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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.
| 39551. |
What is probality method |
| Answer» Has some patience witj lecture on study thr os the equation | |
| 39552. |
Nmberic |
| Answer» | |
| 39553. |
Prove BPT . In detail? |
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Answer» You can find it in ncert textbook page no 124. Dg |
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| 39554. |
2+3=0 prove thid |
| Answer» (-2)+(-1)+3=0 | |
| 39555. |
2/10=2 |
| Answer» On multiplying numerator by 10 and denominater by 1=2 | |
| 39556. |
Prove that the diagonals of trapezium intersects each other in the same ratio? |
| Answer» Given in the quadrilateral ABCD{tex}\\frac{AO}{BO}=\\frac{CO}{DO}{/tex}or,\xa0{tex}\\frac { A O } { C O } = \\frac { B O } { D O }{/tex} ...(i)Draw\xa0EO {tex}\\parallel{/tex} AB onIn\xa0{tex}\\triangle A B D,{/tex}\xa0EO {tex}\\parallel{/tex} AB (By construction){tex}\\therefore {/tex}\xa0{tex}\\frac { A E } { E D } = \\frac { B O } { D O }{/tex}\xa0(By BPT)...(ii)From (i) and (ii) we get\xa0{tex}\\frac{AO}{CO}=\\frac{AE}{ED}{/tex}Hence by converse of BPT in {tex}\\triangle{/tex}ADC\xa0{tex}EO\\|CD{/tex}But\xa0{tex}EO \\|AB {/tex}So {tex}AB\\|CD {/tex}Therefore ABCD is a trapezium | |
| 39557. |
Prove the sss similarity criterion? |
| Answer» | |
| 39558. |
O/O =2 |
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Answer» YouTube 10+10÷(10)square +(10)square |
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| 39559. |
What are +,+= |
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Answer» + + + + |
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| 39560. |
All trignometry identities in class 10 |
| Answer» 1)sin² +cos² =1~ sin²= 1– cos² cos²= 1– sin²2)1+ tan² = sec²~ tan² = sec² – 1 1 = sec² – tan²3)1+ cot² = cosec²~ cot² = cosec² – 1 1 = cosec² – cot² | |
| 39561. |
What are the main real numbers |
| Answer» Real Numbers:\tAll rational and all irrational number makes the collection of real number. It is denoted by the letter R\tWe can represent real numbers on the number line. The square root of any positive real number exists and that also can be represented on number line\tThe sum or difference of a rational number and an irrational number is an irrational number.\tThe product or division of a rational number with an irrational number is an irrational number.\tThis process of visualization of representing a decimal expansion on the number line is known as the process of successive magnification | |
| 39562. |
Value of cot10° |
| Answer» Cot10=5.671282 | |
| 39563. |
Ex-14.4 question no. 3 is right but in app draw graph more than type but written less than type |
| Answer» | |
| 39564. |
Sir 3.4 ka q 1 |
| Answer» Iet x + y= 5 eqn1. and 2x - 3y=4 eqn2.Then multipling the eqn1. from 2 and eqn2. from 1, then you get2x + 2y=10 and 2x - 3y=4 now solve the both eqn you will get the value "y" is 6:5 and "x" is 19:5.... | |
| 39565. |
How many terms of the A.P 9,17,25......must be taken to give a sum of 636 |
| Answer» According to the question,we have,\xa0a=9. Therefore, common difference d =17-9=8let the required number of terms be n.Therefore, Sn=636{tex}\\Rightarrow{/tex}{tex}\\frac{n}{2}{/tex}[2a+(n-1)d]=636{tex}\\Rightarrow{/tex}{tex}\\frac{n}{2}{/tex}[2(9)+(n-1)8]=636{tex}\\Rightarrow{/tex}n[18+8n-8]=1272{tex}\\Rightarrow{/tex}8n2+10n-1272=0{tex}\\Rightarrow{/tex}4n2+5n-636=0{tex}\\Rightarrow{/tex}4n2+53n-48n-636=0{tex}\\Rightarrow{/tex}n(4n+53)-12(4n+53)=0{tex}\\Rightarrow{/tex}(4n+53)(n-12)=0{tex}\\Rightarrow{/tex}4n+53=0 or n-12=0{tex}\\Rightarrow{/tex}n={tex}\\frac{{ - 53}}{4}{/tex}\xa0or n=12Since number of terms cannot neither be negative nor fraction, n=12hence, the required number of terms is 12. | |
| 39566. |
135 and 225 find the HCF |
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Answer» HCF is 45 Its ans is 45 It is 45 3 |
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| 39567. |
What is the sa1 portion of math 2018 - 19 |
| Answer» I think 1to 9 chapters. | |
| 39568. |
2x +3y=14x+6y=4 show that the system has no solution |
| Answer» 8x+12y=4 8x+12y=8_ _. _ y= 4 | |
| 39569. |
Amar eat 2out of 8slices of a pizza . What fraction of the pizza was left over? |
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Answer» 1/4 2/8=1/4 1/4 8/2 =4/1 3/4 |
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| 39570. |
Chapter 5 k all examples |
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| 39571. |
If one zero of the polynomial xpower2 + a is -3 then find other zero? |
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Answer» P(x)=x^2+aGiven that one zero or root is -3Then put -3 instead of x(-3)^2+a=09+a=0a=-9 x2 +a+3=0a=xx2+x+3=0 D = 1-4×1×1=3X = -1+√3/2 & -1-√3/2 Putting x= -3 we geta= -9After factoring it i.e.(x-3)(x+3)We get +3 is another zero |
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| 39572. |
Alzebra quastion |
| Answer» | |
| 39573. |
How can i solve Q.2(iii) of Exercise 8.2 of Chapter 8 in another way? |
| Answer» LHS = sin 2 (0) = sin 0 = 0RHS =2 sin 0 =2 × 0 = 0 A = 0 | |
| 39574. |
59 |
| Answer» | |
| 39575. |
3 x square - K root 3 X + 4 is equal to zero |
| Answer» K=+4 and -4 | |
| 39576. |
Solve 2xy/x+y=3/2 and xy/2x-y=-3/10,x+y is not =0,2x-y is not =0 |
| Answer» x=1/2,y=-3/2 | |
| 39577. |
If alpha, bita are the zeros of two polynomial f(d) |
| Answer» X^2-(alpha+beta)+alpha×beta | |
| 39578. |
What is marking scheme for all chapters |
| Answer» Check marking scheme here :\xa0https://mycbseguide.com/cbse-syllabus.html | |
| 39579. |
What is elevtion |
| Answer» Elevation is height above a given level, especially sea level. | |
| 39580. |
Term one mn maths ka portion kya hn kripaya jaldi batane |
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Answer» Mn means 5ki powe x amd multiply by 2 ki power y Tu pagal hai kya tere school ka first term ka portion hame kauae pata hoga |
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| 39581. |
What is the value of tan square10 - cot square 80 |
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Answer» Answer is oviously 0 because tan10-cot(90-80)=0 Bro this higher class concept please do only 10 class concept |
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| 39582. |
I don\'t |
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Answer» Incomplete question What is your question????? Wrong |
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| 39583. |
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60° |
| Answer» 18.8cm^2 | |
| 39584. |
a=5 ,d=2 |
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Answer» What to find 367 |
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| 39585. |
If sum of the squares of zereos of quadratic polynomial f(x)= x2- 8x+k is 40 . Find the value of k |
| Answer» α+β=8αβ=k α^2+β^2=40(α+β)^2-2αβ =408^2-2k=4064 -2k =40-2k =40-64-2k =-24k=12 | |
| 39586. |
find the value of p for which the numbers 2p-1,3p+4,11 are in AP. Hence,find the numbers |
| Answer» 3p+4 -2p +1=11-3p-4 ( common difference is same ) p+5=7-3p4p=2p= 1/2 | |
| 39587. |
Define Thales theorem |
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Answer» If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,the other two sides are divided in the same ratio.MAY THIS HELP U MY FRND!!!!!? When a line cut the opposite two sides of the triangle which is parallel to the third side is divide the lines in the same ratio |
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| 39588. |
Basic concepts of traingle |
| Answer» Check notes from here :\xa0https://mycbseguide.com/cbse-revision-notes.html | |
| 39589. |
I need MCQ worksheet |
| Answer» Sorry cant be posted | |
| 39590. |
Prove that 3+2√5 is irrational |
| Answer» Let us assume ,to the contrary that 3+2√5 is rational.That is , we can find coprimr a&b (b not equal to 0)such that 3+2√5=a/b 2√5=a/b 2√5=a/b - 3/1 2√5=a-3b/b Ir not equal to R3+2√5= Irrational . | |
| 39591. |
Find Sum and product of zeros at t2-15 |
| Answer» Sum = 0,product =-15 | |
| 39592. |
The 8th term of an AP is half of its second term exceeds one third of its fourth by 1.find |
| Answer» a+7d=a+d/22a + 14d = a + da = -13da +10d = 1/3(a +3d) + 1a +10d -1 = 1/3(a + 3d)3(a + 10d -1) = a +3d3a + 30d -3 = a+ 3d2a + 27d =3 2(-13d) + 27d =3(since a = -13d)-26d + 27d = 3 d = 3a= -13d a = -39a + 14d = -39 + 42 hence a15 =3 | |
| 39593. |
What will be the sum of whole numbers from 1-40? |
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Answer» Where n=40, a=1, and d=1. Therefore answer comes 820 We can use the formula sum to nth term =n/2{2a+(n-1)d} |
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| 39594. |
CosA+sinA |
| Answer» cosA+sinA=1 | |
| 39595. |
A cos theta + b sin theta = cProve that : a sin theta - b cos theta =√a square + b square - c square |
| Answer» | |
| 39596. |
simplify : 1/2a + b+ 2x = 1/2a + 1/b + 1/2x |
| Answer» X=1/2b | |
| 39597. |
How to find hcf in euclid lemma division |
| Answer» | |
| 39598. |
What is the syllabus of claas 10 |
| Answer» Download \'cbse guide\' app for detail syllabus notes of each chapter of all subject | |
| 39599. |
If tan(A+B)= √3 and tan(A-B)=1/√3;0° |
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Answer» What nonsense is this \' Ayush yelpale\' .? 15 and 45 |
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| 39600. |
In a quadrilateral△ABC D is a pount on side BC such that 4BD =BC. prove that 16ADsquare =13BC square |
| Answer» In equilateral {tex}\\triangle{/tex}ABC. 4BD = BCConstruction: Draw AE\xa0{tex}\\perp{/tex} BC.\xa0{tex}\\therefore{/tex} BE = {tex}\\frac{1}{2}{/tex}BC.In right {tex}\\triangle{/tex}AED, AD2 = DE2 + AE2\xa0{tex}\\Rightarrow{/tex} AE2 = AD2 - DE2 ..(i)In right {tex}\\triangle{/tex}AEB, AB2 = AE2 + BE2{tex}\\Rightarrow{/tex}\xa0AB2 = AD2 - DE2 + BE2 [using (i)]{tex}\\Rightarrow{/tex}\xa0AB2 + DE2 - BE2 = AD2{tex}\\Rightarrow{/tex}\xa0AB2 + (BE - BD)2 - BE2 = AD2{tex}\\Rightarrow{/tex}\xa0AB2\xa0+ BE2\xa0+ BD2\xa0- 2BE.BD - BE2\xa0= AD2{tex}\\Rightarrow{/tex}\xa0AB2 + ({tex}\\frac{1}{2}{/tex}BC)2 - {tex}2 \\times \\frac{1}{2}{/tex}BC{tex}\\times \\frac{1}{4}{/tex}BC = AD2{tex}\\Rightarrow{/tex}\xa0AB2\xa0+\xa0{tex}\\frac{1}{16}{/tex}BC2\xa0-\xa0{tex}\\frac{1}{4}{/tex}BC2\xa0= AD2{tex}\\Rightarrow{/tex}\xa0BC2\xa0- {tex}\\frac{3}{16}{/tex}BC2\xa0= AD2\xa0[{tex}\\because{/tex} AB = BC]{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { 13 \\mathrm { BC } ^ { 2 } } { 16 }{/tex} = AD2{tex}\\Rightarrow{/tex}\xa013BC2\xa0= 16AD2 | |