For how many days are rations expected to last in a camp of 100 soldie

Let \(t\) be total number of rations initially and \(r\) be daily consumption per soldier

Statement 1 : (1) If 100 additional soldiers join the camp after 13 days, each consuming the same quantity of ration as the initial soldiers, the rations would last another 12 days.

Rations consumed during first 13 days = \(13 * 100 * r\)
so \((t - 1300r)/113r = 12\)
rearranging, \(t = 2656r\)

days rations can last = \(2656 r / 100 r\) (daily consumption by all soldiers) = 27 days

Sufficient

Statement 2: If 20 soldiers leave the camp in the very beginning, the rations would last for 25 percent more number of days.

Let t be number of days it can last.

as per this statement, \(100 * r * t = 80 * r * (5/4) t\) => \(100 * r * t = 100 * r * r\) => this statement does not say anything new => InSuff

Answer (A)

@Bunnel chetan2u
can you provide the official explanation?
I think i am doing wrong here in formulating the equation

Let the total time taken to be taken by 100 soldiers be: T
Statement 1
let time after 13 days time taken by 200 soldiers to complete the remaining ration = a
13*100+ 200a = 100T+12*200
1300+200a = 100(a+13)+ 2400
200a=100a+2400
100a=2400
a=24
hence total time = 24+13 =37
sufficient
Statement B
100T = 80*1.25T
Hence insufficient
Answer A
Kindly clarify

For how many days are rations expected to last in a camp of 100 soldiers? Assume that each soldier consumes equal quantities of rations per day.

(1) If 100 additional soldiers join the camp after 13 days, each consuming the same quantity of ration as the initial soldiers, the rations would last another 12 days.
a) logically...
since they eat same amount of food, if 200 eat some quantity for 12 days, 100 will eat same quantity for 12*2=24
so the ration will last for 13+24=37 days
b) algebraic method
total days if counted for 100 initially and then 200 = \(13*100+200*12=1300+2400=3700\)
total days when counted on 100 soldiers taking them as x days = 100x
\(100x=3700.........x=37\)
Raksat algebraic should be on these lines.
sufficient

(2) If 20 soldiers leave the camp in the very beginning, the rations would last for 25 percent more number of days.
let x be the days initialy so 100x but with 20 moving out remaining use it for 1.25x
so 100x=1.25x*80......100x=100x
it will always be true....
reduction in 20 out of 100 or 20% would always increase days by 25%
insuff

A

Let the consumption rate per soldier be 1 unit of ration per day.

Hence 100 soldiers will consume 100 units of ration per day.

Let n be the # of days the ration can last for 100 soldiers. Hence Total ration for 100 soldiers is 100n

Statement 1: For 13 days ration consumed by 100 soldiers is 13*100 = 1300 units

Remaining ration = 100n - 1300

100 additional soldiers, along with 100 original consume the remaining ration in 12 days.

Hence 200*12 = 100n -1300

n = 27

Statement 1 is Sufficient.


Statement 2: 20 soldiers leave the camp at the beginning. Hence only 80 soldiers left to consumer 100n units of ration.

80 soldiers consume 80 units per day, therefore # of days the ration lasts for them = 100n/80 =5n/4 days

the # of days given is 25% more, which is 1.25n = 5n/4 days

we get same information, No additional information provided.

Statement 2 is Not Sufficient.



Answer A.



Thanks,
GyM

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