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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.
| 19251. |
Represent the number line root 4.2 |
| Answer» ???? | |
| 19252. |
Who wants to take a break..... |
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| 19253. |
Solve by cross multiplication method???2x +3y +5=05x _3y +9=0 |
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Answer» X= - 2. Y= -3????? X=-2 , Y=-3 |
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| 19254. |
X2-(1+√2)x+√2=0 |
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| 19255. |
What is the next term of AP root 8 root 18 root 32 |
| Answer» _/50 (root 50) | |
| 19256. |
1/2x-1/y=-11/x+1/2y=8 |
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| 19257. |
Which term of the sequence 20, 19 1/4, 18 1/2, 17 3/4 is the first negative term |
| Answer» 28th term | |
| 19258. |
Find the area of the shaded region in the given figure |
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Answer» Where is the figure Vikas Figure to do Kha h figure |
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| 19259. |
Way sin²ø+cos²=1 |
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Answer» Yes i want It has identity Do you want to prove it....???? |
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| 19260. |
Surface area of hemisphere |
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Answer» 3πrsquare Total surface area of the hemisphere– While calculating the total surface area of a hemisphere, we need to consider the base of the hemisphere which is circular in shape. Thus the total surface area of a hemisphere is equal tototal surface area= curved surface area + area of the base circle= |
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| 19261. |
Is secA + tanA=p then find in terms of cosecA |
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| 19262. |
find the sum of the first 50 odd natural numbers by using arithmetic progression |
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Answer» 1,3,5 is correct in place 1,2,3 1,2,3.......a=1,d=2,n=50 S=n/2(2a+(n-1)d)S=50/2(2(1)+(50-1)2S=25(2+(49)2)S=25(2+98)S=25(100)S=2500 The first 50 natural odd numbers = 1,3,5,7... (upto 50 terms) which is an AP.Now, sn = n/2 [2a + (n-1)d]S50 = 50/2 [2x1 + (50-1)2]= 25 (2 + 98)= 25 x 100= 2500 |
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| 19263. |
What is formula of Area of segment of a circle |
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Answer» The circular region enclosed between a chord and the corresponding arc is called the segment of a circle.Minor segment : If the boundary of a segment is a minor arc of a circle, then the corresponding segment is called a minor segment.Major segment: A segment corresponding a major arc of a circle is called as major segment.Area of Minor segment = θ/360 πr2Area of major segment = Area if circle - Area of Minor segment Area of segment of a circle =area of the corresponding sector-area of the corresponding triangle |
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| 19264. |
If x=2 & x=3 are the roots of the equation 3x2-2mx+2n=0 then find the value of m and n |
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Answer» 3x2 - 2mx + 2n = 0x = 22 (2)2 - 2m(2) + 2n = 02(4) - 4m +2n = 08 - 4m + 2n = 04- 2m + n = 02m - n = 4.... (i)x = 32 (3)2 - 2m(3) + 2n = 02(9) - 6m +2n = 018 - 6m + 2n = 03m - n = 9 ...... (ii)After solving m = 5 and n = 6.\xa0 M=15/2N=9 |
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| 19265. |
Find the area of minor segment of circle of radius 42 cm. If length of corresponding arc is 44cm. |
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Answer» Length of arc = θ/360 2πr = 44 cmPutting r = 42cm, we get θ= 60°Now, Area of minor segment= Area of minor sector-Area of ∆Since = 60°, so the triangle formed will be an equilateral∴ ∆ Area of minor segment= Area of minor sector - Area of equilateral ∆i.e. Area of minor segment = θ/360 πr2 -\xa0√3/4 a2= 924 - 441√3 cm2 76 approx |
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| 19266. |
How find similar triangle |
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Answer» By any rule but most aa rule By AA rulw By AA RULE |
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| 19267. |
Kya koi online hai......... |
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Answer» I m Yes |
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| 19268. |
Prove. SinA upon 1 minus cos A is equals to 1 + sec a |
| Answer» Please give answer of my question | |
| 19269. |
Is anyone having sites for latest cbse sample papers 2019 |
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Answer» I know Search on cbsesqp 2018-19 (CBSE Sample Question Paper) |
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| 19270. |
When is board exam of 10 th |
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Answer» CBSE is not announced yet. 6 march From 5march to 4 april In March |
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| 19271. |
If nth term of an A.P. is 2n+1, then find the sum of first 3 terms of A.P. |
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Answer» 15 15 15. |
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| 19272. |
Solve the following quadratic equation for x : x² - b² = a(2x-a). |
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Answer» X = a+b Aur X= a-b x=a-b and x = a+b |
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| 19273. |
Show that 7-root 5 is irrational given root 5 is rational |
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| 19274. |
what is fundamental theorum of arithmetic |
| Answer» THE FUNDAMENTAL THEOREM OF ARITHMETICS STATES THAT EACH AND EVERY COMPOSITE NUMBER CAN BE WRITTEN AS THE PRODUCT OF PRIME NUMBERS IN ONE AND ONLY ONE WAY EXCEPT THE ORDER OF NUMBERS. | |
| 19275. |
Is baar kon kon top karenge???\u200d??\u200d? |
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Answer» Gall ni kadni Confidence is good but overconfidence is not good , anyways wishing all of you best of luck for your exams...do your best???? In sha allah I think it\'s me ?? ? Inshallah we all try our best.. |
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| 19276. |
Prove the value of sin 45 geometrically |
| Answer» See in ncert book page no.182 | |
| 19277. |
If the mean of 1, 2, 3,., x is, 6x/11 find the value of x. |
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| 19278. |
If secA + tanA = x then find CosecA |
| Answer» Samyak it should be done likeSec2A-Tan2A=1SecA+TanA)(SecA-TanA)=1SecA-TanA=1/xBy this you formed the linear equation solve it to get the answer | |
| 19279. |
Sin 30 +cos 60 |
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Answer» The answer is √3+√3/2 √3/2 + √3/2 = √3/2 |
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| 19280. |
Ap , if the common difference (d)=4 and the seven therm (a7)is 4 then find the first them ? |
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Answer» Ohh sorry? 4=a+6×4. 4=a+24. a=20 Really sorry for a silly mistake? a=-20 a7=a+6d =a +6×4=a+24=0 a=-24 then first term is -24 a=-20. Hope it helps you...... |
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| 19281. |
TheHCF of smallest prime number and the smallest composite number ? |
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Answer» The smalest number= 2The composite numbers =42=2×14=2×2The HCF is 2 The require number 2and 4 Hcf of 2 and 4 is 2 . The smallest prime number is 2 and composite number is 4,so the h.c.f(2,4) =2. 2 |
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| 19282. |
The distance of a point p (x , y)from the origin |
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Answer» Op =[x2+y2) Nice Square root of x^2+y^2 |
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| 19283. |
A cistern |
| Answer» GivenVolume of cistern = {tex}150 \\times 120 \\times 110 \\mathrm { cm } ^ { 3 } = 1980000 \\mathrm { cm } ^ { 3 }{/tex}Volume of water =\xa0129600 cm3Volume of one brick =\xa0{tex}22.5 \\times 7.5 \\times 6.5 \\mathrm { cm } ^ { 3 } = 1096.875 \\mathrm { cm } ^ { 3 }{/tex}Each brick absorbs one - seventeenth of its volume of waterVolume of water absorbed by one brick = {tex}\\frac { 1 } { 17 } \\times{/tex}\xa0volume of brick = {tex}\\frac { 1 } { 17 } \\times 1096.875 \\mathrm { cm } ^ { 3 }{/tex} = 64.52 cm3Let n be the total number of bricks which can be put in the cistern without water overflowing. Then,Volume of water absorbed by n bricks =\xa0{tex}n \\times \\frac { 1 } { 17 } \\times 1096.875 \\mathrm { cm } ^ { 3 }{/tex}{tex}\\therefore{/tex} Volume of water left in cistern =\xa0{tex}= \\left( 129600 - \\frac { n } { 17 } \\times 1096.875 \\right) \\mathrm { cm } ^ { 3 }{/tex}Since the cistern is filled upto the brim.Therefore, Volume of the cistern =\xa0Volume of water left in the cistern + Volume of bricks\xa0{tex} 1980000{/tex} =\xa0{tex}129600 - \\frac { n } { 17 } \\times 1096.875 + n \\times 1096.875 {/tex}{tex}n \\times 1096.875 - \\frac { n } { 17 } \\times 1096.875 = 1980000 - 129600{/tex}{tex}1096.875 \\times \\left( n - \\frac { n } { 17 } \\right) = 1850400{/tex}{tex}1096.875 \\times \\frac { 16 n } { 17 } = 1850400{/tex}{tex}17550 \\times \\frac { n } { 17 } = 1850400 \\Rightarrow n = \\frac { 1850400 \\times 17 } { 17550 } = 1792.41 {/tex}since the number of bricks cannot be in decimalstherefore, required number of bricks = 1792 | |
| 19284. |
For Ø=30°;Verify that:sin 2ø=2tanø/1+tan2ø |
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| 19285. |
Radha made a picture of an aeroplane with coloured paper .find the total area of paper used. |
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Answer» Plz tell me the measurment of this que. Sorry but she? only made picture What are the measurements |
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| 19286. |
How to learn trigonometry formulas on tips? |
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Answer» PBP= pandit badri prasad? HHB= Har Har bholey?? Some people have=(some?sin theta)=(people?perpendicular)/(have?height),sinΦ=P/H.........Curly brown hair=(curly?cos theta)=(brown?base)/(hair?height),cosΦ=B/H.......Tightly pulled back=(tightly?tan theta)=(pulled?perpendicular)/(back?base),tanΦ=P/B.... PBP/HHB Pakistan bhooka pyasa Hindustan hara bhara Like in serieswise p/h =sin B/H =cos P/B = tan and h/p = cosec h/b = sec b/p = cotPbp/hhbPandit badri prasad har har boleSIN COS TAN COSEC SEC COT Just learn formula of sin,cos . Tan can be solved by dividing them . And cot,sec,cosec are there opposite. Like sin30° ÷ cos30°= 1/2 ÷ √3/2 = 1/2 = tan 30° |
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| 19287. |
A man is 5 times old as his son and sum of the square of their ages ages is 2106 find the ages |
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| 19288. |
Show that 7-√5is irrational give that √5is irrational |
| Answer» Seven underroot 5Put equal to a/b so Now underroot 5 is equal to a/7bTherefore a/7b is rationl no So5 also is a rational no But this contradict that underroot 5 is irrrational So ,our assumption is wrong and 7 underroot 5 is irrational no | |
| 19289. |
SinA+casA=p,prove that -cosecA=? |
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Answer» Om,bhai question 100%galat hai vishwas nhi hai toh jahan se dekh kar pucha hai na waha se ek baar correction kar lo ho skta gai ki galat type ho gya ho..hmm Ye question worng haii q ki (Cas= ka koi maan hi nahi hota haii)☺️☺️☺️ 99.95℅ it is wrong Is question is correct. |
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| 19290. |
Similar quadrilateral |
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| 19291. |
Guys Kya maths ke do paper h means standard level or higher level plz reply. |
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Answer» Yes, this year 2018-19 Yes for next year I also want to know..... Maine suna tha ki 2020 ke board mein maths ke easy or difficult ke papers ayenge, par ab mujhe bhi nahi pata?? Vo next year se hai this year aisa kuch nai hai I think? but it is dam sure? |
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| 19292. |
(Sin^3A+cos^3A/sinA+cosA) + sinAcosA |
| Answer» {tex}\\begin{array}{l}\\frac{\\sin\\;^3A\\;+\\cos^3A}{\\sin A+\\cos A}+\\;\\sin AcosA\\\\={\\textstyle\\frac{\\textstyle(\\sin A\\;+\\;\\cos A)(\\sin^2A+\\cos^2A-\\sin A\\cos A)}{\\sin{\\textstyle A}{\\textstyle\\;}{\\textstyle+}{\\textstyle\\cos}{\\textstyle A}}}\\;+\\;\\sin AcosA\\\\=\\sin^2A+\\cos^2A-\\sin A\\cos A+\\;\\sin AcosA\\\\=1\\\\\\end{array}{/tex} | |
| 19293. |
express the trigonometric ratios sinA , secA and tanA in terms of cotA. |
| Answer» For sin A,By using identity {tex}cosec ^ { 2 } A - \\cot ^ { 2 } A = 1 \\Rightarrow \\cos e c ^ { 2 } A = 1 + \\cot ^ { 2 } A{/tex}{tex}\\Rightarrow \\frac { 1 } { \\sin ^ { 2 } A } = 1 + \\cot ^ { 2 } A{/tex}{tex}\\Rightarrow \\sin A = \\frac { 1 } { \\sqrt { 1 + \\cot ^ { 2 } A } }{/tex}For secA,\xa0By using identity {tex}\\sec ^ { 2 } A - \\tan ^ { 2 } A = 1 \\Rightarrow \\sec ^ { 2 } A = 1 + \\tan ^ { 2 } A{/tex}{tex}\\Rightarrow \\sec ^ { 2 } A = 1 + \\frac { 1 } { \\cot ^ { 2 } A } = \\frac { \\cot ^ { 2 } A + 1 } { \\cot ^ { 2 } A } \\Rightarrow \\sec ^ { 2 } A = \\frac { 1 + \\cot ^ { 2 } A } { \\cot ^ { 2 } A }{/tex}{tex}\\Rightarrow \\sec A = \\frac { \\sqrt { 1 + \\cot ^ { 2 } A } } { \\cot A }{/tex}For tanA,{tex}\\tan A = \\frac { 1 } { \\cot A }{/tex} | |
| 19294. |
Can anybody tell me , ki board exams February me honge ya March mein....?? |
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Answer» Possibility is March but we can\'t say anything about CBSE\'s mood. Thanxxxxxxx??????? Vocational subjects in last of Feb and other subjects will be in March. |
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| 19295. |
What is the formula of midpont of line? |
| Answer» X+Y/2 | |
| 19296. |
Where we can use CSA and TSA while solving a question? |
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| 19297. |
Fine the value of k, when the distance between the points(3, 2k) and (4,1) is root 10 unit |
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Answer» By using distance formula. 4 |
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| 19298. |
What is sin 2x formula |
| Answer» sin 2x = 2 sinxcosx\xa0 | |
| 19299. |
For some integer m,every even integer is of the form |
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| 19300. |
Please help me to understand not to confuse about abscissa and ordinate???????????? |
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Answer» Thanks! See DearAbscissa is nothing but the horizontal line called the x-axisAnd ordinate is nothing but the vertical line called y-axis. |
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