Question

An aeroplane left 30 minutes later than its scheduled time and in order to reach it's destination 1500 km away in time , it had to increase it's speed by 250 km/hr from its usual speed . Determine it's usual speed.

plzzzz solve this

Answer 1

Heya here ,....

Solution is given below ::-
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Let the usual speed be x km /hr.

Actual speed = ( x + 250 ) km/hr.

Time taken at actual speed = (1500/x) hr.

Different between the two times taken = 1/2 hr.

$= > \frac{1500}{x} - \frac{1500}{x + 250} = \frac{1}{2} \\ \\ = > \frac{1}{x} - \frac{1}{(x + 250)} = \frac{1}{3000} \\ \\ = > \frac{(x + 250) - x}{x(x + 250)} = \frac{1}{3000} \\ \\ = > \frac{250}{( {x}^{2} + 250x) } = \frac{1}{3000} \\ \\ = > {x}^{2} + 250x - 750000 = 0 \\ = > {x}^{2} + 1000x - 750x - 750000 = 0 \\ = > x(x + 1000) - 750(x + 1000) = 0 \\ = > (x + 1000)(x - 750) = 0 \\ = > x + 1000 = 0 \: \: or \: \: x - 750 = 0 \\ = > x = - 1000 \: \: or \: \: x = 750$
=> x = 750 【 speed cannot be negative】

Hence , the usual speed of the aeroplane was 750 km / hr.
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Hope it's helps you.
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Answer 2

Let,

Usual speed = x km/hr
Distance/time = x
1500/time = x
1500/x= time
×××××××××××××××××××××××

Speed at that time = (x + 250) km/hr
Distance/time = x + 250
1500/time = x + 250
1500/(x + 250) = time


Then, According to the question,



1500/x - 1500/(x+250) = 30× 1/60

1500(1/x - 1/x+250) = 1/2

3000×[ (x + 250-x)/(x²+250)] = 1

3000{250} = x² + 250

0 = x² + 250x - 750000

0 = x² + (1000-750)x - 750000

0 = x² + 1000x - 750x - 750000

0 = x(x + 1000) - 750(x + 1000)

0 = (x + 1000) (x - 750)

By zero product rule,

x = -1000 or x = 750


Usual speed = x = 750 km /hr




I hope this will help you

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