**Question 1. A Can Do A Work In 60 Days And B Can Do The Same Work In 40 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 A and B in 12 days is = 12 * 5/120 = 1/2

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

B does 1/40 work in one day

Therefore B does 1/2 work in 40*1/2 =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 = 42 [M/D] =2/1 and Solving the above equations we get D = 28 yrs.

**Question 3. The Price Of Sugar Is Increased By 25%.find By How Much Percent The Consumption Of Sugar Be Decreased So As Not To Increase The Expenditure?****Answer :**Using the formula to calculate % decrease as [R/(100+R)]x100 where R = percentage increase in price, we get

Required % decrease in consumption = 25/125 MULTIPLIED 100 = 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 know, Average speed = Total distance travelled / Total time taken Average = [45+50+25]/ [3+2+ {25/15}] = 18 kmph

**Question 5. A Car Travels A Distance Of 170 Km In 2 Hours Partly At A Speed Of 100 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 100 kmph

So he covers 170-x at 50 kmph

So {X/100}+{170-X}/50=2

Solving this equation, we get x = 140.

So he covers 30km at 50 kmph.

**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 value is

400 = 320 (1 + profit/100),

So profit is 25%

**Question 7. The Ratio Of The Present Ages Of A And B 9: 5. Five Years Earlier The Ratio Of Their Was 2 : 1. What Is The Average Of Their Present Ages?****Answer :**A/b=9/5; i.e. 5a = 9B………. (i)

A-5/B-5 = 2/1; i.e. A-5 = 2B ………10

∴A – 2B = -5………………. (ii)

From equation (i) and (ii), A = 45, B = 25

∴ Average =45+25/2 = 70/2 = 35.

**Question 8. 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 as well as girls, Therefore

[20+X]/ [32+X] =3/4

Solving this we get x = 16

So total will 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=0 And The Greater Root Of The Equation X2+14x+45=0.**Answer :**Roots of x2−12x+35=0 are 5 and 7, lesser root is 5

Roots of x2+14x+45=0 are -5 and -9, greater root is -5

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

**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 the work in 10 days, he/she will complete the work in 20 days.

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

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

So, they can complete the entire work in 80/7 days

Therefore, they can complete the remaining 50% of the work in (1/2) x 80/7 = 40/7 days.

**Question 11. Two Trains Of Length 150 M And 200 M Respectively, Are Travelling In Opposite Directions At A Speed Of 54 Km/hr And 72 Km/hr. What Is The Total Time Taken By Them To Cross Each Other?****Answer :**54 km/hr = 54 x (5/18) = 15m/s

72 km/hr = 72 x (5/18) = 20 m/s

Total distance = 150 + 200 = 350 m

Relative speed = 15 + 20 = 35 m/s

Total time = Total distance/Relative speed

= 350/35 = 10s

**Question 12. The Sum Of 3rd And 6th 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

**Question 13. What Is The Probability Of Finding A Red Face Card In A Deck Of Cards?****Answer :**A deck of card has 52 cards of which 26 are black and the other 26 are red.

The number of face cards is 12, but only 6 of them are red.

So, required probability 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 group the letters that need to come together (A & E) and consider them as a single letter. So, here the letters are O, R, C, L, and AE.

Number of ways A & E can be arranged is 2!

So, the total number of ways in which the words can be formed so that all vowels are together is 5! x 2! = 240 ways.

**Question 15. On Selling 100 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 each article be Re. 1

So, SP of 100 articles =Rs. 100

Implies profit = Rs. 50

So, CP = 100 – 50 = Rs. 50

Therefore, profit percentage of the shopkeeper is 50/50 × 100 = 100%

**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) = 8 cm.

Volume of cone = (1/3) x π x r2 x h= (1/3) x π x 82 x 15

Radius of hemisphere=radius of cone

Volume of hemisphere = (2/3) x π x r3 = (2/3) x π x 83

Total volume of figure = 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: 9. How Many Litres Of Liquid A Were Contained By The Can Initially?****Answer :**Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively

Quantity of An in mixture left

= (7x – 7/12 x 9) litres = (7x – 21/4) litres.

Quantity of B in mixture left

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

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

=› 112x = 336 =’ x = 3.

So, the can contained 21 litres of A.

**Question 18. A Vessel Is Filled With Liquid, 3 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 initially contains 8 litres of liquid. Let x litters of this liquid be replaced with water.

Quantity of water in new mixture = (3 – 3x/8 + x) litres.

Quantity of syrup in new mixture = (5 – 5x/8) litres.

(3 – 3x/8 + x) = (5 – 5x/8) = 5x + 24 = 40 – 5x

=› 10x = 16 =› x = 8/5

So, part of the mixture replaced = (8/5 x 1/8) = 1/5.

**Question 19. A Rectangular Parking Space Is Marked Out By Painting Three Of Its Sides. If The Length Of The Unpainted Side Is 9 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) sq.ft = 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, length = (x + 20) metres.

Perimeter = (5300 / 26.50) m

= 200m

**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 40 Kg And That Of Sections B Is 35 Kg. Find The Average Weight Of The Whole Class?****Answer :**Total weight of (36+44) Students = (36×40+44×35) Kg = 2980 kg.

Average weight of the whole class = (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 17th Inning?****Answer :**Let the average after 17th inning = x. Then, average after 16th inning = (x – 3)

Average =16 (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 = (100×72 / 12×1800)

= 1/3 year = 4 months.

**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 100 / 12 x 1)

= Rs.50.

**Question 25. A Boat Can Travel With A Speed Of 13 Km/hr In Still Water. If The Speed Of The Stream Is 4 Km/hr. Find The Time Taken By The Boat To Go 68 Km Downstream?****Answer :**Speed Downstream = (13 + 4) km/hr

= 17 km/hr.

Time taken to travel 68 km downstream = (68 / 17) hrs

= 4 hrs.

**Question 26. The Speed Of A Boat In Still Water Is 15 Km/hr And The Rate Of Current Is 3 Km/hr. The Distance Travelled Downstream In 12 Minutes Is?****Answer :**Speed Downstream = (15 + 3) km/hr

= 18 km/hr.

Distance travelled = (18 x 12/60) hrs

= 3.6km.

**Question 27. An Accurate Clock Shows 8 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 hour hand in

5 hrs 10 min. = (360/12 x 6) °

= 180°.

**Question 28. A Clock Is Set At 5 A.m. The Clock Loses 16 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 10 p.m.on 4th day = 89 hours.

Now 23 hrs 44 min. of this clock = 24 hours of correct clock.

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

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

= 90 hrs

So, the correct time is 11 p.m.

**Question 29. Find Compound Interest On Rs. 7500 At 4% Per Annum For 2 Years, Compounded Annually?****Answer :**Amount = Rs [7500x (1+4/100)²]

= Rs.(7500 x 26/25×26/25Rs)

= Rs.8112.

C.I = Rs (8112 – 7500)

= Rs.612.

**Question 30. Find The Compound Interest On Rs.16, 000 At 20% Per Annum For 9 Months, Compounded Quarterly?****Answer :**Principal = Rs.16, 000;

Time=9 months = 3 quarters;

Amount = Rs. [16000x (1+5/100)³]

= [16000×21/20×21/20×21/20]

= 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 must have been given. So, the data is inadequate.

**Question 33. If Log 2 = 0.30103, The Number Of Digits In 5**^{20}Is?**Answer :**Log 5

^{20}=20 log 5=20 × [log (10/2)]

=20 (log 10 – log 2)

=20 (1 – 0.3010)

=20×0.6990

=13.9800.

Characteristics = 13.

**Question 34. The Value Of Log**_{2}16 Is?**Answer :**Let log

_{2}16 = n.Then, 2

^{n}= 16 = 24‹=› n=4.

**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

‹=›x

^{2}– 28x + 192 = 0.‹=›(x – 16) (x – 12) = 0

‹=›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+4) + 3

‹=›x = 11

Third integer = x + 4 = 15.

**Question 37. Suganya And Surya Are Partners In A Business. Suganya Invests Rs. 35,000 For 8 Months And Surya Invests Rs.42, 000 For 10 Months. Out Of A Profit Of Rs.31, 570. Suganya’s Share Is?****Answer :**Ratio of their shares = (35000×8): (42000×10)

= 2: 3.

Suganya’s share = 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 = 85% 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. (100 / 75×34.80)

= Rs.46.40.

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