 17 }{/tex}Therefore, the solution is {tex}x = 5,\\ y = 9{/tex} | |
| 37018. |
Find the zeroes of the quadratic polynomialY2+92y+1920 |
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Answer» y2 + 92y + 1920= y2 + 32y + 60y + 1920= y( y+ 32 ) + 60(y + 32)= (y+ 32) (y + 60)=> y = -32 and y = -60 Y= -32 and -60 |
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| 37019. |
Prove that there are infinite no of prime no |
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| 37020. |
Solve the following system of equation: x+y=a-b and ax-by=a^2+b^2 |
| Answer» The given system of equations may be written asx + y = a - bSo, x + y -(a - b)=0 ......... (i)and ax - by = a2 + b2So, ax\xa0- by -(a2 + b2)=0 ........ (ii)By cross-multiplication, using (i) and (ii) , we have{tex}\\frac { x } { - \\left( a ^ { 2 } + b ^ { 2 } \\right) - b ( a - b ) } = \\frac { y } { - a ( a - b ) + \\left( a ^ { 2 } + b ^ { 2 } \\right) } = \\frac { 1 } { - b - a }{/tex}{tex}\\Rightarrow \\quad \\frac { x } { - a ^ { 2 } - a b } = \\frac { y } { a b + b ^ { 2 } } = \\frac { 1 } { - b - a }{/tex}{tex}\\Rightarrow \\quad \\frac { x } { - a ( a + b ) } = \\frac { y } { b ( a + b ) } = \\frac { 1 } { - ( a + b ) }{/tex}{tex}\\Rightarrow \\quad x = \\frac { - a ( a + b ) } { - ( a + b ) } = a \\text { and } y = \\frac { b ( a + b ) } { - ( a + b ) } = - b{/tex}Hence, x = a, y = -b is the solution of the given system of equations. | |
| 37021. |
(A-B)^2 identity |
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Answer» Thankd (A-B)^2=A^2+B^2-2AB ? A^2+B^2-2AB A^2 -2ab+b^2 |
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| 37022. |
Mark √8.2 on no. Line |
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| 37023. |
Find the hcf and lcm of 10211,2517 |
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| 37024. |
Squre |
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| 37025. |
They are perfect aap |
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| 37026. |
Find the zeroes of the polynomial p(x)=x3-12x2+39x-28,if it is given that the zeroes are a-d, a, a+d |
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Answer» -5,4and 13 are zeroes of the given polynomial. First find alpha+bita+gama and then use the value of a in alpha*bita*gama then put the value of a and d in the given zeroes. |
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| 37027. |
What are real no.s. |
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Answer» Hii gud to see u using app for study purpose ppz dont use it for chating Real no include rational number and irrational now \tAll rational and all irrational number makes the collection of real numbers. It is denoted by the letter R\tWe can represent real numbers on the number line. The square root of any positive real number exists and that also can be represented on number line\tThe sum or difference of a rational number and an irrational number is an irrational number.\tThe product or division of a rational number with an irrational number is an irrational number. Real number are a mixture of rational and irrat numbers. Real numbers include all types of number |
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| 37028. |
3x-y-5=06x-2y-p=0Find p when given lines are parellel |
| Answer» The given equations are3x - y - 5 = 0 ......... (i)6x - 2y + k = 0 .......... (ii)The system of linear equations\xa0a1x + b1y + c1\xa0= 0a2x + b2y + c2\xa0= 0Compare (i) and (ii), we geta1 = 3, b1 = -1, c1 = -5and a2 = 6, b2 = -2 , c2 = kThe equations has no solution if\xa0{tex}\\frac { a _ { 1 } } { a _ { 2 } } = \\frac { b _ { 1 } } { b _ { 2 } } \\neq \\frac { c _ { 1 } } { c _ { 2 } }{/tex}{tex}\\Rightarrow \\frac { 3 } { 6 } = \\frac { - 1 } { - 2 } \\neq \\frac { - 5 } { k }{/tex}So,{tex}- \\mathrm { k } \\neq 10{/tex}{tex} \\Rightarrow \\mathrm { k } \\neq - 10{/tex} | |
| 37029. |
Write the prime factorisation : 5005 |
| Answer» So, 5005 = 5 {tex}\\times{/tex}\xa07{tex}\\times{/tex}\xa011 {tex}\\times{/tex}\xa013. | |
| 37030. |
Write the prime factorisation of : 72 , 5005 |
| Answer» 72 = 8 x = 23\xa0x 33So, 5005 = 5 {tex}\\times{/tex}\xa07{tex}\\times{/tex}\xa011 {tex}\\times{/tex}\xa013. | |
| 37031. |
Find the sum of all number from 1 to 20 |
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Answer» 120 Sn=n/2(2a+(n-1)d) a=1d=1n÷2(2a + n-1× d)n=20 |
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| 37032. |
In an AP , 2 , 7 , 12 , 17 , .........find a5 and a10 |
| Answer» AP , 2 , 7 , 12 , 17 , .........a = 2 and d = 7 -2 = 5a5 = a + 4d = 2 + 4(5) = 2 + 20 = 22a10 = a + 9d = 2 + 9(5) = 2 + 45 = 47\xa0 | |
| 37033. |
Answer of ch 1 real number |
| Answer» Get NCERT solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 37034. |
prove root 10 is irrational |
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| 37035. |
Solve pair of linear equation by substitution method1-....6x-5y=1 2x+3y=5 |
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Answer» Sorry,the value of x is (1+5y)÷6 not 3 Taking equation 16x-5y=1Here x=(1+5y)÷3Put the value of x in equation 2, 2x+3y=52[(1+5y)÷6]+3y=5(1+5y)÷3+3y=55y÷3+3y=5-1÷3(5y+9y)÷3=(15-1)÷314y=14Therefore,y=1Now,put the value of y in equation 16x-5(1)=16x-5=16x=1+56x=6Therefore,x=1The answer is X=1 and Y=1 x=1, y=1. |
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| 37036. |
Mathematics me ncert ke alawa aur kaun si book lu rs or rd |
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Answer» R d sharma nhi Exempler Rs Ncert exampler Rs |
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| 37037. |
Find the HCF and LCM of 12, 15 and 21 by prime factorisation method |
| Answer» L.c.m of( 12,15,21)= 2×3×2×5×7 = 6×10×7 =420 l.c.m H.c.f =3 | |
| 37038. |
4u^2+8u |
| Answer» 4usquare + 8u =4u(u+2)U=0 and also u =-2So the zeroes are o and -2 Case 1... (a+b)=0+(-2) = -(-8)/4 =(-b/a)-2 = 8/4Case 2... (a*b) = (0)(-2) = 0/4= (c/a)0=0/4 Hence verified...Hope sooo ...it would help u..? | |
| 37039. |
I am so weak in mathematics |
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Answer» Bass practice Karo Bhai aap Do practice on daily basis |
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| 37040. |
Solve by elimination method...3x+2y=42x-3y=7 |
| Answer» 3x+2y=4 ........ (i)2x-3y=7 ....... (ii)Multiply (i) by 3 and (ii) by 29x + 6y = 12......... (iii)4x - 6y = 14 ......... (iv)Add (iii) and (iv)13x = 26x = 26/13 = 2Put x = 2 in (i) , we get3(2) + 2y = 46 + 2y = 42y = 4 - 62y = -2y = -1 | |
| 37041. |
Find the polynomial whose zeroes are 2 and -3 |
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| 37042. |
Solve by elimination method...x=2y-1y=5-3x |
| Answer» x=2y-1x - 2y = -1 ......... (i)y=5-3x3x + y = 5 ......... (ii)multiply (ii) by 26x + 2y = 10 ......... (iii)Adding (i) and (iii)x - 2y + 6x + 2y = -1 + 107x = 9x = 9/7Put x = 9/7 in y = 5 -3xy = 5 - 3(9/7)y = 5 - 27/7= (35 - 27)/7= 8/7\xa0 | |
| 37043. |
Pkease told me maths 10class rs aggarwal ka chapter no. 1real no. |
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| 37044. |
If Sin(1-tan)+Cos(1+cot)=Sec+Cosec |
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| 37045. |
Five terminating decimals between root 2 and root 3? |
| Answer» We knowRoot 2 = 1.414Root 3 = 1.732Now, we can write n number of rational number between these. That is just greater than 1.414 and less than 1.732 and it should be terminating or not terminating but repeating.For example1.415659756, 1.416893, 1.715644, ... | |
| 37046. |
Ask me any Question from ch1 real no. |
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| 37047. |
Solve the linear equation i.e. 5x-4y+8=0;7x+6y-9=0 |
| Answer» By substitution method | |
| 37048. |
Explain why 7×11×13+13 and 7×6×5×4×3×2×1+5are composite no. |
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Answer» We have,7×11×13+13 =13(7×11×1+1)=13×78=13×3×2×13Hence, it is a composite number .We have,7×6×5×4×3×2×1+5 =5(7×6×4×3×2×1+1)=5(1008+1)=5×1009Hence, it is a composite number . 7*11*13+13Let\'s take common 13(7*11*1+1) *1 More then two factors then it is composite no. I HOPE YOU WILL DO SECOND QUESTION SELF Plz slove this |
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| 37049. |
what is linear equation |
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Answer» Linear equation means equation in which highest degree of term is one Equation jiski highest power 1 hiti hai |
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| 37050. |
x-3y=7&3x-3y=15 |
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Answer» ? X=4 and Y= -1 |
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