If speed of train increased by 6km/hrs it takes 4 hrs less and if the speed is decreased by 6km/hrs it takes 6 hrs more.Find distance travelled.
Answers
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- Let the actual speed of the train be x km/hr and the actual time taken be y hours. Then,
Distance covered =(xy)km ..(i) [∴ Distance = Speed × Time]
If the speed is increased by 6 km/hr, then time of journey is reduced by 4 hours i.e., when speed is (x+6)km/hr, time of journey is (y−4) hours.
∴ Distance covered =(x+6)(y−4)
⇒xy=(x+6)(y−4) [Using (i)]
⇒−4x+6y−24=0
⇒−2x+3y−12=0 ..(ii)
When the speed is reduced by 6 km/hr, then the time of journey is increased by 6 hours i.e., when speed is (x−6) km/hr, time of journey is (y−6) hours.
∴ Distance covered =(x−6)(y+6)
⇒xy=(x−6)(y+6) [Using (i)]
⇒6x−6y−36=0
⇒x−y−6=0 (iii)
Thus, we obtain the following system of equations:
−2x+3y−12=0
x−y−6=0
By using cross-multiplication, we have,
3×−6−(−1)×−12x = −2×−6−1×−12−y
= −2×−1−1×3
Let the actual speed of the train be x km/hr and the actual time taken be y hours. Then,
Distance covered =(xy)km ..(i) [∴ Distance = Speed × Time]
If the speed is increased by 6 km/hr, then time of journey is reduced by 4 hours i.e., when speed is (x+6)km/hr, time of journey is (y−4) hours.
∴ Distance covered =(x+6)(y−4)
⇒xy=(x+6)(y−4) [Using (i)]
⇒−4x+6y−24=0
⇒−2x+3y−12=0 ..(ii)
When the speed is reduced by 6 km/hr, then the time of journey is increased by 6 hours i.e., when speed is (x−6) km/hr, time of journey is (y−6) hours.
∴ Distance covered =(x−6)(y+6)
⇒xy=(x−6)(y+6) [Using (i)]
⇒6x−6y−36=0
⇒x−y−6=0 (iii)
Thus, we obtain the following system of eey is 720km
Step-by-step explanation:
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