Algebra
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Q.

A transport agency has 5 carriers, each of capacity 15 tonnes. The carriers are scheduled such that the first carrier makes a trip every day, the second carrier makes a trip every second. The third makes a trip every third day and so on. Find the maximum number of times in a year that it is possible to dispatch a total shipment of 75 tonnes in a single day. The operation starts on the 7th January 2010 and continue till the end of the year (31st December 2010) without any holiday.

 A.

7

 B.

6

 C.

5

 D.

8

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Solution:
Option(B) is correct

The first carrier makes a trip in one day and the second carrier makes one trip in 2 days and so on.

Hence, all the trucks will leave on the same day once every $X$ days where:

$X= LCM (1,2,3,4 \text{ and } 5)=60$

If on the 7th of Jan all the trucks left together than after every 60 days they will leave together (i.e. on the same day).

There will be at most $366−6=360$ days including 7th Jan in that year.

In 360 days there can be at most 6 occasions like that.


(4) Comment(s)


Shubham Goel
 ()

Please update answer to 5



RAHUL
 ()

No 2010 is not a leap year so Nikhil Answer's is correct .l



Nikhil
 ()

2010 is not a leap year.

365-6=359days

so 5 occasion

answer is 5



Ravish
 ()

i dont think 2010 was a leap year..