Shiptheory Interview Questions, Process, and Tips

Total Pages = 42+28 = 70 pages
for 70 pages we give Rs.20/-
and for 42 pages we give x
thus x = (42×20)/70
x = 12
The Answer = 12/-

6.25

=> 3s = d
=> 5(s-9) = d

therefore
5s-45=3s
2s = 45
s=22.5

hence
d= 67.5

avg sal of 3 is $950
first person earns=$1150
second person earns=$650
let thirdperson =x
avg=(1150+650+x)/3
950*3=1800+x
x=2850-1800
x=$1050

R
T,P,R,Q,S

2years

500 = Total+50
Total(450) = only one paper(p) + 29+20+35 + all three (g)
285+212+127 = p + 2( 29+20+35 )+ 3g
solve above .. to get g = 45 …
( small corrctn .. i think .. questn shud be 20 read ONLY
hindu and
times of India and 29 read ONLY hindu and Indian express
and 35
read ONLY times of India and Indian express)

( b ) 216.90

If n = no. of revolutions for back wheel
9n = 7n + 10*7
n=315

18

C. 96

volume of cylinder=volume of plane
pi*r*R*H=L*B*H
16PI*H=176
SO h= length=176/16*pi
ANser is 44pi

Triveni Sangam of the river
Ganga, Yamuna, and Saraswati at Allahabad.

(c)UW

Quick

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