An aeroplane left 50 minutes later than its scheduled time, and in order to reach the destination, 1250 km away, in time, it had to increase its speed by 250 km/hr from its usual speed. Find its usual speed.
Answers
SOLUTION :
Let the usual speed of the aeroplane be x km/h.
and the increase speed be (x + 250) km/h
Time taken to cover 1250 km with the usual speed of x km/h = 1250 / x hrs
Time taken to cover 1500 km with the increase speed of (x + 250) km/h = 1250 / (x + 250) hrs
A.T.Q
1250 / x = 1250 / (x + 250) + 50/60
1250 / x - 1250 / (x + 250) = 5/6
1250(x + 250) - 1250x / x(x + 250) = 5/6
[By taking LCM]
1250x + 1250 × 250 - 1250x / x² + 250x = ⅚
312500 / x² + 250x = ⅚
6(312500 ) = 5(x² + 250x )
[By cross multiplication ]
1875000 = 5x² + 1250x
5x² + 1250x - 1875000 = 0
5(x² + 250x - 375000) = 0
x² + 250x - 375000 = 0
x² - 500x + 750x - 375000 = 0
[By middle term splitting]
x(x - 500) + 750(x - 500)= 0
(x - 500) (x + 750) = 0
(x - 500) = 0 or (x + 750) = 0
x = 500 or x = - 750
Since, Speed can’t be negative, so x ≠ - 750
Hence, the usual speed of an aeroplane is 500 km/h.
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Answer:
500 km/hr
Step-by-step explanation :
Let the usual speed of aeroplane be x km/hr.
And, the increased speed of aeroplane be (x + 250) km/hr.
Time taken to reach the destination of 1250 km ( T₁ ) = 1250 / x km/ hr.
Time taken to reach the destination of 1250 km (T₂) = 1250 / x + 250 km/hr.
Given that ;
T₁ - T₂ = 50 minutes
⇒ 1250 / x - 1250 / x + 250 = 50 /60
⇒ 1250 ( x + 250 ) - 1250x / x ( x + 250) = 5 / 6
⇒ 1250x + 312,500 - 1250x/ x² + 250x = 5 / 6
⇒ 312,500 / x² + 250x = 5 / 6
[ On cross multiplying ]
⇒ 312,500 * 6 = 5 ( x² + 250x )
⇒ 1,875,000 / 5 = x² + 250x
⇒ x² + 250x = 375,000
⇒ x² + 250x - 375,000 = 0
⇒ x² + 750x - 500x - 375,000 = 0
⇒ x ( x + 750 ) - 500 ( x + 750 ) = 0
⇒ ( x + 750 ) ( x - 500 ) = 0
⇒ x + 750 = 0 or x - 500 = 0
⇒ x = - 750 or x = 500
Since, speed cannot be negative, answer is 500 km/hr.
Hence, the usual speed of the aeroplane is 500 k/hr.