Saved Bookmarks
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.
| 39301. |
The sum of three consecutive numbers are divisible by 6 |
| Answer» There are 2,1,3 | |
| 39302. |
If the point P(x,y) is equidistant from the points A(5,1) and B(1,5), prove that x=y. |
| Answer» PA = PB (Given){tex}\\therefore {/tex}\xa0PA2 = PB2{tex} \\Rightarrow {/tex}\xa0(5 - x)2 + (1 - y)2 = (1 - x)2 + (5 - y)2{tex} \\Rightarrow {/tex}\xa025 + x2 - 10x + 1 + y2 - 2y = 1 + x2 - 2x + 25 + y2 - 10y{tex} \\Rightarrow {/tex}\xa0-8x = -10y + 2y{tex} \\Rightarrow {/tex}\xa0-8x = -8y{tex} \\Rightarrow {/tex}\xa0x = y\xa0 | |
| 39303. |
Is ncert sufficient to get 90+marks |
|
Answer» Mistaken, its GOOD question Yes, but do all ncert book. examples also... Including examples and activity. It is not necessary to refer only the ncert. If you need to increase your knowledge, you are free to use/refer any other books related to your syllabus.??By the way, goid question!! Sss |
|
| 39304. |
Excise 10.2 questions number 6 |
| Answer» | |
| 39305. |
What is average means? |
| Answer» Sum of no. /total no. | |
| 39306. |
Prove that 2-3Root 5 |
| Answer» Let, 2-3√5 be a rational number.2-3√3=a/bWhere \'a\' and \'b\' are coprimes-3√3=a/b-2-3√3=a-2b/b √3=a-2b/b×(-3) √3=a-2b/-3b √3=(integer)-2(integer)/-3(integer) √3=rational numberBut,comtradiction √3 is a irrational number.Thus, Our assumption is wrong.Therefore, 2-3√5 is a irrational number. | |
| 39307. |
What is four digital |
| Answer» | |
| 39308. |
a^2x^2 |
| Answer» 10 | |
| 39309. |
formative assesment 1 syllabus |
| Answer» | |
| 39310. |
When 2 die tossed simultaneously. What is the probability of sum is prime no? |
| Answer» 14/36 = 7/18 | |
| 39311. |
Prove--10/10 + 10/10 =2 |
|
Answer» 10/10 + 10/10 =21+1=22=2 10/10 + 10/10 ON CANCELLING 1+1=2 Wrong question. |
|
| 39312. |
Represent root 9.3 on the number line |
| Answer» Mark the distance 9.3 units from a fixed point A on a given line to obtain a point B such that AB = 9.3 units. From B mark a distance of 1 unit and call the new point as C. Find the mid-point of AC and call that point as O. Draw a semi-circle with centre O and radius OC = 5.15 units. Draw a line perpendicular to AC passing through B cutting the semi-circle at D. Then BD = {tex}\\sqrt {9.3} .{/tex} | |
| 39313. |
Why is sin0 =0 |
| Answer» | |
| 39314. |
Trignometry chapter me se ex. 7.1Q6 |
| Answer» 1st is a Square2nd is not a Quadrilateral 3rd is a Parallelogram [email\xa0protected]Hope it helps you!!!??? | |
| 39315. |
Sa 1 class 9 syllabus 2018 |
| Answer» | |
| 39316. |
What is pair of linear eq in to variables |
| Answer» The common solution of linear equation is called pair of linear equation | |
| 39317. |
Find the volume of sphere is 154cm square? |
| Answer» I think some part of que is miss | |
| 39318. |
21 |
| Answer» | |
| 39319. |
Solve the equation 6x+2x=6xy 4x+3y=5xy |
| Answer» The given pair of equation is6x + 3y = 6xy{tex}\\Rightarrow \\quad \\frac { 6 x } { x y } + \\frac { 3 y } { x y } = \\frac { 6 x y } { x y }{/tex}..............Dividing throughout by xy{tex}\\Rightarrow \\quad \\frac { 6 } { y } + \\frac { 3 } { x } = 6{/tex}.........................(1)2x + 4y = 5xy{tex}\\Rightarrow \\quad \\frac { 2 x } { x y } + \\frac { 4 y } { x y } = \\frac { 5 x y } { x y }{/tex} ..................Dividing throughout by xy{tex}\\Rightarrow \\quad \\frac { 2 } { y } + \\frac { 4 } { x } = 5{/tex}..........................(2)Put\xa0{tex}\\frac { 1 } { x } = u{/tex} .....................(3)And\xa0{tex}\\frac { 1 } { y } = v{/tex}..........................(4)Then, the equation (1) and (2) can be written as:{tex}6 v + 3 u = 6{/tex}..................(5)2 v + 4 u = 5 ..................(6)Multiplying equation (6) by 3, we get6 v + 12 u = 15 ..........................(7)Subtracting equation (5) from equation(7), we get 9u = 9\xa0{tex}\\Rightarrow \\quad u = \\frac { 9 } { 9 } = 1{/tex}{tex}\\Rightarrow \\quad \\frac { 1 } { x } = 1{/tex}.......................using (3){tex}\\Rightarrow \\quad x = 1{/tex}Substituting this value of u in equation (5), we get 6v + 3 X 1 = 6{tex}\\Rightarrow \\quad 6 v + 3 = 6{/tex}{tex}\\Rightarrow 6 v = 6 - 3 = 3{/tex}{tex}\\Rightarrow \\quad v = \\frac { 3 } { 6 } = \\frac { 1 } { 2 }{/tex}{tex}\\Rightarrow \\quad \\frac { 1 } { y } = \\frac { 1 } { 2 }{/tex}......................using (4){tex}\\Rightarrow y =2{/tex}Hence, the solution of the given pair of the equation is x =1, y =2Verification: Substituting x =1, y =2We find that both the equation (1) and (2) are satisfied as shown below:{tex}\\frac { 6 } { y } + \\frac { 3 } { x } - \\frac { 6 } { 2 } + \\frac { 3 } { 1 } - 3 + 3 - 6{/tex}{tex}\\frac { 2 } { y } + \\frac { 4 } { x } - \\frac { 2 } { 2 } + \\frac { 4 } { 1 } = 1 + 4 = 5{/tex}Hence, the solution is correct. | |
| 39320. |
Find radius of a sphere whose surface Area is 154cm^2 |
| Answer» area of sphere=4 pi r3so ,154=4×22/7×r2154=88/7×r2154/88/7=r2154/88/7=r2so r=root of 49/4so aftr removing root we get =7/2 | |
| 39321. |
Exercise 14.4 |
| Answer» | |
| 39322. |
If P is a prime number then P is which number |
| Answer» P is not divisible by any number ecept 1 and itself. | |
| 39323. |
Which is the largest 6 digit number that is exactly divisible by 18 , 24 and 36 |
| Answer» | |
| 39324. |
In school 4/7 of the boys If there are 210 girls find the number of boys in the school |
|
Answer» Substitute this value of x for 4x then we get 4*30 that is 120 therefore the no of boys is 120 Then we get tje value of x as 30 Then equate 7x equal to 210 Take the ratios as 4x and 7x |
|
| 39325. |
A piggy bank contaons |
| Answer» Number 50 p coins in the piggy bank = 100Number of Re. 1 coins in the piggy bank = 50Number of Rs. 2 coins in the piggy bank = 20Number of Rs. 5 coins in the piggy bank = 10∴ Total number of coins in the piggy bank = 100 + 50 + 20 -10 = 180∴ Number of all possible outcomes = 180\tNumber of favourable outcomes to the event that the coin will be a 50 p coin = 100\t∴ Probability that the coin will be a 50 p coin\xa0{tex}\\frac{Number\\;of\\;favourable\\;outcomes\\;to\\;the\\;event\\;that\\;the\\;coin\\;will\\;be\\;a\\;50\\;p\\;coin}{\\;Number\\;of\\;all\\;possible\\;outcomes}{/tex}{tex}\\frac{100}{180}=\\frac59{/tex}\tNumber of favourable outcomes to the event that the coin will not be a Rs. 5 coin\t= 100 + 50 + 20 = 170\t∴ Probability that the coin will not be Rs. 5 coin\t{tex}\\frac{Total \\ no.\\;of\\;favourable\\;outcomes}{Total\\;number\\;of\\;possible\\;outcomes}{/tex}{tex}\\frac{170}{180}=\\frac{17}{18}{/tex} | |
| 39326. |
Obtain all the zeros of x^4+6x^3+x^2-24x-30 if 2 zeros are 2 and -5 |
| Answer» As x = 2 and -5 are the zeroes of x4 + 6x3 + x2- 24x - 20.{tex}\\Rightarrow{/tex}\xa0(x - 2) and (x + 5) are two factors of x4 + 6x3 + x2\xa0-24x - 20{tex}\\Rightarrow{/tex}\xa0product of factors is (x - 2) (x + 5) = x2 + 3x - 10Dividing x{tex}^4{/tex} + 6{tex}x^3{/tex}+\xa0{tex}x^2{/tex} - 24x - 20 by\xa0{tex}x^2{/tex} + 3x - 10Dividend = divisor {tex}\\times{/tex}\xa0quotient + remainder{tex}\\Rightarrow{/tex}\xa0x4 + 6x3 + x2 - 24x - 20 = (x2 + 3x -10) (x2 + 3x + 2)= (x - 2) (x + 5) (x + 2) (x + 1)Hence, other two zeroes are -2 and -1. | |
| 39327. |
Ncert solution 8.4 |
| Answer» Check NCERT solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 39328. |
Difference between the two no. is 1. Difference of their reciprocal is 1/6. Find the numbers. |
| Answer» x-y=1. x=1+y-----(1)1/x - 1/y =1/6( y-x)/xy =1/6xy=6y-6x-----(2)(1) in (2),y^2 +y= 6y^2+y-6=0Middle Term Splitting,y^2+3y-2y-6=0y(y+3)-2(y+3)=0(y-2)(y+3)=0Therefore, y=2 or y=-3. | |
| 39329. |
Name of quadrilateral whose all sides are not equal........... |
| Answer» Trapezium | |
| 39330. |
Find rational numbers between root 3 and root 5. |
| Answer» | |
| 39331. |
What is meant by corresponding sides |
| Answer» It means that the sides which contains a common vertex and an angle that pair of sides are commonly known as corresponding sides. | |
| 39332. |
Ex 5.2 solutions |
| Answer» Open the Google Chrome and write exercise 5.2 class 10th open the my CBSE guide app or open the dronstudy you get the answers and in the CBSE guide of all the solutions of exercise 5.2 present | |
| 39333. |
Name a quadrilateral whose sides and diagonal are not equal |
|
Answer» Trapezium Parellelogram |
|
| 39334. |
Solve by using only & only factorisation:a/x-a + b/x-b = 2c/x-c |
| Answer» The given equation is; (x - a)(x - b)+(x - b)(x - c)+(x - c)(x - a) =0\xa0{tex}\\Rightarrow{/tex}\xa0(x2\xa0- ax - bx + ab) + (x2- bx - cx + bc) +(x2\xa0- cx - ax + ac) = 0{tex}\\Rightarrow{/tex}3x2\xa0- 2x(a + b + c) + (ab + bc + ca) = 0.....(1).\xa0Discriminant \'D\' of quadratic equation (1) is given by;{tex}\\therefore{/tex}\xa0D = 4(a + b + c)2\xa0- 12(ab + bc\xa0+ ca)= 4[(a + b + c)2\xa0- 3(ab + bc + ca)]= 4(a2 + b2 + c2\xa0- ab - bc\xa0- ca)= 2(2a2\xa0+ 2b2 + 2c2 -\xa02ab - 2bc\xa0- 2ca)= 2 [(a - b)2 + (b - c)2 + (c - a)2] ≥ 0[{tex}\\because{/tex}\xa0(a - b)2\xa0≥ 0, (b - c)2\xa0{tex}{/tex}\xa0≥ 0 and (c - a)2\xa0{tex}{/tex}\xa0≥ 0].This shows that both the roots of the given equation are real.For equal roots, we must have D = 0.Now, D = 0 {tex}\\Rightarrow{/tex}\xa0(a - b)2\xa0+ (b - c)2\xa0+ (c - a)2= 0{tex}\\Rightarrow{/tex}\xa0(a - b) = 0, (b - c) = 0 and (c - a) = 0 (sum of squares can be zero only if they all are equal to 0){tex}\\Rightarrow{/tex}\xa0a = b = c.Hence, the roots are equal only when a = b = c | |
| 39335. |
If angle APO=33degree then calculate angle AOB |
| Answer» | |
| 39336. |
Ch 15 class 10 ex 15.1 |
| Answer» Q. No. | |
| 39337. |
Find the roots of the equation using quadratic formula5x²_6x_2=0 |
| Answer» 5x^2-6x-2=0Here-a=5,b=-6,c=-2By using quadratic formulax=-b+-√b^2-4ac/2ax=-(-6)+-√(-6)^2-4×5×(-2)/2×5x=6+-√36+40/10x=6+-√76/10x=6+-2√19/10x=2(3+-√19)/10x=3+-√19/5x=3+√19/5 and 3-√19/5 | |
| 39338. |
Find the roots of the equation useing quadratic formula |
| Answer» | |
| 39339. |
The angles of right angle triangle are in AP show that they are in ratio 3:4:5 |
| Answer» | |
| 39340. |
Prove that √2 is an irrational no |
|
Answer» Let assume root 2 is rationalRoot 2=a/b(where a and b are co -prime and b is not equal to 0)Root2b=a--(1)Squaring on both sides(Root 2b)^2=a^2--(2)2 divide a^2}2 divide a}a=2c(where c is some integer)Put a in (2)(Root 2b)^2=(2c)^22b^2=4c^2b^2=4c^2/2b^2=2c^2So a and b have atleast 2 as common factor.But this condradictss that a and b have no,common factor other than 1 and this aris due to incorrect assumption that root 2 is rational.Therefore root 2 is irrational. Hence proved Let is assume that root 2 is rational.Let a and b is a positive integer. Root 2= a/bSo root 2 is rational number But this contradicts the fact that root 2 is rationalBut this contradiction our assumption is wrong so root 2 is rational number. |
|
| 39341. |
Prashant chaudhry are you online i want to ask some questions. |
| Answer» | |
| 39342. |
4x+y=27 and 2x+2y=21 |
| Answer» X=11\\2 y=5 | |
| 39343. |
Solve 2/x+3/y=13 5/x+3/y=2 |
| Answer» Putting {tex}\\frac 1x{/tex}= u and {tex}\\frac 1y{/tex}\xa0= v, the given equations become{tex}2u+ 3v\xa0= 13{/tex} .... (i){tex}5u - 4v = -2{/tex} .......(ii){tex}\\text{Multiplying (i) by 4 and (ii) by 3 and adding the results, we get}{/tex}{tex}8u\xa0+ 15u\xa0= 52 - 6{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}23u = 46{/tex}{tex}\\Rightarrow \\quad u = \\frac { 46 } { 23 } = 2{/tex}{tex}\\text{Putting u\xa0= 2 in (i), we get}{/tex}(2 {tex}\\times {/tex}\xa02) + 3v\xa0= 13 {tex}\\Rightarrow{/tex}\xa03v\xa0= 13 - 4 = 9 {tex}\\Rightarrow{/tex}\xa0v\xa0= 3.Now,u\xa0= 2{tex}\\Rightarrow \\frac { 1 } { x } = 2 \\Rightarrow 2 x = 1 \\Rightarrow x = \\frac { 1 } { 2 }{/tex}And, v\xa0= 3 {tex}\\Rightarrow \\frac { 1 } { y } = 3 \\Rightarrow 3 y = 1 \\Rightarrow y = \\frac { 1 } { 3 }{/tex}Hence,\xa0{tex}x = \\frac { 1 } { 2 } \\text { and } y = \\frac { 1 } { 3 }{/tex} | |
| 39344. |
If HCF(a,b)=12 and a×b = 1800 then find LCM (a,b) |
|
Answer» 1800/12 = 150 LCM X HCF = Product of two no. LCM X 12 = 1800LCM = 1800/12LCM =150 |
|
| 39345. |
x2 ₹ |
| Answer» | |
| 39346. |
pls explain tomorrow\'s is my exam Trigonometric ratios of some specific angles |
|
Answer» All the best gor your exams! !!!!!!????????????✏✒ Just go through the table given in the ncert maths book . Page number 185 Sorry but i dont know? ??? |
|
| 39347. |
Full form of Bpt in maths? Triangles |
|
Answer» BPT= basic proportional theorem Basic proportnality theorom Basic Proportionality Theorem |
|
| 39348. |
Hiii priya good morning |
| Answer» | |
| 39349. |
Write quadratic polynomial the sum and product of whose zeros are 3 and -2 respectively |
|
Answer» X square minus 3x minus 2 x2 - 3x - 2 |
|
| 39350. |
Can i send a pic of problem |
| Answer» Yes you can?? | |