Question

A car after travelling a distance of 110 km develops a problem in the engine and proceeds at 53/99 of its former speed and arrives at the destination 1 hour 55 min late. Had the problem developed 63 km further on, the car would have arrived only 1 hour 10 minutes late. Find the total distance travelled by car.
a)241 km (b)301 km (c)161 km (d)271 km

Answer 1

Answer:

D = 261.47 km

Step-by-step explanation:

Time saved if break down after 63 km

is 1 hr 55 min - 1 hr 10 min  = 45 min

= 3/4 hrs

Let say  Speed = S

reduced Speed  = 53S/99

63/S   + 3/4  = 63 /(53S/99)

=>63/S   + 3/4  = 63*99 /53S

=>63*53   + 3*53S/4  = 63*99

=>63*53*4   + 3*53S  = 63*99*4

=> 169S = 63*4*46

=> S = 63*4*46/169 = 11592/169

D/S + 23/12= 110/S  + (D-110)/(53S/99)  -

=> 169D/11592  + 23/12 =  110*169/11592   + (D-110)*99*169/(53*11592)

=> 169D*53  + 23*966*53  = 110*169*53  + 169(D-110)*99

=> 1177554 - 985270 + 1840410 = 169D(99-53)

=> 2032694 = D *169 * 46

=> D = 261.47 km