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.
| 10601. |
Show that one and only one out of n, n+2,or n +4 is divisible by 3, where n is any positive integer |
| Answer» On dividing n by 3, let q be the quotient and r be the remainder.Then, {tex}n = 3q + r{/tex}, where {tex}0 \\leq r < 3{/tex}{tex}\\Rightarrow\\;n = 3q + r{/tex} , where r = 0,1 or 2{tex}\\Rightarrow{/tex}{tex}n = 3q \\;or \\;n = (3q + 1) \\;or\\; n = (3q + 2){/tex}.Case I If n = 3q then n is clearly divisible by 3.Case II If\xa0{tex}\\;n = (3q + 1)\\; {/tex} then {tex} (n + 2)= (3q + 1 + 2) = (3q + 3) = 3(q + 1){/tex}, which is clearly divisible by 3.In this case, {tex}(n + 2){/tex} is divisible by 3.Case III If n = {tex}(3q + 2){/tex} then {tex}(n + 1) = (3q + 2 + 1) = (3q + 3) = 3(q + 1){/tex}, which is clearly divisible by 3.In this case,{tex} (n + 1){/tex} is divisible by 3.Hence, one and only one out of {tex}n, (n + 1){/tex} and {tex}(n + 2){/tex} is divisible by 3. | |
| 10602. |
CosA=CosB prove prove that A=B |
|
Answer» CosA=CosB, B/H=B/H, AC/AB=BC/AB, AC=AB, /_A= /_B (angle opposite to equal sides are equal) cosA=cosBcos(90-A)=cos(90-B)90-A=90-B-A=-BA=B |
|
| 10603. |
COT |
| Answer» Cotangent (cot)= base / perpendicular | |
| 10604. |
Prove that p+q th term and p-q th term of an Ap is same as double of its p th term |
| Answer» | |
| 10605. |
How many term are there in the Ap 41 38 35 .......8? |
|
Answer» 12 correct 12 terms This does not comes in AP 12 term |
|
| 10606. |
Sin theta=a²-b²%a²+b² |
| Answer» | |
| 10607. |
6x+2√5+5 |
| Answer» | |
| 10608. |
SinA+cosA=2sin (90-A) |
| Answer» | |
| 10609. |
Define Euclid\'s division lemma |
| Answer» It is an proven statement __________ | |
| 10610. |
What is thm of the triangle chip? |
| Answer» | |
| 10611. |
If Sn denotes the sum of nth term, and S2n=3Sn, find S2n/Sn |
| Answer» S2n=3Snthen S2n/Sn=3 | |
| 10612. |
tanA.cotA |
| Answer» 1 | |
| 10613. |
117,124,131........which will be the first AP negative term? |
| Answer» No answer | |
| 10614. |
A line touches a circle of radius 4cm.another line |
| Answer» The question is incomplete | |
| 10615. |
Solve 2x+3y=11 and 2x -4y=-24 and hence find the value of m for which y= mx+3 |
| Answer» Y=5 x=-2 m=-1 | |
| 10616. |
find the value ofsec 60` geometrecally |
| Answer» Ncert uthaker dekho bhai sabkuch ans mil jayega introduction me sari cheez hai | |
| 10617. |
area of triangle |
|
Answer» 1/2×b×h Tumhari life me is ki jaroorat ka hain muje batao daily life mein daily life mein |
|
| 10618. |
What is Integer ? |
| Answer» A number which is not a fraction; a whole number is known as integer. | |
| 10619. |
if cos(9a)=sin(a) and a |
| Answer» Cos(9a)=cos(90-a)So 9a=90-a=> 10a=90a=9Tan(5a)= tan45= 1 :) | |
| 10620. |
what type of triangle |
| Answer» | |
| 10621. |
x+y25 |
| Answer» X + y =25 | |
| 10622. |
Trignometary |
| Answer» | |
| 10623. |
The three vertex of a parallelogram are (3,4, (3,8)and (9,8)find the fourth vertex? |
| Answer» let ABCD is a IIgram with A(3,4) , B(3,8) , C(9,8), D(X,Y)AC and BD are diagonals which bisect each other at mid point O\xa0thus mid point of AC=(3+9)/2, (4+8)/2 = 6, 6mid point of BD = (3+x)/2 , (8+y)/2mid point of AC = mid point of BD6, 6 = (3+x)/2 , (8+y)/2(3+x)/2=6 3+x=12 , x=9(8+y)/2=6 8+y=12 , y = 4the fourth vertex is (9 , 4)check ur answer 3+9/2 = 12/2 , 8+4/2 =12/2 6,6 ok\xa0 | |
| 10624. |
sin 60 value |
| Answer» √3/2 | |
| 10625. |
Cos/1-cos + cos/1+ sin =4 |
| Answer» | |
| 10626. |
An aeroplane when flying at a height of 5000 m |
| Answer» | |
| 10627. |
State Euclid\'s division algorithm. |
| Answer» Given positive integers a and b , there exist unique integers q and r satisfying a=bq+r , r is greater than or equal to zero but less then b. | |
| 10628. |
Prove that: 1/sinA-cosA = Sina+ cosA |
| Answer» | |
| 10629. |
How will be the level of questions in board exam |
|
Answer» 60 no. Ka aasani se kerdoge magar 20 no. Ka hila dega in case of mathematics in science value based chemistry me common name jaroor yad kerlena s.st me previous years paper ke kuch question yad ker lena Very tough Easy if you study |
|
| 10630. |
In maths board exam question from rd sharma will come or only NCERT questions will come??? |
|
Answer» No any questions will come from NCERT all question \'s will come from outside 60% from ncert and 40% other resources ncert is more good and it\'s basic should be clear then you should prefer rdsharma it is better for securing good marks in the exam . In CBSE BOARD exam questions will be from both books and more from rdsharma . BEST OF LUCK ............ 80%ncert and 20%other |
|
| 10631. |
If 2sina=1 and a is an acute angle, then find the value of (3 cosa-4cos*cos*cosa) |
| Answer» | |
| 10632. |
4/x+5/y=7 and x=-9/3 |
|
Answer» Eazzzzzzzzy Use substitution youll get the answers easily |
|
| 10633. |
A triangle dpb and PBA ,angledpa45 Angle. Bpa30find db when pb perpendicular da and PBA 10m |
| Answer» | |
| 10634. |
What are the identities of trigonometry |
|
Answer» Are the koi questions hai identies to maths e ch trigonometry Mai hai Refer to book |
|
| 10635. |
what is the consistency |
|
Answer» In maths, consistency means having at least one solution. In regard of quadratic equation there are two cases-(i) unique solution (ii) infinitely many solutions. Consistency matab ek jagah jiske sare vote ek jagah pade aur be vote ek list mo ho |
|
| 10636. |
How to find cf of less and more than graph |
| Answer» | |
| 10637. |
The sum of two number is 8. Determine the numbers if the sum of their reciprocal is 8/15. |
| Answer» let nos be x and 8 - x1/x + 1/8 -x\xa0=8/15(8-x+x)15=8(xx(8-x)120=64x - 8x28x2 - 64x +120=0x2-8x+15=0x2-5x-3x+15=0x(x-5) - 3(x-5)=0x=5 , 3\xa0 | |
| 10638. |
If tan theta is equal to root 2 minus 1 show that 4 sin theta cos theta is equal to root 2 |
| Answer» | |
| 10639. |
Derivation of s-d formula |
| Answer» | |
| 10640. |
34+54 |
|
Answer» 88 Bhai ix class kahan se kiya hame bhi bata dete ham bhi bahi se karlete 88 88 88 |
|
| 10641. |
A mixed fractions is :7/25 |
| Answer» | |
| 10642. |
5+3+-25-32 |
|
Answer» Deepak bhai app x ke student ko ye sab batane ki jagah 7ke st. Ko samjaho aur inse pusho ku ix kahan se kiyahai 8-57 = -49 5+3+-25-32 (Here - - becomes+)5+3+(-25-32)5+3+(-57) (here + - becomes -)8+ (-57)-49 is solution |
|
| 10643. |
2222222222222*2222222 |
| Answer» Answer s.m. hi aye ga kerke dekhlo | |
| 10644. |
if the sum of the first 14 terms of an AP is 1050 and first term is 10 then find the 20th term |
|
Answer» s14=14/2(2a+13d)1050=7(20+13d)1050=140+91d910 = 91dd=10a=10 (given)a20=a+(19d) = 10+190 = 20020th term = 200 ATQ...... 20TH TERM=200A=10D10 |
|
| 10645. |
if the sum of first 14 terms of an AP is 1050 and its first term is 10, find the 20th term. |
| Answer» Given, a = 10, and S14\xa0= 1050Let the common difference of the A.P. be d{tex}\\therefore \\quad S _ { n } = \\frac { n } { 2 } [ 2 a + ( n - 1 ) d ]{/tex}{tex}\\therefore \\quad S _ { 14 } = \\frac { 14 } { 2 } [ 2 \\times 10 + ( 14 - 1 ) d ]{/tex}\xa01050 = {tex}7 (20 + 13 d ){/tex}{tex}20 + 13d{/tex} =\xa0{tex}\\frac{1050}{7}{/tex}\xa0{tex}20 + 13d{/tex} = 150{tex}13d{/tex} = 150 - 20{tex}13d{/tex} = 130{tex}d{/tex} =\xa0{tex}\\frac{130}{13}{/tex}\xa0= 10{tex}a_{20} = a + (n - 1)d{/tex}= 10 + (20 - 1) 10= 10 + 19\xa0{tex}\\times{/tex}\xa010= 10 + 190= 200Hence, a20\xa0= 200 | |
| 10646. |
If tanA=cotB, prove that A+B=90 |
|
Answer» ok merko nhi aata tan A = cot Bi.e tan A = tan ( 90 - B )A = 90 - BA + B = 90 |
|
| 10647. |
Prove that 1+7=8 person will be awarded with rupees 100000000000000000000000000....... |
| Answer» | |
| 10648. |
(tanA+cosecB)^2—(cotB–secA)^2=2tanAcotB(cosecA+secB) |
| Answer» We have,LHS = (tanA + cosec B)2 - (cotB - sec A)2{tex}\\Rightarrow{/tex}\xa0LHS = (tan2A + cosec2B + 2tanA cosecB ) -\xa0(cot2B + sec2A\xa0- 2cotB secA){tex}\\Rightarrow{/tex}\xa0LHS = (tan2A - sec2A)+(cosec2B - cot2B)+2tanA cosecB +2cotB secABut, Sec2A - tan2A =1 & cosec2A - cot2\xa0A = 1{tex}\\therefore{/tex}\xa0LHS = -1 + 1 + 2 tanA cosecB + 2cotB secA{tex}\\Rightarrow{/tex}\xa0LHS = 2 (tanA cosecB + cotB secA){tex}\\Rightarrow{/tex}\xa0LHS = 2 tanA cotB{tex}\\left( \\frac { cosec\\: B } { \\cot B } + \\frac { \\sec A } { \\tan A } \\right){/tex}\xa0[Dividing and multiplying by tanA cotB]{tex}\\Rightarrow{/tex}\xa0LHS = 2tan A cotB{tex}\\left\\{ \\frac { \\frac { 1 } { \\sin B } } { \\frac { \\cos B } { \\sin B } } + \\frac { \\frac { 1 } { \\cos A } } { \\frac { \\sin A } { \\cos A } } \\right\\}{/tex}\xa0[Since, CosecA.SinA =1 , SecA.cosA =1, (sinA/cosA)= tanA & (cosA/SinA) =cotA ]{tex}\\Rightarrow{/tex}\xa0LHS = 2 tanA cotB{tex}\\left( \\frac { 1 } { \\cos B } + \\frac { 1 } { \\sin A } \\right){/tex}\xa0= 2tanA cotB ( secB + cosecA ) = RHS. Hence, proved. | |
| 10649. |
value |
| Answer» | |
| 10650. |
prove the valu of sin45 |
| Answer» | |