Latest TCS Aptitude Question SOLUTION: truck a = 10kg/min truck b= 131/3kg/min truck c = 5kg/min if work simuntaneosuly how min it take to load 2.4tons load

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truck a = 10kg/min
truck b= 131/3kg/min
truck c = 5kg/min
if work simuntaneosuly
how min it take to load 2.4tons load

Asked In TCS gaurav kothari (11 years ago)

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praveen kumar  (11 years ago)

clearly 2.4 tons=24oo kg
A/q,truck is loaded simulaneously,
thetrefore,in 1 mins,we can load=10+5+131/3=58.6kg
so 2400kg can be loaded in=2400/58.6= 40.95 minutes

Is this solution Helpfull? Yes (29) | No (1)  

aahana  (11 years ago)

a loaded in 1 min =10kg
b loaded in 1 min=40/3kg
c unloaded in 1 min=5kg
so a,b,c are loaded in 1 min=55/3kg
so 1 kg loaded by a,b,c in time =3/55min
now, 2.4 tons loaded by a,b,c in time=(3/55)*2.4*1000=131min

Is this solution Helpfull? Yes (4) | No (3)  

Keshav Gupta  (11 years ago)

Approx 41 is answer...
10+131/3+5=176/3=58.66
so 58.66*41=2400

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KRISHNA KANTH YENUMULA  (11 years ago)

Here it is mentioned clearly that one minute work of truck a ---10kg,one minute work of truck b--131/3,one minute work of truck c--5kg.
let 'x' be the total time taken
=>10x+(131/3)x+5x=2.4*10^3(since 10x-- denotes number of kg's transported)
solving 58.667=2.4*10^3
=>x=40.90minutes

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rohit  (11 years ago)

450/11 MINS

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