Explore topic-wise InterviewSolutions in .

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.

7551.

Find value of k for the following quadratc equation, so that they have two equal roots. 3x²-k√3x+4

Answer»
7552.

Find 3 numbers in AP, whose sum is 15 and whose product is 105.

Answer» And is 3,5,7
a-d. , a, a+d. are the three numbers
7553.

For final board the text book questions only will come?

Answer» ?
No man, only few questions will appear from your book in board exams
No
7554.

Determine if the points (1,5), (2,3) and (-2,-11) are collinear

Answer» If area =0
7555.

Is the paper going to be ncert based

Answer» No it is not complsery that paper from ncert but exam will not hard according to anouncement by cbse
not totoly but about 65% from ncert and rest from RS agarwal
7556.

A card has 3,4,5........50 numbers .find the probability of perfect square number.

Answer» Sorry it\'s not 3/50 it\'s 1/16
Probability is 3/50
7557.

State and prove converse of Pythagoras theorem

Answer» It is very simple go on youtube and find aise likhne se samaj nhi aayega
7558.

The hcf of two numbers is 145 and their lcm is 2175 if one number is 725 then find the other number

Answer» Hcf x lcm = a x b145 x 2175 = 725 x bb = 435
7559.

Distance between the centre of the circle of radius 3 cm and 8 cm

Answer»
7560.

Solve eqation ax+by=b-a

Answer» x\\y = a\\b
7561.

Number of question come on each chapter

Answer»
7562.

bhh

Answer»
7563.

IncontinenceHello

Answer»
7564.

What is hcf

Answer» Highest Common Factor(HCF) of two or more numbers is the greatest number which divides each of them exactly. Greatest Common Measure(GCM) and Greatest Common Divisor(GCD) are the other terms used to refer HCF.
7565.

Ajagsjhsishig

Answer»
7566.

ST line segment joining the points (4,2)&(-6,4)&(-10,5)&(8,1) bisect each other...

Answer»
7567.

If roots of qdrtc eqn (a-b)+(b-c)+(c-a)=0 are equal, PT 2a=b+c..

Answer» We have (a-b)x2 + (b-c)x + (c-a) = 0Here A\xa0= (a-b), B\xa0= (b-c), C= (c-a){tex}\\therefore D = B^2 - 4AC = (b-c)^2 - 4(a-b) (c-a){/tex}For equal roots, D = 0{tex}\\implies{/tex}(b-c)2 - 4 (a-b) (c-a) = 0b2+c2-2bc -4(ac-a2-bc+ab) =0b2+c2-2bc -4ac+4a2+4bc-4ab=04a2+b2+c2+2bc-4ab-4ac=0(2a-b-c)2=0i.e. 2a-b-c =0Hence, b + c = 2a or 2a = b + c
7568.

2x+4x

Answer» 2x+4x=6x
7569.

Kya kar rahe ho

Answer» Kuch ni
7570.

3square

Answer» 9
7571.

Can we answer from last question in board

Answer» Yes
7572.

What is circle

Answer» A closed figure formed by infinite points
7573.

(A^2 + b^2 )^2

Answer»
7574.

5x-4y+8=0 , 7x+6y-9

Answer»
7575.

CH No 6 Triangle 10 class

Answer»
7576.

Greatest integer

Answer»
7577.

Problem

Answer» Solution
7578.

Is upgradation for all subjects?

Answer»
7579.

How to form the quadratic eqn by seeing the question

Answer»
7580.

How many numbers lie between 10 and 300 which divided by 4 leave a remainder 3

Answer» The numbers lying between 10 and 300, which when divided by 4 leave a remainder 3 are11, 15, 19............,299This is an A.P. with a = 11, d = 4 and l = 299Let the number of terms be n.then, an = 299{tex}\\Rightarrow{/tex}\xa0a + (n - 1)d = 299{tex}\\Rightarrow{/tex}\xa011 + (n - 1)4 = 299{tex}\\Rightarrow{/tex}\xa0(n - 1)4 = 288{tex}\\Rightarrow{/tex}\xa0n - 1 = 72{tex}\\Rightarrow{/tex}\xa0n = 73Thus, the required number of terms are 73.
7581.

□+□+□=30With the number (1,3,5,7,9,11,13,15)you can repeat the numbers

Answer» (3+5) +(7+11) +(3+1). Thanks for asking
❎ ? question. Galat hia
11+9+11
7582.

4565+5675×56+56

Answer» 727421
7583.

Pls help me for ogive graph

Answer»
7584.

How to understand the circle chapter example 1

Answer»
7585.

What is cf

Answer» Cumulative frequency
7586.

1/(2a+b+2x)=1/2a+1/b+1/2x

Answer» {tex}\\frac{1}{2a + b + 2x}{/tex}\xa0=\xa0{tex}\\frac{1}{2a}{/tex}\xa0+\xa0{tex}\\frac{1}{b}{/tex}\xa0+\xa0{tex}\\frac{1}{2x}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{1}{2a + b + 2x}{/tex}\xa0-\xa0{tex}\\frac{1}{2x}{/tex}\xa0=\xa0{tex}\\frac{1}{2a}{/tex}\xa0+\xa0{tex}\\frac{1}{b}{/tex}\xa0{tex}\\Rightarrow{/tex}{tex}\\frac { 2 x - 2 a - b - 2 x } { ( 2 a + b + 2 x ) ( 2 x ) }{/tex}\xa0=\xa0{tex}\\frac{b + 2a}{2a \\times b}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { - ( 2 a + b ) } { ( 2 a + b + 2 x ) 2 x }{/tex}\xa0=\xa0{tex}\\frac{b + 2a}{2ab}{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac { - 1 } { 4 a x + 2 b x + 4 x ^ { 2 } }{/tex}\xa0=\xa0{tex}\\frac{1}{2ab}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}4x^2 + 2bx + 4ax = -2ab{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}4x^2 + 2bx + 4ax + 2ab = 0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}2x(2x + b) + 2a(2x + b) = 0{/tex}{tex}\\Rightarrow{/tex}\xa0(2x + b)(2x + 2a) = 0{tex}\\Rightarrow{/tex}\xa0x = -{tex}\\frac{b}{2}{/tex} or x = -a
7587.

What is the probability of getting t3 Sunday\'s match leap year

Answer» If you mean 53rd sunday , then the probability is 2/7
7588.

If the mth term of an AP is 1/n and it\'s nth term be 1/m then show that it\'s mnth 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
7589.

Syllabus for board exams

Answer» It\'s clearly given in mycbseguide.So u can refer this.
7590.

If x=cot^ + cosec^ find the value of 1+cos^/1-cos^

Answer» And : x
7591.

A boiler is in the form of a cylinder 2 m long with hemispherical ends of 2 m find volume????

Answer» According to the question,we are given that,Diameter of common base = 2 mThen, radius of common base =\xa0{tex}\\frac { 2 } { 2 }{/tex}\xa0= 1 mHeight of the cylinder = 2 mVolume of boiler = Volume of cylinder + 2(Volume of hemisphere){tex}= \\pi r ^ { 2 } h + 2 \\times \\frac { 2 } { 3 } \\pi r ^ { 3 }{/tex}{tex}= \\frac { 22 } { 7 } \\times 1 \\times 1 \\times 2 + 2 \\times \\frac { 2 } { 3 } \\times \\frac { 22 } { 7 } \\times 1 \\times 1 \\times 1{/tex}{tex}= \\frac { 44 } { 7 } + \\frac { 88 } { 21 }{/tex}{tex}= \\frac { 132 + 88 } { 21 }{/tex}{tex}= \\frac { 220 } { 21 } \\mathrm { m } ^ { 3 }{/tex}
7592.

5+5

Answer» 10
7593.

4 term of an ap is zero prove that the 25th term of the AP 3 times its 11 term

Answer» 4th term of AP is 0 This implies that 4th term = a+ (4-1)d. 0=a+3d. a=-3d11th term = a+(11-1)d. 11th =-3d+10d. 11th=7d. equation-----(1). 25th term=a+(25-1)d. 25th= -3d+24d25th =21d. equation--------(2)On comparing equation--(1) & equation--(2) ,we get3 times of 11th term =25th term
7594.

ABCD IS A SQUARE OF SIDE14CM AMB &CMD ARE SEMICIRCLES FIND THE PERIMETER OF THE SHADED REGION

Answer» 44cm
42cm
7595.

What is locus

Answer» Locus is\xa0a curve or other figure formed by all the points satisfying a particular equation of the relation between coordinates, or by a point, line, or surface moving according to mathematically defined conditions.
7596.

The sum of the interior angle of a pentegon

Answer» 540
540
7597.

Xraised to the power y=y raised to the power z=z raised to the power x ,then x.y.z= ???

Answer»
7598.

1/secA =

Answer» CosA
7599.

Last 10 years question papers of class 10

Answer» Check last year papers here :\xa0https://mycbseguide.com/cbse-question-papers.html
7600.

Cot^-cos^/cot^+cos^=2-√3/2+√3

Answer» 60°