Quote:
Originally Posted by bhupi3149
Answer the questions on the basis of the information given below.
545 containers were shifted by a certain number of men and the job took 5 working days to complete.
Everyday after the first day 6 more men were put on the job as a result of which each man had to shift 5 fewer containers than what he shifted on the previous day.
What was the total number of containers shifted on the 3rd day?
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let there be x men on day1 and let each of these men shift y containers on day1.
based on info above on second day there were (x+6) men and each of them shifted (y-5) containers. carrying on the same way we get total number of containers shifted=
xy + (x+6)(y-5) + (x+12)(y-10) +(x+18 )(y-15) + (x+24)(y-20) =545
now x and y must be +ve integral value, hence we observe that y>20 (or else last term would become -ve) this gives a lower bound on value of y---------------(i)
simplifying the above expression we get
xy+12y-10x=289
note that both x and y must be odd or else LHS will be an even value where as RHS is odd------------------------(ii)
rearranging the terms we get
x(y-10)+12y=289
note that 12y cannot be greater than 289 because in that case x would be -ve hence y must be < 25 , this puts an upper bound on y-----------(iii)
combine (i), (ii) and (iii) to get that y can be only 21 or 23
for y=23 x=1
hence there was 1 man and he shifted 23 containers on day 1,
hence on third day
(1+12)*(23-10)=169 containers shifted on 3rd day
anytomdickandhary
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(y-15) + (x+24)(y-20) =545
Linear Mode
