Question

A plane left 30 minutes later than the
scheduled time and in order to reach
its destination 1500 km away it has to
increase its speed by
O by
250 km / hour from its usual speed
Find the usual speed.​

Answer 1

Given :

• Distance is 1500 km.
• Left 30 min late.
• Speed increase by 250km/h.

Find :

• The Usual Speed.

Solution :

• Let usual speed = x

Using Formula :

• Time = Distance/Speed

➩ 1500/x

• Now speed increased by 250km/h.
• New speed = (x + 250)

Then,

• Time = 1500/(x + 250)

So we can write :

➥ 1500/x - ½ = 1500/x + 250

➥ 1500/x - 1500/x - 250 = ½

➥ 375000/x(x + 250) = ½

➥ x² + 250x - 75000 = 0

• Now find two numbers who's sums = +250 and multiple = -750000.

☆1000, -750

➨ x² + 1000x - 750x - 750000

➨ x(x + 1000) - 750 (x + 1000)

➨ (x + 1000) (x - 750)

➨ x + 1000 = 0 or x - 750

➨ x = - 1000 or x = 750

❖ Speed can't be negative :

• So, the speed is 750.

Therefore,

• Hence, the Usual Speed is 750km/h.

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Answer 2

Given : A plane left 30 minutes or 1/2 hours later than the scheduled time.

To Find : Find the usual speed ?

_________________________

Solution : Let the original speed of train be x km/hr.

$~$

• ${\sf\leadsto{New ~speed~=~\bigg(x~+~250\bigg)~km/hr}}$

$~$

$\underline{\frak{As ~we~ know ~that~:}}$

• $\boxed{\sf\pink{Time~=~\dfrac{Distance}{Speed}}}$★

$~$

$\pmb{\sf{\underline{According ~to ~the~ Given~ Question~:}}}$

$~$

$\qquad{\sf:\implies{\dfrac{1500}{x}~-~\dfrac{1500}{\bigg(x~+~250\bigg)}~=~\dfrac{1}{2}}}$

$\qquad{\sf:\implies{\dfrac{\bigg(1500x~+~37500~-~1500x\bigg)}{x\bigg(x~+~250\bigg)}~=~\dfrac{1}{2}}}$

$\qquad{\sf:\implies{2\bigg(37500\bigg)~=~x\bigg(x~+~250\bigg)}}$

$\qquad{\sf:\implies{75000~=~x^2~+~250x}}$

$\qquad{\sf:\implies{x^2~+~250x~-~75000~=~0}}$

$\qquad{\sf:\implies{x^2~+~1000x~-~750x~-~75000~=~0}}$

$\qquad{\sf:\implies{x\bigg(x~+~1000\bigg)~-~750\bigg(x~+~1000\bigg)~=~0}}$

$\qquad{\sf:\implies{\bigg(x~-~750\bigg)~\bigg(x~+~1000\bigg)~=~0}}$

$\qquad:\implies{\underline{\boxed{\frak{\pink{x~=~750~~~or~~~x~=~- 1000}}}}}$★

$~$

$\therefore{\sf{x~≠~- 1000~(Because ~speed ~can't ~be ~negative)}}$

$~$

Hence,

$\therefore\underline{\sf{The ~usual ~speed~ is~\bf{\underline{750~km/hr}}}}$