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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.
| 1851. |
How to derive the relation between side and radius of a circle inscribed in a triangle |
| Answer» | |
| 1852. |
the sum of first 14 terms of an ap is 1505 amd first term is 10 then find 25 term |
| Answer» 370 | |
| 1853. |
Find the sum of integers between 100 and 200,which are divisible by 7. |
| Answer» | |
| 1854. |
NCERT CHAPTER 15 PROBABILITY EXERCISE 15.1 Q NO. 15 |
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Answer» Nice Khushi Out of five cards_the ten,Jack,queen, king and ace of diamonds, one card can be drawn in 5 ways.So,total number of elementary events =51.there is only onw queenFavourable number of elementary events =1Hence,P (the queen) =1/52.after keeping the queen card aside,we are left with 4 cards. So,total number of elementary events now=4.(a) there is only one aceFavourable number of elementary events =1Hence,P (an ace)=1/4(b).there is no cards as queen. Favourable number of elementary events =0Hence,P (the queen) =0/4=0.. HENCE PROVED ? |
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| 1855. |
Sir |
| Answer» Hay | |
| 1856. |
About circles |
| Answer» You can get expiation in NCERT book | |
| 1857. |
Formula for middle term of an AP |
| Answer» 1/2( nterm+1) | |
| 1858. |
If radius of circle is 6cm and length of arc is 12cm then find the area of sector |
| Answer» Length of an arc= 12 cm and radius = 6 cmArea of the sector =\xa0{tex}\\frac { 1 } { 2 } \\times{/tex}(length of the corresponding arc) {tex}\\times {/tex}\xa0radius{tex}= \\frac { 1 } { 2 } \\times l \\times r{/tex}{tex} = \\frac { 1 } { 2 } \\times 12 \\times 6{/tex}{tex}= 6 \\times 6{/tex}= 36 cm2 | |
| 1859. |
How to solve median |
| Answer» Solve median by formula L+(n/2-cf/f)h | |
| 1860. |
What is the exact definition of curved and total surface area with examples |
| Answer» I don\'t know what is the correct definition TSA mean whole surface area and curved mean open any surface | |
| 1861. |
If the sum of first m terms of an AP is am sq. + bm ,find its common diffrence |
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| 1862. |
Find the sum of first 15 multiples of 8. |
| Answer» n=15 d=8 a=8Sn=n/2(2a+(n-1)d) =15/2(2×8+14×8) =15/2×128 =15×64=960 | |
| 1863. |
Last 6 year questions |
| Answer» Check last year papers here :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 1864. |
Chapter 8 Trigonometry exercise 8.4 problem how to change sine Chane all trigonometry ratio |
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| 1865. |
For every natural number n 6n always ends with |
| Answer» 6 | |
| 1866. |
Formula of surface area of cube cuboid cylinder sphere hemisphere and cone |
| Answer» | |
| 1867. |
Define tagent? |
| Answer» A tangent to a circle is a line that intersects the circle at only one point,,, the word tangent comes from latin word\'tangere\' which neans to touch | |
| 1868. |
I have some problem in important question of 3 marks in 1 sum 2 equation |
| Answer» | |
| 1869. |
Is official date sheet of board released?? |
| Answer» Not yet | |
| 1870. |
Probability of not getting event |
| Answer» 1-P(of an event) | |
| 1871. |
For what value of k Will k+9,2k+7 are the Conservative terms of an A.P? |
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Answer» K= -7/2 Take take a = k+9 and d= ,2 k +7 -( k +9) |
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| 1872. |
Find the Angle subtended at the center of a circle of radius 5cm by an arc of length (5π/3)cm. |
| Answer» Radius of circle = 5cmLength of arc\xa0{tex} = \\frac{{5\\pi }}{3}cm{/tex}{tex}\\therefore {/tex}\xa0length of arc\xa0{tex} = \\frac{\\theta }{{360^\\circ }} \\times 2\\pi r{/tex}{tex} \\Rightarrow \\frac{{5\\pi }}{3} = \\frac{\\theta }{{360^\\circ }} \\times 2\\pi \\times 5{/tex}{tex} \\Rightarrow \\theta = \\frac{{5\\pi \\times 360^\\circ }}{{3 \\times 2\\pi \\times 5}}{/tex}{tex} \\Rightarrow \\theta = 60^\\circ {/tex} | |
| 1873. |
sin30=? |
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Answer» 1/ 2 1/2 |
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| 1874. |
solve a2u2+2abcdu - (1+2c)(b2d2)=0 by completing the square method |
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| 1875. |
Important question ..jo test m kaafi baar aate h |
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| 1876. |
cncuhrnrjfuf |
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| 1877. |
The difference of two natural no. Is 5. If the difference in their reciprocal is 1/10. Find the no. |
| Answer» Let two numbers be x and y. So x-y= 5 ...(1)1/y-1/x= 1/10x-y/xy = 1/105/xy = 1/10xy = 50From eq (1), we have x= 5+y(5+y)y = 50y2+5y = 50y2\xa0+5y -50=0y2\xa0+10y -5y -50=0y(y+10) -5(y+10) =0(y+10)(y-5) = 0y = 5 or -10y= 5 because -10 can not be a natural numberx = 5+5 =10One number is 5 and other number is 10.\xa0 | |
| 1878. |
Find the sum of fifty positive integers which gives a remainder 2 when divided by 5 |
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| 1879. |
The difference of two numbers is 4 and sum of their reciprocal is 4/21 find the numbers |
| Answer» Let the two no. be x and y.Difference:X-Y=4......Eq.1Reciprocal:1/x+1/Y=4/21X+Y=21/4......Eq.2Addind Eq.1 and 2.2x=37/4.X=37/8Similarly by putting value of X in eq.2 or 1 U will get value of Y. | |
| 1880. |
Find a point on the x axis which is equidistant from the point 7,6and -3,4 |
| Answer» Any point P on x axis is of the form of (x,0). Lwe A=(7,6) and B =(-3,4)It is given that PA= PB√(7-x)2\xa0+(6-0)2\xa0= √(-3-x)2\xa0+ (4-0)2√(7-x)2\xa0+(6)2\xa0= √(-3-x)2+(4)2√49+x2\xa0-14x+36 = √9+x2\xa0+16+6x√x2\xa0-14x+85= √x2+6x+25Squaring both sides , we getx2-14x+85= x2\xa0+6x +25-14x-6x = 25-85-20x = -60x = 3So point P = (3,0)\xa0 | |
| 1881. |
Prove phytogorous theorm |
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Answer» H²=p²+b² I know but I can\'t tell u |
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| 1882. |
what is the formula of equilateral traingle |
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Answer» √3/4 side^2 Root3/4(side)² |
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| 1883. |
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73 |
| Answer» a11\xa0= a+(n-1)d38= a+(11-1)d38=a+10d ...(1)a16\xa0= a+(n-1)d73 = a+(16-1)d73= a+15d ....(2)Subtracting (1) from (2), we get5d= 35d= 35/5 = 7substituting value of d in (1), we get38= a+ 10*738= a+70a= 38-70a= -32So 31 st term a31\xa0= a+(n-1)da31\xa0= -32+(31-1)*7a31\xa0= -32+30*7a31\xa0= -32+210a31 = 178\xa0 | |
| 1884. |
33344% = ?marks |
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| 1885. |
Find the coordinates of the point on y- axis which is nearest to the point [ - 2, 5] |
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Answer» Perpendicular distance from point (-2,5) on y axis is the nearest (0 , 5) -1 , 4 -1,4 |
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| 1886. |
If the sum of squres of the zeroes of polynomual 6x^2+x+k is 25/36 find the value of k |
| Answer» | |
| 1887. |
(1+cotA+tan) ( sinA-cisA) = sinAtanA -cot A cosA |
| Answer» L.H.S\xa0= (1 + cotA + tanA) (sinA - cosA)= sinA - cosA + cotA sinA - cotA cosA + sinA tanA - tanA cosA{tex}= \\sin A - \\cos A + \\frac{{\\cos A}}{{\\sin A}} \\times \\sin A - \\cot A\\cos A + \\sin A\\;\\tan A - \\frac{{\\sin A}}{{\\cos A}} \\times \\cos A{/tex}= sinA - cosA + cosA - cotA cosA + sinA tanA - sinA= sinA tanA - cotA cosA........(1)Now taking ;{tex}\\quad \\frac{{\\sec A}}{{\\cos e{c^2}A}} - \\frac{{\\cos ecA}}{{{{\\sec }^2}A}}{/tex}{tex} = \\frac{{\\frac{1}{{\\cos A}}}}{{\\frac{1}{{{{\\sin }^2}A}}}} - \\frac{{\\frac{1}{{\\sin A}}}}{{\\frac{1}{{{{\\cos }^2}A}}}}{/tex}{tex} = \\frac{{{{\\sin }^2}A}}{{\\cos A}} - \\frac{{{{\\cos }^2}A}}{{\\sin A}}{/tex}{tex} = \\sin A \\times \\frac{{\\sin A}}{{\\cos A}} - \\cos A \\times \\frac{{\\cos A}}{{\\sin A}}{/tex}{tex} = \\sin A \\times \\tan A - \\cos A \\times \\cot A{/tex}.......(2)From (1) & (2),(1 + cotA + tanA) (sinA - cosA) =\xa0{tex}\\frac { \\sec A } { cosec ^ { 2 } A } - \\frac { cosec A } { \\sec ^ { 2 } A }{/tex}\xa0= sinA.tanA - cosA.cotA\xa0Hence, Proved. | |
| 1888. |
What is similar |
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Answer» In two figures if 2 or 3 angles or sides etc are equal. when two figures are of same shape\xa0but of different sizewhen their corresponduing sides are proportional\xa0when their corresponding angles are equal Means same |
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| 1889. |
What is ratio |
| Answer» The quantitative relation between two amounts showing the number of times one value contains or is contained within the other. | |
| 1890. |
If HCF (a,b)=12 and a*b=1800. Then find LCM (a,b) |
| Answer» Hcf*LCM= a*b so 12*LCM= 1800, LCM = 1800/12= 150 | |
| 1891. |
If cot a = 12 upon 5 then find the value of Sin A + Cos A into cosec a |
| Answer» Given:cot A =\xa0{tex}\\frac { 12 } { 5 }{/tex}Now, (sin A + cos A)\xa0{tex}\\times{/tex}\xa0cosec A= (sin A + cos A)\xa0{tex}\\frac { 1 } { \\sin A }{/tex}= sin A\xa0{tex}\\times \\frac { 1 } { \\sin A }{/tex}\xa0+ cos A\xa0{tex}\\times \\frac { 1 } { \\sin A }{/tex}= 1 + cot A = 1 +\xa0{tex}\\frac { 12 } { 5 } = \\frac { 17 } { 5 }{/tex} | |
| 1892. |
Kya koi bata sakta hai ki ch 14 NCERT example 3 me h ko 20 kaise liya |
| Answer» | |
| 1893. |
Find solution 8/2x+3y +6/3x+2y=10 |
| Answer» | |
| 1894. |
Give me a sample paper of mathematics in which latest pattern of CBSE done. |
| Answer» Check sample papers here :\xa0https://mycbseguide.com/cbse-sample-papers.html | |
| 1895. |
Prove that n^3-n is divisible by 6 for any positive integer n. |
| Answer» n3\xa0- n = n (n2\xa0- 1) = n (n - 1) (n + 1)\xa0Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.If n = 3p, then n is divisible by 3.If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.⇒ n (n – 1) (n + 1) is divisible by 3.Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.∴ n = 2q or 2q + 1, where q is some integer.If n = 2q, then n is divisible by 2.If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.⇒ n (n – 1) (n + 1) is divisible by 2.Since, n (n – 1) (n + 1) is divisible by 2 and 3.∴ n (n-1) (n+1) = n3\xa0- n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)\xa0 | |
| 1896. |
Determine k so that (3k-2),(4k-6) and (k+2) are three consecutive terms of an. Ap. |
| Answer» a1= (3k-2)a2= (4k-6)a3= (k+2)Now,a2-a1=a3-a2(4k-6)-(3k-2)=k+2-(4k-6)4k-6-3k+2=k+2-4k+6k-4=-3k+8K+3k=8+44k=12K=12/4K=3 | |
| 1897. |
Pythagorus theorm solved |
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Answer» Tnkx In a right triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides.We are given a triangle ABC right angled at BWe need to prove that AC^2=AB^2+BC^2Let us draw BD perpendicular AC∆ABC~∆ADC. (Theorem)AD\\AB=AB\\AC. (Sides are proportional)AD.AC=AB^2. (1)Also, ∆BDC~∆ABCCD\\BC=BC\\ACCD.AC.=BC^2. (2)Adding (1)&(2),AD.AC+CD.AC=AB^2+BC^2or, AC(AD+CD)=AB^2+BC^2or, AC.AC=AB^2+BC^2or, AC^2=AB^2+BC^2Thank U By ARMAN INAMDAR |
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| 1898. |
I m having maths exam this monday any suggestions please? |
| Answer» Jisa la Kacha katt | |
| 1899. |
Sample paper of class 10 2017 _2018 |
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Answer» Download from cbse.academic.in Download it from google |
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| 1900. |
Sin2-cos2. Proof |
| Answer» | |