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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.
| 40751. |
Prove that in |
| Answer» Nbhhhh | |
| 40752. |
What is the product of 345×104 |
| Answer» 35880 | |
| 40753. |
What are composite no. |
| Answer» A composite number is a positive integer which is not prime (i.e., which has factors other than 1 and itself). | |
| 40754. |
Find the perpendicular distance of the point p(4,2) from the x-axis |
| Answer» It is 2 unit | |
| 40755. |
Ankita please ek tum sa jaruri kam h please talk me |
| Answer» | |
| 40756. |
proof that if a number is doubled then its square becomes four times the square of given number |
| Answer» Let the number be : aIts square is : a^2Then, double it 2aAnd square of 2a will be 4a^2 And a^2 was the square of that number soHence proved | |
| 40757. |
Proved 2+2=5 |
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Answer» Yes but you are intelligent Gi 20-20=25-255×4-5×4=5×5-5×54(5-5)=5(5-5)4=52+2=5 Its very difficult to write its solution here ........ You may check it on google ..... |
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| 40758. |
Surface area of solids |
| Answer» | |
| 40759. |
Plzzz, give the answer fast..... If Sin x -2cos x=1Then prove, 2sin x - cos x=2 |
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Answer» But on google solution is not there sushma You should search it\'s solution on google you will find a brief solution there.? |
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| 40760. |
(X/x+1)2-5x/(x+1)+6=0 |
| Answer» | |
| 40761. |
what is value of2+(3×5÷5×3)-11+45÷2×2options(1) 0 (2) 44 (3) 45 (4) non of these |
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Answer» None of these None of these |
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| 40762. |
Show that sum and product of two irrational numbers 7+ |
| Answer» | |
| 40763. |
koi samja skta hai circle ka sylabus plz help me |
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Answer» Q bhai B m patil Kaun se school me padhte ho yadav g |
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| 40764. |
Kx²_14x+8=0 is 2 |
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Answer» kx2 - 14x + 8 = 0Put x = 2k(2)2 - 14(2) + 8 = 04k - 28 + 8 = 04k - 20 = 04k = 20k = 20/4k = 5 If x=2 . K(2)square-14*2+8=0. 4k-28+8=0. 4k-20=0. K=5 I can\'t understand what you are saying.......?????? |
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| 40765. |
Plzz tell all formula of AP and when we use these formula |
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Answer» I know formula but when what formula used plz tell When u have to find tn then tn=a+(n-1)d Sn=n/2 (a+l)Where,a is first term of an APl=last term of an APd= common difference between second and first term Basic terms is used in A. P 1 a=first term2 d=common difference 3 formula of an= a+(n-1)d4 formula of sn=n/2(2a+(n-1)d) an=a+(n-1)ds=n/2[2a+(n-1)d] |
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| 40766. |
Using the formula, tan2A=2tanA/1-tan²A |
| Answer» Find the value of tan60,it the being given that tan30=1/√3 | |
| 40767. |
(1+sinA) (1- sinA) /(1+cosA) (1-cosA) |
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Answer» My answer is cos^2A/sin^2A What\'s the correct answer 1 |
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| 40768. |
How many tangent draw from an external point |
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Answer» 2 We can draw only two tangents from an external point From an external point only 2 tangents can be drown on the circle Infinite |
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| 40769. |
If 7 times the 7th term of an AP is equal to 11 times of 11th term. Find its 18th term |
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Answer» What ??? Abc |
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| 40770. |
ROOT3 ×ROO2upon 3 |
| Answer» | |
| 40771. |
How i study math before 2 month before from exam plzz help me friend |
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Answer» 2 month bahut hai Do only NCERT Questions only solve last year solved papers and sample papers for better resulr |
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| 40772. |
Prove Pythagoras theorem postulates |
| Answer» Abhi batau ga Mam , you can search in Advisor,Rd,Rs and also on my fav Youtube. | |
| 40773. |
What is the value of √2. |
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Answer» Value of root 2 is1.414. 1.41 |
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| 40774. |
What is corresponig arc and segments |
| Answer» Corresponding ka mtlb hota hai = area of circle - area of triangle.And when we write segment and sector it means it is minor segment And we know that minor segment = area of triangle - area of sector | |
| 40775. |
What is the 9th term of the following A.P 6,8,10.... |
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Answer» We find a9We have a=6 d=a2-a1=8-6=2 n=9So, a9=a+(n-1)d a9=6+(9-1)2 =6+(8)2 =6+16 =22 22 22is the right answer a+8d sha ho kha |
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| 40776. |
Sino+coso=x,prove that sin6o+cos6o=4-3(x2-1)2\\4 |
| Answer» No needed These r not xam based question | |
| 40777. |
Find the eleventh term from the last term of the AP 27,23,19,....,-65 |
| Answer» we use this formula l-(n -1)d where l = -65 , d = -4 , n=11 -65 - ( 10 ) (-4) -65 - (-40) -65 + 40 -25 ans the eleventh term from the last term is -25 | |
| 40778. |
Can you explain sum of first n terms of an AP |
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Answer» Thanks Let Sn be the sum of n terms with \'a\' first step and \'d\' is common differencethen Sn= a + (a+d )+ ................+a+(n-2)*d.+a+(n-1)*dor Sn=(a+(n-1)*d).+(a+(n-2)*d ) ............+(a+d ) + a ( Taking Last term first)2Sn=2a +(n-1)*d +(a+d +a +(n-2)*d +...........\xa02Sn=(a+a +(n-1)*d )+(a+a +(n-1)*d +...........n terms2Sn= n*(a+a +(n-1)*d )Sn=n*(2a +(n-1)*d ) / 2 Aaa |
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| 40779. |
which term of the ap 121 117 113 .... is its negative term? (hint; find n for an |
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Answer» Given AP is:121, 117, 113, ................First term a = 121Common difference d = 117-121 = -4Let nth term is the first negative term in APNow nth tern in AP = a + (n-1)× d = 121 + (n - 1) × (-4) = 121 - 4n + 4 = 125 - 4nNow, we have to find the suitable value of n so that the value of (125 - 4n) is negative.⇒ 125 - 4n < 0⇒ 125 < 4n⇒ 4n > 125⇒ n > 125/4⇒ n > 31.25Since the term can not be in decimal form (term is always taken a positive integer)\xa0So, n = 32\xa0Now, by taking value of n = 32, we getnth term = 125 - 4×32 = 125 - 128 = -3This is the first negative term in the series.So, n = 32Hence, the\xa032nd\xa0term is the first negative term in AP. suppose n th term is 0so 121+(n-1)(-4)=0121-4n+4=0n=125/4 =31.25so 32 nd term is negativeas S32=121-31*4=-3\xa0\xa0\xa0 |
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| 40780. |
Find the sum of three-digit multipleis natural number , which is divisible by 11 |
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Answer» Sorry there was some mistaken990=110+(n-1)*11880=11n-1111n=891 so n=81Sn= 81/2( 2*110+80*11)=81/2 * 1100=44550 110,121,..........990Than A=110D=121-110=11An=990An =a+(n-1)d990=110+(n-1)11990=110+11n-11990=99+11nOr 11n=990-99n=891/11=81 Here first term a= 110, d=11 and last term =990suppose no. of terms =nthen 990= 110 +(n-1) x 11 (nth term of AP)\xa0n-1=8 so n=9sum of all the numbers= 9/2x( 2x110 + 8x11)=9/2 x ( 220+88) = 1386\xa0 |
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| 40781. |
Hlo everyone. Will anyone is there who will top in cbse |
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Answer» Har koi sirf mehnat hee Kar Sakta h sab ☝ upper wale ke marzi... But Agar kisi NE mehnat Kia aur ni v Aaayya toh v apne aap par confidence rakhe aage bohot se mouke milenge.... HO SAKTA HAI AAPKE LIE BHAGWAN NE KUCH BOHOT HEE SPECIAL SOCHA HOGA... AGAR KAVI AESA LAGE KI MUJHSE NI ho paega then main suggest krti Hu ki aap motivational speaker SANDEEP MAHESHWARI KE VIDEOS DEKHE you\'ll feel better and... ALL the best???☺️☺️☺️ Every Hardworker is a topper i think my some friends have the probability to top in exam they r so good everyone tries his/ her best. but the fruit is achieved by the deserving and the most hard working Each and every of us will try our best.....?? |
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| 40782. |
Find a relation between x and y such that the point (x,y) is equidstant from the points (7,1) (3,5). |
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Answer» X - y - 2 = 0 Use distance formulae O(x, y) |
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| 40783. |
Has anyone prepared whole course? |
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Answer» Except construction Yup Not at all 7/8 Yes Yes |
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| 40784. |
Find the roots 1/* |
| Answer» | |
| 40785. |
7x-2y=5xy, 8x+7y=15xy. Solve the linear pair of equation. |
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Answer» {tex}\\begin{array}{l}7x-2y=5xy\\;\\;----(1)\\\\8x+7y=15xy\\;----(2)\\\\from\\;1)\\;and\\;(2)\\\\8x+7y=3(7x-2y)\\\\8x+7y=21x\\;-6y\\\\13y\\;=13x\\;so\\;x=y\\\\Now\\;from(1)\\\\7x-2x=5x^{2\\;}\\;or\\;\\;5x=5x^2\\\\x=1\\;hence\\;y=1\\\\\\\\\\\\\\end{array}{/tex} Sorry, x=1 and y=1 X=1 and y=2 |
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| 40786. |
If the equation x^2-2x(1+3k)+7(3+2k)=0 has eqaul roots, then find the valur of k. |
| Answer» Given, x2 - 2x(1 + 3k) + 7(3 + 2k) = 0Here, a = 1, b = -2(1 + 3k) and c = 7(3 + 2k)The given equation will have equal roots, ifD = 0{tex}\\Rightarrow{/tex}\xa0b2\xa0- 4ac = 0\xa0{tex}\\Rightarrow{/tex}4(1 + 3k)2 - 4\xa0{tex}\\times{/tex} 1\xa0{tex}\\times{/tex} 7(3 + 2k)\xa0= 0{tex} \\Rightarrow{/tex}4(3k\xa0+ 1)2 - 4\xa0{tex}\\times{/tex} 1\xa0{tex}\\times{/tex} 7(3 + 2k) = 0{tex}\\Rightarrow{/tex}\xa04(9k2\xa0+ 1 +6k) - 4(21 + 14k) = 0{tex}\\Rightarrow{/tex}\xa04(9k2\xa0+1 +6k) - 84 - 56k = 0{tex}\\Rightarrow 36k^2+4+24k-84-56k=0 {/tex}{tex} \\Rightarrow{/tex}\xa04(9k2 - 8k- 20) = 0{tex} \\Rightarrow{/tex}\xa09k2 - 8k -20 = 0{tex} \\Rightarrow{/tex}\xa09k2 - 18k + 10k\xa0- 20 = 0{tex} \\Rightarrow \\quad ( k - 2 ) ( 9 k + 10 ) = 0 \\Rightarrow k - 2 = 0 \\text { or, } 9 k + 10 = 0 \\Rightarrow k = 2 \\text { or, } k = - \\frac { 10 } { 9 }{/tex} | |
| 40787. |
What is the HCF smallest prime number and the smallest composite number. |
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Answer» The smallest prime no is 2 and smallest composite no is 4 so the hcf of the no is 2 Maxium gyuies is rong so improve your self Right answer is 2 is prime and 4 is composite any excuise to anyone check any website 2 is the HCF of this eqution 1 is the HCF of smallest prime and composit no. Smallest prime no.= 2Smallest composite no.= 4So, their HCF is = 2 HCF of smallest prime no. and smallest composite no. is 1 |
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| 40788. |
trignometry ratios table?? |
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Answer» Maths book rs aggarwal class 10 chapter,11 Open a ncert book mathematics chapter 6 page.no.185 |
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| 40789. |
Xki power 2 - 2ax-(4b ki pwer 2 - a ki power 2 ) =0Solve the following quadratic equation for X |
| Answer» {tex}x^2\xa0- 2ax - (4b^2\xa0- a^2) = 0{/tex}{tex} \\Rightarrow{/tex}\xa0{tex}x^2\xa0- 2ax + (a^2\xa0- 4b^2) = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}x^2\xa0- 2ax + (a - 2b)(a + 2b) = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}x^2\xa0- (a - 2b)x - ( a+ 2b)x + (a - 2b) (a + 2b) = 0{/tex}{tex}[2ax=(a-2b)x+(a+2b)x]{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}x[x- (a - 2b )] - (a + 2b) [x - (a - 2b)] = 0{/tex}{tex}\\Rightarrow{/tex}\xa0[x - (a - 2b)] [x - (a + 2b)] = 0{tex}\\Rightarrow{/tex}\xa0x - (a - 2b) = 0 or x - (a + 2b) = 0{tex}\\Rightarrow{/tex}\xa0x = a -2b or x = a + 2b | |
| 40790. |
Sir l am very weak in maths so plz help me |
| Answer» | |
| 40791. |
Please tell some book where cbse paper question s came? |
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Answer» Thank u Last 10 years sample paper ki practise kr lo Ncert |
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| 40792. |
If HCF of two numbers are 78 and 13 then expressed in linear combination |
| Answer» Bhai khud nikal lo | |
| 40793. |
In the figure CD||LA and DE||AC find the length of CL if BE = 4cm and EC = 2cm |
| Answer» 10 cm | |
| 40794. |
If sin A=x°and sec A=y find the value of tanA |
| Answer» Given,SinA=x&SecA=y=1/cosA=y=CosA=1/yWe know,TanA=sinA/cosA =x/1/1/y =x.y | |
| 40795. |
Writr trignometri ratio of sin A in terms of cot A |
| Answer» 1/Underoot1+cot²q | |
| 40796. |
Cos(60+2a)÷cos2a-√3sin^2a |
| Answer» | |
| 40797. |
Board marking scheme |
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Answer» You can see marking scheme by this app only.. Of which subject |
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| 40798. |
Any one from patna |
| Answer» From Kanpur | |
| 40799. |
Ch 2 extra questions |
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Answer» You can get by this app only Check out this app u will get some important extra question of chapter 2nd. |
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| 40800. |
Any is ready to help me to solve mathematicsPlz...... |
| Answer» Yes of course | |