4 years
860
B
consider the tank capacity as 90 litres.
to fill the tank in 3 hrs first pipe must flow at a rate of 30 litres/hr
2nd pipe has to flow at a rate of 45 litres/hr
if two pipes are opened at a time the flow rate will become 75 litres/hr
90/75= 6/5 = 1.2= (i.e) 1hr and 12 minutes 1 1/5 hr
b
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
This is more logical …
Let the 1st flag 1 placed at the origin ….
in crossing 8 flags he traveled 7 distances….
s=d/t
=7/8
time for 4 flags t=(d/s)=4/(7/8)=(4*8)/7=4.5714285714285714285714285714286
4
2, 5, 10, 50, 500, 5000
5*2=10
10*5=50
50*10=500
500*50=25000
So odd one is 5000
Given:
In a group of 15 students,
7 have studied Latin,
8 have studied Greek,
3 have not studied either.
To find:
The number of students who studied both Latin and Greek.
Solution:
In a group of 15 students, have studied Latin, 8 have studied Greek, 3 have not studied either.
Therefore,
n(A∪B) = 15 – 3
n(A∪B) = 12
7 have studied Latin,
n(A) = 7
8 have studied Greek,
n(B) = 8
n(A∩B) is the number of students who studied both Latin and Greek.
n(A∩B) = n(A) + n(B) – n(A∪B)
n(A∩B) = 7 + 8 – 12
n(A∩B) = 15 – 12
n(A∩B) = 3
The number of students who studied both Latin and Greek is 3
Final answer:
3 of them studied both Latin and Greek.
Thus, the correct answer .3
I think speed of printing per minute has nothing do to with how long printer being at work.
If “given day” = 8 hours.
8h = 480m
x = 176400 / 480 = 367.5 lines per minute.
B.25
The last number should be 0.
and the rest of the number to be divisible by 8. The x should be 6
So, sum is 6.
CASE 1: First we should take six balls divided equally and
then it is placed on the two pans.three on one and three on
other..
if the two pans are balanced then the defective ball is not
in the six..then we should the two and keep them one ball
on each.
CASE2: Again We should take any of the six balls and
divided equally and then it is placed on the two pans.. if
any of the pan weighs less than the other.. We should take
the three balls seperately..Now from that three we should
take any two and placed one on each.. fi both the pan
balances the ball which is left over is the defective.. if
one ball weighes less than the other,while keeping one on
each,then it is the defective one….
I will try to find that person
Or give that bag to police ..