Fastrack Interview Questions, Process, and Tips

( b ) P

let’s say the Cost Price is 1000x. (CP)
The selling price is also the same as the Cost price. So, here SP=CP (But he sells 950gm instead of 1000gm)

Instead of 1000gm, he is selling 950 grams at the CP. So he sells 050 gms @1000x price. So his net profit is 50gm.
Now 1000gm is 1000x Rs
So, 50gms is 50x rs. [Apllied Unitary Method]

So his profit percentage is:
Profit Percentage Formula: {(Profit/CP)*100%}
So, Here profit is 50x;
CP is 1000x;
so putting the value in the formula we get, Profit Percentage is: (50x/1000x)*100% =(5000x/1000x)%=5%.

4

7/5 * 5/q = 7/10
35/5q = 7/10
q=10

while(true)
{
new Object();
}

length of train = 150 m

let square be x (squares are 4 sides)
i.e., X+X+X+X=4X
4X+3=1460
4X=1460-3
4X=1457
X=364.25

4

Niece

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

1 ball – 4 runs (1st batter – 94+4=98)
2 ball – 3 runs and one run is short run during running between the wicket (1st batter- 98+3-1 =100)and strike changes for 2nd batter
3 ball – 6 runs ( 2nd batter 94+6=100)

30 Sec

Ans is 10%

Printer price = x;
Computer price = 3x
Total overall price = 20x + 60 x 3x = 200x

Printer percent = (20x/200x ) x 100 = 10%

Both conclusion follows