Problem 1 :
A passenger train takes 3 hours less than a slow train for journey of 600 km. If the speed of the slow train is 10 km/hr less than that of the passenger train, find the speed of two trains.
Solution :
Let x be the speed of the of the passenger train
Speed of the slow train is 10 km/hr less than that of the passenger train
So x-10 be the speed of the slow train
Distance has to be covered = 600 km
Time = Distance/speed
Let T1 be the time taken by passenger train
Let T2 be the time taken by the slow train
The differences of time taken by both trains are 3 hours
T1 = 600/x
T2 = 600/(x-0)
T2–T1 = 3 hours
(600/(x–10)) – (600/x) = 3
600[(1/(x-10)-(1/x)] = 3
x-(x-10)/(x2-10x) = 1/200
(x-x+10)/(x2-10x) = 1/200
2000 = x2-10x
x2-10x-2000 = 0
x2–50x+40x-2000 = 0
x(x–50)+40(x–50) = 0
(x+40) (x–50) = 0
By solving, we get x = -40 and x = 50
Therefore the speed of passenger train = 50 km/hr
Speed of slow train = 40 m/hr.
Problem 2 :
The distance between two stations A and B is 192 km. Traveling by a fast train takes 48 minutes less than another train. Calculate the speed of the fast train if the speeds of two trains differ by 20 km/hr
Solution :
Distance between two stations A and B = 192 km
Fast train takes 48 minutes less then the time taken by the slow train.
Let x be the speed taken by the fast train
Speeds of two trains differ by 20 km/hr
So speed of slow train is x – 20.
Time = Distance/speed
Let T1 be the time taken by the fast train
Let T2 be the time taken by the slow train
T1 = 192/x
T2 = 192/(x–20)
48/60 = 4/5 hours
T1 – T2 = 4/5
[192/(x-20)-192/x] = 4/5
192[(x–x+20)/x(x - 20)] = 4/5
192(20)/x2–20 x = 4/5
3840 (5) = 4(x2–20 x)
19200 = 4x2 – 80 x
4800 = x2 – 20 x
x2–20x–4800 = 0
x2–60x+40x-4800 = 0
(x–60) (x+40) = 0
x = 60 and x = -40
Speed of fast train is 60 km/hr.
Problem 3 :
A train covers a distance of 300 km at a certain average speed. If its speed was decreased by 10km/hr, the journey would take 1 hour longer. What is the average speed.
Solution :
Let x be the average speed of the train
So x–10 be the decreased speed
Time = Distance/Speed
T1 and T2 be the time taken by the train to cover the distance with speed of x km/hr and (x-10) km/hr respectively.
T1 = 300/x
T2 = 300/(x–10)
T1 – T2 = 1
[300/x] - [300/(x-10)] = 1
3000/(x2 – 10x) = 1
3000 = x2 – 10 x
x2 – 10x = 3000
x2–10x–3000 = 0
x2–60x+50x–3000 = 0
x (x – 60) + 50 (x – 60) = 0
(x + 50)(x – 60) = 0
By solving, we get
x = -50 and x = 60
So speed of the 60 km/hr.
Problem 4 :
The time taken by a train to travel a distance of 250 km was reduced by 5/4 hours when average speed was increased by 10km/hr. Calculate the average speed.
Solution :
Distance to be covered = 250 km
Let x be the required average speed.
If the average speed was increased by 10 km/hr
x+10 be the increased speed
Let T1 be the time taken to cover the distance in the average speed of x km/hr
Let T2 be the time taken to cover the distance in the average speed of (x + 10) km/hr
Time = Distance/Speed
T1 = 250/x
T2 = 250/(x+10)
T1 – T2 = 5/4
250/x – 250/(x + 10) = 5/4
250 [(x+10–x)/x(x+10)] = 5/4
2500/(x2 + 10x) = 5/4
2500 (4) = 5(x2+10x)
10000 = 5x2+10 x
Now we are going to divide the whole equation by 5, so we get
x2+10x = 2000
x2+10x–2000 = 0
x2+ 50x-40x-2000 = 0
x(x+50)–40(x+50) = 0
(x–40) (x+50) = 0
By solving, we get
x = 40 and x = -50
Therefore the required average speed = 40 km/hr
Increased speed = (40+10)
= 50 km/hr
Problem 5 :
An express train makes run 240 km t a certain speed. Another train whose speed is 12 km/hr less takes an hour longer to make the same trip. Find the speed of the express train.
Solution :
Let x be speed of express train
So x–12 be the speed of another train
Distance to be covered = 240 km
Let T1 be the time taken by the train to cover the distance 240 km at the speed of x km/hr
Let T2 be the time taken by the train to cover the distance 240 km at the speed of (x + 12) km/hr
Time = Distance /speed
T1 = 240/x
T2 = 240/(x - 12)
T2 - T1 = 1 hour
[240/(x- 12)] - [240/x] = 1
240[(1/(x -12) - 1/x] = 1
240[(x - x + 12)/x(x - 12)] = 1
240[12/(x2 - 12 x)] = 1
2880 = (x2-12 x)
x2-12x-2880 = 0
x2+60x-48x-2880 = 0
x(x+60)- 48 (x+60) = 0
(x-48) (x+60) = 0
x = 48 x = -60
Speed of express train = 48 km/hr
Speed of other train = (x - 12)
= (48-12)
= 36 km/hr
Problem 6 :
A plane traveled a distance of 400 km at an average speed of x km/hr. On the return journey, the speed was increased by 40 km/hr. Write down an expression for the time taken for
(i) the onward journey and
(ii) the return journey. If the journey took 30 minutes less than onward journey, write down an equation in x and its value.
Solution :
Let “x” be average speed of plane
On the return journey, the speed was increased by 40 km/hr
So “x + 40” be the speed of plane
Distance to be covered = 400 km
Let T1 be the time taken for onward journey in the speed of x km/hr
Let T2 be the time taken for downward journey to cover the same distance 400 km at the speed of (x + 40) km/hr
Time = Distance /speed
T1 = 400/x
T2 = 400/(x+40)
T1 - T2 = 30 minutes
[400/x]-[400/(x + 40)] = 30/60
400[(1/x) - 1/(x+40)] = 1/2
400[40/(x2 + 40 x)] = 1/2
16000 (2) = (x2+40 x)
x2+40x-32000 = 0
x2+160x-100x-32000 = 0
(x - 100) (x + 160) = 0
x = 100 x = -160
Speed of the plane = 48 km/hr
Increased speed = (x+40)
= (48+40)
= 88 km/hr
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