Heart of worcestershire college Interview Questions, Process, and Tips

Let ‘N’ is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5.
Required number = (LCM of 13 and 16) – (common difference of divisors and remainders)
= (208) – (11) = 197.

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A

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​

The first 10 odd prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31.

Sum of the odd prime numbers = (3+5+7+11+13+17+19+23+29+31)

= 158

Number of odd prime numbers = 10

We know, Average = (sum of the 10 odd prime numbers ÷ Number of odd

prime numbers)

Average =

= 15.8

∴ The Average of first 10 prime numbers which are odd is 15.8

let d distance and s be speed.
d/7 = x and d/5 = x+12
solving we get d=210.

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

0.255555…..

9828

sunday

answer is 306.04