Shalom Hills International School Interview Questions, Process, and Tips

60 liter

1 km/hr

b

12, 10

One day work = 1 / 20
One mans one day work = 1 / ( 20 * 75)
Now:
No. Of workers = 50
One day work = 50 * 1 / ( 20 * 75)
The total no. of days required to complete the work = (75 * 20) / 50 =
30

Ans: 60 kph

Suppose Person meets the train everyday at 3 PM at Station A.

His speed is 12kph.

So normally he reaches 5 km before the meeting point (pt B) at (5/12 hr = 25 min before) 2:35PM.

But if he is late by 30 min, then he will reach that point (pt B) by 3:05 PM.

Train is traveling at its normal speed so it covers the distance of 5 Km in 5 min starting from Station A and reaches the meeting point (pt B) at 3:05 PM.

So speed of the train is 5KM/5min = 60 kph.

Pearl.. Head cut then earl.. tail cut then pear.. Head and tail cut then ear

Ans is 2(Prime numbers)

Sum of 5 consecutive nos is 35

so X + (X+1) + (X+2) + (X+3)+ (X+4) = 35

5X + 10 = 35

X = 5

So the 5 consecutive numbers are : 5, 6, 7, 8, 9

The Prime numbers are 5 and 7

Answer:
11 days.

Step-by-step explanation:
In the question,

Time taken by Ramesh to finish a piece of work = 20 days

Time taken by Sushil to finish a work = 25 days

Time for which they worked together = 5 days

Sushil left after = 5 days

So,

One day work of Ramesh is,

One day work of Sushil is,

So,

Work done in 5 days is given by,

Therefore, Remaining work is given by,

Now, as the Sushil left the remaining work was done by Ramesh,

Time taken by Ramesh for the remaining work is,

Therefore, the remaining work will be completed in 11 days by Ramesh.

North-east

Tuesday

DISTANCE=TIME *SPEED
D=?
48*(5/18)=13.33m/s
D=9*13.333=120m

6,36

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​

let the total work to be done =1
so a+b+c=1 (say)
then a and b are supposed to do 7/11 th of work
so c= 1- (7/11)
c=4/11
therefore c gets (4/11)*550
=200 is the answer