Innovations for peace and development Interview Questions, Process, and Tips

414 Rs
1.2*37.8=45.36=46 sq.feet
46*9=414 RS

.

Let length of tunnel is x meter
Distance = 800+x meter
Time = 1 minute = 60 seconds

Speed = 78 km/hr = 78*5/18 m/s = 65/3 m/s

Distance = Speed*Time
=>800+x=653∗60=>800+x=20∗65=1300=>x=1300−800=500

225

Total number of pairs is NC2^{N}C_2NC2​. Number of pairs standing next to each other = N. Therefore, number of pairs in question = NC2^{N}C_2NC2​ – N = 28/2 = 14. If N = 7,
7C2​ – 7 = 21 – 7 = 14….
N =7

D = 180m
S = 42 – 6 = 36 * 5/18 = 10
T = D/S = 180/10
T = 18s

45 km/hr

For optimal size of a project team..
the % increase in staff size should be “Zero”
reason:
additional member directly proportional to increase in
staff size..

1.5 hr

D

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

Rs. 1500

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

4400+(97 leap year )=4497

R
T,P,R,Q,S

P+C+M=80
P+C=70
Therefore M=10
Given , P+M=90
If M=10
P=80