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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.
| 3251. |
Sahil mene kuch puchha |
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| 3252. |
Sahil ek baat bolu kya |
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| 3253. |
Sahil aapko kya Hua hai mujhe pata nhi |
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| 3254. |
A cottage industry produces a certain no.of pottery articles |
| Answer» Let cost of production of each article be Rs xWe are given total cost of production on that particular day = Rs 90Therefore, total number of articles produced that day = 90/xAccording to the given conditions,{tex} x = 2 \\left( \\frac { 90 } { x } \\right) + 3{/tex}{tex}\\Rightarrow x = \\frac { 180 } { x } + 3{/tex}{tex}\\Rightarrow x = \\frac { 180 + 3 x } { x }{/tex}{tex}\\Rightarrow x ^ { 2 } = 180 + 3 x{/tex}{tex}\\Rightarrow x ^ { 2 } - 3 x - 180 = 0{/tex}{tex}\\Rightarrow x ^ { 2 } - 15 x + 12 x - 180 = 0{/tex}⇒\xa0x (x − 15) + 12 (x − 15) = 0⇒ (x − 15) (x+ 12) = 0 ⇒\xa0x = 15, −12Cost cannot be in negative, therefore, we discard x = − 12Therefore, x = Rs 15 which is the cost of production of each article.Number of articles produced on that particular day = {tex} \\frac { 90 } { 15 }{/tex}\xa0= 6 | |
| 3255. |
Basic converse therom |
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| 3256. |
xsquare+√x-2=4. |
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| 3257. |
cos45*/sec30*+cosec30*...evaluate?? |
| Answer» 4+√3/8 | |
| 3258. |
The sum of first n terms of an AP is (3n²+6n).find the nth term &find the 15th termif this AP. |
| Answer» Here it is given that sum of first n terms\xa0is equal to Sn\xa0= 3n2\xa0+ 6n\xa0Now,\xa0Sn-1\xa0= 3(n - 1)2\xa0+ 6(n - 1)= 3(n2\xa0- 2n + 1) + 6n - 6= 3n2\xa0- 6n + 3 + 6n - 6= 3n2\xa0- 3Therefore nth term, Tn\xa0= Sn\xa0- Sn-1\xa0= 3n2\xa0+ 6n - 3n2\xa0+ 3= 6n + 3And, 15th term T15\xa0=\xa06(15) + 3= 90 + 3= 93\xa0Therefore nth\xa0term is 6n+3 and 15th\xa0term is 93 | |
| 3259. |
47x+31y=63 ,31x+47y=15 find the value of xandy |
| Answer» X=2 & y=-1 | |
| 3260. |
Good morning sahil ji |
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| 3261. |
25 perfect square |
| Answer» 5 | |
| 3262. |
In an application the sum of first n terms is 3n^2/2+13n/2.find it\'s 25 term. |
| Answer» According to the question,Given Sum of n terms {tex}\\left( {{S_n}} \\right) = \\frac{{3{n^2}}}{2} + \\frac{{13}}{2}n{/tex}Put n = 24, {tex}{S_{24}} = \\frac{{3 \\times 24 \\times 24}}{2} + \\frac{{13 \\times 24}}{2}{/tex}= 864 + 156= 1020Put n = 25, {tex}{S_{25}} = \\frac{{3 \\times 25 \\times 25}}{2} + \\frac{{13 \\times 25}}{2}{/tex}{tex} = \\frac{{1875}}{2} + \\frac{{325}}{2}{/tex}{tex} = \\frac{{2200}}{2} = 1100{/tex}{tex}\\therefore{/tex} 25th term (a{tex}_{25}{/tex}) = S25 - S24= 1100 - 1020= 80 | |
| 3263. |
Kaisa ho sab |
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| 3264. |
By what number should 1365 be divided to get 31 as quotient and 32 as remainder? |
| Answer» 43 | |
| 3265. |
If sinA+2cosA=1 then prove that 2sinA-cosA=2 |
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| 3266. |
Explain why 3×5×7+7 is a composite number |
| Answer» We have, (3\xa0× 5\xa0× 7) + 7 = 105 + 7 = 112Prime factors of 112 = 2\xa0× 2\xa0× 2\xa0× 2\xa0× 7 = 24\xa0{tex} \\times {/tex}\xa07So, it is the product of prime factors 2 and 7, i.e. it has factors other than 1 and itself. Hence, it is a composite number. | |
| 3267. |
Find x if the distance between the points (x,2) (3,4) be √8 units |
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| 3268. |
Hlovfrienss how r u all |
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| 3269. |
Wat is area of segment |
| Answer» The area bounded by chord and the arc is called area of segment.It can be found by subtracting area of triangle from area of sector. | |
| 3270. |
RIMSHA kahan chali gyi aap |
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| 3271. |
Which samplepaper is best oswal or i-succed of arihant |
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Answer» No...... Need of samplepaper only study carefully Ideal is the best but I suggest u all in one . |
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| 3272. |
1/x+1/y=1/a+1/b |
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| 3273. |
ishi r u here |
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| 3274. |
Science me problem |
| Answer» Magnesium is third reactive metal | |
| 3275. |
the ratio of n terms of 2 ap is 7n+1 |
| Answer» Let a1\xa0and a2 be the first terms and d1\xa0and d2 be the common difference of the two APs respectively.Let Sn and S\'n be the sums of the first n terms of the two APs and Tn and T\'n\xa0be their nth terms respectively.Then,\xa0{tex}\\frac { S _ { n } } { S _ { n } ^ { \\prime } } = \\frac { 7 n + 1 } { 4 n + 27 } \\Rightarrow \\frac { \\frac { n } { 2 } \\left[ 2 a _ { 1 } + ( n - 1 ) d _ { 1 } \\right] } { \\frac { n } { 2 } \\left[ 2 a _ { 2 } + ( n - 1 ) d _ { 2 } \\right] } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}{tex}\\Rightarrow \\frac { 2 a _ { 1 } + ( n - 1 ) d _ { 1 } } { 2 a _ { 2 } + ( n - 1 ) d _ { 2 } } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}\xa0........(i)To find the ratio of mth terms, we replace n by (2m -1) in the above expression.Replacing n by (2 {tex}\\times{/tex}\xa09 -1), i.e., 17 on both sides in (i), we get{tex}\\frac { 2 a _ { 1 } + ( 17 - 1 ) d _ { 1 } } { 2 a _ { 2 } + ( 17 - 1 ) d _ { 2 } } = \\frac { 7 \\times 17 + 1 } { 4 \\times 17 + 27 } \\Rightarrow \\frac { 2 a _ { 1 } + 16 d _ { 1 } } { 2 a _ { 2 } + 16 d _ { 2 } } = \\frac { 120 } { 95 }{/tex}{tex}\\Rightarrow \\frac { a _ { 1 } + 8 d _ { 1 } } { a _ { 2 } + 8 d _ { 2 } } = \\frac { 24 } { 19 }{/tex}{tex}\\Rightarrow \\frac { a _ { 1 } + ( 9 - 1 ) d _ { 1 } } { a _ { 2 } + ( 9 - 1 ) d _ { 2 } } = \\frac { 24 } { 19 }{/tex}{tex}\\Rightarrow \\frac { a _ { 1 } + ( 9 - 1 ) d _ { 1 } } { a _ { 2 } + ( 9 - 1 ) d _ { 2 } } = \\frac { 24 } { 19 }{/tex}{tex}\\Rightarrow \\frac { T _ { 9 } } { T _ { 9 } ^ { \\prime } } = \\frac { 24 } { 19 }{/tex}{tex}\\therefore{/tex}\xa0required ratio = 24:19. | |
| 3276. |
Tan theta |
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| 3277. |
1/cot. = tan |
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| 3278. |
If the total surface area of a solid hemisphere is 462 cm square ,find its volume. |
| Answer» Let r cm be the radius of the base and h cm be the height of the cylinder. Then, total surface area of cylinder {tex}= 2\\pi r(r + h){/tex}Curved surface area of cylinder {tex}= 2\\pi rh{/tex}We have,Curved surface area = {tex}\\frac{1}{3}{/tex}(Total surface area) = {tex}\\frac{1}{3}{/tex}{tex}\\times{/tex}\xa0{tex}462 cm^2 = 154 cm^2\xa0{/tex}{tex}\\Rightarrow{/tex}{tex}2\\pi rh = 154{/tex}Also, {tex}2\\pi rh + 2\\pi r^2 = 462 {/tex}{tex}\\Rightarrow{/tex}154 + 2{tex}\\pi{/tex}{tex}r^2 = 462{/tex}{tex}\\Rightarrow{/tex}2{tex}\\pi{/tex}{tex}r^2 = 462 - 154 = 308 cm^2{/tex}2 {tex}\\times{/tex}{tex}\\frac{22}{7}{/tex}{tex}\\times{/tex}{tex}r^2 = 308\xa0{/tex}{tex}\\Rightarrow{/tex}r2 ={tex}\\frac{308 \\times 7}{2 \\times22}{/tex} = 72{tex}\\Rightarrow{/tex}\xa0{tex}r = 7 cm{/tex}Again 2{tex}\\pi{/tex}rh = 154 {tex}\\Rightarrow{/tex}\xa02{tex}\\times{/tex}{tex}\\frac{22}{7}{/tex}{tex}\\times{/tex}7{tex}\\times{/tex}{tex}h = 154{/tex}{tex}\\Rightarrow{/tex}\xa0h = {tex}\\frac{154}{2 \\times22}{/tex}={tex}\\frac{7}{2}{/tex}cmVolume of the cylinder = {tex}\\pi{/tex}r2h = {tex}\\frac{22}{7}{/tex}{tex}\\times{/tex}\xa07 {tex}\\times{/tex}\xa07 {tex}\\times{/tex}\xa0{tex}\\frac{7}{2}{/tex}\xa0{tex}= 539 cm^3{/tex} | |
| 3279. |
How to do learn maths formulae |
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| 3280. |
Find highest value of 15 sin theta +16cos theta |
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| 3281. |
Sahil you only like me |
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| 3282. |
Find the sum of all 3-digits natural numbers which are multiples of 11 |
| Answer» 44550 | |
| 3283. |
Explain whether 3×12×101+4 is a prime or composite number (2 marks) |
| Answer» {tex}\\begin{array}{l}=3\\times12\\times101+4\\\\=4(3\\times3\\times101+1)\\\\=4(909+1)\\\\=4\\times910=2^3\\times5\\times7\\times13\\\\\\\\\\\\\\end{array}{/tex}= a composite number[{tex}\\because{/tex}\xa0Product of more than two prime factors] | |
| 3284. |
425 which. Squra |
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| 3285. |
An=3n-5 |
| Answer» n=5\\3 I THINK SO | |
| 3286. |
234×2 |
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Answer» 468 and hai 468 |
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| 3287. |
If\xa0Cos A - Sin A =1\xa0then prove that\xa0Cos A + Sin A=+1.\xa0 |
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| 3288. |
RIMSHA KYA AAP ONLINE HO |
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| 3289. |
Happy new to all my friends ??????????????????? |
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| 3290. |
What is square root of 35 |
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| 3291. |
Sahil illiana kon hai |
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| 3292. |
Byeeeee frnds |
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| 3293. |
Happy New Year to all of you❤️ |
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| 3294. |
RIMSHA plzz come yrr |
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| 3295. |
Sahil mujhe sachhi nhi pata aap boldo |
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| 3296. |
7×6×5×4×3×2×1+5 is composite no. |
| Answer» yes ☺☺☺ | |
| 3297. |
Sahil ye kya bol rhe ho |
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| 3298. |
(A + B) whole 3 |
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Answer» a3+b3+3a2b+3ab2 (a) ka whole 3 + (b)Ka whole 3+ 3ab(a+b) A3+ b3+ 3*a*b(a+b) |
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| 3299. |
Only few hours left for new year |
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| 3300. |
Sahil Bolo am really serious agar Gaye then I\'ll not come back |
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