Astellas pharma Interview Questions, Process, and Tips

1.75-1.68=0.07

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

6/11 or 54.54%

P’s one min work=1/18
Q’s one min work=1/24
P’s x min work+Q’s 12 min work = tank fully filled ( 1 )
x(1/18)+12(1/24)=1
x=9 min

let no. of boys not participating be x
then the no. of girls not participating = x+5
no. of boys : girls participating = 3:2
given no.of boys participating = 15
therefore, the ratio is now 15:y(say)
then 3:2 = 15 : x
on solving 3/2 =15/y ie.., 3y =30 we get y =10
hence no. of girls participating =10
therefore total no of students paricipating = 15+10=25
total no of students in class =60 given
hence no. of students not participating = 60-25=35
therefore x+(x+5)=35
2x=30
x=15
therfore no of girls not participating =15+5=20
therefore total no of girls in class = no of girls
participating + no of girls not participating
=10+20
=30 is the answer

ans : 101 bcoz after every two matches one team is eliminated…so to eliminate 50 team there will be require 100 matches…+ 1 for deciding winner…

419

The first number is 36

Step-by-step explanation:
let their common factor be ‘x’

hence,

the ratio be 3x/5x

hence, by condition;

(3x-9)/(5x-9)=9/17 ch upu

9(5x-9)=17(3x-9)

45x-81=51x-153

153-81=51x-45x

72=6x

x=72/6

x=12

the first number is 3x=3×12=36

Let and A and B are playing and A has 3$ and B won 3 times.

Total money earn by A = no of games A won – no of games A loss
if A loss game means B won.
Therefore
Total money earn by A = no of games A won – no of games B won
3 = no of games A won -3
no of games A won=6
So
A won = 6 games
B won = 3 games

Hence total no of games is 9

32
Solution:

life as a boy = 1/4
life as a youth = 1/8
life as an active man = 1/2
sum of life as boy, youth and active man = 1/4 + 1/8 + 1/2 = 7/8
life as an old man = 1 − 7/8 = 1/8
1/8 Wrinkle’s life (as an old man) is 8 years.
and 1/2 = 1/8 *4
So, 1/2 Wrinkle’s Age (as active man) = 8*4 = 32years.

3/7

38 / 2 = 19 – 5 = 14 years old

Accounting and Finance

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

64

I’m interested in this job because I can see that, in this role, my skills could help solve this problem within your company. I also see an opportunity for me to learn and grow these skills, so we both would benefit personally, professionally, and financially.