InterviewSolution
Saved Bookmarks
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.
| 36251. |
Write the quardratic polynomial whose zeroes are 4,-1 |
| Answer» | |
| 36252. |
X*x-\' |
| Answer» | |
| 36253. |
Write the equation representing y.axis |
| Answer» 0+y | |
| 36254. |
How to get square root of 15 |
|
Answer» √3×53√5 divesion mathed se |
|
| 36255. |
If one root of the polynomial p(x)=5x |
|
Answer» Complete ur question Please tell it properly Tell the question properly |
|
| 36256. |
What is the use of triangle like chapters of maths in real life |
| Answer» And what is the use of trigonometry like chapters in real life???????? | |
| 36257. |
Properties of quadilaterals |
| Answer» They have four sides and the sum of their interior angles is 360 degree | |
| 36258. |
If one root of the quadratic equation 6x2-x-k=0 is 2/3 |
|
Answer» Sorry properly Write the question prope8 |
|
| 36259. |
In the following APs,find the missing term in box |
| Answer» | |
| 36260. |
slove using by quadratic formula,x2-35x+10=0 |
| Answer» D = b2 -4ac=. 352 -4*1*10=. | |
| 36261. |
In an AP,a=-18.9 , d=2.5, an=3.6. Than find n=? |
|
Answer» an= a+(n-1)d n=10 |
|
| 36262. |
Rs aggarwaal solution |
| Answer» | |
| 36263. |
There are 15 cards numbered from 1 to 30 Find the probability of numbers not divisible by 3 |
|
Answer» I am not sure but the answer may be right Total Cards - 15 No. Divisible by 3 - 6, 9, 12, 15, 18, 21, 24, 27, 30 No. Not divisible by 3 - 1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26, 28, 29. Probability of no. Not divible by 3 =20/30 = 2/3 Sorry 30 cards are there |
|
| 36264. |
Any trick to learn t ratios |
| Answer» NnooooOooo | |
| 36265. |
3x-7y=8 ,5x+4y=9 |
| Answer» 3x-7y=8 ...........1st5x+4y=9 .............2ndBy elimination method. -5 ×(3x-7y=83 ×(5x+4y=9We get.-15x+35y=-4015x+12y=27On solving them like addition we get.47y=-13 Then y=-13/47Put the value of y in any equation you will get the value of x.Value of x and y will be answer. | |
| 36266. |
(x+4)(x_4)=0 |
| Answer» | |
| 36267. |
Tell me an easy way to revise chapter 1to 9 |
| Answer» Learn only thier formulas nothing else.Exercises will get easierBecauseall questions are based on formula only.. | |
| 36268. |
Triangle ka important question |
|
Answer» And example que.. With example que. All questions |
|
| 36269. |
Find the sum of the following series: 10+14+18+22+...........104 |
| Answer» | |
| 36270. |
Under root 51 approximate value |
|
Answer» 7.1414284285 7.14(approx) |
|
| 36271. |
Chapter 1st 2 Mark\'s questions |
|
Answer» Ohk Whatever be the chapter of any subject , just go through this app. |
|
| 36272. |
Rs aggarwal chapter AP ex 11-b question no 9 |
|
Answer» Sorry, I don\'t have Rs aggarwal. Tell |
|
| 36273. |
11x+15y+23=0,7x-2y-20=0 solve it elimination method |
|
Answer» The given system of equations is11x + 15y + 23 = 0 ....(1)7x - 2y - 20 = 0 ....(2)To solve the equations (1) and (2) by cross multiplication method,we draw the diagram below:Then,{tex}\\Rightarrow \\;\\frac{x}{{(15)( - 20) - ( - 2)(23)}} = \\frac{y}{{(23)(7) - ( - 20)(11)}}{/tex}{tex}= \\frac{1}{{(11)( - 2) - (7)(15)}}{/tex}{tex}\\Rightarrow \\;\\frac{x}{{ - 300 + 46}} = \\frac{y}{{161 + 220}} = \\frac{1}{{ - 22 - 105}}{/tex}{tex}\\Rightarrow \\;\\frac{x}{{ - 254}} = \\frac{y}{{381}} = \\frac{1}{{ - 127}}{/tex}{tex}\\Rightarrow \\;x=\\frac{{ - 254}}{{ - 127}} = 2{/tex} and {tex}y = \\frac{{381}}{{ - 127}} = - 3{/tex}Hence, the required solution of the given pair of equations isx = 2, y = -3Verification : substituting x = 2, y = -3,We find that both the equations (1) and (2) are satisfied as shown below:11x + 15y + 23 = 11(2) + 15(-3) + 23= 22 - 45 + 23 = 07x - 2y - 20 = 7(2) - 2(-3) - 20= 14 + 6 - 20 = 0Hence, the solution we have got is correct. 14+6-20=0 |
|
| 36274. |
Subtract of z from y |
| Answer» Y-z | |
| 36275. |
Find a.p. whose 2 term is 10 and 6term exceeds the 4th term |
| Answer» | |
| 36276. |
Please give syllabus of maths with each chapter weightage |
| Answer» Sorry, I don\'t know | |
| 36277. |
Q.1). Solve for x and y? 1)6(ax+by)=3a+2b 6(bx-ay) =3b-2a |
| Answer» 6(ax + by) = 3a + 2b6ax + 6by = 3a + 2b.........(i)6(bx - ay) = 3b - 2a6bx - 6ay = 3b - 2a.........(ii)Multiplying (i) by a and (2) by b,So, 6a2x + 6aby = 3a2 + 2ab .......... (iii)And 6b2x - 6aby = 3b2 - 2ab ......... (iv)Add (iii) and (iv), we get6a2x + 6aby + 6b2x - 6aby = 3a2 + 2ab + 3b2 - 2ab⇒ 6a2x + 6b2x = 3a2\xa0+ 3b2⇒6 (a2x + b2x) = 3(a2\xa0+ b2)⇒{tex}x = \\frac { 3 \\left( a ^ { 2 } + b ^ { 2 } \\right) } { 6 \\left( a ^ { 2 } + b ^ { 2 } \\right) } = \\frac { 3 } { 6 } = \\frac { 1 } { 2 }{/tex}Substituting\xa0{tex}x = \\frac 12{/tex}\xa0in (i),we get{tex}6 a \\times \\frac { 1 } { 2 } + 6 b y = 3 a + 2 b{/tex}⇒ 3a + 6by = 3a + 2b⇒ 6by =3a + 2b -3a⇒ 6by = 2b⇒ {tex}y = \\frac { 2 b } { 6 b } = \\frac { 1 } { 3 }{/tex}Hence, the solution is\xa0{tex}x = \\frac { 1 } { 2 } , y = \\frac { 1 } { 3 }{/tex} | |
| 36278. |
Prove that √5 is irrational. |
| Answer» Because this is not form in p/q | |
| 36279. |
I want chapter co-ordinate geometry of RD Sharma |
| Answer» You should download the app from play store . | |
| 36280. |
What are equally likely outcomes? |
| Answer» | |
| 36281. |
Divide 19 into two parts such that the sum of their squares is 193 |
| Answer» 19 should be divided into (12)² + (7)² . It would be equal to 144+49=193. | |
| 36282. |
Find the nature of roots . If real roots exist find them :-(1).2×square -6×square +3=0 |
| Answer» The given quadratic equation is\xa02x2\xa0- 6x + 3= 0Here, a = 2, b = -6, c = 3Therefore, discriminant = b2\xa0- 4ac{tex}= ( - 6 ) ^ { 2 } - 4 ( 2 ) ( 3 ) = 36 - 24{/tex}= 12 > 0So, the given quadratic equation has two distinct real rootsSolving the quadratic equation\xa0{tex}2 x ^ { 2 } - 6 x + 3 = 0{/tex} ,\xa0by the quadratic formula,\xa0{tex}x = \\frac { - b \\pm \\sqrt { b ^ { 2 } - 4 a c } } { 2 a }{/tex}we get {tex}= \\frac { - ( - 6 ) \\pm \\sqrt { 12 } } { 2 ( 2 ) } = \\frac { 6 \\pm 2 \\sqrt { 3 } } { 4 } = \\frac { 3 \\pm \\sqrt { 3 } } { 2 }{/tex}Therefore, the roots are\xa0{tex}\\frac { 3 \\pm \\sqrt { 3 } } { 2 } , \\text { i.e. } \\frac { 3 + \\sqrt { 3 } } { 2 } \\text { and } \\frac { 3 - \\sqrt { 3 } } { 2 }{/tex}\u200b\u200b\u200b\u200b\u200b\u200b | |
| 36283. |
How many circle of a pizza bounded by three line segment |
| Answer» | |
| 36284. |
Acute angle theorem proof |
| Answer» Ask your doubts not something that is given in the book | |
| 36285. |
if the tan theta + cos theta is equal to 3 .find the value of tan squared theta + cos squared theta |
|
Answer» =9 Ans 9 =9 ans =9 |
|
| 36286. |
Find HCF of 2 and 4 |
|
Answer» 2 2 2 Answer is 2 HCF of 2 and 4 is 2 |
|
| 36287. |
If d=HCF(870,255),write the value of d. |
|
Answer» 15 HCF(870,225) = 15 so the value of d is 15. |
|
| 36288. |
When we get rd sharma full solutions |
| Answer» From app | |
| 36289. |
How to prove that the sum of angles of a triangle is 180 degree. |
| Answer» Let the triangle\'s of angle is x, y and zwe have to provex + y + z = 180 ..........(i)we know that sum of two interior opposite angle is equal to exterior angle.so, x + y = 180 - z ....... (ii)By linear pair exterior angle is 180 - z when triangle\'s angle is z.By (i) also we getx + y = 180 - z ..........(iii)Hence proved(ii) and (iii) are the same, so we can say that the sum of the three angles triangle is 180o. | |
| 36290. |
Find the value of b for which the polynomial 2x+3 is a factor of 2x^3+9x^2-x-b |
| Answer» -3/4 | |
| 36291. |
If sin teta =4/5 , find the value of 4tan teta -5cos teta/sec teta + 4cot teta |
|
Answer» 19/56 19/56 is the answer of your question |
|
| 36292. |
6=3,2=12 how? |
| Answer» Let\'s suppose 6 = 3Then,6 × 0 = 3 × 0 [multiplying 0 both the sides]0 = 0 Since 0 = 0 is true so our assumption that 6 = 3 is also true.?? | |
| 36293. |
Prove that area of rhombus is equal to half of product of its diagonals |
| Answer» Let ABCD be a rhombus whose diagonals are AC and BD.Then, Area of rhombus ABCD= Area of {tex}\\bigtriangleup{/tex}ABD + Area of {tex}\\bigtriangleup{/tex}CBD{tex}={(BD)(AO)\\over2}+{(BD)(OC)\\over2}{/tex} . .[Diagonals of a rhombus are perpendiculars to each other]{tex}={(BD)\\over2}(AO+OC)={(BD)(AC)\\over2}{/tex}={tex}1\\over2{/tex} product of the lengths of its diagonals. | |
| 36294. |
Is the following numbers perfect square ?5398 |
|
Answer» No No |
|
| 36295. |
Exercise 5.3 11th question |
| Answer» | |
| 36296. |
If p is a prime then p is an ______number |
| Answer» | |
| 36297. |
If x=2 then find y as x+y=7 |
|
Answer» X=2,x+y=7,2+y=7,y=7-2;y=5 Y= 5 5 X=2X+y=72+y=7(x=2)Y=7-2Y=5ans Y-5 |
|
| 36298. |
Who chapters must study first term exam school to cbse. ? |
| Answer» | |
| 36299. |
Find the qudrite polyonimal/Who Zero are |
| Answer» General form of a quadratic polynomial: x2 - (sum of the roots)x + (product of the roots) = 0 | |
| 36300. |
How to make class interval continous |
| Answer» Substract the sufficing value from the lower limit and add the same value to the upper limit.hit the thanks button? | |