# Top 38 Oracle Aptitude Interview Questions You Must Prepare 03.Oct.2023

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.

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

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)

= 1/3 year = 4 months.

Amount = Rs [7500x (1+4/100)²]

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

= Rs.8112.

C.I         = Rs (8112 - 7500)

= Rs.612.

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.

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

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%

Speed Downstream = (15 + 3) km/hr

= 18 km/hr.

Distance travelled = (18 x 12/60) hrs

= 3.6km.

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

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%

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

x (28 - x) = 192

‹=›x2 - 28x + 192 = 0.

‹=›(x - 16) (x - 12) = 0

‹=›x = 16 or x = 12.

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.

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 = 7/8@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.

We know, Average speed = Total distance travelled / Total time taken Average = [45+50+25]/ [3+2+ {25/15}] = 18 kmph

C.P = Rs. (100 / 75×34.80)

= Rs.46.40.

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

Then 3x = 2(x+4) + 3

‹=›x = 11

Third integer = x + 4 = 15.

Let log216 = n.

Then, 2n = 16 = 24

‹=› n=4.

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

Average weight of the whole class = (2980 / 80) = 37.25.

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

=12: 15: 16.

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.

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

= Rs. (6 x 100 / 12 x 1)

= Rs.50.

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.

Then, length = (x + 20) metres.

Perimeter = (5300 / 26.50) m

= 200m

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.

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.

S.P = 85% of Rs.1400

= Rs. (85/100×1400)

Rs.1190.

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.

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

Area = (l x b)

= (9 x 14) sq.ft = 126 sq.ft.

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.

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

= 2: 3.

Suganya's share = Rs. (31570 ×2/5)

= Rs.12628.

Principal = Rs.16, 000;

Time=9 months = 3 quarters;

Amount = Rs. [16000x (1+5/100)³]

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

= Rs.18522.

C.I = Rs. (18522 - 16000)

= Rs.2522.

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.

Angle traced by hour hand in

5 hrs 10 min. = (360/12 x 6) °

= 180°.

Required decimal = 1/ 60 x 60

= 1/ 3600

= .00027.

Log 520 =20 log 5

=20 × [log (10/2)]

=20 (log 10 - log 2)

=20 (1 - 0.3010)

=20×0.6990

=13.9800.

Characteristics = 13.

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

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%.

Speed Downstream = (13 + 4) km/hr

= 17 km/hr.

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

= 4 hrs.