Question

please help me ❤️❤️❤️​

Answer 1

hi good morning dear

Let the usual speed be x km/hr.

Then, the speed taken =(x+100) km/hr

Distance =1500 km

Usual time =

usual speed

distance

=

x

1500

hrs

Time taken =

x+100

1500

hrs

It is given that Time taken=Usual time−0.5 hrs

⇒

x

1500

−

x+100

1500

=

2

1

⇒

x(x+100)

1500(x+100)−1500x

=

2

1

⇒x

2

+100x=1500×100×2

⇒x

2

+100x−300000=0

⇒x=

2

−100±

100

2

+4×300000

=

2

−100±1100

=500 or−600

Since speed cannot be negative, x=500 km/hr

∴ The usual speed of the plane is 500 km/hr.

Answer 2

$\orange{\mathfrak{Given}} \\ \\ \tt Distance = 1500 \:km \\ \\ \orange{\mathfrak{To\:find}} \\ \\ \tt Usual\:speed \\ \\ \orange{\mathfrak{Solution}} \\ \\ \tt Let\:the\:usual\:speed\:be\:x\:km/hr. \\ \\ \tt New\:speed = (x + 100)\:km/hr \\ \\ \tt We\:know\: that \\ \\ \boxed{\bold{\underline{\green{\tt speed = \frac{distance}{time} }}}} \\ \\ \implies \tt {t}_{1} = \frac{1500}{x} \\ \\ \tt \implies {t}_{2} = \frac{1500}{x+100} \\ \\ \tt Given\:that\: {t}_{1}-{t}_{2} = 0.5 \\ \\ \tt \implies \frac{1500}{x}-\frac{1500}{x+100} = \frac{1}{2} \\ \\ \tt \implies \frac{1500(x+100) - 1500x}{x(x+100)} = \frac{1}{2} \\ \\ \tt \implies \frac{1500x - 1500x + 150000}{{x}^{2}+100x} = \frac{1}{2} \\ \\ \tt \implies \frac{150000}{{x}^{2}+100x} = \frac{1}{2} \\ \\ \tt \implies 300000 = {x}^{2}+100x \\ \\ \tt \implies {x}^{2}+100x-300000 = 0 \\ \\ \tt \implies {x}^{2}+600x-500x-300000 = 0 \\ \\ \tt \implies x(x+600) - 500(x+600) = 0 \\ \\ \tt \implies (x+600)(x-500)=0 \\ \\ \tt \implies x = 500 \\ \\ \tt \therefore \boxed{\bold{\underline{\red{\tt Its\:usual\:speed\:is\:500\:km/hr}}}}$