when speed of a train increased by 20km/hr it can cover a distance 40 minutes early.if speed of train is decreased by 30km/hr it taked 70minutes more to cover. find distance covered by train?
Answers
Given:
✰ When speed of a train increased by 20 km/hr, it can cover a distance 40 minutes early.
✰ If speed of train is decreased by 30 km/hr, it take 70 minutes more to cover.
To find:
✠ Distance covered by train.
Solution:
Here in this question, first we will assume speed, time and distance, then we will form the requisite equations. By substitution method and doing required calculations, we can easily find out the distance covered by train.
Let us assume:
Total distance be D.
Original speed of a train be S.
When speed of a train increased by 20 km/hr it can cover a distance in 40 minutes early.
i.e., D/S - D/(S + 20) = 40/60 ______(i)
If speed of train is decreased by 30 km/hr it taked 70 minutes more to cover.
i.e., D/(S - 30) - D/S = 70/60 ______(ii)
Finding the original speed:
In equation (i).
⇒ D/S - D/(S + 20) = 40/60
Taking S(S + 20) as LCM.
⇒ (DS + 20D - DS)/(S² + 20S) = 2/3
Cross multiplication.
⇒ 3(20D) = 2(S² + 20S)
⇒ 60D = 2S² + 40S ______(iii)
In equation (ii).
⇒ D/(S - 30) - D/S = 70/60
Taking (S - 30)S as LCM.
⇒ (DS - DS + 30D)/(S² - 30S) = 7/6
Cross multiplication.
⇒ 6(30D) = 7(S² - 30S)
⇒ 180D = 7S² - 210S ______(iv)
Substracting eqⁿ(iii) × 3 from equation (iv).
⇒ 180D - 180D = (7S² - 210S) - (6S² + 120S)
⇒ 0 = 7S² - 210S - 6S² - 120S
⇒ S² = 330S
⇒ S × S = 330 × S
Cancelling S both sides.
⇒ S = 330
∴ Original speed of a train = 330 km/h
Substituting the value of S in eqⁿ(iii).
⇒ 60D = 2 × (330)² + 40 × 330
⇒ 60D = 2 × 108900 + 13200
⇒ 60D = 217800 + 13200
⇒ 60D = 231000
⇒ D = 231000/60
⇒ D = 3850
∴ Total distance = 3850 km
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