St. charles parish school board Interview Questions, Process, and Tips

speed = distance / time.
240/16=15m/sec=54km/hr

252

13. One group of each = 2+3+5 = 10 and 1 remainder for each = 1+1+1 = 3. 10+3 = 13

225

answer can be 1,2,3,4 can’t be determined exact number
just explaining a case:
there are 10 people in the party.
name of people no. of people with they made handshake list of those people(this can vary but showing the possibility)
1 1 9
2 2 8, 9
3 3 7, 8, 9
4 4 6, 7, 8, 9
5 5 6, 7, 8, 9. 10
6 6 4, 5, 7, 8, 9, 10
7 7 3, 4, 5, 6, 8, 9, 10
8 8 2, 3, 4, 5, 6, 7, 9, 10
9 9 1, 2, 3, 4, 5, 6, 7, 8, 10
10 4 5, 6, 7, 8, 9
jack got 9 different answer so jack can be either 4th number or 10th number and jack’s wife know jack very well so she can’t have handshake with jack so if 4th is jack then she can’t be handshake with 6,7,8,9, in this case she can be 1,2,3, 5, 10 and now depending upon which no is jack’s wife she can have hand shake with- 1- 4 people, and if jack is number 10 then she can’t be 5,6,7,8,9 so again depending upon her number she can handshake with people in range of 1-4

Answers as expected

cant say

If c is 5 times faster than a then he can do it in 2 days alone.

the answer is At 9:48 PM
At 1:00 pm the difference between A & B = 8 km
after 2:00 pm ………………. = 11 km (as B’s speed
is 1 and A’s 4 km, then eqv speed=(4-1)=3 km)
After 3:00………………….. = 13 km (as B’s speed 2 km)
After 4:00………………….. = 14 km
after 5:00………………….. = 14 km (A’s speed= B’s
speed)
after 6:00………………….. = 13 km
after 7:00………………….. = 11 km
after 8:00………………….. = 8 km
after 9:00………………….. = 4 km
and now the eqv speed is= (9-4) =5 km/hr;
and the renaming distance is 4 km;
then, time=(60*4)/5=48 min;
then the meeting time is=9:00+48 min=9:48 pm;

30

E. Copper

2:3

Let A, B and C be the three 6-faced dice.
Then, according to the question,
Since two dices has to be equal, that value can be any of the 6 faces, i.e., 6C1​ cases.
Now for each case, 2 equal dices can be selected from 3 dices in 3C2​ i.e., 3 ways.
And for each of the above, the third dice can have any of the 5 remaining faces
The possible outcomes are P(A)=61​,P(B)=61​,P(C)=65​,P(A)=61​,P(B)=65​,P(C)=61​ and P(A)=65​,P(B)=61​,P(C)=61​
Hence the required probability = 61​×61​×65​×6×3=21690​=125​

B, 8mm x 5 mm = 40 mm^2
1000mm x 1000mm = 1,000,000mm^2
= 1,000,000 mm^2/40mm^2
= 25,000