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.
| 39651. |
The nth term of A.P. is 6n+2 find the common difference. Pls answer quickly |
| Answer» 6n+2 written as 2+6n=2+6n-6+6=8+(n-1)6 whic is of form a+(n-1)d. Hence,d=6 | |
| 39652. |
Find the least number that is divisible by first five even numbers |
|
Answer» 1st 5 even no. =0, 2,4,6,8L.C.M. of these no. = 24So the least no. that is divisible by 1st 5 even no. = 24 24 120 |
|
| 39653. |
Hey my friends please help me give me best suggestions for preparation of board exams |
| Answer» Gothrough all ncrt portions and examples and activities in maths and science | |
| 39654. |
3n×n-50n+80=0 |
| Answer» | |
| 39655. |
Questions 3 of chapter 5 exercise 5.2 |
| Answer» Where | |
| 39656. |
Can anyone explain prove of root two by contradiction method in easy way |
| Answer» | |
| 39657. |
How to solve formulas |
|
Answer» Try to make formulas not kust mug them up more you deeive them more properties you will learn and it will be difficult at the end of year to handle huge amount of matter so to ease up derive them on your own you will never forget them Learn and remember all the formulas thoroughly and practice as many as questions as u can..... Geography chapter 2 |
|
| 39658. |
the following distribustion gives the daily income of 50 workers of factory |
| Answer» | |
| 39659. |
prove that ¥11 is irrational number |
| Answer» | |
| 39660. |
Find the hcf of 65 and 117 and express it in the form of 65m+117n |
| Answer» First find the HCF of 65 and 117 by Using Euclid\'s division algorithm,117 = 65{tex}\\times{/tex} 1 + 5265 = 52{tex}\\times{/tex} 1 + 1352 = 13{tex}\\times{/tex} 4 + 0So, HCF of 117 and 65 = 13HCF = {tex}65m + 117n{/tex}For, {tex}m= 2{/tex} and {tex}n = -1{/tex},HCF = 65{tex}\\times{/tex} 2 + 117{tex}\\times{/tex} (-1)= 130 - 117= 13Hence, the integral values of m and\xa0n are 2 and -1 respectively and the HCF of 117 and 65 is 13. | |
| 39661. |
R.D. sharma class 9 mcq question no.37 |
| Answer» | |
| 39662. |
Find zeroes of p(x)=xcube-2x |
| Answer» x(x square - 2)x square - 2= 0x=√2ORX=0 | |
| 39663. |
Find the distance ofa point(x,y) from the origin |
|
Answer» (0-x)^2+(0-y)^2=x^2+y^2 ✓x2+y2 ( pathagores thorem) |
|
| 39664. |
Sin 67° + Cos 75° convert it into trigo rations |
|
Answer» Sin 90-75 Sin (90-67) + cos (90-75)=cos23+sin15 Sin(90°-67°) + Cos (90°-75°)Cos 23° + Sin 15° (ans) |
|
| 39665. |
Middle |
| Answer» | |
| 39666. |
The sum of the first n terms of an AP is (3n2/2+5n/2).Find its nth terms and the 25th term |
| Answer» According to the question,we are given that,\xa0{tex}S _ { n } = \\frac { 3 n ^ { 2 } } { 2 } + \\frac { 5 n } { 2 } = \\frac { 3 n ^ { 2 } + 5 n } { 2 }{/tex}{tex}\\Rightarrow S _ { n - 1 } = \\frac { 3 ( n - 1 ) ^ { 2 } + 5 ( n - 1 ) } { 2 }{/tex}{tex}= \\frac { 3 \\left( n ^ { 2 } - 2 n + 1 \\right) + 5 n - 5 } { 2 }{/tex}{tex}= \\frac { 3 n ^ { 2 } - 6 n + 3 + 5 n - 5 } { 2 }{/tex}{tex}= \\frac { 3 n ^ { 2 } - n - 2 } { 2 }{/tex}Now,nth term = Tn=Sn-Sn-1={tex}= \\frac { 3 n ^ { 2 } + 5 n } { 2 } - \\frac { 3 n ^ { 2 } - n - 2 } { 2 } = \\frac { 3 n ^ { 2 } + 5 n - 3 n ^ { 2 } + n + 2 } { 2 }{/tex}={tex}\\frac{{6n + 2}}{2}{/tex}=3n+125th term=T25=3(25)+1=75+1=76. | |
| 39667. |
find the largest number which divides 615 and 963 leaving remainder 6 in rach one |
| Answer» The required number when divides 615 and 963, leaves remainder 6, this means 615 - 6 = 609 and 963 - 6 = 957 are completely divisible by the number.Therefore,The required number = H.C.F. of 609 and 957.By applying Euclid’s division lemma957 = 609 {tex}\\times{/tex}\xa01+ 348609 = 348 {tex}\\times{/tex}\xa01 + 261348 = 216 {tex}\\times{/tex}\xa01 + 87261 = 87 {tex}\\times{/tex}\xa03 + 0.Therefore, H.C.F. of 957 and 609 is = 87.Hence, the largest number which divides 615 and 963 leaving remainder 6 in each case is 87. | |
| 39668. |
Evaluate sin60° geometrically |
| Answer» We consider any right angled triangle ABDWe have base BD = a,Perpendicular AD = √ 3aHipotenuse AB = 2a andAngle ABD = 60•Sin60• = AD / AB = √3a /2a Sin 60• = √3/2May this will help full for you ....? | |
| 39669. |
Sin48° × sec42° + cos48° × cosec48° = 2 |
|
Answer» Answer :............... ___sin 48• × sec( 90•-48•) + cos ( 90•- 48•) × cosec 48• =........___sin 48• ×cosec 48• + sin 48• × cosec48•= 2............___1+1 =2 :::::Hence LHS = RHS I think there is a mistake in the question.....It can be....... Sin 48•× sec 42• + cos 42•× cosec 48• = 2 |
|
| 39670. |
HsfcagavvHbz |
|
Answer» Hhhhhhhhhhhhhhhhbbbhhhhhhhhh Vsahmkhswnkoak |
|
| 39671. |
3x-4y=7,5x+2y=3 slove graphically |
| Answer» {tex}3x - 4y = 7\\ and\\ 5x + 2y = 3{/tex}The given system of linear equation is {tex}3x - 4y = 7\\ and\\ 5x + 2y = 3{/tex}Now, {tex}3x - 4y = 7{/tex}{tex}y = \\frac { 3 x - 7 } { 4 }{/tex}When x = 1 then, y = -1When x = -3 then y = -4\tx1-3y-1-4\tNow, 5x + 2y = 3{tex}y = \\frac { 3 - 5 x } { 2 }{/tex}When x = 1 then, y = -1When x = 3 then y = -6Thus, we have the following table\tx13y-1-6\tGraph of the given system of equations areClearly the two lines intersect at A(1, -1)Hence, x = 1 and y = -1 is the solution of the given system of equations. | |
| 39672. |
X3-3x2+5x-3÷x2-2 |
| Answer» | |
| 39673. |
What is the solution of sinA=2÷3 |
| Answer» | |
| 39674. |
SA 1 Sample Papers 2018-19 Class 10th |
| Answer» Name the sample paper | |
| 39675. |
Asinx |
| Answer» | |
| 39676. |
All formula of a.p |
|
Answer» S = n/2(a+l) First one is an= a+ (n-1)d second is sn =n/2(2a+(n-1)d |
|
| 39677. |
Find the |
| Answer» | |
| 39678. |
Prove the value of sin30 geometrical |
| Answer» 1/2 | |
| 39679. |
Explain ex 1.1 Q1 (iii) |
| Answer» | |
| 39680. |
Two tangents TP and TQ are drawn to a circle |
| Answer» Let\xa0{tex}\\angle O P Q \\text { be } \\theta{/tex}{tex}\\therefore \\quad \\angle T P Q = \\left( 90 ^ { \\circ } - \\theta \\right){/tex}Since\xa0TP = TQ (Tangents){tex}\\therefore \\quad \\angle T Q P = \\left( 90 ^ { \\circ } - \\theta \\right){/tex}(Opposite angels of equal sides)Now,\xa0{tex}\\angle T P Q + \\angle T Q P + \\angle P T Q{/tex}\xa0= 180o{tex}\\Rightarrow 90 ^ { \\circ } - \\theta + 90 ^ { \\circ } - \\theta + \\angle P T Q{/tex}{tex}= 180 ^ { \\circ }{/tex}{tex}\\Rightarrow \\quad \\angle P T Q = 180 ^ { \\circ } - 180 ^ { \\circ } + 2 \\theta{/tex}\xa0{tex}\\Rightarrow \\quad \\angle P T Q = 2 \\theta{/tex}Hence\xa0{tex}\\angle P T Q = 2 \\angle O P Q{/tex} | |
| 39681. |
Plz any one tell me about ch6 triangle |
|
Answer» Theorems Hahaha????? It is the chapter in which we read that triangle has three sides What you want to know |
|
| 39682. |
If a=5. D=18. N=21. Find an |
|
Answer» an=a+(n-1)dan=5+(21-1)18an=5+20*18an=5+360an=365 use formula an=a+(n-1)d. given- a= 5 ,d=18,n=21an=a+(n-1)d =5+(21-1)18 =5+(20)18 =5+360 =365 365 Using formula an = a + ( n-1 ) d an = 5 + (21+1) 18an = 5 + 22× 18= 5 + 396 =401Hope it helps.... |
|
| 39683. |
If a=3 n=4 añ=6 then find d |
|
Answer» an=a+(n-1)*d then 6=3+4-1*d So d=1 1 an=a+(n-1)dSo,6=3+3d3=3dD=1 |
|
| 39684. |
What is value - cos 30\' and tan 60\' |
|
Answer» Cos30°=√3/2 and tan60°=√3 cos 30 is =under root 3/2 and tan60 equal = root3 Value of cos 30=√3/2tan 60=√3 Cos 30 is root3/2 and tan 60 is root3 |
|
| 39685. |
Find the center of circle passing through the points (6,-6),(3,-7),(3,3). |
| Answer» 3.3 | |
| 39686. |
if alpha &betaare roots of ax2-bx+c=0 than find alpha +beta |
|
Answer» Jatin is wrong 1/2 _b /2a You can find the value of alpha and beeta by splitting middle term |
|
| 39687. |
Mth ap =1/n, nth ap =1/m prove that = mnth term =1 |
| Answer» Let a and d be the first term and common difference respectively of the given A.P. Thenan = a + (n - 1)d{tex}\\frac { 1 } { n } ={/tex}\xa0mth term\xa0{tex}\\Rightarrow \\frac { 1 } { n } {/tex}= a + ( m - 1 ) d\xa0...(i){tex}\\frac { 1 } { m }{/tex}= nth term{tex}\\Rightarrow \\frac { 1 } { m } {/tex}= a + ( n - 1 ) d\xa0...(ii)On subtracting equation (ii) from equation (i), we get{tex}\\frac { 1 } { n } - \\frac { 1 } { m } = {/tex} [a+ (m-1) d] -[ a+ (n -1)d]= a + md - d - a - nd + d{tex}= ( m - n ) d{/tex}{tex} \\Rightarrow \\frac { m - n } { m n } = ( m - n ) d {/tex}{tex}\\Rightarrow d = \\frac { 1 } { m n }{/tex}Putting d = {tex}\\frac { 1 } { m n }{/tex}\xa0in equation (i), we get{tex}\\frac { 1 } { n } = a + \\frac { ( m - 1 ) } { m n } {/tex}{tex}\\Rightarrow \\frac { 1 } { n } = a + \\frac { 1 } { n } - \\frac { 1 } { m n } {/tex}{tex}\\Rightarrow a = \\frac { 1 } { m n }{/tex}{tex}\\therefore{/tex}\xa0(mn)th term = a + (mn - 1) d=\xa0{tex}\\frac { 1 } { m n } + ( m n - 1 ) \\frac { 1 } { m n } {/tex}{tex}\\left[ \\because a = \\frac { 1 } { m n } = d \\right]{/tex}= {tex}\\frac { 1 } { m n } + \\frac { mn } { m n } - \\frac { 1 } { m n }{/tex}= 1 | |
| 39688. |
Check whether x =-1/3,2/3 are the roots of 9x^2-3x-2=0 |
| Answer» | |
| 39689. |
class 10 bio note of chapter 7 |
| Answer» bio note of chapter 9 | |
| 39690. |
What is the value of tan 20×tan 70 |
| Answer» 1 | |
| 39691. |
_2x+kx+6 |
| Answer» Science capter 3 | |
| 39692. |
What is modal class?what is the formula of modal class? |
| Answer» It is an answer | |
| 39693. |
Define thales law |
|
Answer» Also known as basic proportionality theorem. If a line are drawn through two points on triangle and line are parallel then corresponding sides and angle are equal It is also known as BPT. |
|
| 39694. |
7a+5a |
|
Answer» 7a+5a will be 12a? a=-12 Is it a joke if not then answer is 12a. 12a..??? 12a 12a |
|
| 39695. |
Important question on square and square root. |
| Answer» | |
| 39696. |
If an AP common diffrence is 3 ,then what is the value of a20- a15 |
|
Answer» A20=19da15=14d then a20-a15=5dAs per sum a20-a15=5Therefor a20-a15=15I HPOE THIS WILL HELP YOU. What is the 1st term |
|
| 39697. |
X-y=6. If y=3 ,find x |
|
Answer» 9 x - y = 6 So, x = 6 + y ------------------AWe know that, y = 3.Put value in equation A . ....... x=6+3...... ....... x=9 x-y=6;y=3 then,x-3=6,x=6+3 now x=9 x-y=6 ; y = 3So, x-3=6 x=6+3 x=9 Hence the value of x is 9... 56 x-y=6x-3=6x=6+3x=9 |
|
| 39698. |
State thales therom. |
|
Answer» Basic proportionality therom If a line is drawn parallel to one side of a triangle intersecting other two sides, then it divides the two sides in the same ratio. |
|
| 39699. |
exercise 14.1 questions and answers |
|
Answer» Sorry app Find in the aap |
|
| 39700. |
150÷14. By long division method |
| Answer» | |