# CLASS-10QUADRATIC EQUATIONS- PROBLEM ON TIME & DISTANCE

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.)