Pershing, a bny mellon company Interview Questions, Process, and Tips

let x be lenghth of candles.

thicker one—- in 1 hr 1/6th will go.
thinner one— in 1 hr 1/4th will go.

let he lit for n hrs.

so in n hrs thicker one goes n/6.
and thinner one goes n/4.

so remainini length are x-n/6 and x-n/4.so x-n/6=2(x-n/4)
implie n=3x. if x=1 n=3hrs.

say the work is w and let no of days taken by b is ‘x’ which
we have to calculate
so work done by a in one day is w/6
work done by b in one day is w/x
a and b together can do work in 4 days ie=(w/6)+(w/x)=(w/4)
solving equation x=12
so no of days taken by b=12

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

77

1/8

friendly environment, supportive colleagues and place where i’ll have to work.

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

20&130

d

10 x 27=270

suppose
pipe:
A -30 hours A’s effeciency (60/30) =2
60( lcm of 30 and 20)
B- 20 hours B’s effeciency (60/20)=3

time taken by both to fill = 60/5 =12 as given in question (effeciencies of both a+b =2+3=5)
time taken by faster pipe i.e b = 60/3 =20

Marks in thrice in english as in science
so if science marks is x
then english marks = 3x
lets maths marks = y
ratio of english and maths marks is
3x/y = 3/5
=> y = 5x

and total marks = 162
3x + x + 5x = 162
x = 18 —-answer