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.
| 40901. |
Can two numbers have 18 as thierHcF and 380 as thier lcm? Give reason |
|
Answer» Yes because all numbers have lcm as well as hcf No because LCM is not the factor of HCF |
|
| 40902. |
Sec A +tan A= p then find the value of cosec A |
| Answer» 1/cosA + sinA/cosA = p1 + sinA = pcosAcosA = p root (1 - sin2A) (1 + sinA)2 = p2(1 - sin2A) D = 4 - 4 (1 + p2)(1 - p2) / 4 - 4(1 - p4) =4p4SinA = - 2 +_root(4p4)/ 2(1 + p2) - 1+_p2/1 + p2 p2-1/p2+1 = -1CosecA = p2-1/p2-1 = - 1 | |
| 40903. |
Which term of the progression 20, 19¼, 18½,. 17¾............ Is the first term negative term. |
|
Answer» An will be less than 0 Aditya apne question kya pucha hai 29th term Can you give me answer of my question Answer please |
|
| 40904. |
1+cos-sin²/sin(1+cos)=cot |
| Answer» 1+cos-sin^2/sin(1+cos)Here we can write sin^2 as 1-cos^2Therefore , 1+cos-(1-cos^2)/sin(1+cos)=1+cos-1+cos^2/sin(1+cos)= cos+cos^2/sin(1+cos)= cos(1+cos)/sin(1+cos)= cos/sin= CotHence Proved | |
| 40905. |
Which is which is important question in 7 chapter maths NCRT books |
| Answer» Ques no . 1,2,3,4,and6 of ex. 7.4( optional) | |
| 40906. |
Koi gkp ka hai |
|
Answer» गोरखपुर Gkp matlab |
|
| 40907. |
Which is importent question in probability |
|
Answer» Card n dice ques More than the cards question try out rd sharma Questions related to card Card question |
|
| 40908. |
2*cos^2 77°+1+2*cos^2 13°/5*sec^2 25°+8-5*cot^2 65 |
| Answer» Change any cos into sin and sec into cosec...... U will get formulae... Apply them.. | |
| 40909. |
Ncert optional questions should be studied or not? |
|
Answer» Yes they should be given as much as importance as the other exercises...they r helpful too?? Yes it also helps us |
|
| 40910. |
Trigonometric identities |
| Answer» Sin2 +cos2=1Cosec2=cot2+1 | |
| 40911. |
find the positive value of k for which the equation 2x-5x-4=0 and 2x -8x + k |
| Answer» I think the question is wrong there is no sitiuation like infinite many solution ,no solution and unique solution so plz check and ur equation is also wrong. | |
| 40912. |
Can you tell me all the formula of class 10 ? Chapterwise |
| Answer» | |
| 40913. |
I want to score above 90% to mare sara course kab tak khatam ho jana chahiye........ |
|
Answer» December tak your whole course should be completed till now and you should practice more in which subjects you think you have to practice more |
|
| 40914. |
If the diameter of a semicircular protector is 42 unit. Then find its perimeter |
| Answer» Diameter is 42 units therefore r is 21 units Perimeter of semicircle=diameter+length of the arc =42+(22÷7×21) =42+66 =108cm | |
| 40915. |
How to solve sin theotta+ cos thetta |
| Answer» Sin theta+cos theta is in the simplest form. It cannot be simplified furthur..... | |
| 40916. |
If secQ + tanQ =.then find the value of cosec Q |
| Answer» | |
| 40917. |
show that an even integer is of the form 6q and 6q+2 for 6q+4 where q is a positive integer |
| Answer» let a be any positive integerthenb= 6a= bq+r0≤r | |
| 40918. |
Find the nature of roots in the quadratic equation x2-4x-3under root 2 |
| Answer» Real and diff. root | |
| 40919. |
If sec theta + tan theta = p , then find cosec theta |
| Answer» Sec=1/cos and tan= sin/cos then cos lcm and 1+sin/cos so, 1+sin/cos=p:1+sin=pcos andsin=pcoss-1andsin=1/cosectocosec=1/pcoss-1 | |
| 40920. |
If nth term of an ap is 2n+1 what is thw sum of its firts three terms |
|
Answer» Pagal ha tu is the only one who can be in a position to help If nth term of an A.P = 2n+1then, first term of A.P = 2(1)+1= 2+1= 3then, second term = 2 (2) + 1》4+1》5third term = 2 (3) +1》6+1》7sum of terms = 3+5+7sum of terms = 15 |
|
| 40921. |
If one zero of polynomial p(x)= (k+4)x2+13 x+3k is reciprocal of the other then find value of k |
|
Answer» Pls give full solution...It will help me understand better☺☺ The image is the only one I could think to do so it is the best I could do to help out the process for this one of those days when we were in our room The answer is 8. |
|
| 40922. |
Root 3 sin theta minus cos theta equals to zero then find the value of theta |
| Answer» | |
| 40923. |
Guys Kya maths ke do paper h means standard level or higher level plz reply plz. |
|
Answer» It is from next year......... Yes, but from next year maths ke do paper hoge higher and lower No it is from next year |
|
| 40924. |
Plzxx tell what re the imp. Questions of chapter 13 |
|
Answer» 1 year Koi bata Sakta hai mujhe ki kisko kitne din hue iss app ko use Karte Sab important questions hai Sab padho tabhi sahi hoga Full chapter is very important |
|
| 40925. |
If the 3rd and 9th term of an Ap is 4 and -8 respectively which term of this Ap is zero |
|
Answer» Please reply me a+2d = 4 …(1)a+8d = -8 …(2)Deduct (1) from (2), to get6d = -12, or d = -2.From (1), a = 4–2d = 4+4 = 8.So the numbers of the AP are- 8, 6, 4, 2, 0, -2, -4, -6, -8.So the 5th term is zero. |
|
| 40926. |
Evaluate: Sin 18° _________ Cos 72° |
|
Answer» Sin 18°upon cos 72°Sin changes to cos So ,sin(90°-18°) upon cos 72°Sin will change to cos 72 °upon cos72°Then cos 72° will cancel with cos72°The answer will be (1) Answer is zero (0) |
|
| 40927. |
Prove the BPT convers theorem |
| Answer» Take a triangle ABC with CD as parallel points on the side AB and AC. We need to prove CD||BC.By the bpt theorem we came to know thatAC/CB = AD/DClet us assume on contradiction that cd is not ll to bc.If CD is not ll to BC then there must be another line which is ll to it. Let us assume that ll line as CE from C. If CE is ll to BC then AC/CB = AE/EC. On substituting bpt theorem and this we get AE/EC = AD/DC. Therefore, CE and CD is same line. Hence proved | |
| 40928. |
Prove that (cosecΦ - sinΦ)(secΦ - cosΦ)(tanΦ + cotΦ) = 1 |
| Answer» (1by sin theta-sin theta) (1by cos theta- cos) (sin by cos +cos by sin) =(1-sin square theta/sintheta) (1-cos square theta/cos theta)[ (sin sq. +cos sq theta) /sin cos] =(cos sq. /sin) (sin sq. Theta/cos theta) (1/sin cos) =1 | |
| 40929. |
Ch triangle state and proof thales theorm |
|
Answer» If there is any other way of sending pic i will send u the proof. Kk Statement: if a line is drwn ll to 1 side of a triangle to intersect the other 2 sides in distinct points the other 2 sides are divided in the same ratio. |
|
| 40930. |
CosecA +cosA =x find the value of cosecA -cosA |
|
Answer» It\'s right Pls check your ques |
|
| 40931. |
Cosis |
|
Answer» Question? What\'s your question Nikhil Please check the question.... |
|
| 40932. |
7th time of 7th term is equal to the 11 time of 11th term find the 18th term |
|
Answer» The 18th term will be equal to 0, Sourabh Because 7a+42d = 11a+110d4a+68d = 0a+17d = 0. ( 18 th term = a+17d 18th term is equal to 0 |
|
| 40933. |
Show that every positive odd integer in of the form (6m+1) or (6m+3) or (6m+5). |
| Answer» Let n be a given positive odd integer.On dividing n by 6, let m be the quotient and r be the remainder.Then, by Euclid\'s division lemma, we have{tex}n = 6m + r{/tex}, where {tex}0 \\leq r < 6{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}n = 6m + r{/tex}, where {tex}r = 0, 1, 2, 3, 4, 5{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}n = 6m\\ or\\ (6m + 1){/tex}{tex}or\\ (6m + 2)\\ or\\ (6m + 3)\\ or\\ (6m + 4)\\ or\\ (6m + 5).{/tex}But, {tex}n = 6m, (6m + 2), (6m + 4){/tex} give even values of n.Thus, when n is odd, it is of the form {tex}(6m + 1)\\ or\\ (6m + 3)\\ or\\ (6m + 5){/tex} for some integer m. | |
| 40934. |
Two no are 18and 24 if hcf is 6find their LCM |
|
Answer» LCM*HCF= A*BSO LCM=18*24/6=72 LCM is equal to the 72 |
|
| 40935. |
Probability chapter solution |
| Answer» You can see the solutions by this app only.. | |
| 40936. |
How to find root of any number |
|
Answer» By adding you asw By prime factorisation method |
|
| 40937. |
Prove that if x and y are odd positive integers, then x2 +y2 is even but not divisible by 4 |
|
Answer» Thanks Let the two odd positive no. be x = 2k + 1 and y = 2p + 1Hence, x2\xa0+ y2 = (2k + 1)2\xa0+(2p + 1)2 = 4k2\xa0+ 4k + 1 + 4p2\xa0+ 4p + 1 = 4k2\xa0+ 4p2\xa0+ 4k + 4p +\xa02 = 4 (k2\xa0+ p2\xa0+ k + p) + 2\xa0clearly, notice that the sum of square is even the no. is not divisible by 4hence, if x and y are odd positive integer, then x2\xa0+ y2\xa0is even but not divisible by four Shut up re |
|
| 40938. |
Find the value of p for which the Equation px2-5x+p=0 has equal roots |
|
Answer» Roots of px2-5x+p=0 are equal\xa0if (-5)2=4p(p)4p2=25p2=25/4\xa0p=5/2 The value of p is 5/2. |
|
| 40939. |
Is number \'7*11*13+13+13*2\' a composite number? |
|
Answer» Yes it is.A composite number is a one which has at least 1 factor other than 1 and itself.Since the given number,i.e.,13(7x11+1+2) is divisible by 13,hence it is a composite number(by definition). Hope it help you.#THANKS! Yes it is a composite number because it is product of primes.As 7×11×13+13+13×2=13(7×11×1+1+1×2)=13(77+1+2)=13(80)=13×2×2×2×2×5 |
|
| 40940. |
How can we solve sin*2 + cos*2 = 1 |
|
Answer» by pythagoras theorembase2 + heigh2=hypo2divide by hypo2base2/hypo2 + height2/hypo2=hypo2/hypo2sin2x+cos2x=1\xa0 By relate trignometery with Pythagoras |
|
| 40941. |
In pallagramPQRS l is mid point ofQR prove mis point of trisection on pr |
| Answer» | |
| 40942. |
If sec p + tan p =yFind the value of cosec p |
| Answer» The value of cosec p = y² + 1/y² - 1. | |
| 40943. |
How to do trignometric proves any trick |
|
Answer» Learn the identities, and if don\'t remember identities than just change the function, like sin into cosec, cos into sec and tan into cot Try to change the identity into sin or cos I think that there are no trics....just remember the concepts.......hope it will help u out...???? There are no tricks just learn the formulas No only learn formula |
|
| 40944. |
Common Kese lete h |
| Answer» Jo number ya vareable same ho usko common lata ha | |
| 40945. |
1(2+5) |
| Answer» 7 | |
| 40946. |
What is the maximum value of sin and minimum value of sin |
| Answer» 1and 1 only | |
| 40947. |
Solve for x x=1/2-1÷2-1/2-x |
| Answer» X is -1 | |
| 40948. |
If d is the hcf of45 and 27 then find x, y satisfying d= 27x+ 45y |
|
Answer» I am right or wrong? Here the value of x and y are 2 and -1 respectively The numbers are :-45 and 27Lets find their HCFBY Euclid\'s division Lemma :-a = bq + r45 = 27 x 1 + 1827 = 18 x 1 + 918 = 9 x 2 + 0HCF = d = 9Given:-d = 27x+45y9 = 27x + 45y9 = 27 - 18 x 118 = 45 - 27 x 1 Therefore ,9 = 27 - [ 45 x 27 x 1 ] x 1= 27 x 2 - 45 9 = 27 x 2 + 45 x [ -1 ]x = 2y = -1 |
|
| 40949. |
if p and q are the two roots of the equation x^2+px-q=0 |
|
Answer» Write ur question properly. Complete your question |
|
| 40950. |
find the polynomial whose zeroes are -5 and 4 |
|
Answer» Sum of zeros = –5 + 4 = – 1,Product of zeros = – 5 × 4 = – 20Required polynomial = x2 + x – 20 Polynomial will be (x square + x - 20) Firstle we find sum and product of zeroes and then we put this in xsquare-(alpha+beeta)x+alpha beeta The equation is (x square +x -20) |
|