A train is going at 1/3 of its usual speed and it takes an extra 30 minutes to reach its destination. Find its usual time to cover the same distance.
I wasn't given the distance, but isn't there any way to solve it?
pop
so the formula is time = distance / speed
im sorry i checked again
its not letting me type anym
isearch mo kaya
To solve this problem, we can use a simple equation to relate the usual speed of the train, the time it takes to cover the distance, and the extra time it takes when going slower.
Let's denote the usual speed of the train as "S" and the usual time it takes to cover the distance as "T". We are given that when the train is going at 1/3 of its usual speed, it takes an extra 30 minutes to reach its destination.
When the train is going at 1/3 of its usual speed, the speed becomes (1/3)S, and the time it takes to cover the same distance is T + 30 minutes.
Using the formula Distance = Speed × Time, we have:
Distance = S × T (Usual distance covered by the train)
Distance = (1/3)S × (T + 30) (Distance covered when going at 1/3 of the usual speed, which takes an extra 30 minutes)
Since both equations represent the same distance, we can set them equal to each other:
S × T = (1/3)S × (T + 30)
Now, we can solve this equation to find the value of T, the usual time it takes to cover the distance.
Cancel out the common factors:
3 × S × T = S × (T + 30)
Expand:
3ST = ST + 30S
Subtract ST from both sides:
2ST = 30S
Divide both sides by 2S:
T = 15
Therefore, the usual time it takes for the train to cover the distance is 15 minutes.
Not enough information:
e.g. suppose the distance is 90 km
and the usual speed is x km/h
then 90/(x/3) - 90/x = 1/2
(270-90)/x = 1/2 ----> x = 360 km/h
but suppose the distance is 60 km
then 60/(x/3) - 60/x = 1/2
(180-60)/x = 1/2 ---> x = 240 km/h
So the speed depends on the distance, and we are not given that