Brothers of Charity Services Interview Questions, Process, and Tips

1. 1g
2. 3g with 1g counter
3. 3g
4. 3g plus 1g
5. 3g plus 1 g plus 1g weighed medicine
6. 9g with 3g counter
7. 9g with 1g of counter and 1g weighed medicine
8. 9g with 1g counter
9. 9g
10. 9g plus 1g
11. 9g plus 3g with 1g on counter
12. 9g plus 3g
13. All 3 weights on one side.

80

13 Kigs & 6 Libs –> 510tors in 10hrs
” –> 51 tors/hr
14 Libs –> 484 tors in 12hrs
” –> 121/3 tors/hr
1 Lib –> 121/42 tors/hr
==> 6 Libs –> 121/7 tors/hr
Now,
13 kigs and 6 Libs –> 51 tors/hr
Only 13 Kigs –> 51 – (121/7) tors/hr
= 236/91 tors/hr

32
Solution:

life as a boy = 1/4
life as a youth = 1/8
life as an active man = 1/2
sum of life as boy, youth and active man = 1/4 + 1/8 + 1/2 = 7/8
life as an old man = 1 − 7/8 = 1/8
1/8 Wrinkle’s life (as an old man) is 8 years.
and 1/2 = 1/8 *4
So, 1/2 Wrinkle’s Age (as active man) = 8*4 = 32years.

192 zeros in 1 to 1000

The last person covered 120.71 meters.
It is given that the platoon and the last person moved with
uniform speed. Also, they both moved for the identical
amount of time. Hence, the ratio of the distance they
covered – while person moving forward and backword – are
equal.
Let’s assume that when the last person reached the first
person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X)
meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-
X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total
distance covered by the last person is 100 meters, as he
ran the total lenght of the platoon (50 meters) twice.
TRUE, but that’s the relative distance covered by the last
person i.e. assuming that the platoon is stationary.

3 hours ago.

Thin candle melts 3/4 in 3 hours leaving 1/4
Where as in the same time thick candle melts 3/6 leaving 3/6 which is 1/2. Now thick candle is exactly twice than the thin candle.

Or via modeling:

We need to find time at which the length of the thin candle is half the thick candle. Let x be the time. Thin candle melts at 1/4 an hour and thick candle melts at 1/6 an hour. In x hours they melt at x/4 and x/6 respectively. What’s left will be 1 – x/4 and 1 – x/6. We need to find x at which :

2 * (1 – (x/4)) = 1 – (x/6)

This equation results in x = 3

Its 2 times faster than the other train
v1*t=v2
v2*t=4*v1
solving these two,we get
v2/v1=2

Joe is fencing in a square area of 576 square feet.
Fence posts, which are needed every three feet, cost 32.00 each.
The fencing cost 4.50 per foot. What is the total cost of the fencing materials?

Solution:-
Area of Square is 576 sq ft.
so, the length of side = root(576) = 24 ft

Since the fencing would be done on the perimeter, we would need that
formula is 4*side = 96 ft.

Number of Fence post needed = 96/3 = 32

1 Fence post cost = 32
32 Fence Post cost = 32 * 32 = 1024

Fencing cost = Perimeter * fencing cost = 96 * 4.5 = 432

Total Cost = 432 + 1024 = 1456

Answer: C

Explanation:

900 — 100
100 — ? => 100/900*100 => 11.11%

China

A