The midcounties co-operative Interview Questions, Process, and Tips

– While the train is moving, the jogger will also be running in the same direction.
– for the head(engine) of the train to get to the current position of the jogger 240m away, it will take:
45km/hr => 12.5m/s => 240/12.5 = 19.2 seconds.

– But in the same period of time, the jogger will still be running and will have moved to a new location by: 9km/hr => 2.5m/s => 2.5 * 19.2 = 48m

To get to the new location at the speed of 12.5m/s will take the train:
48/12.5 = 3.84sec

In this additional time, the jogger will move forward by:
3.84 * 2.5 = 9.6m

at a speed of 12.5m/s, it will take the train less than a second to cover the additional 9.6m

If we add the distance the jogger will cover in 1 second to 9.6, it is still less than what the train can cover per second. let us see (9.6 + 2.5 = 12.1)

Therefore, the head of the train will pass the runner at approximately: 19.2 + 3.84 + 1 => 24.04 seconds.

For the train to completely pass the runner, it will need its whole length of 120m to be in front of the runner.

This will take an additional (9.6 + 2) seconds.

Therefore for the length of the train to be ahead of the runner it will take approx. 35.65 (24.04 + 9.6 + 2) seconds

Here is the solution to the given version of the puzzle (9 balls, one is heavier, need to identify oddball), where we label the balls A, B, …, I:

1. Weigh ABC versus DEF.

Scenario a: If these (1) balance, then we know the oddball is one of G, H, I.

2. Weigh G versus H.

Scenario a.i: If these (2) balance, the oddball is I.

Scenario a.ii: If these (2) do not balance, the heavier one is the oddball.

Scenario b: If these (1) do not balance, then the oddball is on the heavier side. For simplicity, assume the ABC side is heavier, so the oddball is one of A, B, C.

2. Weigh A versus B.

Scenario b.i: If these (2) balance, the oddball is C.

Scenario b.ii: If these (2) do not balance, the heavier one is the oddball.

answer is maximum of 2.

length of the base=3 times height
b=3h;
h=b/3;
so, 1/2bh=24
1/2b(b/3)=24
(b^2)/6=24
b^2=144
b=12

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​

A

X-340 = X x 4 x 8/ 100

100X – 34000 = 32X
68X = 34000
X = 34000/68
X= 500.

C-2896

324

the sum of all = 10
Accept
324 = 3+2+4 =/=10 (NOT EQUAL TO 10)

cube root of 5 is 125 so 128 is nearer to 125 and hence 5+5+5=15 and the nearest answer amongst options is 17 so 17 is correct answer:) yipppy……..

( b ) P

avg of 10nos.=23==>23*10=230
if each no increased by 4 ==> 4*10=40
then new avg is giveen by : 230+40=270
270/10=27
hence the new avg =27

3 days

12