Problem on Time & Distance –
Example.1) A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/hr. more, the time taken for the journey would have been 1 hr. 40 minutes less. Find the original speed of the car.
Ans.) Let the original speed of the car be x km/hr.
400
Times taken to cover 400 km at original speed = -------- hrs.
x
400
Time taken to cover 400 km at the new speed = ---------- hrs.
(x + 1)
Now, as per the given condition –
400 400 100
So, --------- - ---------- = ---------
x (x + 12) 60
400 {(x + 12) – x} 100
=> -------------------- = ---------
x (x + 12) 60
48 1
=> ------------ = -------
(x² + 12x) 60
=> (60 X 48) = x² + 12x
=> x² + 12x – 2880 = 0
=> x² + (60 – 48)x – 2880 = 0
=> x² + 60x – 48x – 2880 = 0
=> x (x + 60) – 48 (x + 60) = 0
=> (x + 60) (x – 48) = 0
=> (x + 60) = 0 or (x – 48) = 0
=> x = - 60 or x = 48
=> x = 48 [As we know that speed cannot be negative, x ≠ - 60]
Hence the original speed of the car 48 km/hr. (Ans.)
Example.2) An aeroplane travelled a distance of 400 km at 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 –
1) The onward journey, 2) the return journey
If the return journey took 30 minutes less than the onwards journey, write an equation in ‘x’ and find the value of x
Ans.) Speed of the aeroplane on onwards journey = x km/hr
Distance covered = 400 km
Distance 400
Time taken for onward journey = ----------- = -------- hrs.…………..(i)
Speed x
Speed of the aeroplane on the return journey = (x + 40) km/hr
Distance covered = 400 km
Distance 400
Time taken for return journey = ------------ = ----------- hrs
Speed (x + 40)
As per the given condition, we have –
400 400 30
--------- - ----------- = --------
x (x + 40) 60
400 {(x + 40) – x} 1
=> -------------------- = -------
x (x + 40) 2
=> 800 X 40 = x (x + 40)
=> x² + 40x – 32000 = 0
=> x² + (200 – 160) x - 32000 = 0
=> x² + 200x – 160x – 32000 = 0
=> x (x + 200) – 160 (x + 200) = 0
=> (x + 200) (x – 160) = 0
=> (x + 200) = 0 or (x – 160) = 0
=> x = 160 [x ≠ - 200, because speed cannot be negative]
Hence, x = 160 km/hrs. (Ans.)
Example.3) A motor boat whose speed is 15 km/hr. in still water goes 36 km upstream and comes back to the starting point in 5 hours. Find the speed of the stream.
Ans.) Let the speed of the stream be x km/hr. Then,
Speed upstream = (15 – x) km/hr.
Speed downstream = (15 + x) km/hr.
36
Time taken to cover 36 km downstream = ---------- hrs.
(15 – x)
36
Time taken to cover 36 km upstream = ---------- hrs.
(15 + x)
Now, as per the given condition we have –
36 36
----------- + ----------- = 5
(15 – x) (15 + x)
1 1 5
=> ---------- + ---------- = --------
(15 – x) (15 + x) 36
15 + x + 15 – x 5
=> ------------------- = --------
(15 + x) (15 – x) 36
=> (30 X 36) = 5 (15² - x²)
=> 1080 = 1125 – 5 x²
=> 5x² = 1125 – 1080
=> 5x² = 45
=> x² = 9 = 3²
=> x = ± 3
=> x = 3, or x = -3
=> x = 3 [x ≠ - 3, because speed cannot be negative]
Hence the speed of the stream is 3 km/hr. (Ans.)
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