University of florence Interview Questions, Process, and Tips

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Answer: Option D

Explanation:

As Bob is sitting between Charles and Elena and Douglas is sitting to the left of Elena. We get the following arrangement.

Andy is opposite Bob. Therefore, the final arrangement is as follows.

Fred will be sitting right of Charles if each person interchange his/her place with the person sitting opposite them.

When none of the digits are repeated:

The hundred’s place can be filled by any of the digits: 2, 3, 5, 6, 7 or 8 except the one which has already been used at the thousand’s place, so it can be filled in 5 ways.

Similarly tens’ place can be filled in 4 ways: only those 4 numbers which have not been use either at hundred’s or thousand’s place.
Unit’s place can be filled in only 3 ways. So, total number of nos. Possible =4×5×4×3= 240

NO but i hope to you hire me

Cousin

18th

4

G.c.d=num1 *num2/lcm
Num1=20
Gcd=5
Lcm=60
5=(20*num2)/60
Num2= 5*60/20
Num2=15

144

the ratio of present age of x:y is 5:6
And,we have to find the present age of x.
so,the present age of X is definitely will be multiple of 5.
X=35
so,35 is the answer.

the answer is 3:10
explanation:

let the first number be x
and the second number be y
(7/8)((3/7)x)=(25/100)((45/100)y) given
ie.., 21x/56= 9y/80
on solving we get x:y= 3:10
is the answer

Total 55
Manisha is a girl name so 54 boys
1 girl

Answer: 3121 gold coins
Let total no of coins be M
Let the disbursement D to each son:

D1 = 1 + (M – 1)/5 = (M + 4)/5
D2 = 1 + ( M – D1 -1)/5 = (D1) * 4/5
D3= (D2) * 4/5
D4= (D3) * 4/5
D5= (D4) * 4/5
Total disbursements to sons=
= ∑D= (M+4)*1/5[ 1+4/5+(4/5)(4/5)+ (4/5)(4/5)(4/5)+(4/5)(4/5)(4/5)(4/5) ]
= (2101/3125)*(M+4)
Thus balance left for daughters =M-{(2101/3125)*(M+4)}
=(1024M-8404)/3125
This balance should be a positive integer ( assuming M and all disbursements are full coins )
Thus 1024M-8404 should be a multiple of 3125….so….
1024M – 8404 = N*3125 where N is an integer
Using Python code:
n=int(input(“Enter num n: “))
X=int()
a=int()
a=0
X=’ ‘
for a in range(0,n+1):
a=a+1
X= (3125*a + 8404)/1024
if (3125*a + 8404)% 1024== 0:
print(X,a)
Enter num n: 10000
3121.0 1020
6246.0 2044
9371.0 3068
12496.0 4092
15621.0 5116
18746.0 6140
21871.0 7164
24996.0 8188
28121.0 9212
We get minimum value of N = 1021 and M = 3121 gold coins

C. Every real number is a rational number.