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InterviewSolution
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.
| 36851. |
prove that 22/7 is a rational number |
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Answer» To prove - 22/7 is a rational number.So first recall the definition of a rational number that is-" A rational number can be represented in the form of p/q where p and q are integers and q#0".Also-"Decimal expansion of rational no . is either terminatinating or repeating" .So here in this case 22 and 7 are integers .7#0 And also -Decimal expansion of 22/7 is 3.14285714 that is nothing but terminating .Hence , by both the ways 22/7 is a rational number. hemce proved |
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| 36852. |
Plz give me all formulas in class 10 maths |
| Answer» You can check the formulae in the notes :\xa0https://mycbseguide.com/cbse-revision-notes.html | |
| 36853. |
Aaj kya world shyayari day hai |
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| 36854. |
I am going in class 10 so how to manage the syllabus |
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Answer» Sorry fir spelling error !!! Books ke liye udhar udhar mat bhatakna ....sirf NCERT rat lena ....extra ke liye NCERT exemplar ......bass isse jayada ki jarurat nahi hiti hau First need to solve ncert ?? ... then baad me karna kuch |
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| 36855. |
What is the books name |
| Answer» Jaat कै तैने भी हद कर दी..... ??? | |
| 36856. |
What is the LCM of 412 & 540 |
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| 36857. |
Civics political parties |
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| 36858. |
After how many digits the decimal expansion of 1/7 should repeats? |
| Answer» The given number is\xa0{tex}\\frac 17{/tex}Now let us divide 1 by 7 using long division method.Now we see that remainder 1 is returned so same pattern\xa0will repeat in the quotient, so we can say that{tex}\\frac{1}{7}{/tex}\xa0= 0.142857142857............Hence\xa0{tex}\\frac{1}{7}{/tex}\xa0=\xa0{tex}0. \\overline {142857}{/tex} | |
| 36859. |
For what value of p (-4) is a zero of the polynomial x2-2x-(7p+3)? |
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| 36860. |
Someone tell me about 10 class |
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| 36861. |
7×11×13×15+15 is a |
| Answer» 15030 | |
| 36862. |
What is algebraic method |
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| 36863. |
Show that there exists no positive integer x, √x-1+ √x+1 |
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| 36864. |
(a+b)2:hole square???? |
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Answer» Pattta hai mai aise hi puch raha thha a square + b square + 2ab..... ye bhi ni ata apko???? |
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| 36865. |
Show that HCF of (867,255) is in the form of 867m + 255n |
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| 36866. |
Prove theroem lemma |
| Answer» Check in this app? for more information.. | |
| 36867. |
What are irrational numbers ? |
| Answer» A number which cannot be written in the form a/b , where a and b are integers and b ≠ 0 is called a irrational number. Irrational numbers which have non-terminating and non-repeating decimal representation.The sum or difference of a two irrational numbers is also rational or an irrational number. | |
| 36868. |
Factorise x+2√2x-6 |
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| 36869. |
Tan60° |
| Answer» √3 | |
| 36870. |
P:payara,,,A:anokha,G:gora,A:akela,L:ladka,,,,,,,PAGAL |
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| 36871. |
Show root 2 as an irrational number |
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Answer» Given √2 is irrational number.Let √2 = a / b wher a,b are integers b ≠ 0we also suppose that a / b is written in the simplest formNow √2 = a / b⇒ 2 = a2 / b2\xa0⇒ 2b2 = a2∴ 2b2 is divisible by 2⇒ a2 is divisible by 2 ⇒ a is divisible by 2 ∴ let a = 2ca2 = 4c2⇒ 2b2 = 4c2 ⇒\xa0b2 = 2c2∴ 2c2 is divisible by 2∴ b2 is divisible by 2∴ b is divisible by 2∴ a are b are divisible by 2 .this contradicts our supposition that a/b is written in the simplest formHence our supposition is wrong∴ √2 is irrational number. You will find your answer in maths NCERT Book........It is very... difficult to type here Yup u will find answer in ur ncert book..... It is difficult to type here.... Bohat bada h..yaha nhi bta skte...ncert me dekho Refer to your ncert book.. |
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| 36872. |
what is irrational no. |
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Answer» A number which cannot be written in the form , where a and b are integers and b ≠ 0 is called a irrational number. Irrational numbers which have non-terminating and non-repeating decimal representation. The sum or difference of a two irrational numbers is also rational or an irrational number. Irrational number are those number that are not written in the form of p by q ,where q not equal to zero. |
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| 36873. |
13113111311113 is................. |
| Answer» It is a non terminating repeating decimal expansion | |
| 36874. |
Show that every integer is polygon |
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| 36875. |
In 3rd chapter of Exercise 3.1 (1st ) sumanswer is what and what are their agesOf intersecting point |
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| 36876. |
A lemma is an axiom used to preving |
| Answer» Other statements | |
| 36877. |
√5 is ratinal |
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Answer» Yes by contradiction method keep √5 as rational and a/b are co primes Yes |
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| 36878. |
Identify √49/√147 as rational or irrational |
| Answer» √49/√147\xa0=7/√49x3=7/7√3=1/√3=√3/3 is a irrational( as\xa0√3 is irrational) | |
| 36879. |
Paper kaisa gaya भाइयों / बहनो |
| Answer» Mast | |
| 36880. |
Find circumference of circle whose area is 6.16cm² |
| Answer» πr^²=6.16 is givenr^²/7=6.16/22r^²=(0.28)(7)r^²=1.96r=1.4So r=1.4Circumference of circle=2πr=(2)(22/7)(1.4)=(44)(0.2)=8.8 | |
| 36881. |
Volume calculations |
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| 36882. |
Please solve 3x square+21x+18 |
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| 36883. |
If 2 is 3 then 4 is ? |
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Answer» Good answer 5 |
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| 36884. |
12xsquare-4+19x |
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| 36885. |
What is the HCF of 81 and 237 |
| Answer» By Euclid\'s Division Algorithm,237 = 81(2) + (75)81 = 75(1) + (6)75 = 6(12) + (3)6 = 3(2) + (0)Hcf = 3\xa0 | |
| 36886. |
Show that one and only one out of n( n+1) and (x+2) is divisible by 3 |
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Answer» We applied Euclid Division algorithm on n and 3.a = bq +r on putting a = n and b = 3n = 3q +r , 0 | |
| 36887. |
Any one here online |
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| 36888. |
Check whether 6n can end with the digit 0 for any natural number N |
| Answer» If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5Prime factorisation of 6n = (2 ×3)nIt can be observed that 5 is not in the prime factorisation of 6n.Hence, for any value of n, 6n will not be divisible by 5.Therefore, 6n cannot end with the digit 0 for any natural number n. | |
| 36889. |
How to find hcf please 1quesrion solved me |
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| 36890. |
Show that the sqare of any positive integers cannot be of the form 6m+2 or 6m+5 for any integer m |
| Answer» Let a be the positive integer and b = 6.Then, by Euclid’s algorithm, a = 6q + r for some integer q ≥ 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 5.So,\xa0a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5.(6q)2 = 36q2 = 6(6q2)= 6m, where m is any integer.(6q + 1)2 = 36q2 + 12q + 1= 6(6q2 + 2q) + 1= 6m + 1, where m is any integer.(6q + 2)2 = 36q2 + 24q + 4= 6(6q2 + 4q) + 4= 6m + 4, where m is any integer.(6q + 3)2 = 36q2 + 36q + 9= 6(6q2 + 6q + 1) + 3= 6m + 3, where m is any integer.(6q + 4)2 = 36q2 + 48q + 16= 6(6q2 + 7q + 2) + 4= 6m + 4, where m is any integer.(6q + 5)2 = 36q2 + 60q + 25= 6(6q2 + 10q + 4) + 1= 6m + 1, where m is any integer.Hence, The square of any positive integer is of the form 6m, 6m + 1, 6m + 3, 6m + 4 and cannot be of the form 6m + 2 or 6m + 5 for any integer m. | |
| 36891. |
Use Euclid division lemma find which of the following pairs are coprime:1. 160, 247 2. 231, 396 |
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| 36892. |
given triangle ABC is similar to triangle PQR, if AB/PQ= 1/3, then find ar(ABC)/ar(PQR) |
| Answer» ΔABC\xa0~ ΔPQRAB / PQ = BC / QR = AC / PR = 1 / 3and AB / PQ = BC / PQ = AC / PR = 1 / 3ar\xa0ΔABC / ar\xa0ΔPQR = (1 /\xa03)2\xa0= 1 / 9ar\xa0ΔABC / ar\xa0ΔPQR = 1 / 9 | |
| 36893. |
How many two digit numbers are divisible by 3 |
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Answer» Let two digit numbers are 10, 11, 12 , -------------99which are divisible by 3 are 12, 15 ...............,99.Here a = 12 and common dofference = 3 , an\xa0= l = 99nth\xa0term (tn) = a + ( n - 1) d⇒ 99 = 12 + ( n - 1) 3⇒ (n - 1)3 = 99 - 12 = 87⇒ ( n -1) = 87 / 3⇒ n - 1 = 29∴ n = 30∴ 30 two digit numbers which are divisible by 3. 33 |
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| 36894. |
Find quadratic polynomial whose zeros are 7+root3 and 7-root3 |
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Answer» Let the zeroes of the polynomial be a and b.Given Zeroes of the Quadratic polynomial are\xa0Therefore the required quadratic polynomial is:\xa0 x^2-14x+46 |
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| 36895. |
1 upon root 2 |
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Answer» 1\\√2*√2/√2. Answer √2/2 Root2 upon 2 |
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| 36896. |
Stata euclids division lemma |
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Answer» For a pair of given positive integers ‘a’ and ‘b’, there exist unique integers ‘q’ and ‘r’ such thata = bq+r , where 0 ≤r < bExplanation:Thus, for any pair of two positive integers a and b; the relationa = bq + r , where 0≤r |
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| 36897. |
If tan x =3cot x . Find the value of x=? |
| Answer» tanx = 3cotxBut cotx = 1/tanx⇒ tanx = 3/tanxMultiply both side by tanx=> tan 2x = 3tanx = ±√3X = 60 when tan x is +3X = 120 when tan x is -3\xa0 | |
| 36898. |
Show that 3underoot2 is irrational. |
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Answer» Bhai app me h...example section me.. Book me example me diya hai bro bhai abhi bhi Maths.. |
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| 36899. |
The decimal representation of 136/1400 will be ?? |
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| 36900. |
Find the roots of the equation 5x square - 6x - 2=0 by the method of completing the squate |
| Answer» 5x2 - 6x - 2 = 0Multiplying the above equation by 1/5{tex} \\Rightarrow {x^2} - \\frac{6}{5}x - \\frac{2}{5} = 0{/tex}{tex}\\Rightarrow x ^ { 2 } - \\frac { 6 } { 5 } x + \\left( \\frac { 3 } { 5 } \\right) ^ { 2 } - \\left( \\frac { 3 } { 5 } \\right) ^ { 2 } - \\frac { 2 } { 5 } = 0{/tex}{tex}\\Rightarrow \\left( x - \\frac { 3 } { 5 } \\right) ^ { 2 } = \\frac { 9 } { 25 } + \\frac { 2 } { 5 }{/tex}{tex}\\Rightarrow \\left( x - \\frac { 3 } { 5 } \\right) ^ { 2 } = \\frac { 9 + 10 } { 25 }{/tex}{tex}\\Rightarrow \\left( x - \\frac { 3 } { 5 } \\right) ^ { 2 } = \\frac { 19 } { 25 }{/tex}{tex}\\Rightarrow x - \\frac { 3 } { 5 } = \\pm \\frac { \\sqrt { 19 } } { 5 }{/tex}{tex}\\Rightarrow x = \\frac { 3 } { 5 } \\pm \\frac { \\sqrt { 19 } } { 5 }{/tex}{tex}\\Rightarrow x = \\frac { 3 + \\sqrt { 19 } } { 5 } \\text { or } x = \\frac { 3 - \\sqrt { 19 } } { 5 }{/tex} | |