A lorry, a van and a car set off at the same time travelling at a constant speed of 60km, 80km/h and 120km/h respectively. The lorry and the van were travelling from town G to Town H while the car was travelling from Town H to Town G. THe car passed the lorry 2 minutes after passing the van.
a) find the ratio of the distance travelled by the lorry to the van to the car at the moment when the car passed the van?
b) find the distance between town G to Town H?
2007-09-28 04:22:50 · 5 answers · asked by Speed??? 1 in Science & Mathematics ➔ Mathematics
since the car and the others are moving in opposite direction, the rate of the car adds up to the rate of the others...
d = r*t
note: 60km/h = 60km/h(1h/60min) = 1km/min : lorry
80km/h = 4/3 km/min : van
120km/h = 2km/min : car
total distance = d = (2+4/3)t = (2+1)(t+2) ... where t denotes the time the car will meet the first opposing object...
10/3t = 3t + 6
1/3t = 6
t = 18 minutes... the car and the van met after 18 minutes...
a. distance of lorry: 18 km
distance of van: 24 km
distance of car: 36 km
ratios... 18:24:36 which is the same as 3:4:6
b. d = 3*20 = 60 km.
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2007-09-28 04:56:19 · answer #1 · answered by Alam Ko Iyan 7 · 0⤊ 0⤋
a) The ratio is trivial...60:80:120. Since they never change speed, it doesn't matter just when the ratio is calculated. Simplified, that's 3:4:6.
b) Ah, the fun part. I'll swap km/h for km/minute, since 2 minutes is 1/30 hours, and I'd prefer to not convert 1/30 to decimal.
The lorry goes 1 km/minute, the van 4/3 km/minute, and the car 2 km/minute. Let T1 be the time in minutes at which the car passes the van, and T2 be the time at which the car passes the lorry. And let X be the distance between Town G and Town H.
We know that at time T1, the van has traveled 4/3*T1 km, and that the car has traveled 2*T1 km. We also know that the distance between Town G and Town H, X = 4/3 *T1 + 2*T1, or (4/3+2)*T1 km.
We also know that at time T2, the lorry has traveled 1*T2 km, and the car has traveled 2*T2 km. Also, X=1*T2+2*T2, or 3*T2 km.
And, we also know that T2=T1+2, since it takes 2 minutes for the car to get from the van to the lorry. We have a system of equations!
(4/3+2)*T1=X
3*T2=X
Substituting in T2=T1+2,
(4/3+6/3)*T1=X
3*(T1+2)=X
((4+6)/3)*T1=X
3*T1+3*2=X
10/3*T1=X
3*T1+6=X
Multiplying the first equation by 3 (to clear the fraction),
10*T1=3*X
3*T1+6=X
And multiplying the second equation by 3 so that we can set the two equations equal to each other:
10*T1=3*X
9*T1+18=3*X
Now, setting them equal to each other:
10*T1=9*T1+18
Subtracting 9*T1 from both sides...
T1=18 minutes.
We already know that T2=T1+2, so T2 is 20 minutes. It takes the car 18 minutes to reach the van, and 20 minutes total to reach the lorry. But how far apart are the two towns?
10/3*T1=X
10/3*18=X
10*6=X
60 = X, so Town G is 60 km from Town H. Let's check:
3*T2=X
3*20=X
60 = X
60 = 60, so we must have done it right.
2007-09-28 05:03:58 · answer #2 · answered by El Jefe 7 · 0⤊ 0⤋
let the car passed the van after t hrs journey
the time taken by car to pass the lorry = t +2/60 hrs = t+1/30
=(30t +1)/30 hrs
distance travelleed by car when it passed van = speed * time
= 120t km
distance travelled by van during this time = 80 t km
distance between G and H = 120t + 80t = 200t
distance travelleed by car when it passed lorry =
= 120[(30t+1)/30]
= 120t + 4 km
distance traveleed by lorry during this= 60((30t+1)/30)
= 60t + 2
total distance = 120t + 4 + 60t + 2 = 180 t + 6
Both the distances are equal to the distance between towns
200t = 180 t + 6
20 t = 6
t = 6/20 = 3/10 hrs
a)
distance travelled by car when it passed van= 120(3/10) = 36 km
distance travelled by van = 80(3/10) = 24 km
distance travelled by lorry = 60((3/10) = 18 km
ratio of distances by lorry : van: car = 18: 24:36
3:4:6
The distance between towns = 200(3/10)
= 60 km
2007-09-28 05:19:24 · answer #3 · answered by mohanrao d 7 · 0⤊ 0⤋
suitable and shortest path is from Bangalore to Belgaum to Kohlapur to Satara to Pune to Nasik to Valsad to Surat to Baroda and ultimately to Udaipur. I even have not travelled in this path yet i think via fact the roads on those trunk routes are NHs so danger is that majority of the sectiond are good Motorable Roads. happy experience.
2016-12-28 06:12:32 · answer #4 · answered by ? 3 · 0⤊ 0⤋
I dont want to sound rude but i am sooooooo busy. Ill get bacl to you. The important thing here is TIME. you need to set up equations and equate the time somehow - ill work it out for you later - hope that helps for now
2007-09-28 04:36:21 · answer #5 · answered by uberifrit 2 · 0⤊ 1⤋