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.
| 41401. |
if curved surface area of cone is 88 cm then find volume of cone |
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| 41402. |
Blueprint{chapterwise marking} for class 10th Boards 2019.pls somebody help |
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| 41403. |
Find two numbers whose sum is 27 and product is 182 solve through quadratic eqn |
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Answer» Yup.....bt ye question ncert me hai na Take 1st no. as 27-2nd no. and then use other info of the ques. 14 and 13 13 and 14..... 13 anx 14 |
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| 41404. |
What is the value of (cos2 67°-sin2 23°)? |
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Answer» Zero Cos 67 -sin(90-67)Cos67-cos67=0 0 Zeroooooiiiiioo |
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| 41405. |
For a ogive ,we have to join the line to origin.. |
| Answer» No for both the ogives | |
| 41406. |
Find sum of first 50 multiples of 3 |
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Answer» 3825? Sn=3825 3825 |
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| 41407. |
0/0 is what? |
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Answer» No 0 |
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| 41408. |
The angle between the line x=2 and y=10 is................. .. |
| Answer» 90° | |
| 41409. |
The line x=my, where m‹0, lies in which quadrants |
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| 41410. |
If the points A(4,3) and B(x,5) are on the circle with centre O(2,3) , find the value of x |
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Answer» Is it x=0 or x=4 Using distance formula solve forAO^2 =BO^2 |
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| 41411. |
in triangle ABC,AD perpendicular to BC and 2DB=3CD . Prove that 5AB.AB=5AC.AC+BC.BC |
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| 41412. |
Minimum and maximize value of cosec |
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| 41413. |
100से कम तथा 4से विभाजित सभी प्राकृत संख्यायो का योग ज्ञात किजीए। |
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Answer» Ur answer is correct but please fully solve it. Please 1200 |
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| 41414. |
3a+2b is rational numbee |
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| 41415. |
For any positive integer n ,prove that n³-n is divisible by 6 |
| Answer» n3\xa0- n = n (n2\xa0- 1) = n (n - 1) (n + 1)\xa0Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.If n = 3p, then n is divisible by 3.If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.⇒ n (n – 1) (n + 1) is divisible by 3.\xa0Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.∴ n = 2q or 2q + 1, where q is some integer.If n = 2q, then n is divisible by 2.If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.⇒ n (n – 1) (n + 1) is divisible by 2.Since, n (n – 1) (n + 1) is divisible by 2 and 3.∴ n (n-1) (n+1) = n3\xa0- n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)\xa0 | |
| 41416. |
Find all the zero\'s of 2x⁴-9x³+5x²+3x-1 if two of it\'s zeroes are (2+√3) & (2-√3) |
| Answer» All zeroes are (2+√3) (2-√3) (1) (-1/2) | |
| 41417. |
For what value of p , x=b is a zero of the polynomial X×X-(a+b)X+a(p-2b) |
| Answer» If p= 3b then x=b is a zero of the given polynomil | |
| 41418. |
Find the real no. whose product with irrational number, 4-√3 is 39 |
| Answer» {tex}\\begin{array}{l}Let\\;the\\;number\\;is\\;x\\\\so\\;x(4-\\sqrt3)=39\\\\x=\\frac{39}{(4-\\sqrt3)}=\\frac{39(4+\\sqrt3)}{(4-\\sqrt3)(4+\\sqrt{3)}}=\\frac{39(4+\\sqrt3)}{16-3}\\\\=\\frac{39(4+\\sqrt3)}{13}=3(4+\\sqrt3)\\\\\\end{array}{/tex} | |
| 41419. |
Sin theta upon cos theta + cosec theta equals to 2 + sin theta upon cot theta minus cosec theta |
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| 41420. |
Solve the X 11 upon X + 4 minus one upon x minus 7 equals to 11 by 30 X is not equals to - 4 7 |
| Answer» I have answered this question just now. | |
| 41421. |
(a,2) (-3,-a) (7,-5) these are the coordinates of a line find a |
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Answer» Equidistant hi to nhi he Coordinates should be placed on equ.idistant Question correct hai ki nahi |
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| 41422. |
How many seconds in in 5 minutes |
| Answer» 5×60= 300 sec | |
| 41423. |
How to easily solve trigonometry questions? |
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Answer» Learn all the identities.. And practice well First u should write all formula and remember them and then practice questions Practice, practice and alot of practice..... |
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| 41424. |
2019 ka math ,hindi b, communicative English,sst and science ke board ke paper bhej do |
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Answer» Its provided in this app Or u cn get frm this app also Its available on google |
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| 41425. |
Proof of converse of bpt theorm |
| Answer» Given in ncert...u can prefer | |
| 41426. |
Find the area of sector with radius 6cm if angle of the sector is 60 |
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Answer» Use formula ø/360 π r square 132/7 cm square |
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| 41427. |
what is Pythagorean triplet |
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Answer» In a triangle, if sum of two sides is greater than the third side,then these sides can make a triangle otherwise not.so the three sides are calles pythagorean triplet. 3,4,5 also. All no.s are not arranged like that to follow the Pythagoras Theorem. Like25,24and7 follow the Pythagoras theorem.This is called Pythagorean triplet |
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| 41428. |
Show that there are infinitely many primes of the form 4n+3 |
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| 41429. |
Formula for median.. Mode.. Mean |
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Answer» Mode= l+(f1-f0)/(2f1-f0-f2)h Mean=£fixi/fiMedian=l+(n/2+cf/2)hPlz follow ncert Plz follow ncert book Open ncert maths book page no.263, 274, 282 |
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| 41430. |
The sum of two number is 15. If the sum of their reciprocal is 3/10.find the number |
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Answer» Let the two numbers be A and B=>A+B=15 .............(1)and\xa01/A +1/B=3/10⇒(A+B)/AB = 3/10from (1)(A+B)/AB = 15/AB⇒15/AB = 3/10⇒AB = 50 ............(2)solving (1) and (2)A = 10 and B = 5orA = 5 and B = 10 10and5 |
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| 41431. |
in figure DE parallel to BC then find value of x given AD=2cm DB=3cm DE=4cm and BC=x |
| Answer» Given That AD=2 cm , DB=3 cm DE= 4 cm BC=x{tex}\\begin{array}{l}As\\;DE\\parallel\\;BC\\\\So\\\\\\frac{AD}{AB}=\\frac{DE}{BC}\\\\\\frac2{2+3}=\\frac4x\\\\2x=4\\times5=20\\\\x=10\\\\\\end{array}{/tex}\xa0 | |
| 41432. |
Area of circlr |
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Answer» πr square Pie r square units pie r square Pie r square πrsquare |
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| 41433. |
If sinA +sin2A =1 ,prove that cos2A +cos4A =1 |
| Answer» Given that sinA + sin2A =1Then, sin A= 1 - sin2ATherefore, sin A = cos2A (since [1 - sin2A] is Cos2A by identity)Now substitute this in cos2A + cos4A = 1Instead of cos2A substitute (1-sin2A)Equating it, 1 - sin2A + cos4A = 1Now instead of sin A substitute cos2A , then sin2A = cos4A1- cos4A + Cos4A = 1. Cos4A gets striked outThus 1=1.Hence proved.... | |
| 41434. |
If m*cotA=n, then find {(m*sinA)-(n*cosA)}/{(n*cosA)+(m*sinA)} |
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| 41435. |
Find the 12 th term from the end of the A.P, -2,-4,-6,.............,-100 |
| Answer» -100,-98.........-2Now find common difference a2-a1=-98-(-100)-98+100d=2Now find the 12th term a+(n-1)d -100+(12-1)2-100+22-78 answer | |
| 41436. |
All chapter formulas |
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| 41437. |
9f. (X-2) (x+3) hcf of polynomial P(x) (2x-3x+2) (4+7x+a) Q(x) (2x+4x+) (6-7x+b) Find a &b |
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| 41438. |
How do we come to know that which median class should be taken aftwr getting n/2 |
| Answer» i also have a doubt in dis pls guys help me.... | |
| 41439. |
Prove that length of tangent drawn from an external point to a circle are equal |
| Answer» You can prove it by congruency rule i.e.Given in ncert page no. 211 | |
| 41440. |
I have my matjs pre board tommorow give me tip to solve the questions of trignometry |
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Answer» Pls dont study at the last time as it will make you confeuse and ?all the best? Just do practice...... In proving identities questions first read the question carefully and cjoose what step you should taken . Change all th value in sin ans cos in mostly qusetions As the whole concept is primarily based on NCERT so u must practise ncert questions quite well and thus learn the application of identities. BEST OF LUCK btw ?. |
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| 41441. |
(7\'4) or (-1\'8) distance |
| Answer» √(-1-7)² + (8-4)²√64+16√804√5 | |
| 41442. |
(2-5-5-5-8-7-4) |
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Answer» - 32 is the answer -32 |
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| 41443. |
Ch2. Polynomial k extra formulas bta do |
| Answer» Check notes for formulae :\xa0https://mycbseguide.com/cbse-revision-notes.html | |
| 41444. |
Tell me the all maths class 10 ncert formula |
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Answer» In the mathematics section , apko jo bhi chapter ka solution chahiye, choose the chapter and then choose ncert solutions............ Iss app pr mil jayege.. ??? solid wale formulae which help u in ur formula learning process of all chapters We r not free check from net Available in this app......... |
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| 41445. |
If cos A- sin A=1, prove that cos A+sin A=+-1 |
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Answer» Sorry for wrong answer Cos A minus sin a is equal to 1 square in both side cos square A + sin square A |
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| 41446. |
Easy method to learn ch 6 of maths |
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Answer» You should try question from sample paper You must learn all theorms and proof then it will become to easier to solve |
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| 41447. |
What would be the value of n for n²-1 divisible by 8 |
| Answer» n=3,5,7,9..... So on, any odd no. Except 1 | |
| 41448. |
Cot(90-0)/tan0+cosec(90-0)Sin0/tan(90-0)=sec,20 |
| Answer» Bhai u 0(zero) hai ya theta | |
| 41449. |
CosA/1+sinA + 1+SinA/cosA= 2secA |
| Answer» CosA 1+sinA---------- + ----------- = 2secA1+ SinA cos Acos^2 + (1+sinA)^2--------------------------(1+sinA)(cosA)Cos^2 + 1+ sin^2 +2sinA----------------------------------------(1+sinA)(cosA)2(1 + SinA)-------------------(+sinA)(cosA)=2/cosA=2secA♡♡♡ | |
| 41450. |
1+secA/SecA = sin²A/1-cosA prove |
| Answer» I don\'t now the answer | |