Qvantel Interview Questions, Process, and Tips

In the example first we have to find the area of the field

we have given the following values ,

cost of fencing per meter = Rs 1.50

Diameter of circular field = 28

We have to find the area , a = ?

We know that,

Area = 2 π r

= 2 x 22/7 x 14

= 2 x 22 x 2

= 44 x 2

= 88 sq. m

So, area of circle is 88 sq.m

Now just multiply it with 1.50

cost of fencing = 88 x 1.50

= Rs. 132

So, the cost of fencing the circular field is Rs. 132

a

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

24 times

47

297

8 games They played
A own 3 games = 18 Rs
B loss 3 Rs and own 1 games i,e 9(A own 3 Games)-6(B own
1 game)=3
c own 12 Rs and loss 4 games(A own 3+b own 1) and own 4
games i.e 24(own)-12(loss)
so totally A own 3 games
B own 1 game
c own 4 games
=8

C. 96

Divide 30*60 seconds by LCM of Numbers =15

Given y/x = 1/3, x+2y =10
3y=x
Then substitute x=3y in x+2y=10

3y+2y=10
5y=10
y=2

Then substitute y=2 in x=3y
x=3*2
x=6
: x=6, y=2

Ans-44
4x+8=6Y
5X=5Y
X=Y
FROM EQ 1
2X=8
X=4
6Y+5Y=11*4=44

length of the base=3 times height
b=3h;
h=b/3;
so, 1/2bh=24
1/2b(b/3)=24
(b^2)/6=24
b^2=144
b=12