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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.
| 10201. |
Find the 9th term of A.P : 3, 2.5 , 2 , ... |
| Answer» a = 3, d = 2.5 - 3 = -0.5, n = 9.an = a + (n - 1) d. a9 = a + 8d = 3 + 8(-0.5) = 3 + (-4) = -1.// | |
| 10202. |
Sn=4095, r=4, n=6, a=? |
| Answer» Sn = n/2 (a + r). 6/2 (a + 4) = 4095. 3 (a + 4) = 4095. a + 4 = 4095/3. a + 4 = 1365. a = 1361.// | |
| 10203. |
Find quotient and remainder on dividing x2 + 1 by x – 1. |
| Answer» Quotient = x+1Remainder = 2Hope u got the answer Plz reply.. | |
| 10204. |
Explain why 7×11×13+13 and 7×6×5×4×3×2×1+5 are composite |
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Answer» 7×11×13+13 = 7×11×26•26 is a composite number. So, 7×11×13+13 is a composite number. 7×6×5×4×3×2×1+5. In this, number 6,4 etc., are composite numbers. Also, +5 will become a composite number. 7×6×5×4×3×2×1+5 is a composite number. The number 7 * 11 * 13 + 13 can also be written as 13 (7 * 11 + 1). This means that the number has two other factors 13 and 78 while not considering 1 and the number itself. The case is similar for the number 7 * 6 * 5 * 4 * 3 * 2 * 1 + 5 which can also be written as 5 (7 * 6 * 4 * 3 * 2 * 1 + 1). Therefore, both these numbers are COMPOSITE in nature. |
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| 10205. |
Find the zeroes of p(x)= 2x2-x-6 and verify the relationship of zeroes with these co-efficients |
| Answer» 2x^2 - x - 6 = 0 (S = -1, P = -12, No. = -4, 3). 2x^2 - 4x + 3x - 6 = 0. 2x (x - 2) + 3 (x - 2) = 0. (2x + 3) (x - 2) = 0. Therefore, x = -3/2, 2.// | |
| 10206. |
The lcm of two number is 1200. What of the following can\'t be hcf ?A 600B 500C 400 D. 200 |
| Answer» 500 | |
| 10207. |
Solve for x 2x^-6√3-6 |
| Answer» 2x^2 - 6√3 - 6 = 0. Discriminant = b^2 - 4ac = (-6√3)^2 - 4 * -6 * 2 = 108 + 48 = 156.//Therefore, x = (-b _+_ √D)/2a = 6√3 + √156/4, 6√3 - √156/4 = 6√3 + 2√39/4, 6√3 - 2√39/4// | |
| 10208. |
8.4 ka 5(7) question solve kar do |
| Answer» | |
| 10209. |
Show that for odd positive integer to be a perfect square, it should be of the form 8k + 1 |
| Answer» | |
| 10210. |
Solve the pair of linera equation:- root2 x-root3 x=0 and root5 x +root2 y=0 by substition method |
| Answer» | |
| 10211. |
Show that any integer is of the form 5q,5q+1,5q+2,5q+3,5q+4Where q is some integer |
| Answer» | |
| 10212. |
HCF if 2,7,12 by division method |
| Answer» HCF (2, 7, 12) = 1.// | |
| 10213. |
How to prove 2+√3 is an irrational number? |
| Answer» First prove that √3 is irrational. Then, since the sum of a rational and an irrational number is always irrational, 2 + √3 will also be IRRATIONAL in nature. | |
| 10214. |
Draw the graph for the polynomial p(x) =3x-17 |
| Answer» Take p(x)=03x-17=0or, x=17/3 So x=17/3 is a solution of p(x)So draw a straight line on x axis at x=17/3. [NOTE: The graph will be not be accurate it will be approximate] | |
| 10215. |
Express 17017 as a prime factor |
| Answer» The prime factor of 17017 = 7*11*13*17 | |
| 10216. |
Find the lcm and hcf of the following integer by applying the prime factorisation method (1)12,15,21 |
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Answer» LCM(12, 15, 21) = 420 LCM (12, 15, 21) = 420. HCF (12, 15, 21) = 3. |
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| 10217. |
If the product of zeroes of the polynomial a * x ^ 2 - 6x - 6 is 4, find the value of "a". |
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Answer» p (x) = ax^2 - 6x - 6 = 0. Now, according to the relationship between zeroes and coefficients in a quadratic equations, (if α and β are the zeroes of p (x)), αβ = C/A. 4 = -6/a. Therefore, a = -6/4 = -1.5// Chapter no, 12 |
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| 10218. |
Find the HCF of two numbers experimentally based on Euclid’s division lemma. |
| Answer» Hope this helps you | |
| 10219. |
Prove that op=p/q |
| Answer» Kindly provide more clarity. Is there any figure? | |
| 10220. |
Are the following pair of linear equations consistence? Justify your answer 3/5 x-y=1/2 |
| Answer» Dk | |
| 10221. |
2xy+3xy-2ab |
| Answer» 2xy + 3xy - 2ab = 5xy - 2ab// | |
| 10222. |
Solve the following pairs of linear equation by elimination method5x-4y=-8 , 7x+6y=9 |
| Answer» [(5x - 4y = -8) * 3] + [(7x + 6y = 9)] * 2 ==> (15x - 12y = -24.) + (14x + 12y = 18.) ==> 29x = -6. ==> x = -6/29// Therefore, y = 9 - 7x /6 = 9 + 42/29 / 6 = (261 + 42/29) / 6 = 303 / 174 = 101/58// | |
| 10223. |
Find a quadratic polynomial, the sum and product of whose zeroes are – 5and 3, respectively. |
| Answer» S = -5, P = 3. f (x) = k (x^2 - sx + p) = k (x^2 + 5x + 3)// | |
| 10224. |
Assignment chapter 1 class 10 |
| Answer» 12(($ | |
| 10225. |
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. | |
| 10226. |
State and prove Basic Proportionality Theorem (BPT) |
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Answer» Proof: refer class 10 triangles chapter theorem6.1 State: if a line is drawn parallel to one side of the triangle then it intersects the other 2 sides in same ratio |
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| 10227. |
1/4(cat4 30-cosec4 60) |
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Answer» 1/4 (cot^4 30° - cosec^4 60°) = 1/4 ((√3)^4 - (2/√3)^4) = 1/4 (9 - 16/9) = 1/4 ((81 - 16)/9) = 1/4 (65/9) = 65/36// cot 30 = √3. cosec 60 = 2/√3. Cot to the Power 4hai? |
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| 10228. |
If the graph of the equation 4x + 3y = 12, has x = k – 3, and y = 5k – 1, find k. |
| Answer» 4x + 3y = 12. 4 (k - 3) + 3 (5k - 1) = 12. 4k - 12 + 15k - 3 = 12. 19k - 15 = 12. 19k = 12 + 15 = 27. k = 27/19// | |
| 10229. |
If the perimeter and the area of a circle are numerically equal then find the radius of the circle |
| Answer» 2πr = πr^2. 2r = r^2. r^2/r = 2. Therefore, r = 2 units. | |
| 10230. |
He |
| Answer» Pardon!!!! Kindly provide more clarity. | |
| 10231. |
If 1+sin^2A=3 sinA cosA, then prove that tanA=1or tanA=1/2 |
| Answer» 1 + sin^2 A = 3 sin A cos A. 1 - 3 sin A cos A + sin^2 A = 0. 1/cos^2 A - 3 sin A cos A /cos^2 A + sin^2 A/cos^2 A = 0. sec^2 A - 3 tan A + tan^2 A = 0. 1 + tan^2 A + tan^2 A - 3 tan A = 0. 2 tan^2 A - 3 tan A + 1 = 0. (S = -3, P = 2, No. = -2, -1.) 2 tan^2 A - 2 tan A - tan A + 1 = 0. 2 tan A (tan A - 1) - 1 (tan A - 1) = 0. (2 tan A - 1) (tan A - 1) = 0. Therefore, tan A = 1, 1/2// | |
| 10232. |
SinA + cosA =x, find the value of sin^6A + cos^6A |
| Answer» | |
| 10233. |
Write the zeroes of the polynomial x^2-x-6 |
| Answer» x^2 - x - 6 = 0. (S = -1, P = -6, No. = -3, 2) x^2 - 3x + 2x - 6 = 0. x (x - 3) + 2 (x - 3) = 0. (x + 2) (x - 3) = 0. Therefore, x = -2, 3. | |
| 10234. |
Plz vote for Mathematics Basic or Standard of ur choice by writing Basic or Standard |
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Answer» If u wanna opt for science you need to choose standard and if don\'t wanna go for science opt for Basic..... Standard* Stands Standard Standee |
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| 10235. |
If zero of the polynimial 3x cube -x square -3x+1 is 1 , then find all the other zeroes . |
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Answer» sin A = 3/4. Now, sin^2 A + cos^2 A = 1. cos A = √1 - sin^2 A = √1 - 9/16 = √7/16 = √7/4. Therefore, sec A = 4/√7. Let p(x) =3x^3 -x^2 -3x + 1.One zero of p of x is given 1. Therefore ( x -1) is a factor of p of x. Now. Divide p of x with (x - 1).we get. 3x^2 +2x -1. Now factorise it. We get, Therefore. Other zero of p of x is -1 and 1/3 . ???? ?? ???? ???? ?. if sin A=3/4, calculate sec A |
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| 10236. |
Determine the nature of the roots of the equation 2x^2+x-1 |
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Answer» D=b^2-4ac =(1)^2-4 (2)(-1) =1+8 =9D <9.: there exist real and distinct roots... Discriminant = b^2 - 4ac = (1) ^2 - 4 * -1 * 2 = 1 + 8 = 9. Therefore, the roots are REAL and DISTINCT in nature. |
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| 10237. |
Express 59/2raises4 5raises3 |
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| 10238. |
Ex 1.2 |
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Answer» Ask question with complete information.. ??... :) Please check the question before writing. ?? |
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| 10239. |
Show that 5-√3 is irrational. Anyone please help |
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Answer» Thanku adithya dev a Prove √3 is irrational. It is given in the text. And then, since the difference between a rational and an irrational number is always irrational, 5 - √3 is irrational. |
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| 10240. |
Prove that √2 is an irrational number? |
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Answer» Given √2To prove: √2 is an irrational number.Proof:Let us assume that √2 is a rational number.So it can be expressed in the form p/q where p, q are co-prime integers and q≠0√2 = p/qHere p and q are coprime numbers and q ≠ 0Solving√2 = p/qOn squaring both the side we get,=>2 = (p/q)2=> 2q2 = p2……………………………..(1)p2/2 = q2So 2 divides p and p is a multiple of 2.⇒ p = 2m⇒ p² = 4m² ………………………………..(2)From equations (1) and (2), we get,2q² = 4m²⇒ q² = 2m²⇒ q² is a multiple of 2⇒ q is a multiple of 2Hence, p, q have a common factor 2. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number√2 is an irrational number. Any root of imperfect square is an \'irrational number\'.As \'2\' is not a perfect square(i.e.4,9,16....) so root of it is surely irrational. |
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| 10241. |
-1, -3/2, -2, 5/2 .... from an AP as a2-a1=a3-a2 |
| Answer» -3/2 + 1 = -2 + 3/2.(-3 + 2)/2 = (-4 + 3)/2.-1/2 = -1/2. Hence, the above sequence is an A.P. | |
| 10242. |
Form cubic polynomial whose zeroes are 1, 1, 2. |
| Answer» S = 1 + 1 + 2 = 4.P = 1 * 1 * 2 = 2.PZ = 1*1 + 1*2 + 1*2 = 1 + 2 + 2 = 5. p (x) = k(x^3- sx^2 + pz x - p)= k(x^3 - 4x^2 + 5x - 2)// | |
| 10243. |
What can we say about a²-b².If a>b and a&b are two odd prime numbers. |
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Answer» Considering the following condition, we can arrive to a conclusion that a^2 - b^2 would always be multiple of 8. Pls explain Ftutf |
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| 10244. |
Show that 6^n-5^n always end with digit 1. |
| Answer» 6^n, where n is any positive integer, always ends with 6. Eg :- 6^2 = 36, 6^3 = 216, etc. Whereas, 5^n , for any value of n, always ends in 5. Eg :- 5^2 = 25, 5^4 = 625, etc. Therefore, since 6 - 5 is always equal to one, the no 6^n - 5^n always ends with one. | |
| 10245. |
11th |
| Answer» Please ask question with complete information. | |
| 10246. |
Is 0 a rational number or an irrational number? |
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Answer» It is a rational number 0 is a rational number. We know that rational number is in the for of p/q . |
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| 10247. |
Find the value of X and Y 3x+2y=-9/23x+4y=5/2 |
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Answer» x=-13/6, y=1 No the above one is wrong the right answer is X=-23/6 Y=7/2 X=1, y=-8/6 |
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| 10248. |
Express the trigonometry ratio sin A,sec and tan A in terms of cot A |
| Answer» tan A = 1/cot A. sec A = √1 + 1/cot^2 A. sin A = cos A/cot A. | |
| 10249. |
Check whether 6^n can end with the digit 0 for any natural number n . |
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Answer» It never end with the digit 0, as it\'s prime factorization is 2 × 3 = 6. It doesn\'t contain 5, so it can never end with 0. Never because it doesn\'t consist of prime factor as multiple of 2×5 i. e6^n=(2×3)^n |
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| 10250. |
Find the greatest number of 6 digits exactly divisible by 18, 24 and 36 |
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Answer» No mention, Kajal. Thanku Chapter 1 in hindi LCM (18, 24, 36) = 72. Largest number of 6 digits = 999999. Dividing 999999 by 72, we get a remainder 63. Therefore, the greatest number of 6 digits exactly divisible by 18, 24, 36 is 999999 - 63 = 999936// |
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