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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.
| 37351. |
Show that an odd integer is of the form 5q, 5q+2 or 5q+4 ,where q is some positive integer |
| Answer» Any integer can be written in the form 5m , 5m+1, 5m+2 (where m is any integer)(5m)² = 25m² = 5 × 5m² = 5q (where q = 5m²)(5m+1)² = (5m)² + 2 ×5m × 1 + 1² [by using identity- (a+b)²= (a²+ 2ab +b² )] = 25m² + 10 m + 1 = 5 (5m²+ 2m) +1 = 5q +1 ( where q= 5m²+2m)(5m+2)²= (5m)² + 2× 5m × 2 +2² [by using identity- (a+b)²= (a²+ 2ab +b² )] = 25m² + 20 m +4 = 5 (5m²+4m ) + 4 = 5q+4 (where q= 5m²+4m)Hence proved | |
| 37352. |
for what value of n where n is natural number is(19)^n is divisible by 9\u200b |
| Answer» n=8 | |
| 37353. |
What is sin,cos,and tan? |
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Answer» sin ,cos and tan isa thheta part of trigonometry. बदमाश Please give correct answer Rakshash..?? |
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| 37354. |
If (x+a)is a factor of the polynomialxsquare+px+q and x square +mx+n then prove that a=n-q/m-p |
| Answer» Since x + a is a factor of x2+ px + qthen (-a)2\xa0- pa + q = 0{tex}\\Rightarrow{/tex}a2 = pa -q ..........(i)also (x + a) is a factor of x2+ mx + n then we get(-a )2\xa0- am + n = 0{tex}\\Rightarrow{/tex} a2 - am + n= 0{tex}\\Rightarrow{/tex}a2\xa0=\xa0am - n........(ii)From eq(i) and (ii), we getam - n = ap - q{tex}\\Rightarrow{/tex}am - ap = n - qHence, a =\xa0{tex}\\left[ \\frac { n - q } { m - p } \\right]{/tex} | |
| 37355. |
If a zero of the quadriccpoly. 4xsqaure -8kx -9 is negative of other find the value of k |
| Answer» Let, the two zeroes of the polynomial f(x) = 4x2 - 8kx - 9 be α and -α.Comparing f(x) = 4x2\xa0- 8kx - 9 with ax2+bx+c we geta = 4; b = -8k and c = -9.Sum of the zeroes = α + (-α) ={tex}-\\frac ba=\\;-\\frac{-8k}4 {/tex}0 = 2kk = 0 | |
| 37356. |
Find roots of x×x+x-p(p+1) |
| Answer» {tex}x^2 + x - p(p+1) = 0{/tex}{tex}x^2 + (p+1)x - px - p(p+1) = 0{/tex}{tex}x(x+ p + 1) -p(x + p + 1) = 0{/tex}(x + p + 1)(x - p) = 0\xa0x = - p - 1, p | |
| 37357. |
If the roots of the equation a(b-c)x^2+b(c-a)x+c(a-b)x+c(a-b) are equal then show that 2/b =1/a+1/c |
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| 37358. |
How can we know that In such what problem we so LCM OR HCF? |
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Answer» When question ask you maximum and minimum or largest or smallest number then we find HCF.When question ask you meeting point of ride (run)* then we find LCM. Whenever you are required to find a number that completely divides more than one number, you have to find HCF.Whenever you are required to find a number that is divisible by more than one number, you have to find LCM. |
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| 37359. |
In the following real number which one is non terminating repeating decimal expansion |
| Answer» Where are the following real no.???? | |
| 37360. |
Arthematic progression exercise 1.1 solutions |
| Answer» Check NCERT solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 37361. |
Root 5 irrational |
| Answer» Let\xa0{tex}\\sqrt 5{/tex}\xa0is a rational number.{tex}\\sqrt { 5 } = \\frac { a } { b }{/tex}(a, b are co-primes and b{tex}\\neq{/tex}0)or,\xa0{tex}a = b \\sqrt { 5 }{/tex}On squaring both the sides, we get\xa0a2=5b2 ---------------------------------(1)Hence 5 is a factor of a2so 5 is a factor of aLet a = 5c, (c is some integer){tex}\\therefore{/tex}\xa0a2\xa0= 25c2\xa0From equation(1) putting the value of a2or, 5b2\xa0= 25c2or b2=5c2so 5 is a\xa0factor of b2or 5 is a factor of bHence 5 is a common factor of a and bBut this contradicts the fact that a and b are co-primes.This is because we assumed that\xa0{tex}\\sqrt 5{/tex}\xa0is rational{tex}\\therefore{/tex}\xa0{tex}\\sqrt 5{/tex}\xa0is irrational. | |
| 37362. |
N³-N divisible by6? |
| Answer» n3\xa0- n = n (n2\xa0- 1) = n (n - 1) (n + 1)\xa0Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.∴ 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, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.If n = 3p + 2, then n + 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 + 1 is always divisible by 3.⇒ n (n – 1) (n + 1) is divisible by 3.\xa0Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.∴ n = 2q or 2q + 1, where q is some integer.If n = 2q, then n is divisible by 2.If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 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 + 1 is always divisible by 2.⇒ n (n – 1) (n + 1) is divisible by 2.Since, n (n – 1) (n + 1) is divisible by 2 and 3.∴ n (n-1) (n+1) = n3\xa0- n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6) | |
| 37363. |
Who are tenthers here? |
| Answer» मैं हूं रानी मिश्रा दसवीं कक्षा में पढ़ती हूं | |
| 37364. |
Draw the graph of the following pair of linear equations x+3y=6 and 2x-3y=12 |
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Answer» Solution:Plotting the above points and drawing lines joining them, we get the graphs of the equations x + 3y = 6 and 2x - 3y = 12 3y=6-xy=6-x/3 So by the using form then you find the value of x and then you have value of x to put the value in given equation and you have value of y |
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| 37365. |
If -1&-2 are the zeroes of the polinomial xcube -4xsqaure-7x+10 ,find its third zero |
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| 37366. |
Find the HCF of 243 and 711 and express it as a linear combination of 243 and 711 |
| Answer» 711 = 243*2+225 243 = 225*1+18225 = 18*12+918 = 9*2+0HCF = 99 = 225-18*12 = 225-(243-225*1)*12 = 225-243*12+225 *12 = 225*13-243*12 = (711-243*2)*13-243*12 = 711*13-243*26-243*12 = 711*13-243*38 = 711*13+243*(-38) = 711m+243n = m=13 and n=-38 | |
| 37367. |
X+y=35 4x-5y=5 |
| Answer» x + y = 35 ............(1)4x - 5y = 5 ...............(2)From (1) , x = 35 - yPutting x = 35 - y in (2) we get,4 ( 35 - y)- 5y = 5 140- 4y - 5y = 5 - 9y = 5 - 140 y = -135/-9 y = 15 x = 35 - 15 = 20 | |
| 37368. |
Q .no. 15 ex. 9.1 |
| Answer» Check NCERT solutions here : https://mycbseguide.com/ncert-solutions.html | |
| 37369. |
Solve it (a-b)x+(a+b)y=asquare-bsqure and. (a+b)+(x+y)=a square +b square |
| Answer» x = a + b{tex}\\therefore{/tex} (a - b)(a + b) + (a + b)y = a2 - b2 - 2aba2 - b2 + (a + b)y = a2 - b2 - 2ab{tex}y = \\frac{{ - 2ab}}{{a + b}}{/tex} | |
| 37370. |
Trigonomatric identies |
| Answer» Check revision notes for the formulae :\xa0https://mycbseguide.com/cbse-revision-notes.html | |
| 37371. |
Define general lok sabha election |
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Answer» Thnx Member of lok Sabha ( house of the people) or the lower house of India is parliament and elected by being voted upon by all adult citizen of India from a set of candidates who stand in their respective constituencies. Every adult citizen of India can vote only in their constituency. |
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| 37372. |
To obtain the condition for consistency of system of linear equation in two variables. |
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| 37373. |
To draw the graph of quadratic polynomial and observe it . |
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Answer» Mera 9th m 83% aaya h 10th m to saayd 80 > hi rhega bina tution Thanx bro .Maths m kaafi tej ho hm to weak hote jaa rhe bcoz dont have tution in my village ..Or town bhi ghr se 8km dur h beech m ghana jangal ? Question:To draw the graph of a quadratic polynomial and observe(i) the shape of the curve when the co-efficient of x square is positive and (ii) same when its negativeSolution:1 ) Draw a quadratic equation when co-efficient of x2\xa0is positive\xa0Lets our quadratic equation is y = x2 ( Here co-efficient of x2\xa0is +1 )Step 1 -\xa0find several points for equation y\xa0= x2 As:\txy = x2-39-24-11002439\tStep 2 -\xa0draw these points on graph and join them by smooth curving line As:2 )\xa0Draw a quadratic equation when co-efficient of\xa0x2\xa0is negative.Let our quardetic equation is As : y = -x2( Here co-efficient of x2\xa0is -1 )Step 1 -\xa0find several points for equation\xa0y\xa0= - x2\xa0As:\txy = -x2\xa0-3\xa0- 9\xa0-2\xa0- 4\xa0- 1\xa0- 1\xa00\xa00\xa01\xa01\xa02\xa04\xa03\xa09\xa04\xa016\tStep 2 -\xa0draw these points on graph and join them by smooth curving line As: |
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| 37374. |
Explain why 17×5×11×3×2+2×11 is a coposite number |
| Answer» 17×5×11×3×2 + 2×11=2×11 × ( 17×5×3 + 1)=2×11 × ( 255+1)=22 × 256=22 ×16 ×16The number is product of more than one number so the number is composite | |
| 37375. |
Why is 2/3x-5/2 a binomial? |
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| 37376. |
Find the value of p for which graphs of 2x - py = 9 and 6x-9y=18 will be parallel |
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| 37377. |
Solve xcube +2xsquare +5x+6 |
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| 37378. |
If the product of root of x2-3 x+ k=10 is -2 what is the value of k |
| Answer» Any one can give me answer of this please | |
| 37379. |
If alpha and beta are two zeroes of x2 - 3x+ 7 then find the value of alpha-1 +beta-1 |
| Answer» Try to imagine the question in a different way.Solution:{tex}\\huge\\implies \\alpha-1+\\beta-1\\\\\\huge\\implies\\alpha+\\beta-1-1\\\\\\huge\\implies(\\alpha+\\beta)-2{/tex}{tex}\\huge\\implies Since,\\alpha+\\beta=-\\frac ba\\\\\\huge\\implies (given)\\space\\space a=1,b=-3\\\\\\huge\\implies Thus,(\\alpha+\\beta)-2\\implies-\\frac{-3}{1}-2\\\\\\huge\\implies 3-2\\implies\\boxed1{/tex} | |
| 37380. |
Square root of 588 |
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Answer» 24.668 14 root 3 24.248 |
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| 37381. |
3 +2√7 i irrational by contradictory method |
| Answer» Let us assume that 3 + 2√7 is rational3+2√7=a/b2√7=a/b-3√7=a-3b/2bSince a and b are integers where b is not equal to 0Therefore a-3b/2b is rationalSo,√7 is also rationalBut this contradicts the fact ghat √7 is irrational.Hence, 3+2√7 is irrational. Proved | |
| 37382. |
cos^2A +1/1+cot^2A =1 |
| Answer» Please learn the trigonometric formulae, this is a very simple question to ask .{tex}\\large Question: \\space To\\space prove:cos^2A +\\frac{1}{1+cot^2A} =1?{/tex}{tex}\\large{Solution\\implies LHS\\implies cos^2A +\\frac1{1+cot^2A}}{/tex}{tex}\\large{\\big(1+cot^2A = cosec^2A \\big)}{/tex}{tex}\\large{\\implies cos^2A +\\frac1{\\underline {1+cot^2A}}\\implies cos^2A+\\frac 1 { cosec^2A}}{/tex}{tex}\\large{\\big(\\frac1{cosec\\space A} = sin\\space A \\big)\\\\\\big(\\frac1{cosec^2A} = sin^2A \\big)}{/tex}{tex}\\large{\\implies cos^2A+sin^2A}{/tex}{tex}\\large{\\big(cos^2A+sin^2A=1 \\big)}{/tex}{tex}\\large{Ans\\implies \\boxed1}{/tex} | |
| 37383. |
2xsquare -5x +3=0 solve this by complitting square method |
| Answer» 2x^2-5x+3=02x^2/2-5x/2=-3/2x^2-5x/2=-3/2x^2-5x/2-5/2=-3/2+5/2(x-5/2)^2=1x-5/2=√1x=+-(1+5/2)x=3.5 or x=(-1.5)Maybe this helps... | |
| 37384. |
Hyyyy ....koi h .... ????? |
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Answer» Hii Hello Hii riya Hlo HLo riyuu |
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| 37385. |
Write the condition for parallel lines |
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Answer» It is a part of graphical method.It is inconsistent.It has no solution, so it is parallel.a1/a2 = b1/b2≠ c1/c2. a1/a2 = b1/b2 ≠ c1/c2 |
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| 37386. |
What is the underoot of 10176 |
| Answer» 100.87616170334793 | |
| 37387. |
3x-y=39x-3y=9Substitution method |
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Answer» It has infinitely many solutions...so you can not find the answer 3x-y=39x-3y=9 - + --3x+2y=-3 Hii Hlo |
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| 37388. |
Using the Formal, tan2=2tanA÷1-tansquareA, Find the value of tan 60, it being given that 30=1÷root3 |
| Answer» Given Formula : tan2A = 2 tan A / 1 - tan²A Also, tan30 = 1/√3Required to find : tan60Now, Let A = 30 , 2A = 60° tan2A = 2tanA / 1 - tan²A tan60 = 2(tan30 ) / 1 - ( tan²30)tan60 = 2(1/√3) / 1 -( 1/√3)²tan60 = 2/√3 / 1 - 1/3tan60 = 2/√3 / 2/3 tan60 = 2/√3 * 3/2 tan60 = √3 Therefore , tan60 = √3 | |
| 37389. |
Find the Lcm and Hcf of 6 by the prime factorisation method |
| Answer» 2,3 | |
| 37390. |
Find the sum of first 10 multiples of 6. |
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Answer» S=330 Step-by-step explanation: 6,12,18.......a=6d=6n=10s=10/2[2*6+(10-1)(6)]s=5[12+54]s=5[66]s=330 |
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| 37391. |
Survey of your favourite chennel of 25 different families and also make pie chart and bar graph |
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| 37392. |
Where are Activities in mathematics book |
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| 37393. |
If tan2A=cot(A-18°) where 2A is an acute angle, find the value of A. |
| Answer» A=36 | |
| 37394. |
if cot? + tan? = x and sec? – cos? = y, prove that (x2y)2/3 – (xy2)2/3 = 1. |
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| 37395. |
1_2+ |
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Answer» ,......?,??? what\'s the question bro ......??....?? What is the ques |
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| 37396. |
Ch 5 AP extra questions |
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| 37397. |
Value of log 32 base2 |
| Answer» 5 | |
| 37398. |
Find the value of log 32 with base 2 |
| Answer» 5 | |
| 37399. |
Show that. Any positive odd integer is form of 6q+1or6q+3or6q+5 whereq is some integer |
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Answer» & Let a be any positive integer and b = 6∴ by Euclid’s division lemmaa = bq + r, 0≤ r and q be any integer q ≥ 0∴ a = 6q + r,where, r = 0, 1, 2, 3, 4, 5If a is even then then remainder by division of 6 is 0,2 or 4Hence r=0,2,or 4or A is of form 6q,6q+2,6q+4As, a = 6q = 2(3q), ora = 6q + 2 = 2(3q + 1), ora = 6q + 4 = 2(3q + 2).If these 3 cases a is an even integer.but if the remainder is 1,3 or 5 then r=1,3 or 5or A is of form 6q+1,,6q+3 or,6q+5Case 1:a = 6q + 1 = 2(3q) + 1 = 2n + 1,\xa0Case 2: a = 6q + 3 = 6q + 2 + 1,= 2(3q + 1) + 1 = 2n + 1,\xa0Case 3: a = 6q + 5 = 6q + 4 + 1= 2(3q + 2) + 1 = 2n + 1This shows that odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer. A=bq+rLet b=6 and r= 0,1,2,3,4,5Putting r=0A=6qPutting r=1A=6q+1Putting r=26q+2Putting r=36q+3Putting r=46q+4Putting r=56q+5We need positive odd integer so 6q+1,6q+3, 6q+5 are the positive odd integer |
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| 37400. |
8x² |
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