maths question?

Question

Can someone please help me with question and show me how to do it?
The excavator at the mine uses a litre of fuel every 3 minutes. The truck uses a litre of fuel every 6 minutes. How long will a 100 litres of fuel last of both machines are being used?

thanx.

Answer 1

Excavator:
3 mins = 1 litre
1 min = 1/3 litre

Truck:
6min = 1 litre
1min = 1/6 litre

Total

1min
= 1/3 +1/6 = 2/6+1/6=3/6
= 1/2 litre per minute.

1/2 litre = 1 minute

1 litre = 2 minutes

100 litre = 200 minutes....or 3 hours 20 minutes.

Answer 2

Find how much fuel is used in one minute.
Excavator: 1 litre in 3 minutes, 1/3 litre in 1 minute
Truck: 1 litre in 6 minutes, 1/6 litre in 1 minute

Total fuel per minute is (1/3 + 1/6) litre = 1/2 litre

so 100 litres lasts 100 /(1/2) = 200 minutes

Answer 3

Excavator uses 1/3 litres / min
Truck uses 1/6 litres / min

Total fuel used per min = 1/3 + 1/6
= 2/6 + 1/6 = 3/6 = 1/2 litres

1/2 litre<-----> 1 min
100 litres<--> 100/(1/2) x 1 min = 200min = 3hours 20 min

Answer 4

both excavator and track consume 1/3 + 1/6 liters of fuel in a minute.

so time = 100/ ( 1/3 +1/6)
= 100 / (1/2)
= 200 mins
= 3 hours and 20 mins

Answer 5

if separated,

excavator
100 litres of fuel x (3 minutes/1litre of fuel) = 300 minutes

truck
100 litres of fuel x (6 minutes/ 1 litre of fuel) = 600 minutes

if both,

1minute x (1 litre/3 minutes) = 1/3 litre or .333
1 minute x (1 litre/6 minutes) = 1/6 litre or .167
---------
1/2 or .497

100 litre x (1 minute/1/2 litre)
or
100 litre x (2/1) = 200 minutes

if hours

200 minutes x (1 hour/60 minutes) = 3.33 hours

3.33-3.00 = .33 x 60 minutes = 19.98 or approx. 20 minutes

therefore, 3 hours and 20 minutes

or exact

19.98-19= .98 x 60 seconds = 59.98 seconds

so 3 hours, 19 minutes and 59.98 seconds

if seconds

3 hours x (60 minutes/1 hour) x (60 seconds/1 minute) = 10800
19 minutes x (60 seconds/ 1 minute) = 1140
+ 59.98 s
---------
11999.98s

Answer 6

3 hours 20 minutes

Answer 7

Given:
Excavator (E) = 1L / 3 min.
Truck (T) = 1L / 6 min.
Total Fuel = 100L
Calculate time = ?

Use of Fuel:
Excavator (E) = 1L / 3 min.
Truck (T) = 1L / 6 min.
-----------------------------------------------------
Total (E) + (T) = 1L /3min + 1L /6min.
Total (E) + (T) = 1L /3min + (½)1L /(½)6min. (÷ 2 part)
Total (E) + (T) = 1L /3min + ½ L/ 3min. (÷ 3)
Total (E) + (T) = (1/3)1L / min + (1/3)½ L/ min. (x 2 part)
Total (E) + (T) = 2 /6L /3min + 1/6 L/ min.
Total (E) + (T) = 3/6 L/ min.
Total (E) + (T) = ½ L/ min.

Full Fuel supply = 100L
Rate of Fuel being used up = 100L / ½L/min.
Rate of Fuel being used up = 100L * 2/1 min./L
Rate of Fuel being used up = 200 min.
Rate of Fuel being used up = 200/ 60 min. = 3hr. 20 min.

Answer 8

EXCAVATER 1 /3 (L/min)
TRUCK 1/6 (L/min)
Both 1/3 + 1/6 = 3/6 = 1/2 (L/min)
Then 100L will last for 100L / 1/2 (L/min)= 200 min
-----------

Answer 9

not sure but,33 mins,with a spare of 1 min?