This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A hall has eight cylindrical pillars. Each pillar has a diameter16 cm and height 2.5 m. Find the cost of painting the pillars atRs 5.00 per square metre. |
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Answer» if any doubt in the above answer comment me |
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| 2. |
Test the convergence of 1+\frac{2}{3} x+\frac{6}{9} x^{2}+\ldots+\frac{2^{n}-2}{2^{r}+1} x^{n-3}+ |
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| 3. |
3"/n10. Test for the convergence of the series ÎŁ.Iln!n-1 T |
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| 4. |
Innd the radius of convergence for the seriesn=1n + 2 |
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| 5. |
Examine the convergence ofthe series Σn=0-->∞n^2 +1/5n |
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Answer» Using ratio convergence test a(n+1)----------- a(n) ((n+1)² +1)(5ⁿ) ------------------------ 5^(n+1) (n²+1) 1/5 < 1So, convergent series |
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| 6. |
ERCISE 1(A)log n1. Test for the convergence of the sequencen= |
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| 7. |
23Ramesh sold two books 480 each 20% profit on one băว1, 20% loss on thebook. Overall check whether he gets loss or profit . By how much? |
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Answer» Cost price of two booksCP = 480*2 = 960 Selling price of first bookSP1 = CP1(1 + 20/100) = 480(6/5) = 96*6 = 576 Selling price of second bookSP2 = CP2(1 - 20/100) = 480(4/5) = 96*4 = 384 Total selling price SP = 576 + 384 = 960 There is no loss no profit as Selling price is equal to cost price of two books |
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| 8. |
There are 4 books belonging to 4 students. The books were put into a box, and each student pulls out abook one after the other. What is the probability that each student gets his or her own book? |
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Answer» ¼ is the probability of a student to get his/her book |
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| 9. |
Check whether the pair of equations x -2y= 0 and 3x + 4y =20coincident lines. Find the solution, if the equations are consistent. (R. V |
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| 10. |
tHan29 Draw a line segment of length 7.6 em and divide it in the ratio 5 :8. Measure the two partsel ground is found to be 45 m longer when sun's altitude is 30° |
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Answer» Length of line segment = 7.6cmLets two parts length be 5x and 8x Then 5x + 8x = 7.6 13x = 7.6 x = 7.6/13 x =.58 Length of two parts are 2.9 and 4.7 cm |
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| 11. |
Express 2310 as a product offactors. Also see how your friends havefactorized the number. Have they doneit as you ? Verify your final productwith your friend's result. Try this for3 or 4 more numbers. What do youconclude?prime(Page No. 7) |
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Answer» Like if you find it useful |
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| 12. |
what is the length of the longer part when a line segment of length 10cm is divided in the ratio |
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Answer» what's the ratio?first tell the ratio and then any one would be able to give answer 2:3 write euclidian division lemma |
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| 13. |
2. Theratio of incomes of two friends Jasmine and Aman is 9:7 and the ratio of their expenditures4:3. If each of them saves 6,000 per month, find their monthly incomes. Also, if each of themdonates 2% of his/her income to a charity working for old age destitutes, find the resulting savingsof each. |
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Answer» Ratio of income = 9 : 7 let us assume inc. = x Jasmin income = 9X Armaan income = 7X Ratio of expenditure = 4:3 jasmine expenditure = 4X Armaan expenditure = 3X Both savings = 6000 Jasmine = SAVINGS= 9X -4X6000= 5X6000/5= XX= 1200 9X = 9 * 12009X = RS.108002 % DonatedSo 2% Of 10800 = 216Final Savings = 6000 - 216= RS 5784 Armaan Savings = 7X - 3X6000= 4X6000/4= 1500X = 15007X = 7 * 15007X= RS 105002% DonatedSo 2% Of 10500 = 210Final Saving= 6000 - 210=RS 5790 |
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| 14. |
12.flogr =logy = logZth'3 2-1, find a=k.and2 |
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Answer» Didn't understand fully |
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| 15. |
Two pillars are 120 ft a part and the height ofone is double that of the other. From the middlepoint of the line joining their feet,an observerfinds that the angular elevations of their topsare complementary. The height of the longertower is feet |
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| 16. |
. The ratio of incomes of two friendsJasmine and Aman is 9: 7 and thetheir expenditures is 4: 3. If each of themsaves 6,000 per month, find theirmonthly incomes. Also, if each of themdonates 2% of her/his income to a charityratio oforking for old age destitutes, find theresulting savings of each. What value iindicated from this action? |
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Answer» Let income of Jasmine is 9xLet income of Aman is 7x Let expenditure of jasmine is 4yLet expenditure of Aman be 3 y Then,9x - 4y = 6000.... (1)x37x - 3y = 6000.....(2)x4 27x - 12y = 18000.....(3)28x - 12y = 24000.....(4) Subract eq(3) from eq(4)x = 6000 Salary of jasmine = 9*6000 = 54000Salary of Aman = 7*6000 = 42000 Put value of x = 6000 in eq(1)54000 - 4y = 60004y = 48000y = 12000 Expenditure of Jasmine = 4*12000 = 48000Expenditure of Aman = 3*12000 = 36000 If they donate 2% of incomeThen, saving of Jamine= 54000 - (48000 + 2/100*54000) = 54000 - 49080= 4920 savinv of Aman= 48000 - (36000+ 2/100*48000) = 48000 - 36960= 11040 |
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| 17. |
If the sum of first 6, terms of an A.P. is 36 and that of the first 16 ferms is 256, find thesum of first 10 terms. |
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Answer» Sum of n terms = n/2{2a + (n -1)d } A/C to question ; S6 = 6/2{ 2a + (6-1)d } 36 = 3(2a + 5d) 12 = 2a + 5d --------------(1) again, S16 = 16/2{ 2a + (16-1)d }256 = 8{ 2a + 15d }32 = 2a + 15d --------(2) solve eqns (1) and (2) 10d = 20 d = 2 put in equation (1) a = 1 now, sum of 10th terms = S10 = 10/2{2a +(10-1)d}= 5{ 2 × 1 + 9 ×2 }= 5 { 2 + 18}= 100 hence, sum of first 10terms = 100 |
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| 18. |
If the sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find thesum of first 10 terms. |
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Answer» Sum of n terms = n/2{2a + (n -1)d } A/C to question ; S6 = 6/2{ 2a + (6-1)d } 36 = 3(2a + 5d) 12 = 2a + 5d --------------(1) again, S16 = 16/2{ 2a + (16-1)d }256 = 8{ 2a + 15d }32 = 2a + 15d --------(2) solve eqns (1) and (2) 10d = 20 d = 2 put in equation (1) a = 1 now, sum of 10th terms = S10 = 10/2{2a +(10-1)d}= 5{ 2 × 1 + 9 ×2 }= 5 { 2 + 18}= 100 hence sum is 100 |
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| 19. |
it if the sunn of first 6 terms of an A.P is 36 and that of the first 16 terms is 256, find thesum of first 10 terms |
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Answer» Sum of n terms = n/2{2a + (n -1)d } A/C to question ; S6 = 6/2{ 2a + (6-1)d } 36 = 3(2a + 5d) 12 = 2a + 5d --------------(1) again, S16 = 16/2{ 2a + (16-1)d }256 = 8{ 2a + 15d }32 = 2a + 15d --------(2) solve eqns (1) and (2) 10d = 20 d = 2 put in equation (1) a = 1 now, sum of 10th terms = S10 = 10/2{2a +(10-1)d}= 5{ 2 × 1 + 9 ×2 }= 5 { 2 + 18}= 100 hence, sum of first 10terms = 100 |
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| 20. |
7. (i) Find 3 G.M. between 3 and 48.(ii) Find 6 G.M. between 2 and 256. |
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Answer» 1)3 , a1, a2, a3 , 4848 = 3r^4r^4 = 16r = 2a1 = 3*2 = 6a2 = 3*2^2 = 12 = 6*2a3 = 3*2^3 = 24 = 12*2 2 , a1, a2, a3 , a4 , a5, a6, 256256 = 2r^7r^7 = 124r = 2a1 = 2*2 = 4a2 = 2*2^2 = 8a3 = 2*2^3 = 16 a4 = 2*2^4 = 32a5 = 2*2^5 = 64a6 = 2*2^6 = 128 |
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| 21. |
\sqrt { 6 \frac { 145 } { 256 } } |
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Answer» root(6 145/256)=root(1681/256)=+-41/16 |
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| 22. |
The mean of eight observations is 25. Itobservation 11 is excluded, find the meanthe remaining observations. |
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Answer» If the mean of 8 observations is 25, its total is 25 * 8 = 200. If the observation 11 is excluded, then the total of 7 observations will be 200 - 11 = 189. Therefore, the mean of 7 observations with a total of 181 is 189/7 = 27. please like the solution 👍 ✔️ |
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| 23. |
81/256 |
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Answer» 81/256=9²/16²=(3²)²/(4²)²=3⁴/4⁴=(3/4)⁴ |
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| 24. |
( 256 ) ^ { 0.16 } \times ( 256 ) ^ { 0.09 } |
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| 25. |
2y = 4 and wh(27+7=th3. Check which of the following are solutions of the equation x-27=4not:(0) (0.2)(1) (20)(iii) (4.0)(iv) (SZ. 412) (1) (1,1) |
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Answer» 3)(i)(0,2); x-2y=4 ==0-2(2)=4;-4=4; (ii)(2,0); x-2y=4; 2-2(0)=4; 2=4,; 1=2; (iii)(4, 0); x-2y=4; (4)-2(0)=4,; 4=4; (iv)(V2, 4V2); x-2y=4; (V2)-2(4V2)=4; V2-8V2=4; V2(1-8)=4; V2(-7)=4;; V2=4/-7; (v)(1,1);; x-2y=4; 1-2(1)=4; 1-2=4; -1=4 |
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| 26. |
Fill in the blanks:(a) Longer line in the symbol for a cell represents its___ |
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Answer» In thesymbolof the electriccell, the longer line representsthe positive terminal and the thicker, shorterline representsthe negative terminal. |
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| 27. |
(1) Find the probability that a leap year selected at random, will contain 53 Sundays |
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| 28. |
What is the probability that a leap year selected at random will contain 53 Fridays? |
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| 29. |
Find the probability that a non-leap year selected at random will have 53 Fridays. |
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Answer» Any non-leap year has 365 days which is divided into 7x52=364 + 1 days. There will be 52 fridays, one for each week except when the first day of the year is a Friday. That probability is 1/7. Also, the probability of non-leap year is 3/4 so overall probability is 1/7*3/4=3/28 |
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| 30. |
12. What is the probability that a leap year selected at random will contain 53 Sundays? 2 |
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| 31. |
. Solve the following questions : (Any Three)(1) Find the probability that a leap year selected at random, will contain 53 Sundays. |
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Answer» tha |
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| 32. |
plify and express the resu3] (v)5the value of.(3° + 4-1)x 2 |
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Answer» (-4)^5 ÷ (-4)^8 = (-4)^5-8 = (-4)^-3 = (-1/4)^3 = -1/64 |
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| 33. |
The number of times a particular observationsoccurs is called5. |
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Answer» The number of times a particular observation occurs is called as frequency. |
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| 34. |
(vii) 2(viii) 6 256Find the square root of the decimals.(i) 2.1025 (ii) 69.5556(u 0.000676(iv) 0.7921 |
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Answer» (1) square root of 2.1025 = 1.45 (2) square root of 69.5556 = 8.34 ER. RAVI KUMAR ROY (3) square root of 0.000676 = 0.026 (4)square root of 0.7921 = 0.89 2(14/25)= 2.56; viii)6(145/256)=6.566; square root i)2.1025=1.45 ii) 69.5556 = 8. 34 |
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| 35. |
[ jee main-2017]A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also 7 friends3 of them are ladies and 4 are men. Assume X and Y have no common friends, thenthe total number of ways in which X and Y can throw a party inviting3 ladies and 3 men so that 3 friends of each X and Y are in this party is(A) 484(B)485(C)468(D) 469 |
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Answer» option (b) is correct |
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| 36. |
19. Two coins are tossed once. What is theprobability of getting at least two T's ? |
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Answer» Whentwo coins are tossedsimultaneously then the possible outcomes obtained are {HH, HT, TH, and TT}. Probablity of getting atleast two t's one possible outcome so probability is1/4 |
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| 37. |
256+256 |
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Answer» the answer is 256 +256 -512 |
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| 38. |
check 20:25 and 5:4 are equivalent |
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Answer» 20: 25 = 20/25 = 4/5 And given ratio is 5:4 that is 5/4So, they are not equivalent |
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| 39. |
oOTestn n(log n) convergernc(-1)nfor convergence and absolute convergencen 2 nlog n)2 |
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| 40. |
Visualize 4.26 on the number line, up to 4 decimal places. |
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| 41. |
test for convergence |
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Answer» Inmathematics,convergence testsare methods of testing for theconvergence,conditional convergence,absolute convergence,interval of convergenceor divergence of aninfinite series{\displaystyle \sum _{n=1}^{\infty }a_{n}}. |
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| 42. |
४ जी mo | a ‘8\‘9,“ = |
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Answer» 8r² = r + 2 In standard form 8r² - r - 2 = 0 Use quadratic formula r = [-(-1) + or - √(-1)² - (-4)× (8) × (-2) ]/16 r = [ 1 + or - √(-63) ]/16 r = [ 1 or - √63 i ] /16 |
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| 43. |
9.) Find the probability that a non-leap yearselected at random will have 53 Tuesdays. |
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Answer» no, according to my answer book answer is 1/7 but I can't understand how answer is 1/7 |
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| 44. |
Q.1 Find the difference between the greatest and least 6 digit number formed by using the digits 3.6.9.7,5 and 2 |
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Answer» ihfyfihohohohigihohohoh Difference between largest number and smallest number is 740853 greatest-least=976532-235679=740853 greatest number formed = 976532least number formed = 235679difference between both = 976532 - 235679 = 740853 ( ANS ) 740853 is the following question answer 740853 is the correct answer of the given question 740853 is the right answer 740853 is the correct one 740853 is the right answer 740853 is the right answer. 740853 is the correct answer |
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| 45. |
Three consecutive integers are such thatwhen they are taken in increasing order andmultiplied by 2, 3 and 4 respectively, theyadd up to 74. Find these numbers |
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| 46. |
, o, ), o, r, s, 4, 9, io,, s, o, s, 4, /, 8, o,.. 3.8, 7.63, e, i ,s10, 7,95. The temperature (in degree Celsius) of Delhi during a particular forthnight in the moof June is given below:38, 40, 42, 43, 45, 39, 40, 37, 38, 41, 44, 44, 40, 42, 40Rearrange the data in descending order and find the range. |
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Answer» Descending order is from high to low. 45, 44, 44, 43, 42, 42, 41, 40, 40, 40, 40, 39, 38, 38, 37 thus the range is from 37 to 45 |
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| 47. |
6. Three consecutive integers are such thatwhen they are taken in increasing order andmultiplied by 2, 3 and 4 respectively, theyadd up to 74. Find these numbers. |
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| 48. |
1 A cmmittee of 5 persons is to be formed out of 6 gents and 4 ladies Find the number of ways inwhich this oan be done, when at least two ladies are inciuded |
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| 49. |
5. Visualise 4.26 on the number line, up to 4 decimal places. |
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| 50. |
2. Visualise 4.26 on the number line, up to 4 decimal places. |
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Answer» Have 4.26 = 4.2626 This number lies between 4 and 5 the distance between 4 and 5 is divided into 10 equal parts.Then the first mark to the right of 4 will represent 4.1 & second 4.2 and so on. Now 4.2626 lies between 4.2 & 4.3. We divide the distance between 4.2 and 4.3 into 10 equal parts. 4.2626 lies between 4.26 and 4.27. Again we divide the distance between 4.26 & 4.27 into 10 equal parts.The number 4.2626 lies between 4. 262 and 4.263. The distance between 4.262 and 4.263 in again divided into 10 equal parts. Sixth mark from right to the 4.262 is 4.2626 |
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