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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.
| 3401. |
× |
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| 3402. |
2017 2018 exam date for class 10th |
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Answer» Exam is on 5. Datesheet will announced this week but the exam will be in March |
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| 3403. |
Sample paper class 10 maths solutions |
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| 3404. |
Ab |
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| 3405. |
Which branch is better ? Commerce or pcb |
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| 3406. |
If A=2n+13 and B=n+7, where n is a natural number than find HCF of A and B. |
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Answer» answer HCF(A,B)=1, since A and B are coprimes. Try |
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| 3407. |
Hi shrushti are u there ???? |
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| 3408. |
Find the probability that the month June may have five Monday in a 1.leap year 2.non-leap year |
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| 3409. |
If the sum of an A.P is ap2 + 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. | |
| 3410. |
Koi h ya yaha PE bhi kisi Ko mere se baat nhi karni??? |
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| 3411. |
Yahaaaa |
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| 3412. |
0 ➗ 0 karne par proof karo ans 0,1,2or infinity ye sare ans ane chaiye |
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Answer» 0/0=01/0=02/0=0 0÷0 = infinity |
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| 3413. |
Hmm sahil ab bolo kya bol rhe the |
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| 3414. |
Hlo.... Yaha mil |
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| 3415. |
Koi mera bhai nd frnd banega |
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| 3416. |
Can I start writing answers from Section-D in Board Exam? |
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| 3417. |
What is anodising ???? |
| Answer» Anodising is an electrolytic passivation process used to increase the thickness of the natural oxide layer on the surface of metal parts. | |
| 3418. |
Galvanization ???? |
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Answer» Right prateek Coating iron and steel with a thin layer of zinc |
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| 3419. |
What is amalgam ???? From chemistry |
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Answer» A solution of a metal in mercury( hg). Metal +mercury |
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| 3420. |
What is tatti |
| Answer» Bhai maths mein tatti kab se aane lagi. | |
| 3421. |
Find the value of tan15°, tan75° |
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| 3422. |
Sahil I am going know so |
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| 3423. |
Sahil are you there here |
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| 3424. |
Area of a quadrilateral ABCD whose vertices are A(3,-1),B(9,-5),C(14,0),D(9,19) |
| Answer» Join A and CNow, Area of quadrilateral ABCD = Area of\xa0{tex} \\Delta{/tex}ABC + Area of\xa0{tex}\\Delta{/tex}ACDArea of\xa0{tex} \\Delta A B C = \\frac { 1 } { 2 } | 3 ( - 5 - 0 ) + 9 ( 0 + 1 ) + 14 ( - 1 + 5 ) |{/tex}{tex}= \\frac { 1 } { 2 } | - 15 + 9 + 56 |{/tex}{tex}= \\frac { 1 } { 2 } \\times 50{/tex}= 25 sq. unitsArea of\xa0{tex}\\Delta ACD = \\frac{1}{2}|3(0 - 19) + 14(19 + 1) + 9( - 1 - 0)|{/tex}{tex}= \\frac { 1 } { 2 } | - 57 + 280 - 9 |{/tex}{tex}= \\frac { 1 } { 2 } \\times 214{/tex}= 107 sq. unitsArea of quad. ABCD = Area of\xa0{tex}\\Delta{/tex}ABC + Area of\xa0{tex}\\Delta{/tex}ACD= (25 + 107) sq. units=132 sq. units | |
| 3425. |
2x^+ax-a^=0 . Find the value of x |
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| 3426. |
States Euclid\'s Division LEMMA |
| Answer» Euclid’s Division Lemma: If we have two positive integers a and b,then there exists unique integers\xa0q\xa0and\xa0r\xa0which satisfies the conditiona = bq + r (where 0\xa0≤ r ≤ b)Example: If we have two integers a=27 b=4Then {tex}27 = 4 × 6 + 3,{/tex}Here q = 6 and r = 3 | |
| 3427. |
2018-19 ka syllabus |
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| 3428. |
Sir 10 class CBSE me date sheet bata do |
| Answer» Yet date sheet not release | |
| 3429. |
What is trignometric ratios |
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| 3430. |
X^2+3|X|+2=3 ,find the no. Of real roots |
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| 3431. |
Aditi sharam whatspp pe aao |
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| 3432. |
Find the root of the equation of 16x-10/x=27 |
| Answer» -10/11 i think | |
| 3433. |
For board exam we have to prepare optional exercise also or not? |
| Answer» Nothing is optional bhai. Vese vo thofe high lvl ke hote hai u can leave it | |
| 3434. |
Show graphically that the system of equation 3x-y=2,9x-3y=6 |
| Answer» Graph of {tex}3x - y = 2{/tex}:We have, {tex}3x - y = 2\\Rightarrow y = 3x - 2{/tex}When x = 2, we havey = 3 {tex}\\times{/tex}\xa02 - 2 = 4When x = 1, we havey = 3 {tex}\\times{/tex}\xa01 - 2 = 1\xa0\tx21y41\tPlotting the points {tex}A(2, 4)\\ and\\ B(1, 1){/tex} onthe graph paper and drawing a linepassing through A and B, we obtain thegraph of {tex}3x - y = 2{/tex} as shown in Fig.Graph of {tex}9x - 3 y = 6{/tex} :We have, {tex}9x - 3y = 6{/tex}{tex}\\Rightarrow \\quad y = 9 x - 6{/tex}{tex}\\Rightarrow \\quad y = \\frac { 9 x - 6 } { 3 }{/tex}When, x = 0, We\xa0have{tex}y = \\frac { 9 \\times 0 - 6 } { 3 } = - 2{/tex}When x = -1, we have{tex}y = \\frac { 9 \\times - 1 - 6 } { 3 } = - 5{/tex}\tx0-1y-2-5\tPlotting the points {tex}C(0,-2) and D (-1, - 5){/tex} on the graph paper and drawing a line passing through these two points on the same graph paper we obtain the graph of{tex}9x - 3y = 6{/tex}. We find the C and D both lie on the graph of {tex}3x - y = 2{/tex}. Thus, the graphs of the two equations are coincident.\xa0Hence, the system of equations has infinitely many solutions. | |
| 3435. |
Similarity theoram in maths |
| Answer» Bpt theorem | |
| 3436. |
The end points of a diameter of a circle are P (-3,8) and Q (7,4) . Find the centre of circle . |
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| 3437. |
How we can prepare different materials Ron different books in a better way for board exam |
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| 3438. |
How we can prepare in a better way from different books for our |
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| 3439. |
Tan (45-theta) - cot(45+theta |
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| 3440. |
Hii everyone where is Akshita |
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| 3441. |
Using the quadratic equation ,solve the equation:a×a×b×bx×x-(4b×b×b×b-3a×a×a×a) x-12a×a×b×b |
| Answer» The given quadratic equation isa2b2x2 - (4b4 - 3a4)x - 12a2b2 = 0.Comparing with Ax2 + Bx + C = 0, we getA = a2b2, B = -(4b4 - 3a4), C = -12a2b2Using the quadratic formula, {tex}x = \\frac { - B \\pm \\sqrt { B ^ { 2 } - 4 A C } } { 2 A }{/tex}we get{tex}= \\frac { \\left\\{ \\left( 4 b ^ { 4 } - 3 a ^ { 4 } \\right) \\right\\} \\pm \\sqrt { \\left( - \\left( 4 b ^ { 4 } - 3 a ^ { 4 } \\right) \\right\\} ^ { 2 } - 4 \\left( a ^ { 2 } b ^ { 2 } \\right) \\left( - 12 a ^ { 2 } b ^ { 2 } \\right) } } { 2 a ^ { 2 } b ^ { 2 } }{/tex}{tex}= \\frac { 4 b ^ { 4 } - 3 a ^ { 4 } \\pm \\sqrt { 16 b ^ { 8 } + 9 a ^ { 8 } - 24 a ^ { 4 } b ^ { 4 } + 48 a ^ { 4 } b ^ { 4 } } } { 2 a ^ { 2 } b ^ { 2 } }{/tex}{tex}= \\frac { 4 b ^ { 4 } - 3 a ^ { 4 } \\pm \\sqrt { 16 b ^ { 8 } + 9 a ^ { 8 } + 24 a ^ { 4 } b ^ { 4 } } } { 2 a ^ { 2 } b ^ { 2 } }{/tex}{tex}= \\frac { 4 b ^ { 4 } - 3 a ^ { 4 } \\pm \\sqrt { \\left( 4 b ^ { 4 } + 3 a ^ { 4 } \\right) ^ { 2 } } } { 2 a ^ { 2 } b ^ { 2 } }{/tex}{tex}= \\frac { 4 b ^ { 4 } - 3 a ^ { 4 } \\pm \\left( 4 b ^ { 4 } + 3 a ^ { 4 } \\right) } { 2 a ^ { 2 } b ^ { 2 } }{/tex}{tex}\\frac { 4 b ^ { 4 } - 3 a ^ { 4 } + 4 b ^ { 4 } + 3 a ^ { 4 } } { 2 a ^ { 2 } b ^ { 2 } } , \\frac { 4 b ^ { 4 } - 3 a ^ { 4 } - 4 b ^ { 4 } - 3 a ^ { 4 } } { 2 a ^ { 2 } b ^ { 2 } }{/tex}{tex}= \\frac { 8 b ^ { 4 } } { 2 a ^ { 2 } b ^ { 2 } } , \\frac { - 6 a ^ { 4 } } { 2 a ^ { 2 } b ^ { 2 } } = \\frac { 4 b ^ { 2 } } { a ^ { 2 } } , - \\frac { 3 a ^ { 2 } } { b ^ { 2 } }{/tex}{tex}\\therefore{/tex} the solutions of equation are {tex}\\frac { 4 b ^ { 2 } } { a ^ { 2 } } \\text { and } \\frac { - 3 a ^ { 2 } } { b ^ { 2 } }{/tex}. | |
| 3442. |
If tan A=cot B, then find A+B. |
| Answer» {tex}\\tan A = \\cot B{/tex}{tex}\\Rightarrow \\tan A = \\tan \\left( 90 ^ { \\circ } - B \\right) \\quad \\because \\tan \\left( 90 ^ { \\circ } - \\theta \\right) = \\cot \\theta{/tex}{tex}\\Rightarrow A = 90 ^ { \\circ } - B \\quad \\quad \\quad \\because A \\text { and } 90 ^ { \\circ } -B{/tex}\xa0are both acute angles{tex}\\Rightarrow A + B = 90 ^ { \\circ }{/tex} | |
| 3443. |
Sahil I am confused please real wala.reply karo |
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| 3444. |
Sahil so Gaye kya |
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| 3445. |
Hy amaira?? |
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| 3446. |
prove that the parallelogram circumscribe a circle is a rhombus |
| Answer» Given ABCD is a parallelogram in which all the sides touch a given circleTo prove:- ABCD is a rhombusProof:-{tex}\\because{/tex}\xa0ABCD is a parallelogram{tex}\\therefore{/tex}\xa0AB = DC and AD = BCAgain AP, AQ are tangents to the circle from the point A{tex}\\therefore{/tex}\xa0AP = AQSimilarly, BR = BQCR = CSDP = DS{tex}\\therefore{/tex}(AP + DP) + (BR + CR) = AQ + DS + BQ + CS = (AQ + BQ) + (CS + DS){tex}\\Rightarrow{/tex}\xa0AD + BC = AB + DC{tex}\\Rightarrow{/tex}\xa0BC + BC = AB + AB [{tex}\\because{/tex} AB = DC, AD = BC]{tex}\\Rightarrow{/tex}\xa02BC = 2AB{tex}\\Rightarrow{/tex}\xa0BC = ABHence, parallelogram ABCD is a rhombus | |
| 3447. |
intregation |
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| 3448. |
(a+b) ka cube = ? |
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Answer» a3+b3+3a2b+3ab2 A3+B3+3a2b+3ab3 |
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| 3449. |
If xsin^3theta+ycos^3theta=sinthetacostheta And xsintheta=ycostheta . To prove:x^2+y^2=1 |
| Answer» We have,xsin3{tex}\\theta{/tex}\xa0+ ycos3{tex}\\theta{/tex}\xa0= sin{tex}\\theta{/tex}\xa0cos{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0(xsin{tex}\\theta{/tex}) sin2{tex}\\theta{/tex}\xa0+ (ycos{tex}\\theta{/tex}) cos2{tex}\\theta{/tex}\xa0= sin{tex}\\theta{/tex}\xa0cos{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0x sin{tex}\\theta{/tex}\xa0(sin2{tex}\\theta{/tex}) + (x sin{tex}\\theta{/tex}) cos2{tex}\\theta{/tex}\xa0= sin{tex}\\theta{/tex}\xa0cos{tex}\\theta{/tex}\xa0[{tex}\\because{/tex}\xa0x sin{tex}\\theta{/tex}\xa0= y cos{tex}\\theta{/tex}]{tex}\\Rightarrow{/tex}\xa0x sin{tex}\\theta{/tex}\xa0(sin2{tex}\\theta{/tex}\xa0+ cos2{tex}\\theta{/tex}) = sin{tex}\\theta{/tex}\xa0cos{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0x sin{tex}\\theta{/tex}\xa0= sin{tex}\\theta{/tex}\xa0cos{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0x = cos{tex}\\theta{/tex}Now, xsin {tex}\\theta{/tex}\xa0= ycos{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0cos{tex}\\theta{/tex}\xa0sin{tex}\\theta{/tex}\xa0= y cos{tex}\\theta{/tex}\xa0[{tex}\\because{/tex} x = cos{tex}\\theta{/tex}]{tex}\\Rightarrow{/tex}\xa0y = sin{tex}\\theta{/tex}Hence, x2 + y2 = cos2{tex}\\theta{/tex}\xa0+ sin2{tex}\\theta{/tex}\xa0= 1 | |
| 3450. |
Find the perpendicular distance of A(5,12) from the Y axis |
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Answer» 12 12 |
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