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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. |
49P A 641649(A) 35(C) 36(B) 41(D) 40 |
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Answer» 1-4 = -34 -9= -59-16= -716-25= -9therefore 25-x=-11x=25+11=36 |
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| 2. |
2. A shopkeeper buys 30 chocolate bars at * 15 each and sold all of them for525. Find the profit per cent he made. |
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Answer» total cost price -15x30 is 450total selling price -525profit percent -525-450) /450 x 100that is 16.666667 total price=15x30=450, total selling price=525, profit percent=525-450/450x100=16.666 |
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| 3. |
5457+456-34+3344-3453224 |
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Answer» -3444001 is the answer -3444001 is the answer 5913-34+3344-34532245913+3344-34-34532249257-34-34532249223-3453224=-3444001 ans.if helpful like it -3444001 is the correct answer -3444001 is the right answer. |
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| 4. |
Th.H.T O4562 31 5 |
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Answer» 2+5=76+1=75+3=84+2=6 henceanswer is 6877 |
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| 5. |
(a) 30°(b) 60°(c) 90°(d) 120°(a) 10°(b) 15°(c)20°ノ(d) 30° |
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Answer» 2cos3theta= 1/cos3theta= 1/2cos3theta= cos60°3theta= 60° theta = 60°/3= 20° |
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| 6. |
23. The ratio of the areas of two circle is 64: 81, then the ratio of their cirumference is(a) 8:9(b) 64: 81(c) 16:25(d) none of these |
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| 7. |
50P<3230 230 B 30 2500SA 4C30 |
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| 8. |
:11. A shopkeeper buys paper foră50 per ream. At what price per quire should he sell it to gain |
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| 9. |
30.0There are 100 families in a society, 40 families buy newspaper A, 30families buy newspaper B, 30 families buy newspaper C, 10 buy A and B,8buy B and C, 5 buy A and C, 3 families buy newspapers A, B and C, thenthumber of families who do not buy my newpepec is |
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Answer» so the answer is 20 |
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| 10. |
A shopkeeper buys paper for50 per ream. At what price per quire should he sell it to gain 20%? |
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| 11. |
789-456+56×456÷6775=686×658÷547-4565 |
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| 12. |
By selling a Tan for ( 1200,am1. A shopkeeper buys paper for50 per ream. At what price per quire should he sell it to gain 20%? |
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| 13. |
5. A shopkeeper buys an inductioncooker for 2750 and sells its for2860. Find his gain percentage. |
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Answer» 4 percentage is the right answer |
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| 14. |
A shopkeeper buys an article for Rs 490and sells for Rs 465. Find the loss and lospercentage? |
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| 15. |
A shopkeeper buys 10 music player for rs 26000. If he sells them for profit of 5% find the selling price? |
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| 16. |
Find the least number that should be added to 36,516 so that theresult is exactly divisible by 456.Hint: you have to add (456 - remainder) to 36,516) |
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Answer» By using division algorithm:************************************Dividend = divisor × quotient + remainder *********************************************** According to the problem , Dividend = 36516 Divisor = 456 36516 = 456 × 80 + 36 To find the least number which should be added to dividend We use following rule : least number = divisor - remainder least number = 456 - 36 = 420 Therefore Required least number = 420 420 is the least number which should be added to the 36516 so that the new Dividend is exactly divisible by 456. Verification:***************New dividend = 36516 + 420 = 36936 36936 = 456 × 81 + 0 Remainder = 0 36936 is exactly divisible by 456. |
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| 17. |
456+577 |
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Answer» 456+577 =1,033( basic additon) |
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| 18. |
450*456 |
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Answer» The answer of question450*456=205,200 the total answer is 450*456=205,200 |
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| 19. |
questions, WillyDULIULLUULLIILLILU2. To buy a flat, a man paid 47,250 every month for six years.How much did he pay in all? |
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Answer» each year has 12 months. He paid for 6 years every month so,, 47,250×12×6=3,402,000rs he paid 3,402,000rs. ₹34,02,000 is the answer of the following each year twelve month 6×12×47250= 34,02, 000 |
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| 20. |
24Arunrepayshisloan amount of1,18,000 by paying every month,starting with the first installment of .1000. If he increases theinstallment by .100 every month what amount will be paid as thelast installment of loan?. |
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Answer» Loan Amount = Rs 118000 First installment = 1000Second installment = 1100Third installment = 1200 Thus loan installments form an AP with first digit a = 1000, common difference d = 100 Let number of installments are n to repay loan Then,Sn = (2a + (n-1)d)n/2118000*2 = [2*1000 + (n-1)100]n1180*2 = n(20 + (n-1))2360 = 19n + n^2 n^2 + 19n - 2360 = 0n^2 + 59n - 40n - 2360 = 0n(n + 59) - 40(n + 59) = 0n = 40,-59Negative value not possible So total number of installments are 40 Amount paid for last installment= 1000 + (39)*100= Rs 4900 |
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| 21. |
(2 x 30°) = 2 sin 30° - cos 30° L4 tan 3 3030, |
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Answer» Sin60 ° = sin2*× 30° from LHS sin60° = √3/2 from RHS 2 sin30° ×cos30° = 2×1/2×√3/2 2√3/2×2 √3/2 LHS = RHSNow tan30=1/√32tan30/1+tan^230=2*1/√3/1+1/3=2/√3/4/3=√3/2 |
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| 22. |
what will Rs So0 ameunt to un lonondypdit ima bank uhich pouys cannuneual |
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Answer» compound interestinterest rate is 10%initial deposit 500n = 10 yearsso after 10 years using compound interest amount will be = 500(1+10/100)^10=500*(1.1)^10 = 1296.87 thanks |
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| 23. |
14. Rahul reads 123 pages in 3 hrs. How many hours will he take tocomplete a book containing 287 pages? 42.0 |
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Answer» 123pages in 3 hourshence in 1 hour=123/3=41pageshence in 1 hour 41 pageshence for 287pages =287/41=7 hrs |
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| 24. |
7. Ishita read 2 of a book containing 50 pages while Nikita readof a book containing 80 pages.10Who reads more? |
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Answer» ishita read =(3/5) of 50 pages=3×10=30 pagesNikita read (7/10) of 80 pages=7×8=56 pagesNikita read more |
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| 25. |
hoThree persons can build a small house in 8 days. To build the same house in 6 days,many persons are required? |
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Answer» Number of persons required let y, for building the same on 6 days is equal toy = 3 x 8/6=4 persons |
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| 26. |
Three persons can build a small house in 8 days. To build the same house in 6 days, howmany persons are required?(4) |
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Answer» 3 person can build house in 8 days1 person can build house in 8*3 = 24 daysNo. Of persons require to build house in6 days = 24/6 = 4 4 persons ans |
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| 27. |
Three persons can build a smallhouse in 8 days. To build thesame house in 6 days, how manypersons are required?4. |
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| 28. |
. Three persons can build a small house in 10 daysHow many persons are required to build the same(3 marks)house in 6 days?lution : |
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Answer» 3 men can build house in 10 days so to build house in 6 days men are require=(10×3)/6=5 |
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| 29. |
Seven different coins are to be divided amongst three persons. If no two of the persons receive the samenumber of coins but each receives at least one coin & none is left over, then the number of ways in whichthe division can be made is |
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Answer» Well with three people and no two people getting the same number (and everyone gets at least 1)- the only way is to divide it is 1-2-4,,, Assume person one gets one coin... there are 7 possibilitiesPerson two gets 2 coins from the remaining 6 = 6!/(6-2)!2!=15 (because the order of the coins doesn't matter)Person 3 gets what's left - 4 coins (1 possibility again order does not matter) but there are also 6 ways to arrange the people...7*15*6=630 |
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| 30. |
*10. In a hall there are 960 persons. If the number of womenis two times the number of men and the number of menis three times the number of children, find how manywomen are there? |
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Answer» 96 children288 men576 women |
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| 31. |
If tan x = sin 45° cos 45° + sin 30°, then find the valueS1 |
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Answer» tan x = sin 45°.cos 45° + sin 30°tan x = 1/√2 × 1/√2 + 1/2tan x = 1/2 + 1/2tan x = 2/2tan x = 1tan 45° = 1x = 45° |
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| 32. |
6, Find the value of x iftan 3x=sin 45° cos 45° sin 30. |
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| 33. |
Find the value of x if tan 3x=Sin 45.cos 45+sin 30 |
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| 34. |
( 243 ) ^ { 5 } \div ( 64 ) - \frac { 1 } { 6 } |
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| 35. |
10 Prove that parallelograms on the same base and betwéenthe same parallels are equal in area |
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| 36. |
Sample Paper (Class X Studying)25. If there are three numbers a, b, c such that HCF ofeach pair is 7 and the LCM of all the three numbersis 630, then product of all the three numbers is(1) 61740(2) 20580(3) 82320(4) 30870 |
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Answer» b) is right answer of this question. please like my answer. 30870 is the right ans of this numerical 30870 is correct answer 4 th is the correct option HCF of abc ×LCM of abc = product of abchcf of each pair =7so product of no = 7×7×630= 30870 is the answer 30870 is the answerfor method refer answer of sri sai 30870 is correct answer |
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| 37. |
Three numbers are in the ratio 2:3:4 and the sumof their cubes is 33957. What are those threenumbers ? |
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Answer» Let the numbers be 2x, 3x and 4xsum of cubes = 33957⇒(2x)³ + (3x)³ + (4x)³ = 33957⇒8x³ + 27x³ + 64x³ = 33957⇒99x³ = 33957⇒x³ = 33957/99 = 343⇒x =∛343 = 7 Numbers are:2x = 2×7 = 143x = 3×7 = 214x = 4×7 = 28 Numbers are: 2×=2×7=143×=3×7=214×=4×7=28 correct answer is 14,21,28 |
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| 38. |
If the three positive numbers 2r 1, 2x3 and 4x 3 arc in continuedproportion, then find the three numbers. |
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| 39. |
The sum of three consecutive odd numbers is 129What is the smallest of the three numbers? |
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| 40. |
2| 10|1| 123।15।।चिह्न359 4 5(1) 36(3) 64व्याख्या :-(2) 48(4) 75|{||जिस प्रकार, 105 =2 तथा 2x32=12| 12-4 = 3 तथा x 9 x 1 = 27उसी प्रकार, 153=5 तथा 5x5X3=5ज।।89 155 |
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Answer» dgshbsysnwhsmzyz zy7sbdhsbdbxgsjsbzvzgsjbszhsvszbsd right answer is 75 .... |
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| 41. |
1.4 अतेक्:*2 o 51 (== XA sm(&—lS”) AT |
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| 42. |
DODHI PUBLIC SCHOOLARAR MORE NEAR KAMARA COLLGOPALGANJ-341428 (BIHAR)OFFICE No - 8210621421IDENTITY CARDCLASSV-APradeep KumarSambhu YadKundi DewF NAMEM NAMEROLL.NOD.O.BMOB NO |
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Answer» Name is sambhu yadav NAME:PRADEEP KUMARCLASS:V ASCHOOL: DELHI PUBLIC SCHOOL your name is Pradeep Kumar name pradeep kumar class 5 a name Sambhu Yadavclass V Aschool Delhi public schoolroll no~26 |
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| 43. |
पा का मान क्या है, (6 % )* Whatsm (61" =(6]) |
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Answer» -7 Bharat ke purv mein sthit sthan raja mizoram nagaland aur tripura ke sath ke naam se Jana jata hai Inka kshetrafal 250000 laga hua hai is kshetra ki jansankhya Bharat ki kul jansankhya ka lagbhag 8.8 pratishat bhag hai iske Seema Bhutan Chin aur se lagti hai yah kshetra pichhle hone ke Karan swatantra rajya ki mang kar rahe hain |
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| 44. |
IfPm = Sn for some. A.P. then prove that Sm+n-ă |
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Answer» thku |
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| 45. |
(0) IFind.tnu1ome aMrimi no |
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Answer» First ten prime numbers are = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 Mean total sum/ total elements = (2+3+5+7+11+13+17+19+23+29)/10= 129/10 = 12.9 |
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| 46. |
thelethuxdixgnolAwd aceen ahoww umSm |
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Answer» we have taken the sides same because it is a regular octagon |
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| 47. |
K2. Express each of the following in power notation:25(ii) -273664243(1)(iil) -32 |
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| 48. |
The angle made by the straight line y x with the pasitive direction of x - axis is-a)-45°, b) 135, c) 45°, d)90°14. |
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Answer» y = -x has slope = -1 also, the slope is tan∅ = -1 and ∅ for which tan∅ = -1 is 135° |
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| 49. |
Gin 12(ii)1535(1) 25 33(1) 1028 39* 283627 65 562. Simplify:3 44 35(iv) 17*22* 153. Find:1 of 242,3324x7x19 52(vi) 3-19x73x125(11)of 32( of 45(vi) 3 of 220meno(vl 3 of 1920 |
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Answer» i |
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| 50. |
6) Find the unknown angles in these figures(i) AB | | CD45"92x 4530*70"(ii) x+y 9060* |
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Answer» (i)AB || CD so y =92°.x + y + 45 = 180X = 43. (ii) y + 70 + 45 = 180 y = 65.z + y = 180, z = 115.x + z + 30 = 180, x =35. |
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