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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.
| 4401. |
Length |
| Answer» | |
| 4402. |
How to consruct 105° angle |
| Answer» Very silmple1)Draw angle of 90 degree .2)Then,draw angle of 120dgre.3)take 90 as a center draw an arc and take 120 as a center cut the arc. | |
| 4403. |
3 tan,Q=4, evaluate 3 sinQ+2sinQ/3sinQ-2cosQ |
| Answer» 10/3 | |
| 4404. |
If the sum of first p terms of AP is ap^2+bp find its common difference |
| Answer» We are given that sum\xa0of first p terms of this APSp= ap2 + bpLet the first term= x and common difference = d{tex}{\\mathrm S}_1=\\mathrm a(1)^2+\\mathrm b(1)=\\mathrm a+\\mathrm b{/tex}So first term x = a+bNow S2 = a(2)2 + b(2)S2 = 4a + 2b,{tex}\\mathrm{So}\\;{\\mathrm s}_2={\\mathrm a}_1+{\\mathrm a}_2=\\mathrm x+\\mathrm x+\\mathrm d=4\\mathrm a+2\\mathrm b{/tex}2x + d = 4a+2b2(a+b)+d = 4a+2b2a+2b+d=4a+2bd=4a+2b-2a-2bd = 2aTherefore, the common difference of the given AP is 2a. | |
| 4405. |
Plz tell my exam is there tomorrow |
| Answer» Because Isme diameter pucha h nd multiply Ho raha h | |
| 4406. |
Why in example 10 in NCERT page number 250, 2divide by1800 has done not 1800 divide by 2 |
| Answer» | |
| 4407. |
What is euclid division lama? |
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Answer» a=bq+r For any 2 given +ve integers a and b, there exist unique whole numbers q and r Sorry GiveN positive integer a and b there exists unique integer q and r satisfying a=bq+r,,,0 <_r | |
| 4408. |
Check whether 15n end with the digit 0 for any natural number n |
| Answer» Any number can end with 0 only if this is divisible by 10 hence it is divisible by 2 and 5 bothBut prime factors of (15)n are 3n{tex}\\times{/tex}\xa05n\xa0It is clear that 15n is not divisible with 2Hence by the Fundamental Theorem of Arithmetic, there is no natural number n for which (15)n ends with the digit zero. | |
| 4409. |
In triangle ABC, AD perpendicular BC. If AD2 = BD× DC. Prove that ABC is a right triangle. |
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Answer» But ABC is not a right triangle. We know that if perpendicular is drawn from vertex of right angle of triangle.. The two triangle which formed is similar to each other and to whole triangle |
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| 4410. |
I need questions of arihant for maths class 10 cbse |
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Answer» Buy it for 100 rupees U can get it on this app ncert exemplar Ncert exemplar |
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| 4411. |
SinA+sinB=? |
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Answer» By the way it\'s of std 11th!! sina+_sinb=2sin1/2(a+_b)cos1/2(a+_b) |
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| 4412. |
For what value of k , x=a is a solution of equation x² - (a+b) + k = 0 |
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Answer» Please share the procedure . I think K=-ab K=ab |
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| 4413. |
Do we have to study the optionals exercise too?? |
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Answer» Why not Yes that\'s helps uh better to deal with value based questions 4marker ones |
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| 4414. |
Find the roots of x2-x-p(p+1) |
| Answer» {tex}x^2 + x - p(p+1) = 0{/tex}{tex}x^2 + (p+1)x - px - p(p+1) = 0{/tex}{tex}x(x+ p + 1) -p(x + p + 1) = 0{/tex}(x + p + 1)(x - p) = 0\xa0x = - p - 1, p | |
| 4415. |
Do CBSE Board repeats previous questions ?? |
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Answer» Follow NCERT most questions come from it No very few question will repeat but in different type |
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| 4416. |
If tan A = 3\\4 and A+B=90°, then what is the value of cot B? |
| Answer» tan A =\xa0{tex}\\frac 34{/tex}cot A={tex}\\frac{4}{3}{/tex}A + B = 90o\xa0{tex}\\Rightarrow{/tex}\xa0A=90o-B{tex}\\Rightarrow{/tex}\xa0cot A=cot(90o-B){tex}\\Rightarrow{/tex}\xa0cot A=tan B{tex}\\Rightarrow{/tex}\xa0cotA = tanB =\xa0{tex}\\frac{4}{3}{/tex}{tex}\\Rightarrow{/tex}\xa0cot B =\xa0{tex}\\frac 34{/tex} | |
| 4417. |
Cos38.cosec52/ cot18.cot35.cot60cot55cot72 + sin |
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| 4418. |
Median formula in math |
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Answer» M=l+ n/2-cf/f×h Read |
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| 4419. |
Perimeter of quadrant formula |
| Answer» πr+2r | |
| 4420. |
find the vertex D ,if a[1,2] b[4,3] c[6,6] |
| Answer» Let coordinates of D be\xa0{tex}( \\alpha , \\beta ){/tex}\xa0P is mid-point of AC and BD.{tex}\\therefore \\quad \\left( \\frac { \\alpha + 4 } { 2 } , \\frac { \\beta + 3 } { 2 } \\right){/tex}{tex}= \\left( \\frac { 1 + 6 } { 2 } , \\frac { 2 + 6 } { 2 } \\right){/tex}{tex}\\Rightarrow \\frac { \\alpha + 4 } { 2 }{/tex}{tex}= \\frac { 7 } { 2 } ; \\frac { \\beta + 3 } { 2 } = \\frac { 8 } { 2 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\alpha + 4 = 7 ; \\beta + 3 = 8{/tex}{tex}\\Rightarrow \\quad \\alpha = 3 ; \\beta = 5{/tex}{tex}\\therefore{/tex}\xa0Coordinates of D are (3, 5). | |
| 4421. |
Write 0.32 is fraction form |
| Answer» 32/100 | |
| 4422. |
New 1060 |
| Answer» Kyaa | |
| 4423. |
Prove that the product of three consecutive integers is divisible by 6 |
| Answer» Easy but quite long so refer Ncert | |
| 4424. |
Find the 20th term from the last term of the Ap:3,8,13,.....,253 |
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Answer» 158 First find term no. Of 253 then use this formula a+( last term no. - required term no.) D Ye ncert me 5.2 ka 17 question hai |
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| 4425. |
5 marks |
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| 4426. |
What is the meaning of parallel line |
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Answer» Lines which never meet or intersect each other are parallel line Pair of lines which do not intersect It doesn\'t intersects over a long distance and also equal distance between them |
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| 4427. |
(1+m2)x2+2mcx+(c2-a2)=0 |
| Answer» Mc | |
| 4428. |
Dear algebra plz stop asking about ur \'x\'She has left uAnd don\'t ask \'y\'!!! |
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Answer» ??? Nice Nice ? |
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| 4429. |
If the point p(1,2) lies on the line segment joining the points A(2,4) and b(4,8), then find AP/AB |
| Answer» | |
| 4430. |
Which is very bad prepared? For exam |
| Answer» mee | |
| 4431. |
Whose MATHS is complete? |
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Answer» Meee??? Me Mr nhi hui tumhri ho jaye to batana Haan ho gai almost. mera bhi nhi hua yrrrr I completed RS AGARWAL meri bhi nhi hui Which NCERT or RS agarwal And suraj and ilma and smart infera?? Not completed what about yours ruhi?? |
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| 4432. |
Find in square metres the area of reactangle whose length is 13m and breath is 225 cm. |
| Answer» 97.50m^2 | |
| 4433. |
SinA-B=sinacosb-cosa sinb |
| Answer» | |
| 4434. |
Hi is there any body online |
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Answer» Hi Hiiii Yes I m Hii hii |
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| 4435. |
Solve for x:x2+9x+20=0 |
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Answer» Square This is 2 x or squars |
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| 4436. |
Explain 6n can not end with digit 0 |
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Answer» No it is not a wrong question as n is not multiplied it is the power of 6 6 is expressible in the form 2n3n and for a digit to end with zero it should be of the form 2n5n. See ncertbook example when n=5 ..6*5=30 ..so it ends with zero.. wrong question |
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| 4437. |
What is a prime no |
| Answer» Prime numbers are those numbers who have only 2 factors. For example :- 2,3,5. 2= 2×1, 3= 3×1, 5= 5×1. | |
| 4438. |
Ax⅝ +4000-aby⅜ find roots |
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| 4439. |
A circle inscribed in a triangle ABC in which AB=a BC=b and CA= c. Find the radius of a circle. |
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| 4440. |
The graph of y is equal to p(x) where p(x) is polynomial find numbers of zeros of p(x) |
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Answer» Ya right Depend ob graph... jha pr contact me vo uska zero hai |
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| 4441. |
Solutions of evergreen model test paper of 100% sucess |
| Answer» | |
| 4442. |
Find the value of k for which one root of quadratic equation kx2 -14x+8=0 is six times the other |
| Answer» Let one root = {tex}\\alpha{/tex}Other root = 6{tex}\\alpha{/tex}{tex}\\therefore{/tex}Sum of roots =\xa0{tex}\\alpha{/tex}\xa0+ 6{tex}\\alpha{/tex}\xa0=\xa0{tex}\\frac{14}{k}{/tex}or,\xa0{tex}7 \\alpha = \\frac { 14 } { k } \\text { or } \\alpha = \\frac { 2 } { k }{/tex} ...........(i)Product of roots\xa0{tex}\\alpha ( 6 \\alpha ) = \\frac { 8 } { k }{/tex}or,\xa0{tex}6 \\alpha ^ { 2 } = \\frac { 8 } { k }{/tex}..........(ii)Solving (i) and (ii),{tex}6 \\left( \\frac { 2 } { k } \\right) ^ { 2 } = \\frac { 8 } { k }{/tex}{tex}6 \\times \\frac { 4 } { k ^ { 2 } } = \\frac { 8 } { k }{/tex}or,\xa0{tex}\\frac { 3 } { k ^ { 2 } } = \\frac { 1 } { k }{/tex}or, 3k = k2or, 3k - k2\xa0= 0k[3-k] = 0{tex}\\therefore{/tex}\xa0k = 0 or k = 3k = 0 is not possibleHence, k = 3 | |
| 4443. |
2018 question paper of cbse |
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Answer» On net and also in sample paper Do you have |
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| 4444. |
Sin6 theta+cos6 theta= 1-3sin2 theta cos2 theta |
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| 4445. |
In an equilateral triangle with side \'a\' prove that its area = root 3/4a square |
| Answer» According to the question, we have to prove that in an equilateral triangle with side a, area =\xa0{tex}\\frac { \\sqrt { 3 } } { 4 } a ^ { 2 }{/tex}Let\xa0{tex}\\triangle{/tex}ABC be an equilateral triangle with side a.Then, AB = AC = BC = a.Draw AD{tex}\\perp{/tex}BC.In\xa0{tex}\\triangle{/tex}ADB and\xa0{tex}\\triangle{/tex}ADC, we haveAB = AC ({tex}\\Delta ABC{/tex}\xa0equilateral triangle){tex}\\angle{/tex}ADB=\xa0{tex}\\angle{/tex}ADC = 90° (AD is altitude)\xa0AD\xa0=\xa0AD\xa0(Common){tex}\\therefore \\Delta A D B \\cong \\triangle A D C{/tex}\xa0[ RHS Congurency rule]\xa0{tex}\\therefore{/tex}\xa0BD = DC =\xa0{tex}\\frac { a } { 2 }{/tex}From right\xa0{tex}\\triangle{/tex}ADB, we haveAB2\xa0= AD2\xa0+ BD2\xa0(By Pythagoras theorem){tex}\\Rightarrow{/tex}\xa0AD =\xa0{tex}\\sqrt { A B ^ { 2 } - BD ^ { 2 } }{/tex}{tex}\\Rightarrow A D = \\sqrt { a ^ { 2 } - \\left( \\frac { a } { 2 } \\right) ^ { 2 } }{/tex}{tex}\\Rightarrow A D = \\sqrt { a ^ { 2 } - \\frac { a ^ { 2 } } { 4 } }{/tex}{tex}\\Rightarrow A D = \\sqrt { \\frac { 3 a ^ { 2 } } { 4 } }{/tex}So, the altitude is, AD =\xa0{tex}\\frac { \\sqrt { 3 } a } { 2 }{/tex}Area of {tex}\\triangle{/tex}ABC =\xa0{tex}\\frac { 1 } { 2 } \\times B C \\times A D{/tex}{tex}= \\frac { 1 } { 2 } \\times a \\times \\frac { \\sqrt { 3 } a } { 2 }{/tex}{tex}= \\frac { \\sqrt { 3 } a ^ { 2 } } { 4 }{/tex}Hence proved. | |
| 4446. |
How can we get 40 mark in math exam |
| Answer» Study and study | |
| 4447. |
In figure xp/py=xq/qz=3if the area of triangle xyz is32cm, then find area of qadrilateral pyzq |
| Answer» Given {tex}\\frac{{XP}}{{PY}} = \\frac{{XQ}}{{QZ}}{/tex}\xa0\u200b{tex}\\Rightarrow {/tex}\u200b PQ {tex}\\parallel{/tex} YZ .....(i) [By converse of B.P.T]In \u200b{tex}\\triangle {/tex}\u200b XPQ and \u200b{tex}\\triangle {/tex}\u200b XYZ we have[ \u200b{tex}\\angle{/tex}\u200b XPQ = \u200b{tex}\\angle{/tex}\u200b Y [From (i) corresponding angles]\u200b{tex}\\angle{/tex}\u200b X = \u200b{tex}\\angle{/tex}\u200bX [common]\u200b{tex}\\therefore {/tex}\u200b\u200b{tex}\\triangle {/tex}\u200b XPQ \u200b{tex}\\cong{/tex}\u200b{tex}\\triangle {/tex}XYZ [By AA similarity]\u200b{tex}\\therefore {/tex}\u200b{tex}\\frac{{ar\\left( {\\triangle XYZ} \\right)}}{{ar\\left( {\\triangle XPQ} \\right)}}=\\frac{{X{Y^2}}}{{X{P^2}}}{/tex}{tex}\\frac { P Y } { X P } = \\frac { 1 } { 3 }{/tex}We have\xa0{tex}\\Rightarrow \\frac { P Y } { X P } + 1 = \\frac { 1 } { 3 } + 1{/tex}{tex}\\Rightarrow \\frac { P Y + X P } { X P } = \\frac { 4 } { 3 }{/tex}\xa0\u200b\u200b\u200b{tex}\\Rightarrow {/tex}\u200b{tex}\\frac{{XY}}{{XP}} = \\frac{4}{3}{/tex}Substituting in (i), we get{tex}\\frac{{ar\\left( {\\triangle XYZ} \\right)}}{{ar\\left( {\\triangle XPQ} \\right)}} = {\\left( {\\frac{4}{3}} \\right)^2} = \\frac{{16}}{9}{/tex}\u200b{tex}\\Rightarrow {/tex}\u200b{tex}\\frac{{32}}{{ar\\left( {XPQ} \\right)}} = \\frac{{16}}{9}{/tex}{tex}ar\\left( {XPQ} \\right) = \\frac{{32 \\times 9}}{{16}} = 18c{m^2}{/tex}Area of quadrilateral PYZQ = 32 - 18 = 14 cm2 | |
| 4448. |
Pt and pa |
| Answer» | |
| 4449. |
Value of sec 60 |
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Answer» Value is 2 2 |
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| 4450. |
If 9th term of an A.P is zero prove that it\'s 29th term is double the 19th term |
| Answer» We have,a9\xa0= 0{tex}\\Rightarrow{/tex}\xa0a + (9 - 1)d = 0{tex}\\Rightarrow{/tex}\xa0a + 8d = 0{tex}\\Rightarrow{/tex}\xa0a = -8dTo prove: a29 = 2a19Proof:LHS = a29= a + (29 - 1)d= a + 28d= -8d + 28d= 20dRHS = 2a19= 2 a + (19 - 1)d]= 2[ -8d + 18d]= 2\xa0{tex}\\times{/tex}10d= 20d{tex}\\therefore{/tex} LHS = RHSHence, 29th\xa0term is double the 19th term. | |