Question

A train travels 360km/hr with uniform speed.The speed of the train is increased by 5km/hr,it takes 48 minutes less to cover the same distance.Find the initial speed of the train.​

Answer 1

Answer:

Step-by-step explanation:

let initial speed = x km/h

since Speed = distance / time

then time = distance /speed

                = 360/x

if speed is increased

new speed = x+5 km/h

new time = 360/x+5

new time - old time = 48 minutes = 48/60

= 360/x+5  -  360/x  = 48/60

now i have given you equation please solve accordingly

Good day!

Answer 2

$\bf{\Huge{\boxed{\sf{\green{ANSWER\::}}}}}}$

$\bf{\Large{\underline{\Rm{Given\::}}}}}$

A train travels 360km/hr with uniform speed. The speed of the train is increased by 5km/hr, it takes 48 minutes less to cover the same distance.

$\bf{\Large{\underline{\rm{To\:find\::}}}}}}$

The initial speed of the train.

$\bf{\Large{\underline{\sf{\pink{Explanation\::}}}}}}$

Let the Initial speed of the train be R km/hr.

$\bf{We\:have}\begin{cases}\sf{Distance\:=\:360km/hr}\\ \sf{Speed\:=\:R\:km/hr}\end{cases}}$

We know that Formula of time;

$\leadsto\tt{\orange{Time\:=\:\frac{Distance}{Speed} }}$

Therefore,

$\longmapsto\sf{Time\:to\:travel\:360km=\frac{360}{R} hr}}$

$\longmapsto\sf{Time\:with\:increase\:speed\:(R+5)=\frac{360}{R+5} hr}$

$\longmapsto\sf{48\:minutes\:convert\:hour\:=\:0.8hr.}$

A/q

$\mapsto\tt{\frac{360}{R} -\frac{360}{R+5} =0.8}$

$\mapsto\tt{\frac{360(R+5)-360(R)}{R(R+5)} =0.8}$

$\mapsto\tt{\frac{360(R+5-R)}{R(R+5)} =0.8}$

$\mapsto\tt{\frac{360(\cancel{R}+5\cancel{-R})}{R(R+5)} =0.8}$

$\mapsto\tt{\frac{1800}{R^{2}+5R } =\frac{8}{10} }$

$\mapsto\tt{1800*10=8(R^{2} +5R)}$

$\mapsto\tt{18000=8R^{2} +40R}$

$\mapsto\tt{8R^{2} +40R-18000=0}$

$\mapsto\tt{8(R^{2} +5R-2250)=0}$

$\mapsto\tt{R^{2} +5R-2250=\cancel{\frac{0}{8}} }$

$\mapsto\tt{R^{2} +5R-2250=\cancel{\frac{0}{8}} }$

$\mapsto\tt{R^{2} +5R-2250=0}$

Using Sridharacharya Formula:

$\mapsto\tt{x\:=\:\frac{-b ± \sqrt{b^{2} -4ac} }{2a} }$

• A = 1
• B = 5
• C = -2250

So,

$\mapsto\tt{x=\frac{-5±\sqrt{5^{2}-4*1*-2250 } }{2*1} }$

$\mapsto\tt{x=\frac{-5± \sqrt{25+9000} }{2} }$

$\mapsto\tt{x=\frac{-5± 95}{2} }$

Now,

$\leadsto\rm{x\:=\:\frac{-5+95}{2} \:\:\;\:\;\:\:or\:\:\:\;\:x=\frac{-5-95}{2} }$

$\leadsto\rm{x\:=\:\cancel{\frac{90}{2}} \:\:\:\:\:\:\:\;or\:\:\:\:\:\:\:x=\cancel{\frac{-100}{2}} }$

$\leadsto\rm{\red{x\:=\:45\:\:\;\:\:\:or\:\:\:\:\:x=-50}}$

We know that negative value is not acceptable.

Thus,

$\bf{\large{\boxed{\rm{\orange{The\:initial\:speed\:of\:the\:train\:is\:45km/hr.}}}}}}$