Petronas Interview Questions, Process, and Tips

let t = total no of students.. then
students who passed one or both subjects,
n(e U h) = n(e) + n(h) – n(e intersection h)
=> t = 0.8t + 0.7t – 144
=> t = 1.5t – 144
students who failed both subjects is 10% i.e. 0.1t
=>t-n(e U h) = 0.1t,
=>t -(1.5t – 144) = 0.1t
=>t- 1.5t- 0.1t = -144
=> -0.6t = -144
=>t = 240

b

B. 27

( Number × 2 ) + 1
3 × 2 + 1 = 7
7 × 2 + 1 = 15
15 × 2 + 1 = 31 ( not 27 )
31 × 2 + 1 = 63
63 × 2 + 1 = 127
127 × 2 + 1 = 255

22

s=a+(a+10)+(a+10+10)+(a+10+10+10)
s=4a+60
a=(s-60)/4

x- sum of first 4 values
y – 5th value

x/4 = 45
x=45*4=180

(x+y)/5 = 46
(180+y)/5=46
180+y=230
y=50

( b ) 40 sq cm

C. 4

So we consider the 2nd statement first. We can form an equation out of it.
14x-6=13y+3=9z+3
Using this, we can understand that the multiple of 14 and the multiple of 13 and 9 must have a difference of 9. The easiest way to ensure that is multiplying it by 9
14*9=126
13*9=117
If the 5th farmer gives 3 apples to the 4th farmer, they would have 123 and 120 apples respectively. However, we also know that the 2nd farmer has 117 apples (13*9=117, and this is a multiple of 9) if the 5th farmer gives 3 apples too the 2nd farmer, the 3rd, 4th and 5th farmers would have 120 apples each.
Now that we got 120, we should check if the first part of the question makes sense along with it. The equation would be
7a+1=11b-1=120
We know that 11*11=121 and 7*17=119. When we add 1 to 119 and subtract 1 from 121, we get 120 for each. In this way, all the farmers have 120 apples each.
Therefore, the 3rd farmer had a yield of 11 per tree and the 4th farmer had a yield of 9 per tree.

Ans- 6,6 litres
Let x Litres from can1 (where, water = x/4 litres and milk = 3x/4 litres)
Now
12-x liters from can 2 ( where, water = 6-x/2, milk = 6-x/2)

Now given
water/milk = 3/5
0.25x+6-0.5x / 0.75 +6-0.5x = 3/5
6-0.25x/6+0.25 = 3/5
Solve it you get
x = 6 litres from can1
Then 12-x= 12-6=6 litres from can2

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