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.
| 28701. |
What is the probability that leap year having 53sundays |
|
Answer» A leap year has 366 days ( 52 weeks and 2 days) these 2 days can be Sunday mondayMonday tuesdayTuesday WednesdayWednesday ThursdayThursday fridayFriday SaturdaySaturday sundayAs there are 52 weeks in leap year it would have 52 sundays and the probability of having 1 more sunday i.e 53 sundays areTotal outcomes =7No.of favourable outcomes = 2 ( sunday , monday. Saturday, sunday) Prob.= 2/7 A leap year has 366 days or 52 weeks and 2 more days. The two odd days can be {Sunday,Monday},{Monday,Tuesday},{Tuesday,Wednesday},{Wednesday,Thursday},{Thursday,Friday},{Friday,Saturday},{Saturday,Sunday}. So there are 7 possibilities out of which 2 have a Sunday. So the probability of 53 Sundays = {tex}\\frac27{/tex} 2/7 |
|
| 28702. |
Frnds there is my maths exam...if u know some difficult que plz mention them |
| Answer» 1.Be calm in exam drink water as possible.May be you get stressed because of board,friend no differencein this exam than your 9th class exam2.Attempt easy questions first,Do not waste time for difficult questions first3.Best thing attempt questions first from your interesting topics4.Check solution of all question even some questions are left do not worry5.If possible cross check your answers.6.After doing all above,write some thing possible for unsolved questions.Now try not to leaveany question unanswered.All the best for exam\xa0 | |
| 28703. |
Find the roots of1/x - 1/x-2 = 3 |
|
Answer» First take LCM of the left hand side of the equation. Then transpose the denominator to RHS . u will get a quadratic eq. Solve for roots. I hope i was helpful ? Sorry+2 3x^2-6x-2is the equation solve it |
|
| 28704. |
If P is a prime number then p is a _______number |
| Answer» Odd except 2 | |
| 28705. |
If sec + tan= p then find the value of cosec |
| Answer» {tex}sec\\ \\theta+ tan\\ \\theta = p{/tex} ...(i)Also {tex}sec^2\xa0\\theta - tan^2 \\theta = 1{/tex}{tex}\\Rightarrow{/tex}\xa0(sec\xa0{tex}\\theta{/tex}\xa0- tan\xa0{tex}\\theta{/tex}) (sec\xa0{tex}\\theta{/tex}\xa0+ tan\xa0{tex}\\theta{/tex}) = 1{tex}\\Rightarrow{/tex}\xa0p(sec\xa0{tex}\\theta{/tex}\xa0-\xa0tan\xa0{tex}\\theta{/tex}) = 1[using equation (i)]{tex}\\Rightarrow{/tex}\xa0sec\xa0{tex}\\theta{/tex}\xa0-\xa0tan\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{1}{p}{/tex}\xa0...(ii)(ii) - (i) we get{tex}-2 tan{/tex}\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{1-p^{2}}{p}{/tex}{tex}\\Rightarrow{/tex}- tan\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{1-p^{2}}{2 p}{/tex}{tex}\\Rightarrow{/tex}- cot\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{2 p}{1-p^{2}}{/tex}cot\xa0{tex}\\theta{/tex}\xa0{tex}=\\left(\\frac{2 p}{1-p^{2}}\\right)^{2}{/tex}{tex}=\\frac{-4 p^{2}}{\\left(1-p^{2}\\right)^{2}}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}cosec^2{/tex}\xa0{tex}\\theta{/tex}\xa0- 1\xa0{tex}=\\frac{-4 p^{2}}{\\left(1-p^{2}\\right)^{2}}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}cosec^2{/tex}\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{-4 p^{2}}{\\left(1-p^{2}\\right)^{2}}+1=\\frac{-4 p^{2}+\\left(1-p^{2}\\right)^{2}}{\\left(1-p^{2}\\right)^{2}}{/tex}{tex}cosec^2{/tex}\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{-4 p^{2}+1+p^{4}-2 p^{2}}{\\left(1-p^{2}\\right)^{2}}{/tex}{tex}=\\frac{\\left(1+p^{2}\\right)^{2}}{\\left(1-p^{2}\\right)^{2}}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}cosec\\\xa0\\theta{/tex}\xa0{tex}=\\frac{1+p^{2}}{1-p^{2}}{/tex} | |
| 28706. |
How to solve a problem by trigonometric identities |
|
Answer» Please reply with example But easy way is 2 convert it into sin and cos Depends upon the type of question...... |
|
| 28707. |
Value of cot A in all ternimetory ration |
| Answer» | |
| 28708. |
2+2+24=? |
|
Answer» Whats that? Stupid question 28 28 |
|
| 28709. |
The decimal expressions of 6 / 1250 will terminate after how many place of decimal |
|
Answer» 4 places 4 decimal place 0.0048 .0048 ans.... 4 places... |
|
| 28710. |
If sin theta equal to cos theta then find the value of 2 10 cube theta + sin squared theta |
| Answer» If you write answer in the form of sin theta then it will be 2 + sin square theta or if we write answer in cos theta then it will be 3 - cos squared theta | |
| 28711. |
CosA +SinA =1 |
| Answer» Cos^2a+sin^a=1 | |
| 28712. |
What is other name of pythagorus theorem |
|
Answer» Baudhayana (an Indian scientist) theorem Baudhayana theorem is somewhat like pythagoras theorem Other name of pythagoras is (pythagorean theorem) No it is other name of bpt Thales theorem |
|
| 28713. |
Is it true that 70% ncert will come in main examination. Is ncert sufficient to score good marks. |
|
Answer» Well thanks. Bhai 100% my teacher told me Bai 70%nahi97% Yup , ncert is sufficient!!!? Just go through last years cbse papers along with ncert for above 90%..........this is all my teacher told me...... |
|
| 28714. |
FIND TWO CONSECUTIVE POSITIVE INTEGER SUM OF WHOSE SQUARE 365 |
|
Answer» 13 and 14 13 and 14 |
|
| 28715. |
Is super 20 book is nice for maths anyone used it pls answr me... |
|
Answer» No idea.....?? You used it Yes it is |
|
| 28716. |
Euclids division algorithum find hcf of 455,42 |
| Answer» 455=42×10+35. 42=35×1+7. 35=7x5 +0. :. H.C.F=7 | |
| 28717. |
2×2+3+4÷6 |
|
Answer» =2*2+3+4/6=4+3+4/6=7+0.667=7.667 7.6666.........?? 23/3 ? 23/2 ? Guess11/6 |
|
| 28718. |
If sec = 3x and 3 by X equal to tan find 9 ( x -1 by x square) |
| Answer» 1....... | |
| 28719. |
Sin^6 theta - cos ^6 theta =? |
|
Answer» How ??? 45 diger |
|
| 28720. |
Prove that :, Tan^4 theta + tan^2 theta = sec^4 theta - sec^2 theta. |
| Answer» LHS:-Tan^4 theta +Tan^2 thetaTan^2 theta (Tan^2 theta +1)Tan^2 theta *Sec^2 thetaRHS:-Sec^4 theta - Sec^2 thetaSec^2 theta (Sec^2 theta -1)Sec^2 theta* Tan^2 theta. | |
| 28721. |
Write the value of (256)0.25 |
|
Answer» 64 Hddjeiwkgivusqkfogoewo |
|
| 28722. |
X cos theta/ a+ y sin theta/ b= 1and... |
| Answer» Complete your question | |
| 28723. |
sir plz give me 2017-18 boards sample paper which is released by cbse plzz sir |
| Answer» See from google by writing cbse sample paper | |
| 28724. |
How to convert 8.4hrs into hours |
| Answer» Into hrs??? | |
| 28725. |
Which term of the AP : 121, 117 , 113,....., is its first term negative term? |
|
Answer» Given AP is:121, 117, 113, ................First term a = 121Common difference d = 117-121 = -4Let nth term is the first negative term in APNow nth tern in AP = a + (n-1)*d= 121 + (n - 1)*(-4)= 121 - 4n + 4= 125 - 4nNow, we have to find the suitable value of n so that the value of (125 - 4n) is negative.⇒ 125 - 4n < 0⇒ 125 < 4n⇒ 4n > 125⇒ n > 125/4=> n > 31.25Since the term can not be in decimal form (term is always taken a positive integer)\xa0So, n = 32\xa0Now, by taking value of n = 32, we getnth term = 125 - 4× 32 = 125 - 128 = -3This is the first negative term in the series.So, n = 32Hence, the\xa032nd\xa0term is the first negative term in AP. 32nd term...... |
|
| 28726. |
Find the coordinates of the point which divide the join of (-1,7) and(4-3) in the ratio 2:3 |
|
Answer» Let x1 = -1 , x2 = 4 , y1 = 7 and y2 = -3 , m1 = 2 and m2 = 3\xa0Using Section Formula to find coordinates of point which divides join of (–1, 7) and (4, –3) in the ratio 2:3, we getx = (m1x2 + m2x1)/(m1 + m2) = (8-3)/ 5 = 5/5 = 1y = (m1y2 + m2y1)/(m1 + m2) = (-6 + 21)/ 5 = 15/5 = 3Therefore, the coordinates of point are (1, 3) which divides join of (–1, 7) and (4, –3) in the ratio 2:3. (1,3) the point is (1,3) 1,3 |
|
| 28727. |
Find the sum of the first 15 multiple of 8 |
|
Answer» The positive integers which are multiple of 8 are:8,16,24,................up to 15 termsIt forms an arithmetic series.Now first term a = 8, common difference d = 8, number of terms n = 15Now sum of first 15 terms of AP = (n/2) × {2a + (n-1) × d}= (15/2)× {2× 8 + (15-1)× 8}= (15/2)× (16 + 14× 8)= (15/2) × (16 + 112)= (15 × 128)/2= 15 × 64 (when 64 and 2 is divided by 2) = 960So sum of first 15 terms which are multiple of 8 = 960 960 |
|
| 28728. |
Prove that (a+b)2 |
|
Answer» (a+b)² = a²+2ab+b²(a+b)(a+b) = a²+2ab+b²a(a+b)+b(a+b) = a²+2ab+b²a²+ab+ab+b² = a²+2ab+b²a²+2ab+b² = a²+2ab+b² Question incomplete h |
|
| 28729. |
Pythagoras theoram proof |
|
Answer» SORRY DUE TO SCANNING ERROR I REPEAT AGAIN.YOU CAN COPY AND PRINT IF REQUITED. AC* = AB* + BA* THAT\'S PROOF See from ncert book unit 6 theorm 6.9 See from ncert |
|
| 28730. |
Find the quadratic equation,if x=root5+root5+root 5......infinity and x is natural number |
| Answer» leave first root 5 then take all other root 5 as xthen write x=root 5 +x . Then if yousquare this equatio you can get the answer | |
| 28731. |
TanA÷1-cotA+cotA÷1-tanA=1 +secA×cosecA |
| Answer» We have,{tex}\\mathrm { LHS } = \\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan A } { 1 - \\frac { 1 } { \\tan A } } + \\frac { \\frac { 1 } { \\tan A } } { 1 - \\tan A }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan A } { \\frac { \\tan A -1 } { \\tan A } } + \\frac { 1 } { \\tan A ( 1 - \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A - 1 } + \\frac { 1 } { \\tan A ( 1 - \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A - 1 } - \\frac { 1 } { \\tan A ( \\tan A - 1 ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 3 } A - 1 } { \\tan A ( \\tan A - 1 ) }{/tex}\xa0[Taking LCM]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { ( \\tan A - 1 ) \\left( \\tan ^ { 2 } A + \\tan A + 1 \\right) } { \\tan A ( \\tan A - 1 ) }{/tex}\xa0[{tex}\\because{/tex}\xa0a3\xa0- b3\xa0= ( a - b )(a2\xa0+ ab + b2)]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A + \\tan A + 1 } { \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A } + \\frac { \\tan A } { \\tan A } + \\frac { 1 } { \\tan A }{/tex}{tex}\\Rightarrow{/tex}\xa0LHS = tanA + 1 + cotA [ since (1/tanA) =cotA ].= (1 + tanA + cotA){tex}\\therefore \\quad \\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}\xa0= 1 + tanA + cotA ...........(1)Now, 1 + tanA + cotA = 1 +\xa0{tex}\\frac { \\sin A } { \\cos A } + \\frac { \\cos A } { \\sin A }{/tex}\xa0= 1 +\xa0{tex}\\frac { \\sin ^ { 2 } A + \\cos ^ { 2 } A } { \\sin A \\cos A }{/tex} = 1 +\xa0{tex}\\frac { 1 } { \\sin A \\cos A }{/tex}\xa0[{tex}\\because{/tex}Sin2A + Cos2 A = 1 ]\xa0= 1 + cosecAsecASo, 1 + tanA + cotA = 1+ cosecAsecA.......(2)From (1) and (2), we obtain{tex}\\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}\xa0= 1 + tanA + cotA = 1 + cosecAsecA | |
| 28732. |
Ex 13.26 |
| Answer» | |
| 28733. |
solve rhis question☺4+2 |
|
Answer» 6.......??????? Of course 6 |
|
| 28734. |
The 8th term of ap is 35 and 102th term of ap is 317 find ap |
|
Answer» That is a+7d=35 Ya Ya... Sry I have done mistakes A+7d=35+a+101d=317 solve this a+9d=35a+101d=317Solve it n u wl find the value of a and d |
|
| 28735. |
What is the HCF of prime number and composite number |
|
Answer» Either the answers is 2 or 4?? Write full question abhinav ji |
|
| 28736. |
Find the value of K for which the point A (K+1,2K) ,B(3K,2K+3)&C(5K-1,5K) are collinear |
| Answer» Use area of triangle formula.. U will get the answer | |
| 28737. |
Har bar exam mea ane wala question |
| Answer» google cbse tute | |
| 28738. |
Hamesha ane wala question |
| Answer» Theorem 1 or 3 of circle. | |
| 28739. |
The value of cos^2A= |
|
Answer» 1-sin^2A or Subtract Sin^2A from both sides to get Cos^2A! Since Sin^2A+Cos^2A=1 , Therefore, transpose Sin^2A to the RHS and hence you get your answer... Cos^2A=1-Sin^2A |
|
| 28740. |
what real root |
|
Answer» When d=0it exit real root i.e., real nos. Real roots are roots that are real! |
|
| 28741. |
Obtain all the zeroes of the polynomial x^4-5x^3+2x^2+10x-8 if the zeroes are √2 and -√2 |
|
Answer» The other zeroes are 1 and 4.....☺️☺️ 4 and1 |
|
| 28742. |
How do we find the median clss in median |
|
Answer» Median class is that whose cumilative freq. Is just greater than n/2.Where n=sigma fi Divide no. Of rows by 2Ex if you have 7 rows 7/2=3.5Median class =4th row If you have 6 row median class =3 Which have high c.f.from my side |
|
| 28743. |
If secA+tanA=p show that secA-tanA=1/p hence find the values of sinA and cosA |
|
Answer» Sec A + Tan A =p -equation 1We know thatSec^2 A - Tan^2 A = 1(Sec A -Tan A)(Sec A+ Tan A)=1[a^2 -b^2 =(a+b)(a-b)]From equation 1(Sec A - Tan A)(p) = 1Sec A - Tan A=1÷p -equation 2Adding equations 1 and 2Sec A - Tan A +Sec A+ Tan A=(1+p^2)/p2 Sec A =(1+p^2)/pSec A=(1+p^2)/2pNow you can easily find the value of sin and cos Answer,SecA+tanA=p --------------eq1Now,Sec^2A-tan^2A=1. (a^2-b^2)(SecA+tanA)(secA-tanA)=1 (a+b)(a-b)p(secA-tanA)=1. (using eq1)secA-tanA=1/p PROVED Shift tan to rhs and sq. on both sides and find value of tan in trms of p and put the values in a triangle and then find the t ratios. |
|
| 28744. |
What are minimum passing marks in CBSE? |
|
Answer» 80/27100/33 marks are necessary in board exams 33 only 33/100 |
|
| 28745. |
Find the value of k,for which one root of the quadratic equation kx2-14x+8=2 |
|
Answer» 5 You are right 5 |
|
| 28746. |
If tan a= 5by 12 find the value of ( sin a+ cosa) seca |
|
Answer» 17/12 17/12 17/12 |
|
| 28747. |
How many important chapter in math |
|
Answer» Saare chap se 5 se 6 no ke ques aate h. 1,2,3,7,11,14,15, If you need only passing marks then prepare ch1,14 15 All 15 chapters are important.... ?? Solo??? |
|
| 28748. |
ncube -n is divisible by 6 |
| Answer» | |
| 28749. |
Where we use rational and irrational numbers in our daily life? |
| Answer» | |
| 28750. |
If cos(a+b)=0,then sin(a-b)can be reduced to |
| Answer» | |