InterviewSolution
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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.
| 43301. |
two dice are rolled. What are the possible outcomes |
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Answer» There are total 36 possible outcomes(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)Note that there are 36 possibilities for (a,b). This total number of possibilities can be obtained from the multiplication 6 Probability for rolling two dice with the six sided dots such as 1, 2, 3, 4, 5 and 6 dots in each die. |
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| 43302. |
Hcf &lcm of 2no is 9 & 459 respectivly if one no is 27find the another no |
| Answer» As we know LCM(a,b)×HCF(a,b)=a×b9×459=27×bb=9×459/27b= 459/3=153 Other number is 153. | |
| 43303. |
Math cbse now solutions |
| Answer» Bhai we r not ur Google or elexa? | |
| 43304. |
Ontain all the zero of p(x)2x4+x3-14x2-19x-6 if two of its zeroes are -2 and -1 |
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| 43305. |
How many terms are there in the AP 41,38,35,...,8? |
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Answer» An= a + (n-1)dwhere, a= 41 and d = a2 -a1=38-41=-3now, 8 = 41 + (n-1)-38-41=-3n+3-33 = -3n+3-33-3 =-3n-36=-3nn =12Answer: 12 terms. An= a + (n-1)dwhere, a= 41 and d = a2 -a1=38-41=-3now, 8 = 41 + (n-1)-38-41=-3n+3-33 = -3n+3-33-3 =-3n-36=-3nn =12therefore there are 12 terms. |
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| 43306. |
(x+4y)/(x-5y) |
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| 43307. |
If d is the HCF of 23 and 31find the x,y satisfying d=23x+31y |
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| 43308. |
Find the 20th term from the last term of the ap:3, 8,13.........,253 |
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Answer» Given AP is 3,8,13........253 first term a=3 an=253 common difference =5According to Question Let d=-5 a=253 an=3 n=20a+(n-1)d =an =>253 +(n-1)-5 =3=>(n-1)-5=-230 =>(n-1)=-250 /-5=>n-1=50=>n=50+1=>n=51 Given series is 3, 8, 13, ...... 253here, first term , a = 3common difference , d = 5Let us find total number of terms at first.use\xa0=> 253 = 3 + (n - 1)5=> 250 = 5(n - 1)=> n - 1 = 50=> n = 51so, there are 51 terms in given series.now we know, mth term from last =last term - mth term + 1so, 20th term from last = 51 - 20 + 1 = 32hence, 20th from last = 32th term from firstuse ,\xa0T32= a+(32 -1)d= 3 + 31 × 5 = 3 + 155 = 158hence , 20th term from last = 158\xa0 |
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| 43309. |
Plz see q3 from ch 3 and ex 3.3 |
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Answer» ii) one angle is 81°and other angle is 99° ..... iii) cost of one bat is =Rs. 500 and cost of another ball is= Rs. 50 ..,.. iv) amount paid for traveling 25 km in Rs. 255 ...... v) 7/9(x=7,y=9) .,...... vi) x=40,y=10 i) = one number is 13 and other number is 39 |
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| 43310. |
8.3 ex solution |
| Answer» Available in this app ? | |
| 43311. |
8.1\'s question no 7 |
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| 43312. |
Fan ke niche bathi hu fir bhi grmi lg rhi h |
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Answer» Ache Hiii girls |
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| 43313. |
Gye kya kushi aap?.....? |
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Answer» Hnn bhut grmi h........?? Hlo Hi sona ? Khushi* |
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| 43314. |
Solve for x 1by x+1 + 2byx+2 + 4 byx+4 where x is not equal to-1,-2,-4 |
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| 43315. |
What is the meaning of quadratic equations |
| Answer» A\xa0quadratic\xa0equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is ax² + bx + c = 0 with a, b, and c being constants, or numerical coefficients, and x is an unknown variable. One absolute rule is that the first constant "a" cannot be a zero. | |
| 43316. |
Why the number of zeros of a polynomial depends upon its degree??? |
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| 43317. |
Vgm to all my dear friends....... Th ?? |
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Answer» Boycott attitude girl Gm Gm |
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| 43318. |
Find the zeros of 2 x square - 5 x + 3 |
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Answer» Zero of x=1 and x=3/2 2x2 -5x + 3= 2x2 - 2x -3 x + 3= 2x(x - 1) - 3 ( x - 1)= (x -1) (2x - 3) |
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| 43319. |
Divided x cube minus 3 X square + 3 x minus 5 by x square minus x + 1 find the quotient and reminder |
| Answer» x^3-3x^2+3x-5 by x^2-x+1→Quotient=x-2 & remainder=-3 | |
| 43320. |
Difference between consistent or inconsistent?. |
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Answer» I am not a teacher , i am a student Thanks teacher. Plz tell answer..... I hv to do my wrk also .today is my test of this chapter. A consistent system of equations has at least one solution, and an inconsistent system has no solution |
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| 43321. |
(a-b) x+(a+b)y=a2-2ab-b2(a+b)(x+y)=a2+b2 |
| Answer» Given that (a-b) x + (a + b)y = a2 - 2ab - b2 -------------(1)(a+b) x+ (a+b)y = a2 + b2 -------------(2)\xa0Substracting (2) from (1) we get-2bx = -2ab=2b2 x = a + bSubsitute x in equation (2) we get y = - 2ab / ( a + b). | |
| 43322. |
15x²-41x+14=0 factorisation |
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Answer» Yes you solve write Given :\xa015x² - 41x + 14 = 0\xa0 15x2\xa0- 35x\xa0-\xa06x - 14 = 0 5x (3x-7) + 2\xa0(3x-7) = 0 (5x-2) = 0 (3x-7) = 0x = 2/ 5orx = 7/3 |
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| 43323. |
2x-5 - 3 by 4 is equal to X by 2 - 5 by 6 by multiple method |
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| 43324. |
Hii priyanshu i am neha |
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Answer» Ok , no problem I am fine and what about u Call me..... i feel bore that why i will write this. How are you Ooooo |
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| 43325. |
How many three digit numbers are divisible by seven |
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Answer» The first 3-digit number which is divisible by 7 is 105The last 3-digit number which is divisible by 7 is 994The list of 3-digit numbers divisible by 7 are105, 112, 119,…..994 which forms an A.PNow using the A. P Formula,T(n) = a + (n – 1)dWherea = 105d = 7T(n) = 994994 = 105 + (n – 1)7889 = 7n – 77n = 896n = 128∴ There are 128 3-digits number which are divisible by 7. 128 H |
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| 43326. |
Kvhxvlj FM vkchdaeia hmm nlj |
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Answer» Bhai apki alien language hame samaj nahi a rahi he toh plz aise language use kare Jo hame samaj aye Haven\'t you heard of english? Ky hai ye ? |
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| 43327. |
My name is ghanshyam thakkar |
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Answer» thanks pranali Toh kya kare And i am a human So , what |
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| 43328. |
The positive roots of the equation x²-4=0 |
| Answer» x²-4=0x2 - (2)2 = 0(x - 2) (x + 2) =0\xa0x - 2 = 0 and x + 2 = 0x = 2 and x = -2 | |
| 43329. |
Jaldy bolo sister yes or not i want to chat u in whatappp |
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Answer» Ok by Morning mai Tomorrow aap massage kar na ok bay sister. Thanks Ok |
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| 43330. |
Theorem 6.8 proof |
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Answer» I don\'t have any idea bro Mujhe nhi pta |
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| 43331. |
What is the meaning of do all the words problem of ch 3 |
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| 43332. |
2x+4y=10and3x+6y=12 give graphically solution |
| Answer» Given equations are 2x+4y=10 and 3x+6y=12now in these two equations takea1 =2 and a2 = 3b1 = 4 and b2 = 6c1 =10 and c2 =12by a1/a2 =2/3b1 /b2 = 4/6 =2/3c1/c2 = 10/12 =5/6when a1/a2 = b1/b2 ≠c1/c2 the two equations have no solution | |
| 43333. |
Optional ex.3.7 |
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Answer» Kon sa question Questions |
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| 43334. |
3-(+8)⅞ |
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| 43335. |
CosA/1-TanA+sinA/1-cotA=cosA+sinA |
| Answer» LHS=CosA/1-tanA+sinA/1-cotA→CosA/1-sinA/cosA+sinA/1-cosA/sinA→cosA/cosA-sinA/cosA+sinA/sinA-cosA/sinA→cos^2A/cosA-sinA+sin^2A/sinA-cosA→cos^2A-sin^2A/cosA-sinA→(cosA+sinA)(cosA-sinA)/(cosA-sinA)→cosA+sinA(HENCE PROVED) | |
| 43336. |
12 /70 |
| Answer» 0.17142857142 | |
| 43337. |
What is the fundamental theorem of arithmetic with example |
| Answer» Fundamental Theorem of Arithmetic:Fundamental Theorem of Arithmetic states that every composite number greater than 1 can be expressed or factorised as a unique product of prime numbers except in the order of the prime factors.We can write the prime factorisation of a number in the form of powers of its prime factors.By expressing any two numbers as their prime factors, their highest common factor (HCF) and lowest common multiple (LCM) can be easily calculated.The HCF of two numbers is equal to the product of the terms containing the least powers of common prime factors of the two numbers.The LCM of two numbers is equal to the product of the terms containing the greatest powers of all prime factors of the two numbers. | |
| 43338. |
If (1,2)is solution of equation 3x+Ky=15 then the value of k is? |
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Answer» 3x+Ky=153(1) + k (2) =153 + 2k = 152k = 15 - 3\xa02k = 12k = 12/2k = 6 6 |
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| 43339. |
10th term of AP1,4,7,10. |
| Answer» 10th term in given A.P = 28Step-by-step explanation:Given\xa0A.P\xa0:\xa01,4,7,10,\xa0...First\xa0term\xa0(a)\xa0=\xa01,n\xa0=\xa010,Therefore,10th\xa0term\xa0in\xa0given\xa0A.P\xa0=\xa028 | |
| 43340. |
Find the quadriatic polynomial.1/4,-1 |
| Answer» Let quadratic polynomial be\xa0Let\xa0α\xa0and\xa0β\xa0are two zeroes of above quadratic polynomial.α+β\xa0=\xa0\xa0=\xa0α\xa0×\xa0β\xa0= -1\xa0\xa0\xa0=\xa0\xa0Quadratic polynomial which satisfies above conditions =\xa0 | |
| 43341. |
Ur Instagram id? { O4 GIRLS} ? |
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| 43342. |
Gujffi mkf oh bfj |
| Answer» Kya hai bhai ye | |
| 43343. |
Find thw 21 th term of AP whose 1 st term are -3 and 4 |
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Answer» Bhai aapko answer nhi aata na to koi baat nhi , but plz faltu bakvaas na kr and ?ayese emoji na bhej Dost mai question nahi samajh pa rahi hoon sorry but kya tum bata sakti ho sana kyu nahi aa rahi aap par ?????? ????? Koi to answer doo mere question ka |
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| 43344. |
510 , 92 hcf lcm |
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Answer» 3:4 510 = 2 × 5 × 3 × 1792 = 2 × 2 × 23LCM(510,92) = Product of the greatest power of each factor=2×2×3×5×17×23=23,460HCF(510,92)=Product of lowest power if each common factor=2\xa0 |
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| 43345. |
Given 15 cot A=8, find sin A and sec A. |
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| 43346. |
what is equally likely outcomes |
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Answer» For many experiements it is reasonable to assume that all possible outcomes are equally likely.For example:\tDraw a random sample of size n from a population. The assumption that the sample is drawn at random means that all samples of size n have an equal chance of being chosen (much of statistical analysis depends on the assumption that samples are chosen randomly).\tFlip a fair coin n times and observe the sequence of heads and tails that results.\tRoll n dice, die 1, die 2, die 3, . . . , die n, and observe the ordered sequence of numbers on the uppermost faces. The outcomes of a sample space are called equally likely if all of them have the same chance of occurring |
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| 43347. |
4x²-4a²x+(a⁴-b⁴)=0 |
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| 43348. |
Exercise 3.5 |
| Answer» Available in this app | |
| 43349. |
Factorise:Q n^2-2√5-15 |
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| 43350. |
Prove that 3 is irrational |
| Answer» Let us assume that √3 is a rational number.then, as we know a rational number should be in the form of p/qwhere p and q are co- prime number.So,√3 = p/q { where p and q are co- prime}√3q = pNow, by squaring both the sidewe get,(√3q)² = p²3q² = p² ........ ( i )So,if 3 is the factor of p²then, 3 is also a factor of p ..... ( ii )=> Let p = 3m { where m is any integer }squaring both sidesp² = (3m)²p² = 9m²putting the value of p² in equation ( i )3q² = p²3q² = 9m²q² = 3m²So,if 3 is factor of q²then, 3 is also factor of qSince3 is factor of p & q bothSo, our assumption that p & q are co- prime is wronghence,. √3 is an irrational number | |