Pennecon limited Interview Questions, Process, and Tips

If Vijay gives ‘x’ marbles to Ajay then Vijay and Ajay would have V – x and A + x marbles.
V – x = A + x — (1)
If Ajay gives 2x marbles to Vijay then Ajay and Vijay would have A – 2x and V + 2x marbles.
V + 2x – (A – 2x) = 30 => V – A + 4x = 30 — (2)
From (1) we have V – A = 2x
Substituting V – A = 2x in (2)
6x = 30 => x = 5.

400%

e.g:-

l = 5 b = 2
Area= l*b =10

New after 100% increament
l=10 b = 4
Area = 10*4

16 chapatis* 6 Rs= 96
5 plates of rice * 45 Rs = 225
7 plates of mixed vegetable * 70 Rs = 490
6 ice-cream * ??
The amount that Alok paid the cashier was Rs.961
96+225+490 = 811
961-811= 150
150/6 = 25
The cost of each ice-cream cup is 25 .Rs

Number of persons between Vijay and Jack = 48 – (14 + 17) = 17.
Now, Mary lies in middle of these 17 persons i.e. at the eight position.
So, number of persons between Viji and mary = 7.

Mm

1/24-1/40
(5-3)/120
2/120
1/60
60 minutes

15/46=0.32

I can join your company with in one month that is my notice period time

94

Six men can at a time put mangala sutra (marriage) to their
spouses where six women cant do

75 degrees
30h – 11/2 m = 240 – 165 = 75 ..

a printer produced 176400 lines in 420 minutes
it produces 176400/420 lines per minute = 420
answer is : 420 lines per minutes

1/4, 1/8
4*8=32
32-4=28
28-8=20
28*20/(4^2)=35.00

Ans : 228
7
(7*1)+1=8
(8*2)+2=18
(18*3)+3=57
(57*4)+4=232 [ given 228 ] odd
(232*5)+5=1165
(1165*6)+6=6996

I am looking for a qualified chartered accountant in my profile who is known for his excellence in the field.

$57.30

To determine how many consecutive zeros the product of S will end with, we need to find the highest power of 10 that divides the product. This is equivalent to finding the highest power of 5 that divides the product, since the number of factors of 2 will always be greater than the number of factors of 5.

The primes in S are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}.

There are 24 primes in S, so the product of S is:

2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 x 43 x 47 x 53 x 59 x 61 x 67 x 71 x 73 x 79 x 83 x 89 x 97

We need to find the highest power of 5 that divides this product. To do this, we count the number of factors of 5 in the prime factorization of each number in S.

5 appears once: 5
5 appears once: 25
5 appears once: 35
5 appears once: 55
5 appears once: 65
5 appears once: 85
So, there are six factors of 5 in the product of S. However, we also need to consider the powers of 5 that arise from the factors 25, 35, 55, and 65.

25 = 5 x 5 appears once: 25
35 = 5 x 7 appears once: 35
55 = 5 x 11 appears once: 55
65 = 5 x 13 appears once: 65
Each of these numbers contributes an additional factor of 5 to the product of S. Therefore, there are 6 + 4 = 10 factors of 5 in the product of S.

Since each factor of 5 corresponds to a factor of 10, we know that the product of S will end with 10 zeros. Therefore, the product of S will end with 10 consecutive zeros

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