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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.
| 10151. |
use euclids division algorithm to find hcf of 567 and 693 |
| Answer» 693=567×1+126567=126×4+63126=63×2+0 | |
| 10152. |
Check weather |
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Answer» Note means more typed by mistake Please provide the question with note clarity |
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| 10153. |
IN figer pq and pr are tangent to circle with center a if angel q p a is equal to 27 then find q r |
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| 10154. |
Write emperical formula of central tendency. (Probability, Class-10) |
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Answer» 3 MEDIAN = MODE + 2 MEAN Aap itna aage pahuch gayi class 10 ke syallabus mei |
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| 10155. |
Find the H.C.F of 1260 , 1250 and 1520 by Euclid\'s division lemma |
| Answer» Hcf is 10 | |
| 10156. |
2x²+5x+273 find the zeroes by completing square method |
| Answer» | |
| 10157. |
Prove that under root m + under root n is irrational if under root mn is irrational |
| Answer» Ggg | |
| 10158. |
2x²+3y²+√2x+√3y =0 |
| Answer» | |
| 10159. |
Find the zeroes of x^2+15x? |
| Answer» x^2+15x = x(x+15) Here,x=0x+15=0Therefore, zeroes are 0 , -15I hope it will be helpfull | |
| 10160. |
What is Euclid\'s division |
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Answer» In arithmetic, Euclidean division – or division with remainder – is the process of dividing one integer by another, in a way that produces a quotient and a remainder smaller than the divisor. According to Euclid\'s Division Lemma if we have two positive integers a and b, then there exist unique integers q and r which satisfies the condition a = bq + r where 0 ≤ r < b. That means, on dividing both the integers a and b the remainder is zero. ... |
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| 10161. |
1.1 3 rd questions |
| Answer» Ans:To do it systematically, find HCF of(616 and 32)=So, 616>32 616=32×19+8 32=8×4+0So, HCF of 616 and 32 is 8(common HCF). Thus, the maximum number of columns is 8 in which they can march. | |
| 10162. |
prove that root2+root3 is an irrational number |
| Answer» Obviously as root2+root3 = root 5 which is irrational | |
| 10163. |
Best helpbook for English grammar |
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Answer» Xam IdeaBest book for English grammar for all the grades Number system |
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| 10164. |
Ncert and exemplar is enough to score 100 in maths 10th boards |
| Answer» Yes | |
| 10165. |
Find the HCF and LCM of 11008 and 7344 using fundamental theorem of arithmetic. |
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| 10166. |
Pz give fast my ans I need |
| Answer» Lol | |
| 10167. |
24 +3{16-2 (8-1)} +1 |
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Answer» 31 319 31 |
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| 10168. |
If sec theta =5/4 then verify tan theta/1+tan square theta =sin theta / sec theta |
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| 10169. |
2x^2+x-1=0 |
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| 10170. |
cos(90-theta)*sin(90-theta)/tan(90-theta)=sin^theta |
| Answer» | |
| 10171. |
(X)2+(6)2=14 |
| Answer» (X)2 + 6(2) =14 First of all, we have to divided by 2 both sides (X)2 /2 + 6(2) /2 =14/2 Then we have, X +6 =7X = 7-6X = 1Okay, then check (1)2 + 6(2) = 14 2 + 12 = 14 14 =14 | |
| 10172. |
Ex-3.2 Q-1 |
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Answer» Let number of boys who took part in the quiz = xLet number of girls who took part in the quiz = yAccording to given conditions, we havex + y = 10… (1)And, y = x + 4⇒ x – y = −4 … (2)For equation x + y = 10, we have following points which lie on the lineFor equation x – y = –4, we have following points which lie on the lineWe plot the points for both of the equations to find the solution.We can clearly see that the intersection point of two lines is (3, 7).Therefore, number of boys who took park in the quiz = 3 and, number of girls who took part in the quiz = 7.(ii) Let cost of one pencil = Rs x and Let cost of one pen = Rs yAccording to given conditions, we have5x + 7y = 50… (1)7x + 5y = 46… (2)For equation 5x + 7y = 50, we have following points which lie on the lineFor equation 7x + 5y = 46, we have following points which lie on the lineWe can clearly see that the intersection point of two lines is (3, 5).Therefore, cost of pencil = Rs 3 and, cost of pen = Rs 5 Kindly provide more clarity |
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| 10173. |
Find a quadratic polynomial p (y) with -15 and -7 sem1 of zeros respectively |
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| 10174. |
If-1 is a zero of the polynomial f(x)=x^2-7x-8,then calculate the other zero |
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| 10175. |
Prove that Root3 + root5 is irrational |
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Answer» Let the possible root 3+root5 are rational in the form of p by q and q not equal to 0Root 3+root5=p/q Let as assume on contrary that\xa0:√3 +\xa0√5\xa0is a rational number.\xa0Then there exists co - prime integers p and q such that ⇒ √3\xa0is a rational number .\xa0But this contradicts the fact that\xa0√3 is irrational.So, our assumption is wrong.Hence,√3 +\xa0√5\xa0is irrational. |
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| 10176. |
Check wheather 6 power n Can end with the digit 0 for any natural number n . |
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Answer» No... No because 6^n= (2^n*3^n) hence there are exist no pair of 2^n*5^n factors. |
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| 10177. |
Exampler |
| Answer» Very good,that can be an example for others | |
| 10178. |
A mall is constructing |
| Answer» Please ask question with complete information. | |
| 10179. |
find a quadratic polynomial each with the two zeroes as below :(i) -1,7 |
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Answer» Here is the sol : 1. Sum of the zeroes -1+7=62. Product of zeroes -1×7=-7 So the formula is , X² - (sum of zeroes)x + product of zeroes So the answer is X²-6x-7 Alpha = -b/a = -1Beta = c/a = 7Quadratic polynomial. = x^2 + x + 7 |
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| 10180. |
Draw a line segment of length 7.6cm and divide it into the ratio 5:8. Measures the two parts |
| Answer» Step 1: Draw segment\xa0AB\xa0of length\xa07.6\xa0cm\xa0using a ruler.Step 2: Draw a ray\xa0AC\xa0having an acute angle with line\xa0AB.Step 3: Mark\xa013\xa0equidistant points on ray\xa0AC,\xa0A1\u200b,A2\u200b,A3\u200b,...,A13\u200bStep 4: Join\xa0A13\u200b\xa0to\xa0BStep 5: Draw\xa0A5\u200bP\xa0parallel to\xa0A13\u200bB.Point\xa0P\xa0divides\xa0AB\xa0in the ratio\xa05:8. Measuring the lengths, we get\xa0AP=2.9\xa0cm\xa0and\xa0PB=4.7\xa0cm. | |
| 10181. |
Find all the zeroes of 2x⁴–3x³–3x²+6x–2,if you know that teo of the zeroes are _/2 and –_/2. |
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Answer» 4 zeros We know that √2 and -√2 are the zeroes of p(x) Therefore ...... (x - √2) and (x + √2) are the factors of p(x) It means product of factor also a factor = X² - 2 = g(x) Devide p(x) by g(x) We Get - q(x) = 2x² - 3x +1 By using slitting the middle term method ------ q(x) = 2x² - 3x + 1 = 0 = 2x² - 2x -x + 1 = 0 = 2x( x -1) -1 (x-1) = 0 = (2x-1) (x-1) = 0 Other two zeroes are ----- i) (2x-1) =0 = x = ½ii) (x-1) = 0 x = 1Hope its help you ......... |
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| 10182. |
Kskd |
| Answer» Gfgfd | |
| 10183. |
If one zero are polynomial (a ke power 2+9)x ke power 2+13x+6a is reciprocal other find value of a |
| Answer» | |
| 10184. |
The zeros of the polynomial 4x²+5√2x-3 |
| Answer» ( Root 2x + 3 ) ( 2root2x-1) = 0 | |
| 10185. |
Derive the relationship between zeroes and coefficient |
| Answer» | |
| 10186. |
(4x)-4x-3 |
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Answer» -३ -3 -3 |
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| 10187. |
Find the HCF of225 and745 using Euclid algorithm |
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Answer» 5. Gffjk? 745=225×3+70225=70*3+1570=15*4+1015=10*1+510=5*2+05is HCF of 225and725 |
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| 10188. |
SinA 2/3 find cosA |
| Answer» CosA=√1-4/9 =√5/9 | |
| 10189. |
Prove that 5+3√7 is irrational. |
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| 10190. |
3/x+1+4/x-1=29/4x+1 solve for x |
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| 10191. |
To experimentally find the hcf of 2 numbers based on the euclid division lemma |
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| 10192. |
Find the quadratic polynomial whose sum and product of zeroes are√2+1 and 1/√2+1 |
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Answer» Let the quadratic polynomial be =ax²+bx+calso, $ and & are zeroes of given polynomial so, Since, $+&=-b/a =) √2+1+1/√2+1=-b/a =) 2√2=-b/a also, $•&=c/a =) √2+1×1/√2+1=c/a =) 1 =c/aSo we get a=1 b=-2√2 C=1Hence the required polynomial is X²-2√2x+1=0 ans. S = √2 + 1. P = 1/√2 + 1. Therefore, polynomial p(x) = k (x^2 - Sx + P) = k (x^2 - (√2 + 1)x + 1/√2 + 1 = k [(√2 + 1) x^2 - (3 + 2√2) x + 1]// |
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| 10193. |
The LCM and HCF of two numbers are 675 and 45 if one of the number is 135 then find the other number |
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Answer» Given,H.C.F & L.C.M of two numbers is 675 & 45 If one number = 135 Then, let another number be x Now,H.C.F × L.C.M = PRODUCT OF TWO NUMBERS =675×45 = 135×X =X = 675 × 45 /135 = X = 225 HENCE , the another number is 225...(answer) HCF × LCM = product of two numbers.675 × 45 = 135 × x x = 675 × 45 /135 = 225.The second number is 225. |
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| 10194. |
Prove that only one of n-1, n+1, n+5 is divisible by 3 |
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Answer» Hello shinchan I am Gopika Grggtggg Proofed |
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| 10195. |
If 274 is of the form 5q+4 for an integer q , then find the value of q |
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Answer» 54 q=54 |
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| 10196. |
Check wheather g(x) is a factor of p(x).1.) P(x)= x5 - 4x3 + X2 + 3x + 1, g(x) = X2 - 3x + 1 |
| Answer» Divide p(x) by g(x). If the remainder is 0, then g(x) IS a factor of p(x). | |
| 10197. |
Find the value of k for which the quadratic equation 3x2 + kx + 3 = 0 has real and equal roots |
| Answer» For real and equal roots, D = b^2 - 4ac = 0. (for p (x) = ax^2 + bx + c) Therefore, k^2 = 3 * 3 * 4 = 36. k = √36 = 6.// | |
| 10198. |
Use euclids division algorithm to the hcf of 135 and 225 |
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Answer» 45 is the hcf 225=135×1+90135=90×1+4590=45×2+0So HCF of 135and 225 is 45 ^÷¢°¢¶ |
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| 10199. |
X square + 2 root 2 x minus 6 equals to zero |
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Answer» a = 1 , b = 2√2, c = -6p(x)=-( 2√2) + -√2√2square -4*1*-6/2*1p(x)= 2√2±√8+24/2p(x)= 2√2+-√32/2(+) sign lene pr2√2 +√32/2= 4√2+√32/2(-) sign lene pr4√2 -√32/2Formula which i use = -b ±√b square - 4×a×c /2a = 1 , b = 2√2, c = -6p(x)=-( 2√2) + -√2√2square -4*1*-6/2*1p(x)= 2√2±√8+24/2p(x)= 2√2+-√32/2(+) sign lene pr2√2 +√32/2= 4√2+√32/2(-) sign lene pr4√2 -√32/2Formula which i use = -b ±√b square - 4×a×c /2a |
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| 10200. |
The number of quadratic polynomials having zeros as -2 and 4 is . |
| Answer» One. | |