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.

32501.

LCM of (6,72,120)

Answer» 360
32502.

Rectangles

Answer»
32503.

-3/4 standard form

Answer»
32504.

Find the hcf of 960 and 1575 and expression it in the form of 960m+1575n

Answer»
32505.

Koun si month me 11th class ke form bharne hote hai Plzzzzzzzz tell me???

Answer»
32506.

Q) Rational or Irrational ?? (root 5 - root 2)^10

Answer» real number
Irrational
32507.

Define terminating and non terminating number......

Answer» Pavan partu is wrongThe answer is a number which is stop is called terminating A number which is not stop is called non terminating number
the numbers which r not repeating called terminatingthe num which r repeating called non terminating
32508.

What are trigometry

Answer»
32509.

find the largest no which divide 224, 250 and 302 and least remainder =3 in each case

Answer»
32510.

Prove that root 6 is a irrational number

Answer» If possible, let {tex}\\sqrt { 6 }{/tex} be rational and let its simplest form be {tex}\\frac { a } { b }{/tex} then, a and b are integers having no common factor other than 1, and {tex}b \\neq 0{/tex}.Now, {tex}\\sqrt { 6 } = \\frac { a } { b } {/tex}{tex}\\Rightarrow 6 = \\frac { a ^ { 2 } } { b ^ { 2 } }{/tex} [on squaring both sides]{tex}\\Rightarrow 6b^2 = a^2{/tex} .................(i){tex}\\Rightarrow{/tex} 6 divides {tex}a^2{/tex} [{tex}\\because{/tex} 6 divides {tex}6b^2{/tex}]{tex}\\Rightarrow{/tex} 6 divides {tex}a{/tex}Let {tex}a = 6c{/tex} for some integer {tex}c{/tex}putting {tex} a = 6c{/tex} in (i), we get{tex}a^2 = 36c^2{/tex}{tex}6b^2 = 36c^2 \\;\\;\\;[6b^2 = a^2] {/tex}{tex}\\Rightarrow b^2 = 6c^2{/tex}{tex}\\Rightarrow{/tex} 6 divides {tex}b^2{/tex} [{tex}\\because{/tex} 6 divides {tex}6c^2{/tex}]{tex}\\Rightarrow{/tex} 6 divides {tex}b{/tex} [{tex}\\because{/tex} 6 divides {tex}b^2 = 6{/tex} divides {tex}b{/tex}]Thus, 6 is a common factors of {tex}a{/tex} and {tex}b{/tex}But, this contradicts the fact that {tex}a{/tex} and {tex}b{/tex} have no common factor other than 1The contradiction arises by assuming that {tex}\\sqrt { 6 }{/tex} is rational.Hence {tex}\\sqrt { 6 }{/tex} is irrational.
32511.

If p,qare zeroes of 2xsquare-7x+3,find p square+q square

Answer» 37/4
32512.

What is irrational

Answer» A number which is neither terminates nor repeat or which is not a perfect square
Numbers that cannot be expressed in the form of p/q
32513.

2y -4a-(-4) - (-9y+7a-34

Answer»
32514.

Prove that root 3+3/7root 3 is an irrational number?

Answer»
32515.

1~ How we proof that 2+2= 52~ How we proof that 2× 2 =5

Answer» 2 + 2 = 514 = 14 10 + 4 = 10 + 44 - 4 = 10 - 102^2 - 2^- 2 = 5 ( 2 - 2 )Then we apply identity( 2 + 2 ) ( 2 - 2) = 5 ( 2 - 2)Then we will cancel 2 - 2 by 2 - 2Then it will be2 + 2 = 5# proved#
32516.

Prove under root 3 is itrational

Answer»
32517.

What is co-prime and composite number

Answer» Which come in others table
32518.

Find hot of 256,678

Answer»
32519.

Find a quadratic polynomial, the sum of whose zeroes is 0 and one zero is 5

Answer» Let α=5Given α+β=05+β=0,β=-5So the polynomial f(x)=(x-5)[x-(-5)]=(x-5)(x+5)=x2-25
32520.

How chapters to be read for sa1

Answer»
32521.

(X+5)(x+6)=25/24×24

Answer» According to the question,{tex}(x - 5)(x - 6) = \\frac{{25}}{{{{\\left( {24} \\right)}^2}}}{/tex}{tex}\\Rightarrow x(x - 6) - 5(x - 6) = \\frac{{25}}{{{{(24)}^2}}}{/tex}{tex}\\Rightarrow {x^2} - 6x - 5x + 30 - \\frac{{25}}{{{{(24)}^2}}} = 0{/tex}{tex}\\Rightarrow {x^2} - 11x + 30 - \\frac{{25}}{{{{(24)}^2}}} = 0{/tex}{tex}\\Rightarrow {x^2} - 11x + \\frac{{30 \\times {{24}^2} - 25}}{{{{(24)}^2}}} = 0{/tex}{tex} \\Rightarrow {x^2} - 11x + \\frac{{30 \\times 576 - 25}}{{{{(24)}^2}}} = 0{/tex}{tex} \\Rightarrow {x^2} - 11x + \\frac{{17280 - 25}}{{{{(24)}^2}}} = 0{/tex}{tex}\\Rightarrow {x^2} - \\frac{{264x}}{{24}} + \\frac{{145}}{{24}} \\times \\frac{{119}}{{24}} = 0{/tex}{tex}\\Rightarrow {x^2} - \\left( {\\frac{{145}}{{24}} + \\frac{{119}}{{24}}} \\right)x + \\frac{{145}}{{24}} \\times \\frac{{119}}{{24}} = 0{/tex}{tex}\\Rightarrow {x^2} - \\frac{{145}}{{24}}x - \\frac{{119}}{{24}}x + \\frac{{145}}{{24}} \\times \\frac{{119}}{{24}} = 0{/tex}{tex}\\Rightarrow x\\left( {x - \\frac{{145}}{{24}}} \\right) - \\frac{{119}}{{24}}\\left( {x - \\frac{{145}}{{24}}} \\right) = 0{/tex}{tex}\\Rightarrow \\left( {x - \\frac{{145}}{{24}}} \\right)\\left( {x - \\frac{{119}}{{24}}} \\right) = 0{/tex}{tex}\\Rightarrow x - \\frac{{145}}{{24}} = 0{/tex} or {tex}x - \\frac{{119}}{{24}} = 0{/tex}{tex}\\Rightarrow x = \\frac{{145}}{{24}}{/tex} or {tex}x = \\frac{{119}}{{24}}{/tex}
32522.

Give an example of two irrational whose sum is rational

Answer» Sorry...√3and -√3. Are two irrational no. Which sum is rational no.
2root3 - root3
32523.

1.4

Answer»
32524.

Short solutions ncert exercise

Answer»
32525.

Find the series of the polynomial 8x² - 22x - 21

Answer» We have,{tex}8 x ^ { 2 } - 22 x - 21 = 0{/tex}{tex}\\Rightarrow{/tex}\xa08x2 - 28x + 6x - 21 = 0{tex}\\Rightarrow{/tex}\xa04x(2x - 7) + 3 (2x - 7) = 0{tex}\\Rightarrow{/tex}\xa0(2x - 7) (4x + 3) = 0 {tex}\\Rightarrow{/tex}\xa02x - 7 = 0 or,\xa04x + 3 = 0\xa0{tex}\\Rightarrow{/tex}\xa0{tex}x = \\frac { 7 } { 2 } \\text { or } x = - \\frac { 3 } { 4 }{/tex}Thus,\xa0{tex}x = \\frac { 7 } { 2 } \\text { and } x = - \\frac { 3 } { 4 }{/tex} are two roots of the equation 8x2 - 22x - 21 = 0
32526.

Prove that the product of three consecutive positive integer is divisibel by 6.

Answer» Let three consecutive numbers be x, (x + 1) and (x + 2)Let x = 6q + r 0 {tex}\\leq r < 6{/tex}{tex}\\therefore x = 6 q , 6 q + 1,6 q + 2,6 q + 3,6 q + 4,6 q + 5{/tex}{tex}\\text { product of } x ( x + 1 ) ( x + 2 ) = 6 q ( 6 q + 1 ) ( 6 q + 2 ){/tex}if x = 6q then which is divisible by 6{tex}\\text { if } x = 6 q + 1{/tex}{tex}= ( 6 q + 1 ) ( 6 q + 2 ) ( 6 q + 3 ){/tex}{tex}= 2 ( 3 q + 1 ) .3 ( 2 q + 1 ) ( 6 q + 1 ){/tex}{tex}= 6 ( 3 q + 1 ) \\cdot ( 2 q + 1 ) ( 6 q + 1 ){/tex}which is divisible by 6if x = 6q + 2{tex}= ( 6 q + 2 ) ( 6 q + 3 ) ( 6 q + 4 ){/tex}{tex}= 3 ( 2 q + 1 ) .2 ( 3 q + 1 ) ( 6 q + 4 ){/tex}{tex}= 6 ( 2 q + 1 ) \\cdot ( 3 q + 1 ) ( 6 q + 1 ){/tex}Which is divisible by 6{tex}\\text { if } x = 6 q + 3{/tex}{tex}= ( 6 q + 3 ) ( 6 q + 4 ) ( 6 q + 5 ){/tex}{tex}= 6 ( 2 q + 1 ) ( 3 q + 2 ) ( 6 q + 5 ){/tex}which is divisible by 6{tex}\\text { if } x = 6 q + 4{/tex}{tex}= ( 6 q + 4 ) ( 6 q + 5 ) ( 6 q + 6 ){/tex}{tex}= 6 ( 6 q + 4 ) ( 6 q + 5 ) ( q + 1 ){/tex}which is divisible by 6if x = 6q + 5{tex}= ( 6 q + 5 ) ( 6 q + 6 ) ( 6 q + 7 ){/tex}{tex}= 6 ( 6 q + 5 ) ( q + 1 ) ( 6 q + 7 ){/tex}which is divisible by 6{tex}\\therefore {/tex}\xa0the product of any three natural numbers is divisible by 6.
32527.

RS AGGRAWAL

Answer»
32528.

Rd sharma exercise 5.3 and number 8 solutions

Answer»
32529.

Rat

Answer»
32530.

X2- 2x-8

Answer»
32531.

Find all zeroes of the polynomial p (x)=x^4+x^3-2x-10x^2+4.

Answer»
32532.

Eliminate the following - 4/x+3y=14 3/x-4y=2

Answer» 4/x+3y=14 3/x-4y=2
32533.

Plzz put last year means 2018 question paperPlzz.....Plz.............

Answer» No it is cheating
32534.

What is definition of irrationals

Answer» The numbers that are non terminating and non recurring
The numbers which cannot be written in the form of p÷q where p and q are integers and q is not equal to 0.
32535.

Prove that (n)2-1 is divisible by 8 where n is an odd positive integer

Answer» No
32536.

I am expecting 75 maths, is it good

Answer» Yes it is good but u can score more if u want and1 important thing never compare yourself with others
I got 80 marks out of 80
I have scored 80out of 80in my ninety class and hoping to maintain the position
I got 79 marks out of 80 in class 9th and I think I will do my very best now due to encouragement
I would have scored 80 but due to less time I scored 75
No u should score more. I have scored 76 in class 9 in maths so i can score above 90 and u can also score. Maths of class 10 is very easy as compared to class 9.
32537.

88888888 and 77777777 find the hcf with euclidean division

Answer» 11111111 is the answer
32538.

In chapter 1 find HCF by using Euclid\'s lemma. 1(300,540,890)

Answer»
32539.

For what value of n ,2^n× 5^m ends with digits 5

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.
32540.

2 chapter in hindi

Answer»
32541.

1/4 -1

Answer» Math answer in hindi
32542.

I want the relible of 2018 both termd

Answer»
32543.

3x+4y=18,6x+5y=35

Answer»
32544.

Value of (a – c) [(a – b)2 + (b – c)2 – (a – b)(b – c)] + (c – a)3 is equal to

Answer»
32545.

Find lcm if 12

Answer» The LCM of 12 is 12 itself
32546.

7tm

Answer»
32547.

Example of real number

Answer» 2,1/6, -1and irrational number
32548.

How to solve X3 - 2x2

Answer»
32549.

Write whether 2√45+3√20÷2√5

Answer» {tex}\\frac{2\\sqrt{45\\;\\;}+3\\sqrt{20}}{2\\sqrt5}{/tex}{tex}\\text{=}\\frac{\\displaystyle2\\sqrt{5\\times9}+3\\sqrt{5\\times4}}{2\\sqrt5}{/tex}{tex}=\\frac{6\\sqrt5+6\\sqrt5}{2\\sqrt5}{/tex}{tex}=\\frac{\\displaystyle12\\sqrt5}{2\\sqrt5}{/tex}{tex}=6=\\frac61{/tex}Hence the given number can be written in the form of {tex}\\frac pq{/tex} where p and q are integers.So the given expression is a rational number.
32550.

What is the solution of this equation 6x square-x-8

Answer»