# Oracle Job Preparation for Freshers - Quantitative Aptitude Questions

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## Oracle Recruitment Exam - Aptitude Questions

Searching for an opportunity to seek a career in ORACLE, the American multinational computer technology corporation, headquartered in Redwood Shores, California? Then you need to equip yourself with the knowledge and skills needed to crack the competition.

The questions will range from topics like Algebra, Time & Work, Distance &b Speed, Progression, Probability, Profit & Loss and Geometry.

The paper is in time constraints in so the tips and tricks must be known in order to solve the questions quickly. Below sample questions based on previous year placement papers will help you learn and apply these tricks and ace the exam.

## Oracle Job for Freshers - Aptitude Practice Questions

Question 1

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.

(a) x2-25=0

(b) x2-2x+25=0

(c) 2x2-25=0

(d) x2+4x+45=0

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 2

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?

(a) 54/3

(b) 51/7

(c) 40/7

(d) 75/7

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 3:

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?

(a) 15s

(b) 10s

(c) 12s

(d) 18s

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 4:

The sum of 3rd and 6th term of an A.P is 27. Find the sum of the first 10 terms of the progression.

(a) 135

(b) 150

(c) 81

(d) 360

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 5:

What is the probability of finding a red face card in a deck of cards?

(a) 3/13

(b) 9/52

(c) 3/26

(d) 4/13

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 6:

How many different words can be formed from the word ORACLE so that the vowels always come together?

(a) 200

(b) 720

(c) 240

(d) 120

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, 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 7:

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?

(a) 200% (b) 75%

(c) 100% (d) 50%

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 8:

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?

(a) 4655.98 cu. cm.

(b) 4187.33 cu. cm.

(c) 4233.58 cu. cm.

(d) 4957.67 cu. cm.

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

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.

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