Question 1. A Can Do A Work In 60 Days And B Can Do The Same Work In forty Days. They Work Together For 12 Days And Then 'a' Goes Away. In How Many Days Will 'b' Finish The Remaining Work?

Answer :

Work done by means of A and B in 12 days is = 12 * 5/120 = 1/2

Therefore Remaining work = 1- half = 1/2work

B does 1/forty paintings in in the future

Therefore B does half of paintings in forty*half =20days

Question 2. The Average Age Of A Woman And Her Daughter Is 42 Years. The Ratio Of Their Ages Is 2 : 1 Respectively. What Is The Daughter's Age?

Answer :

Let the age of mother be M and that of her daughter be D Therefore, [M+D]/2 = forty two [M/D] =2/1 and Solving the above equations we get D = 28 yrs.

Oracle 9i Interview Questions

Question three. The Price Of Sugar Is Increased By 25%.Discover By How Much Percent The Consumption Of Sugar Be Decreased So As Not To Increase The Expenditure?

Answer :

Using the components to calculate % lower as [R/(100+R)]x100 where R = percentage boom in fee, we get

Required % lower in consumption = 25/125 MULTIPLIED a hundred = 20%.

Question 4. A Car Travels A Distance Of 45 Km At The Speed Of 15 Km/hr. It Covers The Next 50 Km Of Its Journey At The Speed Of 25km/hr And The Last 25 Km Of Its Journey At The Speed Of 15 Km/hr. What Is The Average Speed Of The Car?

Answer :

We recognise, Average velocity = Total distance travelled / Total time taken Average = [45+50+25]/ [3+2+ 25/15] = 18 kmph

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Question five. A Car Travels A Distance Of one hundred seventy Km In 2 Hours Partly At A Speed Of a hundred Km/h And Partly At 50 Km/h. The Distance Travelled At A Speed Of 50 Km/h Is?

Answer :

Suppose he covers x km at one hundred kmph

So he covers one hundred seventy-x at 50 kmph

So X/one hundred+a hundred and seventy-X/50=2

Solving this equation, we get x = 140.

So he covers 30km at 50 kmph.

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Question 6. Even After Reducing The Marked Price Of A Transistor By Rs. 32, A Shopkeeper Makes A Profit Of 15%. If The Cost Price Be Rs. 320, What Percentage Of Profit Would He Have Made If He Had Sold The Transistor At The Marked Price?

Answer :

Let x be the marked price,

So x - 32 = 320 X 1.15

x = 400.

So required fee is

four hundred = 320 (1 + earnings/a hundred),

So profit is 25%

Question 7. The Ratio Of The Present Ages Of A And B 9: five. Five Years Earlier The Ratio Of Their Was 2 : 1. What Is The Average Of Their Present Ages?

Answer :

A/b=9/five; i.E. 5a = 9B.......... (i)

A-five/B-five = 2/1; i.E. A-five = 2B .........10

∴A - 2B = -5................... (ii)

From equation (i) and (ii), A = forty five, B = 25

∴ Average =45+25/2 = 70/2 = 35.

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Question eight. 20 Boys And 32 Girls Form A Group For Social Work. During Their Membership Drive Same No. Of Boys And Girls Joined The Group. How Many Members Does The Group Have Now, If The Ratio Of Boys To Girls Is 3:4 Respectively?

Answer :

Let x be the new boys in addition to women, Therefore

[20+X]/ [32+X] =3/four

Solving this we get x = sixteen

So general can be 36 + 48 = 84.

Question 9. Find The Quadratic Equation With Roots As The Lesser Root Of The Equation Will Help You Learn And Apply These Tricks And Ace The Exam. X2−12x+35=zero And The Greater Root Of The Equation X2+14x+forty five=zero.

Answer :

Roots of x2−12x+35=zero are five and 7, lesser root is five

Roots of x2+14x+forty five=0 are -five and -9, more root is -five

Required equation = (x-5) (x+five) = x2-25=0

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Question 10. A Completes 50% Of The Work In 10 Days And Then Decides To Take Help From B And C. B Is Half As Efficient As A And Similarly C Is Half As Efficient As B. How Many More Days Will They Take To Complete The Work?

Answer :

As A completes half of of the work in 10 days, he/she will be able to whole the work in 20 days.

As B is half of as green as A and C is half as efficient as B,

They will entire the paintings in 1/20 + 1/40 + 1/80 = 7/80. So 7/80 of the paintings in a single day.

So, they can complete the whole paintings in 80/7 days

Therefore, they are able to complete the last 50% of the work in (half of) x 80/7 = 40/7 days.

Question 11. Two Trains Of Length one hundred fifty M And two hundred M Respectively, Are Travelling In Opposite Directions At A Speed Of fifty four Km/hr And 72 Km/hr. What Is The Total Time Taken By Them To Cross Each Other?

Answer :

fifty four km/hr = 54 x (five/18) = 15m/s

72 km/hr = seventy two x (five/18) = 20 m/s

Total distance = a hundred and fifty + 2 hundred = 350 m

Relative pace = 15 + 20 = 35 m/s

Total time = Total distance/Relative pace

= 350/35 = 10s

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Question 12. The Sum Of third And sixth Term Of An A.P Is 27. Find The Sum Of The First 10 Terms Of The Progression?

Answer :

T3 = a + 3d

T6 = a + 6d

T3 + T6 = 2a + 9d = 27

Sum of first 10 terms of an AP is:

S= (10/2) (2a+9d)

= 5 x 27

= 135

Oracle 9i Interview Questions

Question 13. What Is The Probability Of Finding A Red Face Card In A Deck Of Cards?

Answer :

A deck of card has 52 playing cards of which 26 are black and the other 26 are red.

The number of face playing cards is 12, however handiest 6 of them are crimson.

So, required opportunity is 6/52 = 3/26

Question 14. How Many Different Words Can Be Formed From The Word Oracle So That The Vowels Always Come Together?

Answer :

We will institution the letters that need to come back collectively (A & E) and take into account them as a unmarried letter. So, here the letters are O, R, C, L, and AE.

Number of approaches A & E can be organized is two!

So, the full wide variety of methods in which the words may be formed so that every one vowels are together is 5! X 2! = 240 approaches.

Question 15. On Selling one hundred Articles, A Shopkeeper Earns A Profit Amount Equal To The Selling Price Of 50 Articles. What Is The Profit Percentage Of The Shopkeeper?

Answer :

Let SP of every article be Re. 1

So, SP of one hundred articles =Rs. One hundred

Implies income = Rs. 50

So, CP = a hundred – 50 = Rs. 50

Therefore, income percent of the shopkeeper is 50/50 × 100 = 100%

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Question 16. A Cone Has Vertical Height And Slant Height As 15 Cm And 17 Cm Respectively. A Hemisphere With The Same Radius As The Cone Is Placed On The Face Of The Cone. What Is The Total Volume Of The Figure Formed?

Answer :

Using the Pythagoras theorem, the radius of the cone = √ (172 - 152) = eight cm.

Volume of cone = (1/three) x π x r2 x h= (1/three) x π x 82 x 15

Radius of hemisphere=radius of cone

Volume of hemisphere = (2/three) x π x r3 = (2/3) x π x eighty three

Total volume of determine = Volume of cone + Volume of hemisphere = 4233.58 cu. Cm.

Question 17. A Can Contains A Mixture Of Two Liquids A And B In The Ratio 7: 5. When 9 Litres Of Mixture Are Drawn Off And The Can Is Filled With B, The Ratio Of A And B Becomes 7: nine. How Many Litres Of Liquid A Were Contained By The Can Initially?

Answer :

Suppose the can to start with contains 7x and 5x litres of mixtures A and B respectively

Quantity of An in mixture left

= (7x - 7/12 x 9) litres = (7x - 21/four) litres.

Quantity of B in aggregate left

= (5x - five/12 x 9) litres = (5x - 15/4) litres.

(7x - 21/four) / [(5x - 15/4) +9] = 7/9 = › 28x - 21/20x + 21 = 7/nine =› 252x - 189 = 140x + 147

=› 112x = 336 =’ x = 3.

So, the can contained 21 litres of A.

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Question 18. A Vessel Is Filled With Liquid, three Parts Of Which Are Water And 5 Parts Of Syrup. How Much Of The Mixture Must Be Drawn Off And Replaced With Water So That The Mixture May Be Half Water And Half Syrup?

Answer :

Suppose the vessel to start with carries 8 litres of liquid. Let x litters of this liquid get replaced with water.

Quantity of water in new mixture = (three - 3x/eight + x) litres.

Quantity of syrup in new aggregate = (5 - 5x/eight) litres.

(3 - 3x/eight + x) = (5 - 5x/8) = 5x + 24 = 40 - 5x

=› 10x = 16 =› x = eight/five

So, a part of the combination changed = (8/five x 1/8) = 1/5.

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Question 19. A Rectangular Parking Space Is Marked Out By Painting Three Of Its Sides. If The Length Of The Unpainted Side Is nine Feet, And The Sum Of The Lengths Of The Painted Sides Is 37 Feet, Then What Is The Area Of The Parking Space In Square Feet?

Answer :

Clearly, we have l=9 and l+2b=37

Area = (l x b)

= (9 x 14) squaretoes = 126 sq.Ft.

Question 20. The Length Of A Rectangular Plot Is 20 Metres More Than Its Breadth. If The Cost Of Fencing The Plot @ Rs. 26.50 Per Metre Is Rs. 5300, What Is The Length Of The Plot In Metres?

Answer :

Let breadth = x metres

Then, period = (x + 20) metres.

Perimeter = (5300 / 26.50) m

= 200m

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Question 21. There Are Two Sections A And B Of A Class, Consisting Of 36 And 44 Students Respectively. If The Average Weight Of Sections A Is forty Kg And That Of Sections B Is 35 Kg. Find The Average Weight Of The Whole Class?

Answer :

Total weight of (36+forty four) Students = (36x40+44x35) Kg = 2980 kg.

Average weight of the complete elegance = (2980 / 80) = 37.25.

Question 22. A Batsman Makes A Score Of 87 Runs In The 17th Inning And Thus Increases His Averages By 3. Find His Average After seventeenth Inning?

Answer :

Let the average after seventeenth inning = x. Then, common after sixteenth inning = (x - three)

Average =sixteen (x-3) +87

= 17x or x= (87-48)

= 39.

Question 23. The Banker's Discount On Rs.1800 At 12% Per Annum Is Equal To The True Discount On Rs.1872 For The Same Time At The Same Rate. Find The Time?

Answer :

S.I on Rs.1800 = T.D on Rs.1872.

P.W on Rs.1872 is Rs.1800.

Rs.72 is S.I on Rs. 1800 at 12%.

Time = (100x72 / 12x1800)

= 1/three 12 months = four months.

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Question 24. The Banker's Gain On A Bill Due 1 Year Hence At 12% Per Annum Is Rs.6. The True Discount Is?

Answer :

T.D = [B.G x 100 / R x T]

= Rs. (6 x one hundred / 12 x 1)

= Rs.50.

Oracle 11g Interview Questions

Question 25. A Boat Can Travel With A Speed Of thirteen Km/hr In Still Water. If The Speed Of The Stream Is four Km/hr. Find The Time Taken By The Boat To Go 68 Km Downstream?

Answer :

Speed Downstream = (13 + four) km/hr

= 17 km/hr.

Time taken to journey sixty eight km downstream = (sixty eight / 17) hrs

= four hrs.

Question 26. The Speed Of A Boat In Still Water Is 15 Km/hr And The Rate Of Current Is three Km/hr. The Distance Travelled Downstream In 12 Minutes Is?

Answer :

Speed Downstream = (15 + three) km/hr

= 18 km/hr.

Distance travelled = (18 x 12/60) hrs

= three.6km.

Value Labs Aptitude Interview Questions

Question 27. An Accurate Clock Shows eight O'clock In The Morning. Through How Many Degrees Will The Hour Hand Rotate When The Clock Shows 2 O'clock In The Afternoon?

Answer :

Angle traced by way of hour hand in

five hrs 10 min. = (360/12 x 6) °

= one hundred eighty°.

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Question 28. A Clock Is Set At 5 A.M. The Clock Loses sixteen Minutes In 24 Hours. What Will Be The True Time When The Clock Indicates 10 P.M. On 4th Day?

Answer :

Time from 5 a.M on a day to ten p.M.On 4th day = 89 hours.

Now 23 hrs forty four min. Of this clock = 24 hours of accurate clock.

Therefore 356 / 15 hrs of this clock = 24 hours of accurate clock.

89 hrs of this clock = (24 x 15/356 x 89) hrs

= ninety hrs

So, the best time is eleven p.M.

Question 29. Find Compound Interest On Rs. 7500 At four% Per Annum For 2 Years, Compounded Annually?

Answer :

Amount = Rs [7500x (1+4/100)²]

= Rs.(7500 x 26/25x26/25Rs)

= Rs.8112.

C.I = Rs (8112 - 7500)

= Rs.612.

Abaxis Aptitude Interview Questions

Question 30. Find The Compound Interest On Rs.16, 000 At 20% Per Annum For nine Months, Compounded Quarterly?

Answer :

Principal = Rs.Sixteen, 000;

Time=9 months = three quarters;

Amount = Rs. [16000x (1+5/100)³]

= [16000x21/20x21/20x21/20]

= Rs.18522.

C.I = Rs. (18522 - 16000)

= Rs.2522.

Question 31. What Decimal Of An Hour Is A Second?

Answer :

Required decimal = 1/ 60 x 60

= 1/ 3600

= .00027.

Question 32. A Man Standing At A Point P Is Watching The Top Of A Tower, Which Makes An Angle Of Elevation Of 30° With The Man’s Eye. The Man Walks Some Distance Towards The Tower To Watch Its Top And The Angle Of The Elevation Becomes 60°. What Is The Distance Between The Base Of The Tower And The Point P?

Answer :

One of AB, AD and CD need to had been given. So, the records is insufficient.

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Question 33. If Log 2 = zero.30103, The Number Of Digits In 520 Is?

Answer :

Log 520 =20 log five

=20 × [log (10/2)]

=20 (log 10 - log 2)

=20 (1 - zero.3010)

=20×0.6990

=13.9800.

Characteristics = thirteen.

Oracle BPM Interview Questions

Question 34. The Value Of Log2 16 Is?

Answer :

Let log216 = n.

Then, 2n = 16 = 24

‹=› n=four.

Question 35. The Product Of Two Numbers Is 192 And The Sum Of These Two Numbers Is 28. What Is The Smaller Of These Two Numbers?

Answer :

Let the number be x and (28 - x) = Then,

x (28 - x) = 192

‹=›x2 - 28x + 192 = zero.

‹=›(x - sixteen) (x - 12) = zero

‹=›x = 16 or x = 12.

Question 36. Three Times The First Of Three Consecutive Odd Integers Is 3 More Than Twice The Third. The Third Integer Is?

Answer :

Let the three numbers be x, x+2, x+4

Then 3x = 2(x+four) + 3

‹=›x = eleven

Third integer = x + four = 15.

Oracle GoldenGate Interview Questions

Question 37. Suganya And Surya Are Partners In A Business. Suganya Invests Rs. 35,000 For eight Months And Surya Invests Rs.Forty two, 000 For 10 Months. Out Of A Profit Of Rs.31, 570. Suganya's Share Is?

Answer :

Ratio in their shares = (35000×8): (42000×10)

= 2: 3.

Suganya's percentage = Rs. (31570 ×2/5)

= Rs.12628.

Question 38. Aman Started A Business Investing Rs.70, 000. Rakhi Joined Him After Six Months With An Amount Of Rs.1, 05,000 And Sagar Joined Them With Rs.1.4 Lakhs After Another Six Months. The Amount Of Profit Earned Should Be Distributed In What Ratio Among Aman, Rakhi And Sagar Respectively, 3 Years After Aman Started The Business?

Answer :

Aman : Rakhi : Sagar =(70,000 x 36):(1,05,000 x 30):(1,40,000 x 24)

=12: 15: 16.

Question 39. A Man Buys A Cycle For Rs.1400 And Sells It At A Loss Of 15%. What Is The Selling Price Of The Cycle?

Answer :

S.P = eighty five% of Rs.1400

= Rs. (85/100×1400)

Rs.1190.

Question 40. When A Commodity Is Sold For Rs.34.80, There Is A Loss Of 2%. What Is The Cost Price Of The Commodity?

Answer :

C.P = Rs. (a hundred / 75×34.80)

= Rs.Forty six.40.

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