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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.
| 24401. |
Find the sum of three digit natural no. Which are divisible by 13 |
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| 24402. |
Hey guys......How r u all????????and how ur studies r going on..... |
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| 24403. |
Proove that the length of tangents draw from an external point are equal in length |
| Answer» 1st step: take a point p outside the circle and and name the point A & B where it touches the circle.2nd step: join OA and OB as a radius from the circle.3rd step : join OP.4th step: now congruent the triangles OPA and OPB by RHS criterian. Hence, by c.p.c.t. AP and BP are equal. | |
| 24404. |
CI. f50-52. 2053-55. 12056-58. 10559-61. 12562-64. 30Find the mean of the given data. |
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Answer» Sorry I still can\'t understand because our sir said to convert them into continuous 1st take out class mark of class by upper limit + lower limit/2 Class marks :- 51 ,54,57,60,63 Now multiply corresponding freq. with class marksfx:- 20×51=1020,120×54=6480,105×57=5985, 125×60=7500,30×63=1890.Mean =sum of fx/ sum of freq. = 22875/400=57.1875 |
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| 24405. |
Draw the electronic configration of iron |
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| 24406. |
If 17th term of an AP exceeds it\'s 10th term by 7. Then the common difference is. ? |
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Answer» D =1 A17=a10+7,--------------1A10=a+(10-1)dA10=a+9d-----------2A17=a+16d-----------3Equating equation 1,2,and3a+16d=a+9d+7a-a+16d-9d=77d=7d=1=common difference is 1 |
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| 24407. |
Is -150 a term of an AP 11,8,5,2 |
| Answer» Let an=-150,,, a=11,d=-3,,,,,an=a +(n-1)d,,,,-150=11+(n-1)-3,,,,,-150=11-3n+3,,,,,-150=14-3n,,,,-150-14=-3n,,,-3n=-164,,,3n=164,,,n=164/3,,,,n=54*2/3,,,,,Since, n can\'t be a negative, fraction or come in decimal. Therfore.,it is not a term of an AP(,,,,,,, =second line) | |
| 24408. |
tfefgg |
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Answer» I don\'t know ? sorry |
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| 24409. |
1dm=? |
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| 24410. |
In board exam 2018 which book is generally refear by board examinor nceart book or Rd sharma |
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| 24411. |
3-2 |
| Answer» 1 | |
| 24412. |
What is a arithmatic progression |
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| 24413. |
α+β=2αβ=1Then α-β= ? |
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Answer» 0 0 |
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| 24414. |
Formulas of surface area and volume . |
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| 24415. |
Area of parallelogram |
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| 24416. |
sir maths ka sample paper kar uplod hoga 2017-18 ka |
| Answer» Its Added now check the shop in mycbsegude app | |
| 24417. |
Solve this equation 4nsqare+5n-636 |
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| 24418. |
SinA+cosA= |
| Answer» 1 | |
| 24419. |
7%5=100 |
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| 24420. |
I need some some solved example of find the value of k for which the points are collinear |
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Answer» RD????? See RD sharma.....?????many such questions are there.... |
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| 24421. |
Find sin 30geometrically |
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| 24422. |
In which month board 2018 will held? |
| Answer» I think February | |
| 24423. |
the common diffrence of an ap is -2 find its sum if it\'s 1st term is 100 and last term -10 |
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Answer» a= 100d= -2an= -10an=a+(n-1)d-10 = 100+2-2n2n=112n=56S56=56/2(2a+(55)-2) = 28(200 - 110) = 28 x 90=2520\xa0 Given,d = 2 , a = 1 and l = 10 We have to find value of n To find the value of n the given formula is a+(n-1)d solution = a+(n-1)d = 100+(n-1)-2 = 10 (n-1)-2=10-100 = (n-1)-2= -90 = (n-1)=-90/-2(n-1) =45 = n= 45+1= n=46Now we have to find sum formula is n/2[2a+(n-1)d] 46/2 [2×100(46-1)-2] = 23(200-90) =23×110= 2530 is the sum Its sum is 2565 |
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| 24424. |
What is the HCF of the smallest composite number and smallest prime number |
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| 24425. |
Exercise 7.2 question no. 2 |
| Answer» Dividing a line into 3 equal parts | |
| 24426. |
What is a prime no. |
| Answer» Prime numbers are the numbers which have only or 1as a factor and themselves | |
| 24427. |
If the seventh term of an AP is 1/9 and its ninth term is 1/7 find its 1631st term |
| Answer» a7=a+6d = 1/9.............ia9=a+8d=1/7..... iifrom i and ii2d=1/7 - 1/963 x 2d=9 - 7d=2/63x2=1/63a=19 -6/63=19-2/31=589- 2/31=587/31a1631=587/31+1630 x 1/63 | |
| 24428. |
What are the formula of surface areas and volume |
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| 24429. |
Identities formula |
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Answer» A plus b ka whole square is equals to a square + b square + 2 a b (a+b)2 =a2 +b2+2ab |
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| 24430. |
Please publish marking scheme for maths paper |
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| 24431. |
When does case 10 board exam accur |
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Answer» In CBSE the exams will occur around 15 Feb,2018. In may |
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| 24432. |
SinA(1+tanA)+cosA(1+cot)=(secA+cosecA). prove this. |
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| 24433. |
Cosec=2x cot=2/x find value 2(2x-2/x) |
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| 24434. |
x-y+z=4,x-2y-2z=9,2x+y+3z=1 |
| Answer» x-y+z=4 ...(1) x-2y-2z= 9 ...(2) 2x+y+3z= 1...(3)Multiply eq(1) by 2, we get 2x-2y+2z =4.....(4)Add eq(4) and eq(2), we get, 2x-2y+2z= 8 x-2y-2z=9 ----------------------- 3x-4y= 17 ...(5)Multiply eq(2) by 3 ,we get 3x-6y-6z= 27 ....(6)Multiply eq(3) by 2, we get 4x+2y+6z= 2 ...(7)Adding eq(6) and\xa0eq(7), we get3x-6y-6z= 274x+2y+6z=2-----------------7x-4y=29 ...(8)Subtracting eq(5) from eq(8), we get7x-4y=293x-4y=17--------------------4x=12x=12/4= 3\xa0Substituting the value of x in eq(8), we get7x3- 4y= 2921--4y=29-4y= 29-21-4y=8y=-8/4=-2Substituting the value of x and y in eq (1), we get3-(-2)+z = 4\xa03+2+z=45+z=4z=4-5z=-1So, x= 3, y=-2 and z= -1\xa0\xa0\xa0 | |
| 24435. |
From when our board exam will start.?? |
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Answer» 22 march Mid of Feb or start of march March |
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| 24436. |
tan 3 A /1+tan 2A + cot3 A /1+ cot 2A |
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| 24437. |
Ftgghg tu |
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| 24438. |
tan3 A |
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| 24439. |
Paper pattern of board exam |
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| 24440. |
Mathematics Sample paper is available right now ? |
| Answer» Yes | |
| 24441. |
What is Circle ? |
| Answer» Circle is a figure made from all the points that are at equal distance from a fixed point. | |
| 24442. |
From when the board xam fr class 10 will start...... |
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| 24443. |
What is the probability |
| Answer» Probability means the chances for occurence of an eventP= No. of favourable outcomes/ No. of total outcomes | |
| 24444. |
I need sample papers of classXII science U.P.Board please update me as soon as you could |
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| 24445. |
The sum of the first 16 terms of an AP is 112and the sum of its next 14 terms is 518 find AP |
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| 24446. |
A =12, d=4 find n |
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| 24447. |
Prove that cos20cos40cos60cos80=1/16 |
| Answer» cos20 cos40 1/2 cos80 (cos60 = 1/2)multiply and divide by 21/4 (2 cos20 cos40 cos80)1/4 (cos(20+80)+ cos(20-80)) cos40 (2cosa cosb= cos(a+b) + cos(a-b))1/4 (cos(-60)+ cos(100)) cos401/4(1/2 + cos100)cos401/8 cos40+ 1/4 (cos40 cos100)multiplt nd divide by 22/2(1/8 cos40) + 1/8(2 cos40 cos100)1/8 cos40+ 1/8 (cos140+ cos(-60)) (2cosa cosb= cos(a+b) cos(a-b))1/8 cos40+ 1/8 cos140 + 1/16 (cos60= 1/2)1/8(cos40+cos140) + 1/161/8(2 cos90 cos(-50)) + 1/16cos90= 01/16= r.h.shence proved | |
| 24448. |
Prove that perpendicular at the point of contact to the tangent to a circle passes through a center |
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| 24449. |
Find the distance b/w the points A(-6,6) and B(-2,2) |
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| 24450. |
find common differnce of AP :1/p,1-p/p |
| Answer» Common difference(d) = {tex}n^{th} term-(n-1)^{th} term{/tex}{tex}\\therefore d=a_{2}-a_{1}{/tex}{tex}d =( \\frac{{1 - p}}{p}) - (\\frac{1}{p}) = \\frac{{(1-p)-(1)}}{p} = \\frac{-p}{p} = -1{/tex}d = -1\xa0 | |