InterviewSolution
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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.
| 2301. |
kisi ke boyfriend ya girlfriend h |
| Answer» | |
| 2302. |
Find the volume 1. Radius of the base = 7cm and height = 50cm |
| Answer» if its cylinder then its 22/7×7×7×50 22×7×50 22×350 calculate find out | |
| 2303. |
Natural number definition |
| Answer» Natural numbers are\xa0the positive integers (whole numbers) 1, 2, 3, etc., and sometimes zero as well. | |
| 2304. |
If sin theta=1/3 then find the value of 2cot²theta+2 |
| Answer» 10 ... Okk | |
| 2305. |
Board exam2010 solution |
| Answer» Check it on google | |
| 2306. |
12 solid spheres are made by melting radius 2cm and height is 16cm find the diameter of each sphere |
| Answer» | |
| 2307. |
R.S aggarwal/ Rd shrma/ Vidya book / ncert.Which one I am confused |
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Answer» NCERT R. S. Aggarwal and NCERT. |
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| 2308. |
What ise the value of tan |
| Answer» Sin/ Cos | |
| 2309. |
Show that exactly one number n,n+2 or n+4 is divisible by 3 |
| Answer» Pls see in RD dharma\'s book you will get it | |
| 2310. |
What is area of a minor segment? |
| Answer» {Area of sector- area of triangle }?? | |
| 2311. |
Tignometry |
| Answer» | |
| 2312. |
Tan a-tan a = 2cosec |
| Answer» | |
| 2313. |
lgta h jo koi bhi ye app jb jb chlata h vo pdta km h bate jyada krta h |
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| 2314. |
The sum of first n terms of an ap is 5n^2-3n. Find it\'s nth and 10th terms |
| Answer» Given, Sum of n terms of AP (Sn) = 5n2\xa0- 3nHence, Sum of (n - 1) terms of given AP isSn - 1\xa0= 5(n - 1)2\xa0- 3(n - 1){tex}\\Rightarrow{/tex}\xa0Sn - 1\xa0= 5(n2 - 2n + 1)\xa0- 3n + 3Or, Sn-1\xa0= 5n2\xa0- 10n + 5 - 3n + 3 = 5n2\xa0- 13n + 8We know that,\xa0nth term of AP is given by :-an\xa0= Sn\xa0- Sn - 1\xa0= (5n2\xa0- 3n) - (5n2\xa0- 13n + 8){tex}\\Rightarrow{/tex}\xa0an\xa0= 10n - 8{tex}\\therefore{/tex}\xa01st term of AP = 10\xa0×\xa01 - 8 = 2 2nd term of AP = 10\xa0×\xa02 - 8 = 123rd term of AP = 10\xa0×\xa03 - 8 = 22{tex}\\therefore{/tex}Required AP is 2, 12, 22, 32...........Now, 10th term of A.P isa10\xa0= 10\xa0×\xa010 - 8 = 100 - 8 = 92 | |
| 2315. |
If the circumference & the area of a circle are numerically equal,then diameter of the circle is |
| Answer» Area of circle =\xa0{tex}\\pi r^2{/tex}Perimeter of circle =\xa0{tex}2 \\pi r{/tex}Given , Area of Circle = Perimeter of Circle{tex}\\Rightarrow \\pi r^2= 2\\pi r{/tex}{tex}\\Rightarrow \\frac{r^2}{r}=2{/tex}{tex}\\Rightarrow r=2{/tex}\xa0cmTherefore, diameter of the given circle = 2(r) = 2(2) = 4\xa0cm | |
| 2316. |
222/7 |
| Answer» 31.714285 | |
| 2317. |
Hi Anjali sharma where are you? |
| Answer» | |
| 2318. |
Find the value of sin45°, by using formula ( A+B) = sinA.cosB + sinB.cosA. |
| Answer» | |
| 2319. |
Hi Anjali sharks where are you |
| Answer» | |
| 2320. |
Hi Annalise where are you |
| Answer» | |
| 2321. |
Solve for x=9x2-6ax+(a2+b2) |
| Answer» Given equation, 9x2- 6ax + a2- b2 = 0{tex}\\therefore \\quad x = \\frac { 6 a \\pm \\sqrt { ( - 6 a ) ^ { 2 } - 4 \\times 9 \\times \\left( a ^ { 2 } - b ^ { 2 } \\right) } } { 2 \\times 9 }{/tex}or,\xa0{tex}x = \\frac { 6 a \\pm \\sqrt { 36 a ^ { 2 } - 36 a ^ { 2 } + 36 b ^ { 2 } } } { 18 }{/tex}or,{tex}x = \\frac { 6 a + 6 b } { 18 } , x = \\frac { 6 a - 6 b } { 18 }{/tex}or,\xa0{tex}x = \\frac { 6 ( a + b ) } { 18 } , x = \\frac { 6 ( a - b ) } { 18 }{/tex}or,\xa0{tex}x = \\frac { a + b } { 3 } , x = \\frac { a - b } { 3 }{/tex} | |
| 2322. |
Find the sum of the following .(1-1÷n)+(1-2÷n)+(1-3÷n)+......up to n terms. |
| Answer» -3n | |
| 2323. |
Sec2 + tan2 |
| Answer» Sec2A + Tan2A = 1+Tan2A + Tan2A = 1+2Tan2A | |
| 2324. |
Kab tak mujhe se naraj raho gi ... Soory to bol diye n ..Ab maan bhi jao ??☺☺ |
| Answer» | |
| 2325. |
A man goes 10 m due south and then 24 m due west.How far is he from starting point |
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Answer» 26 m 26m. |
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| 2326. |
Formula of mode |
| Answer» a+【£fiui/€fi】h | |
| 2327. |
What is math actually |
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Answer» It is a very easy game of mind. game of pen and pencil. |
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| 2328. |
Bye Friend |
| Answer» | |
| 2329. |
If cosA+cosB+cosC=2 and A,B,C are angles of a triangle.Then which type of triangle is it? |
| Answer» | |
| 2330. |
Sorry I will never repeat my mistake again |
| Answer» | |
| 2331. |
What is meant by basic proportionality tgeorem |
| Answer» Hello guys....I know you all may not know me | |
| 2332. |
find the greatest no. of 6 digits exactly divisible by 24,15&36 |
| Answer» The greater number of 6 digits is 999999.LCM of 24, 15, and 36 is 360.{tex}999999 = 360 \\times 2777 + 279{/tex}Required number is = 999999 - 279 = 999720\xa0 | |
| 2333. |
for what value of n a natura no. 24 raise to the power n is divisible by 8 |
| Answer» 1 | |
| 2334. |
Define euclidean 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 | |
| 2335. |
X square - 11 x + 80 |
| Answer» Imaginary roots because \'d<0\' | |
| 2336. |
What is lactometer |
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Answer» Lactometer is use to check purity of milk Lactometer is used to check the purity of milk |
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| 2337. |
If the areas of two similar triangles are equal,prove that they are congurent |
| Answer» Solution:\xa0Given: ΔABC ~ ΔPQR. &\xa0ar ΔABC =ar ΔPQRTo Prove: ΔABC ≅ ΔPQRProof: Since, ΔABC ~ ΔPQRar ΔABC =ar ΔPQR. (given)ΔABC / ar ΔPQR = 1⇒ AB²/PQ² = BC²/QR² = CA²/PR² = 1[ USING THEOREM OF AREA OF SIMILAR TRIANGLES]⇒ AB= PQ , BC= QR & CA= PRThus, ΔABC ≅ ΔPQR[BY SSS criterion of congruence]\xa0 | |
| 2338. |
if the number a/b has a terminating decimal expansion ,what is the condition to be satisfied by b |
| Answer» b=2^m * 5^n | |
| 2339. |
Five cubes each of side 6cm are joined end to end .find the surface area of resulting cuboid |
| Answer» | |
| 2340. |
What is the difference between ungrouped and grouped data?? |
| Answer» Unlike\xa0ungrouped data,\xa0grouped data\xa0has been organized into several groups. To create\xa0grouped data, the raw\xa0data\xa0is sorted into groups, and a table showing how many\xa0datapoints occur in each\xa0group\xa0is created. ... This would change the frequency distribution a lot, even though the\xa0ungrouped data\xa0remained the same. | |
| 2341. |
23+32+54+54 |
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Answer» 163 163 163 163 163 |
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| 2342. |
2x+7\\5-3x+11\\2=2x+8\\3-5 |
| Answer» | |
| 2343. |
1 tan_cot Q=1tan2 |
| Answer» | |
| 2344. |
Show that tangent drawn at the end point of a diameter are parallel to each other |
| Answer» Given: PQ is a diameter of a circle with centre O. The lines AB and CD are the tangents at P and Q respectively.\xa0To prove: AB\xa0{tex}\\parallel{/tex} CDProof: AB is a tangent to the circle at P and Op is the radius through the point of contact{tex}\\because{/tex}{tex}\\angle{/tex}OPA = 90o .......(1) [The tangent at any point of a circle is perpendicular to the radius through the point of contact]{tex}\\because{/tex}\xa0CD is a tangent to the circle at Q and OQ is the radius through the point of contact.{tex}\\therefore{/tex}\xa0{tex}\\angle{/tex}OQD = 90o ........(2) [The tangent at any point of a circle is perpendicular to the radius through the point of contact]From (1) and (2),{tex}\\angle{/tex}OPA = {tex}\\angle{/tex}OQDBut these form a pair of equal alternate angles.{tex}\\because{/tex}\xa0AB\xa0{tex}\\parallel{/tex} CD | |
| 2345. |
If p is prime then find the hcf and lcm of p and p+1 |
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| 2346. |
Prove converse of Pythagoras theoem |
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| 2347. |
Find the roots of the quadratic equation 9X-15x+16=0 |
| Answer» Question is wrong | |
| 2348. |
Find the roots using factorisation method 4x^2 + 4root3x +3=0 |
| Answer» x=-√3/2,x=-√3/2 | |
| 2349. |
If one zero of polynomial in 2+√3 find the other zero if polynomial inx^-4x+1 |
| Answer» It is given that one of the zero of the polynomial x2\xa0- 4x + 1 is (2 + {tex}\\sqrt{3}{/tex}), we have to find the other zero.Let the other zero of the polynomial\xa0x2\xa0- 4x + 1 be\xa0{tex} \\alpha{/tex}.Then, sum of roots =\xa0{tex}-\\frac{(-4)}{1}{/tex}{tex}\\therefore{/tex}\xa02 +\xa0{tex}\\sqrt{3}{/tex}\xa0+\xa0{tex}\\alpha{/tex}\xa0= 4{tex}\\Rightarrow{/tex}\xa0{tex}\\alpha{/tex}\xa0= 2 -\xa0{tex}\\sqrt{3}{/tex}Thus, the other zero\xa0is 2 -\xa0{tex}\\sqrt{3}{/tex}. | |
| 2350. |
Au+Bv-(A-B) =0,Bu-Av-(A+B)=0 |
| Answer» | |