Saved Bookmarks
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.
| 16751. |
Find sum 2_12_22? |
| Answer» 60 | |
| 16752. |
If the system of equations kx- 5y=2, 6x+2y=7has no sol. then k= |
|
Answer» What do you mean? Can any one tell question |
|
| 16753. |
CNN you tell some question |
|
Answer» ? Determine graphically the coordinates of the verticesof a triangle whose sides are graphically of the question |
|
| 16754. |
Show that cubic root 3 is irrational |
| Answer» If wecubic is3 than it is 27 | |
| 16755. |
a6=5a and a11=2a5+3 . find a8 |
| Answer» a6_ 5a= 1a | |
| 16756. |
Of alpha and beta are the zeroes f(x) = x square + X + 1, then find 1/alpha + 1/beta. |
| Answer» | |
| 16757. |
if α,β are zeros of ax square-bx+c , find α+β |
| Answer» -b/a | |
| 16758. |
Cos37degreecosec53degree |
| Answer» = cos 37° cosec 53°= cos(90° - 53°) cosec 53°= sin 53° cosec 53°= sin 53° .\xa0{tex}\\frac{1}{sin 53^ \\circ}{/tex}= 1 | |
| 16759. |
Find tansquare 60 +2+tan square 45° |
| Answer» = tan260° + tan245°= ({tex}\\sqrt 3{/tex})2\xa0+ (1)2= 3 + 1= 4 | |
| 16760. |
Ex 3.1 ka 1 question |
| Answer» Go to the solution option in maths subject option........ | |
| 16761. |
Divide 4500 into two parts such that 10%of the first is equal ti 5% of the second part |
| Answer» First part:1500Second part is:3000 | |
| 16762. |
HCF of smallest composite number and smallest prime number |
|
Answer» Smallest composite number 4Smallest prime number 2..So HCF equals 2.. 4 |
|
| 16763. |
Show that n square minus 1 is divisible by 8 if n is an odd positive integer |
| Answer» Let n = 4q + 1 (an odd integer){tex}\\therefore \\quad n ^ { 2 } - 1 = ( 4 q + 1 ) ^ { 2 } - 1{/tex}{tex}= 16 q ^ { 2 } + 1 + 8 q - 1 \\quad \\text { Using Identity } ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }{/tex}{tex}= 16{q^2} + 8q{/tex}{tex}= 8 \\left( 2 q ^ { 2 } + q \\right){/tex}= 8m, which is divisible by 8. | |
| 16764. |
Show that n square minus 1 by 8 |
| Answer» Let n = 4q + 1 (an odd integer){tex}\\therefore \\quad n ^ { 2 } - 1 = ( 4 q + 1 ) ^ { 2 } - 1{/tex}{tex}= 16 q ^ { 2 } + 1 + 8 q - 1 \\quad \\text { Using Identity } ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }{/tex}{tex}= 16{q^2} + 8q{/tex}{tex}= 8 \\left( 2 q ^ { 2 } + q \\right){/tex}= 8m, which is divisible by 8. | |
| 16765. |
Show that 2*3*5*13*17+13 a composite number |
| Answer» As this expression can be simplified as 13(2*3*4*5*17+1) this shows that 13 is a factor of yhis expression and hence it is composite | |
| 16766. |
How to cheack a number is aterm of the list of numbers |
| Answer» | |
| 16767. |
IF -2 is a root of the equation 3x^2 - 5x +2k =0, find the value of k |
| Answer» We have the following equation,\xa0{tex}3x^2-5x+2k=0{/tex}Putting x = -23(-2)2\xa0- 5(-2) + 2k = 0{tex}\\Rightarrow{/tex}\xa012 + 10 + 2k = 0{tex}\\Rightarrow{/tex}\xa022 + 2k = 0{tex}\\Rightarrow{/tex}\xa0k =\xa0{tex}\\frac{-22}{2}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0k = -11 | |
| 16768. |
What is the new syllabus of class 10 math Sessions 2019-2020 |
| Answer» Check syllabus here :\xa0https://mycbseguide.com/cbse-syllabus.html | |
| 16769. |
Table of 7 |
| Answer» 7 × 1 = 77 × 2 = 147 × 3 = 217 × 4 = 287 × 5 = 357 × 6 = 427 × 7 = 497 × 8 = 567 × 9 = 637 × 10 = 70 | |
| 16770. |
4x^2 +5 underoot 2x-3Solve this |
| Answer» | |
| 16771. |
LCM of 18,20,9 |
| Answer» LCM of 18 , 20, 9 is 180. | |
| 16772. |
6x+5y=7x+3y+1=2(x+6y-1) Find the value of x and y. |
|
Answer» x=17and y=10 6x+5y=2(x+6y-1)6x+5y=2x+12y-26x-2x+5y-12y=-24x-7y=-2 ------------------(1)Again,7x+3y+1=2(x+6y-2)7x+3y+1=2x+12y-47x-2x+3y-12y=-4-15x-9y=-5 -------------------(2)Multiplying eqt.(1) by 5 and (2) by 4 20x-35y=-10 20x-36y=-20 ~~+~~~+~ y=10 Put y=10 in any eqt. 4x-7y=-2 4x-7×10=-2 4x=-2+70 x=68/4=17 |
|
| 16773. |
Make polynomials whose zeroes are given2,3 |
|
Answer» I mean (x^2-5x+6) is answer x^2-(2+4)x+(2×4) X^2-5x+6 is answer. Answer |
|
| 16774. |
Using Euclid\'s division algorithm to find the HCF of: 56,9, and404 |
| Answer» 96=56×1+4056=40×1+1640=16×2+816=8×2+0HCF=8404=8×50+48=4×2+0HCF(56,96,404)=4 | |
| 16775. |
Find the centroid of the triangle whose vertices are (3, - 7), (-8,6) & (5,10). |
| Answer» Let A , B and C are three vertices ofthe triangle ( x1 , y1 ) = A ( 3 , - 7 )( x2 , y2 ) = B ( - 8 , 6 )( x3 , y3 ) = C ( 5 , 10 )Centroid ( G ) = ( x1+x2+x3/3 , y1+y2+y3/3 )= ( 3-8+5/3 , -7+6+10/3 )= ( 0 , 9/3 )= ( 0 , 3 )G = ( 0 , 3 ) | |
| 16776. |
9x2-9(a+b)x+(2a2+5ab+2b2)=0 |
| Answer» D = b2 - 4ac{tex}= ( - 9 ( a + b ) ) ^ { 2 } - 4 \\times 9 \\times \\left( 2 a ^ { 2 } + 5 a b + 2 a b ^ { 2 } \\right){/tex}= 81(a + b)2 - 36(2a2 + 5ab + 2b2)= 9[9(a2 + b2 + 2ab - 8a2 - 20ab - 8b2)]= 9[a2 + b2 - 2ab]= 9(a - b)2{tex}x = \\frac { - b \\pm \\sqrt { D } } { 2 a } = \\frac { 9 ( a + b ) \\pm \\sqrt { 9 ( a - b ) ^ { 2 } } } { 2 \\times 9 }{/tex}{tex}{ \\Rightarrow x = 3 \\frac { [ 3 ( a + b ) \\pm ( a - b ) ] } { 2 \\times 9 } }{/tex}{tex}{ \\Rightarrow x = \\frac { ( 3 a + 3 b ) \\pm ( a - b ) } { 6 } }{/tex}{tex}\\Rightarrow x = \\frac { 3 a + 3 b + a - b } { 6 } \\text { or } x = \\frac { 3 a + 3 b - a + b } { 6 }{/tex}{tex}\\Rightarrow x = \\frac { 4 a + 2 b } { 6 } \\text { or } x = \\frac { 2 a + 4 b } { 6 }{/tex}{tex}\\Rightarrow x = \\frac { 2 a + b } { 3 } \\text { or } x = \\frac { a + 2 b } { 3 }{/tex} | |
| 16777. |
Solve by cross multiplication method 5ax + 6by = 28 3ax + 4by = 18 |
| Answer» The systems of equations are:5ax + 6by = 28 .......................(i)3ax + 4by = 18 .......................(ii)From (i) and (ii), we geta1 = 5a, b1 = 6b, c1 = -(28)a2 = 3a, b2 = 4b, c2 = -(18)By cross multiplication method, we get{tex}\\frac{x}{{-108b +112b}} = \\frac{{ - y}}{{ - 90a + 84a }} = \\frac{1}{{20ab - 18ab}}{/tex}{tex}\\frac{x}{{4b}} = \\frac{{ - y}}{{ - 6a}} = \\frac{1}{{2ab}}{/tex}Now, {tex}\\frac{x}{{4b}} = \\frac{1}{{2ab}} {/tex}{tex}⇒ x = \\frac{2}{a}{/tex}And, {tex}\\frac{{ - y}}{{ - 6a}} = \\frac{1}{{2ab}} {/tex}{tex}⇒ y = \\frac{3}{b}{/tex}Therefore the solution of the given system of equation is {tex}\\frac{2}{a}{/tex}\xa0and {tex}\\frac{3}{b}{/tex}. | |
| 16778. |
Solve by cross multiplication method 1. x/a + y/b =a+b x/a2 + y/b2 =2 |
| Answer» The given equations are:{tex}\\frac{x}{a}{/tex}\xa0+\xa0{tex}\\frac{y}{b}{/tex}\xa0{tex}=(a+b){/tex}\xa0{tex}\\Rightarrow{/tex}{tex}bx + ay = ab(a + b){/tex}{tex}\\Rightarrow{/tex}{tex}bx + ay - ab(a + b) = 0{/tex} ....(i)and\xa0{tex}\\frac { x } { a ^ { 2 } } + \\frac { y } { b ^ { 2 } } = 2{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}b^2x + a^2y = 2a^2b^2{/tex}\xa0{tex}\\Rightarrow{/tex}{tex}\xa0b^2x + a^2y -2a^2b^2 = 0{/tex}....(ii)From eq. (i) and (ii), we get{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { x } { - 2 a ^ { 3 } b ^ { 2 } + a ^ { 3 } b ( a + b ) }{/tex}\xa0=\xa0{tex}\\frac { - y } { - 2 a ^ { 2 } b ^ { 3 } + a b ^ { 3 } ( a + b ) }{/tex}\xa0=\xa0{tex}\\frac { 1 } { a ^ { 2 } b - a b ^ { 2 } }{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac { x } { - a ^ { 3 } b ( 2 b - a - b ) }{/tex}\xa0=\xa0{tex}\\frac { - y } { - a b ^ { 3 } ( 2 a - a - b ) }{/tex}\xa0=\xa0{tex}\\frac { 1 } { a b ( a - b ) }{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac { x } { - a ^ { 3 } b ( b - a ) }{/tex}\xa0=\xa0{tex}\\frac { y } { a b ^ { 3 } ( a - b ) }{/tex}\xa0=\xa0{tex}\\frac { 1 } { a b ( a - b ) }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}x = \\frac { a ^ { 3 } b ( a - b ) } { a b ( a - b ) }{/tex}\xa0= a2; {tex}y = \\frac { a b ^ { 3 } ( a - b ) } { a b ( a - b ) } = b ^ { 2 }{/tex}The solution is {tex}x = a^2, y\xa0= b^2{/tex}\xa0. | |
| 16779. |
Compartment sample paper |
| Answer» | |
| 16780. |
×_10×+20=0 |
| Answer» | |
| 16781. |
Compartment exam question sir please |
| Answer» | |
| 16782. |
ax + by = (a + b)\\ 2 ; 3x + by = 5Solve for x and y |
| Answer» I can solve | |
| 16783. |
1/2(x+y) + 5/3(3x-2y)=-3/25/4(x+2y) - 3/5(3x-2y)=61/60 |
| Answer» The given equations are{tex}\\frac { 1 } { 2 ( x + 2 y ) } + \\frac { 5 } { 3 ( 3 x - 2 y ) } = - \\frac { 3 } { 2 }{/tex}.....(1)and\xa0{tex}\\frac { 5 } { 4 ( x + 2 y ) } - \\frac { 3 } { 5 ( 3 x - 2 y ) } = \\frac { 61 } { 60 }{/tex}....(2)Putting\xa0{tex}\\frac 1{x+2y}{/tex}=u and\xa0{tex}\\frac 1{3x-2y}{/tex}= v in equation (1) & equation (2) so that we may get two linear equations in the variables u & v as following:-\xa0{tex}\\frac { 1 } { 2 } u + \\frac { 5 } { 3 } v = - \\frac { 3 } { 2 }{/tex}.....................(1){tex}\\frac { 5 } { 4 } u - \\frac { 3 } { 5 } v = \\frac { 61 } { 60 }{/tex}.................(2)Multiplying (1) by 36 and (2) by 100, we get{tex}{/tex}{tex}{/tex}{tex}18u + 60v = -54{/tex}...............(3){tex}125 u - 60 v = \\frac { 305 } { 3 }{/tex}.............(4)Adding (3) and (4),we get{tex}143 u = \\frac { 305 } { 3 } - 54 = \\frac { 305 - 162 } { 2 } = \\frac { 143 } { 3 }{/tex}{tex}\\therefore \\quad u = \\frac { 1 } { 3 } = \\frac { 1 } { x + 2 y }{/tex}{tex}\\therefore{/tex}{tex} x + 2y = 3{/tex}...........(5)Putting value of u in (3), we get{tex}1 + 10v = -9{/tex} (after dividing by 3){tex}\\therefore 10 \\mathrm { v } = - 10{/tex}\xa0or\xa0{tex}\\mathrm { v } = - 1 {/tex}{tex}\\Rightarrow - 1 = \\frac { 1 } { 3 x - 2 y }{/tex}{tex}\\Rightarrow 3 x - 2 y = - 1{/tex}......(6)Adding (5) and (6), we get{tex}4 x = 2 \\quad{/tex}{tex} \\therefore x = \\frac { 1 } { 2 }{/tex}Putting value of x in (5),{tex}\\frac { 1 } { 2 } + 2 y = 3{/tex}{tex} \\text { or } 2 y = 3 - \\frac { 1 } { 2 } = \\frac { 5 } { 2 }{/tex}{tex}\\therefore \\quad y = \\frac { 5 } { 4 }{/tex}The required solution is\xa0{tex}\\mathrm { x } = \\frac { 1 } { 2 } , \\mathrm { y } = \\frac { 5 } { 4 }{/tex} | |
| 16784. |
Can\'t we get answers of practice papers given here |
|
Answer» Ok No |
|
| 16785. |
Match chapter 1 .2 quetion no 3 |
| Answer» Please specify questions...are you talking about NCERT? | |
| 16786. |
Which of the following is quadratic equation |
| Answer» Where are the quadratic equation [email\xa0protected]? | |
| 16787. |
How to find hcf lab manual work |
| Answer» | |
| 16788. |
Prove root 3 is an irrational |
|
Answer» let root 3 is rational number. root 3=p/q [where,p and q are integers,q is not equal to 0 and p and q are co-primes]. root 3q=p squaring both sides, 3q2 =p2...............[1] 3 divides p2 3 divides p.............[2] p2=3q2 p2=3m Put the value of p2 in eqn [1]. p2=3q2 [3m]2=3q2 9m2=3q2 3m2=q2 q2=3m2 3 divides q2 3 divides q [3] By eqn [2] and [3].... 3 is the common factor of both p and q . This is contradiction that and q are co-primes. Our assumption is wrong. Hence , root 3 is irrational number. \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. |
|
| 16789. |
Prove that cosO ÷ 1- tanO + sinO ÷ 1- cotO = sinO + cosO |
| Answer» | |
| 16790. |
Can anyone give me suggestions for making maths model for class 10 |
|
Answer» There is no competition in our school . Project is for holiday homework What type of math model is there any type of competition in your school? ????? |
|
| 16791. |
9 sec A-9 tan A |
| Answer» | |
| 16792. |
solve these equations 24/(×-y)+28/(×+y)=6 |
| Answer» | |
| 16793. |
SHow that cube of any natural no. Is in the form of 4q,4q+1,4q+3. |
| Answer» | |
| 16794. |
What is the product of the HCF and LCM of the smallest prime number and smallest composite number? |
| Answer» 16 | |
| 16795. |
why is an AP series called so? |
| Answer» It is short form of Arithmetic Progression | |
| 16796. |
3 pairs of numbers whose square of the highest is equal to the other 2 numbers other than 3,4&5 |
| Answer» | |
| 16797. |
Prove that the points A(1,7) B(4,2) C(-1,-1) D(-4,4) are the vertices of a square |
| Answer» Let A(1, 7), B(4, 2), C(-1, -1) and D(-4, 4) be the given points. One way of showing that ABCD is a square is to use the property that all its sides should be equal both its diagonals should also be equal. Now,{tex}A B = \\sqrt { ( 1 - 4 ) ^ { 2 } + ( 7 - 2 ) ^ { 2 } } = \\sqrt { 9 + 25 } = \\sqrt { 34 }{/tex}{tex}B C = \\sqrt { ( 4 + 1 ) ^ { 2 } + ( 2 + 1 ) ^ { 2 } } = \\sqrt { 25 + 9 } = \\sqrt { 34 }{/tex}{tex}C D = \\sqrt { ( - 1 + 4 ) ^ { 2 } + ( - 1 - 4 ) ^ { 2 } } = \\sqrt { 9 + 25 } = \\sqrt { 34 }{/tex}{tex}D A = \\sqrt { ( 1 + 4 ) ^ { 2 } + ( 7 - 4 ) ^ { 2 } } = \\sqrt { 25 + 9 } = \\sqrt { 34 }{/tex}{tex}A C = \\sqrt { ( 1 + 1 ) ^ { 2 } + ( 7 + 1 ) ^ { 2 } } = \\sqrt { 4 + 64 } = \\sqrt { 68 }{/tex}{tex}B D = \\sqrt { ( 4 + 4 ) ^ { 2 } + ( 2 - 4 ) ^ { 2 } } = \\sqrt { 64 + 4 } = \\sqrt { 68 }{/tex}Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral\xa0ABCD are equal and its diagonals\xa0AC and BD are also equal. Therefore, ABCD is a square. | |
| 16798. |
Sec70 sin20 +cos20 cosec70=2 |
| Answer» We have,sec 70° sin 20° +\xa0cos 20° cosec 70°= sec (90° - 20°) sin 20° +\xa0cos 20° cosec (90° - 20°)= cosec20° sin20° +\xa0cos20° sec20° [{tex}\\because{/tex}Sec(90°-A)=CosecA & Cosec(90° - A) = SecA ]{tex}= \\frac { \\sin 20 ^ { \\circ } } { \\sin 20 ^ { \\circ } } + \\frac { \\cos 20 ^ { \\circ } } { \\cos 20 ^ { \\circ } }{/tex}. [{tex}\\because{/tex}Cosec A=(1/SinA) & SecA =(1/CosA) ]= 1 +\xa01 = 2 | |
| 16799. |
Kya aj India match jeet sakthi hai |
|
Answer» India win match by 306 runs Ha ab jeet jayegi May be ... Shubh shubh bolo???? Ya No idea |
|
| 16800. |
All unit test paper 1 solution |
| Answer» | |