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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.
| 39901. |
The no. Which is divisible by no. 1 to 10 all |
| Answer» 5040 | |
| 39902. |
the sum of a natural number and it\'s reciprocal is 3/20 |
| Answer» 8 | |
| 39903. |
Please Anybody can tell what is the formula for (a+b+c)^3 |
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Answer» A (a+b)3+c3 now put the (a+b)3 formula you will get your answer |
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| 39904. |
128×234= |
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Answer» Hi It\'s cbse Hi cbsc |
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| 39905. |
Hyyyyya allBusy buddiesOf.class 10Are your exams over..... |
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Answer» Its from28th No it is near Nupe No ..... Not yet .... Yours ? |
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| 39906. |
Where do u live |
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Answer» Plz send your number to me For asking that question For what Ok i m realy very sorry |
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| 39907. |
X/2+y=0.87/x+y/2=10Find x and y value |
| Answer» {tex}\\frac{x}{2}{/tex}\xa0+ y = 0.8\xa0{tex}\\Rightarrow{/tex}\xa0x + 2y = 1.6{tex}\\Rightarrow{/tex}\xa0x = 1.6 - 2y ...........(i)and\xa0{tex}\\frac{7}{x+\\frac{y}{2}}=10{/tex}{tex} \\Rightarrow \\frac{14}{2 x+y}{/tex}\xa0= 10{tex}\\Rightarrow{/tex}\xa020x + 10y = 14{tex}\\Rightarrow{/tex}\xa010x + 5y = 7 ..........(ii)Put (i) in (ii), we get10(1.6 - 2y) + 5y = 7{tex}\\Rightarrow{/tex}\xa016 - 20y + 5y = 7{tex}\\Rightarrow{/tex}\xa0-15y = -9{tex}\\Rightarrow{/tex}\xa0y =\xa0{tex}\\frac{9}{15}{/tex}\xa0= 0.6Put y = 0.6 in equation (i), we getx = 1.6 - 2(0.6)x = 0.4Hence x = 0.4 and y = 0.6 is the solution of given system of equations. | |
| 39908. |
X2+y3+z |
| Answer» | |
| 39909. |
Prove that tangent is perpendicular to radius |
| Answer» Google srarch | |
| 39910. |
Find the value using suitable properties?7265×7265-7265×265 |
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Answer» but yogita +7265 and -7265 toh cut ho sakte h na 7265 × 7265 - 7265 × 265= 7265(7265 - 265)= 7265(7000)= 50855000 |
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| 39911. |
What is basic professional thoerme in Ch-6 |
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Answer» Its basic propotionality theorm i.e.,A line drawn parallel to one side of triangle intersects to sides in same ratio All therom are good . But 6.1 and 6.8 i think more inportant and use full in many questions |
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| 39912. |
In triangle pqr angle q = 90 and sin r = 3/5 write the value of cos p |
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Answer» Let pq=3x & pr=5x Cos p =adj/hypt. =3/5 R+P=90°=› P=90-R=› cosP=cos(90-R) = sinR = 3/5 |
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| 39913. |
value of sin2A+cos2A |
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Answer» Is it sin2A&cos2A or sin2(sq) A+cos 2(sq)A 1 1 |
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| 39914. |
Sin2x-sin4x+sin6x |
| Answer» sin4x | |
| 39915. |
(a-b)^2 =(a+b )^2- 4ab how |
| Answer» | |
| 39916. |
The largest number which divides 70and 125,leaving remainder 5 and 8 |
| Answer» 13 | |
| 39917. |
determine if the points (1,5) (2,3) and (-2,-11)are collinear?? |
| Answer» arrre vaishnavi area of triangle lga do na according to chapter 7 | |
| 39918. |
If the area of the similar triangles are equal prove that they are congruent. |
| Answer» See in NCERT book | |
| 39919. |
Prove 0plus0=2 |
| Answer» | |
| 39920. |
In two triangle corresponding angles are equal then two triangles are ? what |
| Answer» Similar, by AAA similarity criterian. | |
| 39921. |
2+5root5 |
| Answer» | |
| 39922. |
Prove that 1=2 also 2=3 |
| Answer» Ithink q is incomplete | |
| 39923. |
TanA =root 2-1 show that sinA cosA=root2/4 |
| Answer» | |
| 39924. |
In the adjoining figure pa and pb are tangents from p to a circle o.if angle aob=130 then angle apb |
| Answer» | |
| 39925. |
Completing the square metods i want to unerstand |
| Answer» {tex}x^2 +\xa06x - 16 = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}x^2 + 6x = 16{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}x^2 + 6x + 9 = 16 + 9{/tex} [Adding on both sides square of coefficient of x, i.e. ({tex}\\frac{6}{2}{/tex})2]{tex}\\Rightarrow{/tex}\xa0{tex}(x + 3)^2 = 25{/tex}{tex}\\Rightarrow{/tex}\xa0x + 3 =\xa0{tex}\\pm{/tex}{tex}\\sqrt{25}{/tex}{tex}\\Rightarrow{/tex}\xa0x + 3 =5 or x + 3 = -5{tex}\\Rightarrow{/tex}\xa0x = 2 or x = -8 | |
| 39926. |
If the sum of m termof an ap is the same as the sum of its n terms, then the sum its (m+n)term is |
| Answer» -(m+n) | |
| 39927. |
Mth term of an A.P. equal to n times of its nth term prove that (m+n) th term of A.P is 0 |
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Answer» Its m times of its mth term If you do this please explain this We will be using an formula |
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| 39928. |
IfS1,S2andS3 be the sum of n, 2n, 3n terms respectively of an AP prove that 3S1-3S2+S3=0 |
| Answer» {tex}{S_1} = \\frac{n}{2}\\left[ {2a + (n - 1)d} \\right]{/tex}{tex}{S_2} = \\frac{{2n}}{2}\\left[ {2a + (2n - 1)d} \\right]{/tex}{tex}{S_3} = \\frac{{3n}}{2}\\left[ {2a + (3n - 1)d} \\right]{/tex}R.H.S = 3(S2 - S1){tex} = 3\\left[ {\\frac{{2n}}{2}(2a + (2n - 1)d - \\frac{n}{2}(2a + (n - 1)d)} \\right]{/tex}{tex} = 3\\left[ {\\frac{n}{2}\\left[ {4a + 4nd - 2d - 2a - nd + d} \\right]} \\right]{/tex}{tex} = 3\\left[ {\\frac{n}{2}(2a + 3nd - d)} \\right]{/tex}{tex} = \\frac{{3n}}{2}\\left[ {2a + (3n - 1)d} \\right] = {S_3}{/tex} | |
| 39929. |
1/X+2/y=4 and 3/y -1/x=11 then the value of X and y is ? |
| Answer» | |
| 39930. |
Root13 |
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Answer» 169 3.605555 |
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| 39931. |
Root of 16 |
| Answer» Root of 16 = +4 or -4 | |
| 39932. |
What is rational |
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Answer» These no. Which can be written in the form of p/q Those no are form of p/qand qis smallest from0 are called rational |
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| 39933. |
Solve 3 X + 2 y 1,y>2 the system of inequalites |
| Answer» | |
| 39934. |
Marks distribution of maths of class 10 |
| Answer» | |
| 39935. |
Show that 17×41×43×61+43 is a composite number |
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Answer» 1868174 Hlo Aman....:17×41×43×61+43 (43 as comman ):43×42,517WKT wen a composite no is multiplied with another no then its is said to be a composite no .Thus proved Hope it helps:) |
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| 39936. |
Proof of Pythagoras theorum |
| Answer» See in book | |
| 39937. |
Prove sinx+cosx=1 |
| Answer» Yes sin2ue j cos jwiwu = 1 +0 prooved | |
| 39938. |
Provide me all very imp. Questions |
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Answer» Ch4 class3 mean C-4 of so.study class 3 Which chapter ?? |
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| 39939. |
x+y=14x-y=4 |
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Answer» X+y=14, x-y=4 2x=18, x=9, Y=14-9=5 x+y=14x-y=42x=18x=9x+y=149+y=14y=14-9y=5 |
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| 39940. |
Show that only one of the numbers n, n+2 and n+4 is divisible by 3. |
| Answer» Yes | |
| 39941. |
Derived frustum of cone |
| Answer» | |
| 39942. |
All formulas of ncert book |
| Answer» | |
| 39943. |
Prove that : tanø + secø -1/tanø -secø +1= sinø + 1/cosø |
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Answer» Please solve the question point by point (2+1)²+(-3y-y)=(2-5)²+(-3y-7)² Hindi me de guetion |
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| 39944. |
If theta=45,then find the value of 2cosecsquaretheta+3secsquaretheta |
| Answer» we know that , theta = 45 . then , 2×root 2 ka sqare = 2 + 3× root 2 ka sqare = 2 4+ 6= 10 | |
| 39945. |
In an isosceles triangle ,ABC ,AB =AC and BC is produce to D. prove that AD² - AC² = BC *CD . |
| Answer» I don\'t kn☺☺☺?? | |
| 39946. |
How to prove √3 is irrational |
| Answer» Let us assume that √3 is a rational numberThat is, we can find integers a and b (≠ 0) such that √3 = (a/b)Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.√3b = a⇒\xa03b2=a2 (Squaring on both sides) → (1)Therefore, a2 is divisible by 3Hence \x91a\x92 is also divisible by 3.So, we can write a = 3c for some integer c.Equation (1) becomes,3b2 =(3c)2⇒\xa03b2 = 9c2∴ b2 = 3c2This means that b2 is divisible by 3, and so b is also divisible by 3.Therefore, a and b have at least 3 as a common factor.But this contradicts the fact that a and b are coprime.This contradiction has arisen because of our incorrect assumption that √3 is rational.So, we conclude that √3 is irrational. | |
| 39947. |
The volume of frustum of a cone is equal to the _ of the volume of the two cones |
| Answer» | |
| 39948. |
1+ cos×cos÷cos×cos |
| Answer» | |
| 39949. |
Find the value of k in this 3(2.8)×π |
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Answer» Where is the K in the question? Kaha h k ? K kaha hai Where is k?? |
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| 39950. |
ABCD is a trapezium whose diagonal intersect at zero then prove that OA/OB=OC/OD |
| Answer» Given: A trapezium ABCD, In which AB\xa0{tex}\\parallel{/tex} CD and its diagonals AC and BD intersect at O.To Prove: {tex}\\frac{{AO}}{{OC}} = \\frac{{BO}}{{OD}}{/tex}Construction: Through O draw OE||ABproof: In {tex}\\triangle {/tex}ADC, OE {tex}\\parallel{/tex} DC,Hence {tex}\\frac{{AE}}{{ED}} = \\frac{{AO}}{{OC}}{/tex} .....(i).....[By BPT]Again in {tex}\\triangle {/tex}ABD, we have OE\xa0{tex}\\parallel{/tex} AB,hence {tex}\\frac{{DE}}{{EA}} = \\frac{{DO}}{{OB}}{/tex} ........[By BPT]{tex}\\Rightarrow {/tex}{tex}\\frac{{EA}}{{ED}} = \\frac{{OB}}{{OD}}{/tex} .....(ii)......[Using invertendo]{tex}\\therefore {/tex} From (i) and (ii), we.{tex}\\frac{{OB}}{{OD}} = \\frac{{AO}}{{OC}}{/tex}H once proved | |