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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.
| 36901. |
7(x-2)=5 |
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| 36902. |
Write the first five term of an=3n+2 in sequence whose nth term are |
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| 36903. |
26 and 91 LCM and HCF find |
| Answer» 26 and 9126= 2 × 1391 = 7 × 13HCF = 13LCM = 2 × 7 × 13 = 182Product of two numbers 26 and 91 = 26 × 91 = 2366HCF × LCM = 13 × 182 = 2366Hence, product of two numbers = HCF × LCM | |
| 36904. |
Find zeros |
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Answer» Can be given when you will complete the question Complete answer Complete your question first amd then ask. What Of |
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| 36905. |
prove rhat product of three consecutive integers is divisible by 6... |
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Answer» i dont understand... Let three consecutive positive integers be, n, n + 1 and n + 2.When a number is divided by 3, the remainder obtained is either 0 or 1 or 2.\xa0∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.If n = 3p, then n is divisible by 3.If n = 3p + 1, ⇒\xa0n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.If n = 3p + 2, ⇒\xa0n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.\xa0⇒ n (n + 1) (n + 2) is divisible by 3.Similarly, when a number is divided 2, the remainder obtained is 0 or 1.∴ n = 2q or 2q + 1, where q is some integer.If n = 2q\xa0⇒\xa0n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2.If n = 2q + 1 ⇒\xa0n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2. ⇒ n (n + 1) (n + 2) is divisible by 2.Hence n (n + 1) (n + 2) is divisible by 2 and 3.∴ n (n + 1) (n + 2) is divisible by 6.\xa0
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| 36906. |
Sec²theta+cosec²theta=sec²theta*cosec²theta |
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Answer» bhai answer kaha hai.....nishant jha....paagal saala cosec²theta+sec²theta=cosec²theta×sec²theta |
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| 36907. |
What is trignometry? |
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Answer» Trigonometry (from Greek trigōnon, "triangle" and metron, "measure") is a branch of mathematics that studies relationships involving lengths and angles of triangles.\xa0 Sin thetha,cos thetha....??? |
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| 36908. |
3x+2y=36x+7y=16Chapter 3 pair of linear equ. In two valuable |
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| 36909. |
√5\'0 |
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| 36910. |
:Romeo Akbar Walter :RAW |
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Answer» ... Baad me baat krungi market ghum rhi hu? Okk ? Movie What? |
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| 36911. |
√5+√6 is irrational number |
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| 36912. |
Under root 3 is irrational.prove that |
| Answer» Let us assume that √3 is a rational number.That is, we can find integers\xa0a\xa0and\xa0b\xa0(≠ 0) such that √3 = (a/b)Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.√3b = a⇒\xa03b2=a2\xa0(Squaring on both sides) → (1)Therefore, a2\xa0is divisible by 3Hence ‘a’ is also divisible by 3.So, we can write a = 3c for some integer c.Equation (1) becomes,3b2\xa0=(3c)2⇒\xa03b2\xa0= 9c2∴ b2\xa0= 3c2This means that b2\xa0is divisible by 3, and so b is also divisible by 3.Therefore, a and b have at least 3 as a common factor.But this contradicts the fact that a and b are coprime.This contradiction has arisen because of our incorrect assumption that √3 is rational.So, we conclude that √3 is irrational. | |
| 36913. |
Find hcf of 85,90 |
| Answer» HCF = 5 | |
| 36914. |
What is portion of SA1 in 2019-2020 |
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| 36915. |
Can a number 10^n end with digit 0 |
| Answer» ya | |
| 36916. |
Answer the lesson 1 theorem 1.5,1.6,1.7 |
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| 36917. |
Verify the commutative property of multiplication for the following rational number.(a) 2 and 2÷7 |
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| 36918. |
6x-3y+10=0 |
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| 36919. |
Solve for x and y 5/x-1 + 1/y-2 =06/x-1 - 3/y-2=1 |
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| 36920. |
how can i solve ex - 8.4 easily |
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| 36921. |
Use Euclid\'s division lemma to show that the cube of any positive integer is the form 9m,9m+1or9m+8 |
| Answer» Let x be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.So, we have the following cases :Case I : When x = 3q.then, x3 = (3q)3 = 27q3 = 9 (3q3) = 9m, where m = 3q3.Case II : When x = 3q + 1then, x3 = (3q + 1)3= 27q3 + 27q2 + 9q + 1= 9 q (3q2 + 3q + 1) + 1= 9m + 1, where m = q (3q2 + 3q + 1)Case III. When x = 3q + 2then, x3 = (3q + 2)3= 27 q3 + 54q2 + 36q + 8= 9q (3q2 + 6q + 4) + 8= 9 m + 8, where m = q (3q2 + 6q + 4)Hence, x3 is either of the form 9 m or 9 m + 1 or, 9 m + 8. | |
| 36922. |
Vertex 8 |
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| 36923. |
pie is irrational but 22/7is rational? Why |
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| 36924. |
Prove that sum of three consecutive number is divisible by 6? |
| Answer» Let three consecutive positive integers be, n, n + 1 and n + 2.\'When a number is divided by 3, the remainder obtained is either 0 or 1 or 2.\xa0∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.If n = 3p, then n is divisible by 3.If n = 3p + 1, ⇒\xa0n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.If n = 3p + 2, ⇒\xa0n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.⇒ n (n + 1) (n + 2) is divisible by 3.Similarly, when a number is divided 2, the remainder obtained is 0 or 1.∴ n = 2q or 2q + 1, where q is some integer.If n = 2q\xa0⇒\xa0n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2.If n = 2q + 1 ⇒\xa0n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2.⇒ n (n + 1) (n + 2) is divisible by 2.Hence n (n + 1) (n + 2) is divisible by 2 and 3.∴ n (n + 1) (n + 2) is divisible by 6. | |
| 36925. |
How many zeroes do biquadratic polynomial has? |
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Answer» There r 2 zeroes.... 2 zeroes... Bbiquadratic polynomial has 2 zeroes . |
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| 36926. |
3+589 |
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Answer» 592 592 592 |
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| 36927. |
Find the large number which divides 615 and 963 leaving remainder 6 in each cash |
| Answer» The required number when divides 615 and 963, leaves remainder 6, this means 615 - 6 = 609 and 963 - 6 = 957 are completely divisible by the number.Therefore,The required number = H.C.F. of 609 and 957.By applying Euclid’s division lemma957 = 609 {tex}\\times{/tex}\xa01+ 348609 = 348 {tex}\\times{/tex}\xa01 + 261348 = 216 {tex}\\times{/tex}\xa01 + 87261 = 87 {tex}\\times{/tex}\xa03 + 0.Therefore, H.C.F. of 957 and 609 is = 87.Hence, the largest number which divides 615 and 963 leaving remainder 6 in each case is 87. | |
| 36928. |
If Sin alpha=1/3then evaluate tan alpha.sec alpha+cos alpha. cosec alpha |
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| 36929. |
Prove that 3-2√7 is irrational number |
| Answer» Let if possible 3-2√7 is rationalThese exists two co prime integers p and q such that 3-2√7= p/q2√7=p/q+32√7= p-3q/q√7=p-3q/2q√7=integer/integerWhich will result in rational no.√7 is rationalBut√7 is irrational which is contradiction So,our assumption is wrong Hence 3-2√7 is irrational | |
| 36930. |
What is iota(i) |
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Answer» Which comes under imaginary number(unreal) √-1 |
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| 36931. |
Ncert maths book ka exercise 1.4 |
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| 36932. |
If the h.c.f of a,b=1800 find the product of a,b |
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Answer» Yha koi chatting nhi ho rhi mai apne douts hi clear kr rhi thi jo log chatting krr rhe hai please use bolo l think now you understand I think, Something is missing in this question. What chat is going on here ???. Yeh hamare doubts ke liye hai , so please don\'t do your casual talks here !! |
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| 36933. |
( a+b)ka hole square= ? |
| Answer» (a+b)²=a²+2ab+b² | |
| 36934. |
Any chapter in 12 maths. Which i can do without teachers help easily except matricess |
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Answer» I\'m student\'s of class 10th but likes to study 12th maths I\'m happy people like me exist on earth, I also have such like situation! Integral , differential equation you can\'t do without teachers help! ???? Probability, Determinants, Relations and Functions |
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| 36935. |
find zero of the polynomial x+7x+12 |
| Answer» The zeros of f(x) are given by f(x) = 0{tex} \\Rightarrow {/tex}x2\xa0+ 7x + 12 = 0{tex}\\Rightarrow {/tex}(x + 4)(x + 3) = 0{tex}\\Rightarrow {/tex}\xa0x + 4 = 0 or, x + 3 = 0{tex}\\Rightarrow {/tex}\xa0x = -4 or, -3Thus, the zeroes\xa0of f(x) =\xa0x2 +7x + 12 are\xa0{tex}\\alpha = - 4 \\text { and } \\beta = - 3{/tex} | |
| 36936. |
What is. 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 | |
| 36937. |
How many radii will be drawn in a circle |
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Answer» Unlimited Inifite |
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| 36938. |
X+y=5 , 2x-3y=4 . Solvw in cross multiplication method |
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| 36939. |
Prove that 2-3√3 is an irrational number? |
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| 36940. |
Solve the equation by elimination method x+y=5 and2x-3y=4 |
| Answer» x\xa0+\xa0y\xa0=5 ...\xa0(i)2x\xa0–3y\xa0= 4 ...\xa0(ii)Multiplying equation\xa0(i)\xa0by\xa02, we get2x\xa0+ 2y\xa0= 10 ...\xa0(iii)2x\xa0–3y\xa0= 4 ...\xa0(ii)Subtracting equation\xa0(ii)\xa0from equation\xa0(iii), we get5y\xa0= 6y\xa0= 6/5Putting the value in equation\xa0(i), we getx\xa0= 5 - (6/5) = 19/5Hence,\xa0x\xa0= 19/5 and\xa0y\xa0= 6/5 | |
| 36941. |
Find the prime factorization of 1005 with method |
| Answer» 1005=5*201201=3*671005=5*3*67 | |
| 36942. |
Find the greatest number which are dividing 10669 and 2039 lives remander 8 and 7 respectively |
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| 36943. |
Find equation of polynomial qudratic whose sum of zeroes and product of zeroes are 1/4 and -1 |
| Answer» X*2-(sum)x + productX*2-(1/4)x +(-1)X*2-x-4 | |
| 36944. |
Prove that the following are irrational. 1/√2 |
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| 36945. |
Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468 |
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Answer» Last line isSmallest number which when increased by 17 is exactly divisible by both 520 and 468 = 4680 - 17 = 4663. The given numbers are 520 and 468.The smallest number which when increased by 17 is exactly divisible by both 520 and 468 is obtained by subtracting 17 from the LCM of 520 and 468.Prime factorisation of 520 = 2 × 2 × 2 × 5 × 13Prime factorisation of 468 = 2 × 2 × 3 × 3 × 13LCM of 520 and 468 = 2 × 2 × 2 × 3 × 3 × 5 × 13 = 4680.Smallest number which when increased by 17 is exactly divisible by both 520 and 468 = 4680 × 17 = 4663. |
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| 36946. |
what is mathematics.... |
| Answer» Mathematics is the science that deals with the logic of shape, quantity and arrangement. Math is all around us, in everything we do. It is the building block for everything in our daily lives, including mobile devices, architecture (ancient and modern), art, money, engineering, and even sports. | |
| 36947. |
Express 0.6 as a rational number in the simplest form It\'s urgent |
| Answer» 3/5 | |
| 36948. |
Find the greatest number which exactly divides 280 and 1245 leaving remianders 4and 3 respectively. |
| Answer» To find the largest number which exactly divides 280 and 1245 living remainders 4 and 3 respectively, we subtract 4 and 3 from 280 and 1245.280 - 4 = 2761245 - 3 = 1242276 = 2 x 2 x 3 x 231242 = 2 x 3 x 3 x 3 x 23HCF = 2 x 3 x 23 = 138Therefore, the largest number which exactly devides 280 and 1245 living remainders 4 and 3 respectively is 138. | |
| 36949. |
Show that there are infinitely positive primes. |
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| 36950. |
49÷3×2+8= |
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Answer» 40.66 BODMAS ke tarike se 40.66 hi hai answer. 40.66 |
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