Question

2 Trains travels in 160km track each day. The express travels 10km per hours faster and takes 30min less than normal train. Find the speed of the express train.

Answer 1

Answer:

Let's assume,  

the speed of the normal train be “Sn”

the speed of the express train be “Se” = Sn + 10km/hr

time in hours by the normal train to cover 160 km be “tn” hr

time in hours by the express train to cover 160 km be “te” hr = tn – 30 minutes = tn – (0.5) hr

For the normal train, we have

Time = distance / speed

⇒ tn = 160 / Sn …… (i)

For the express train, we have

(tn – 1/2 ) = 160 / (Sn+10)

Substituting from (i)

160/Sn – 1/2 = 160 / (Sn+10)

⇒ (320-Sn)/2Sn = 160 / (Sn+10)

⇒320 Sn = 320 Sn – Sn² + 3200 – 10 Sn

⇒ Sn² + 10Sn – 3200 = 0

Using x = $\frac{-b+-\sqrt{b^2 - 4ac} }{2a}$

Here, x = Sn, a = 1, b = 10, c = -3200

⇒ Sn = $\frac{-10+- \sqrt{10^2 - 4*1*(-3200)} }{2*1}$

⇒ Sn = $\frac{-10+-\sqrt{100 + 12800} }{2}$

⇒ Sn = $\frac{-10+- \sqrt{12900} }{2}$

Since value of Sn cannot be negative

∴ Sn = $\frac{-10 + 113.57}{2}$ = 51.789 km/hr

Thus,

The speed of the express train, Se = Sn + 10 = 51.789 + 10 = 61.789 km/hr.