Rushcliffe Care Interview Questions, Process, and Tips

c

DISSENTIAL

Sunday
Monday
Tuesday

M
M. L
L

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;

1.5

no

D=5, G=1

x (A to B by car) + y (B to A by cycle) is 7 hours

x (A to B by car) + x (B to A by car) is 7 minus 3 is 4 hours

2x is 4 and hence x is 2

2 + y is 7 and hence y is 5

y (A to B by cycle) + y (B to A by cycle) is 5 + 5 is 10 hours

ans 810 i think

Let:
1. speed of boat in still water = x
2. speed of stream = y;

Then net speed of boat upstream = x – y = 16km / 4 h = 4kmph
Net speed of boat downstream = x + y = 16 / 2 = 8kmph

Now solve for x to get x = 6kmph in still water.

A. 4

Soldier have to move 0.75 miles to West and then 0.375 miles to South to reach the camp.
How:-
Firstly soldier moving 1 mile to East from camp and then 1/2 miles= 0.50 miles to North .

Then it is moving 1/4 miles =0.25 miles to west
And then 1/8 miles =0.125 miles to South

Now we just have to count the difference and minus the value of camp to East with the value of North to west = 1mile – 0.25miles(1/4) = 0.75 miles

Same case with north and south = 0.50 miles(1/2) – 0.125 miles(1/8) = 0.375 miles

Hence proves to return camp the soldier have to move 0.75 miles to west and 0.375 miles to south

28 hrs

( d ) 8404

FLRIHO=cooler

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