TCS
Company
Numerical Ability
Time Distance and Speed
a car travelling with 5/7th of actual speed covers 42km in 1hr40min48sec.find the speed of the car.
Read Solution (Total 10)
-
- let speed be s
(5/7)s=42/(1+ 40/60 + 48/(60*60))
s=35km/hr - 9 years agoHelpfull: Yes(14) No(0)
- 5/7(x)=42*1000/6048
x=1000/144*7/5*18/5
x=35
so ans is 35 KMPH - 9 years agoHelpfull: Yes(4) No(0)
- 35kmph
42*3600*7/(6048*5) - 9 years agoHelpfull: Yes(2) No(0)
- let the actual speed be x
distance= 42km
time= 1hr40min48sec=42/25hr
speed=5x/7
5x/7=42*25/42
5x/7=25
so,x = 35ans:)
- 9 years agoHelpfull: Yes(1) No(6)
- 35 km/h
let x km/h be speed of car
1hr 40(48/60)min
=> 1 hr 204/5 min
=> 1(17/25)hr=(42*7)/(5*x)
=> x=35 - 9 years agoHelpfull: Yes(1) No(0)
- let speed=x
(5/7)x * (1+40/60+48/3600)=42
x=35km/hr - 9 years agoHelpfull: Yes(1) No(0)
- speed of the car=36.75km/hr
- 9 years agoHelpfull: Yes(0) No(3)
- let speed be x
(5/7)x=42/(1+40/60+48/(60*60))
x=35km/hr - 9 years agoHelpfull: Yes(0) No(0)
- Let actual speed be x km/h
New speed is 5x/7
Time taken= 1hr 40min 48secs=126/75
Dist=42km
5x/7*126/75=42
X=35km/hr
Speed of car is 35kmph - 9 years agoHelpfull: Yes(0) No(1)
- total tme =1.6799h
speed=42/1.6799=25
actual speed=25*7/5=35
- 9 years agoHelpfull: Yes(0) No(0)
TCS Other Question