Question

A merchant has three kinds of wine, of the first kind of 403 gallons, of the second 527 gallons and of the third 589 gallons. what is the least number of full casks of equal size in which this can be stored without mixing?​

Answer 1

49 casks of equal size will be required to store the wine.

Step-by-step explanation:

Amount of 1st kind of wine = 403 gallons

Amount of 2nd kind of wine = 527 gallons

Amount of 3rd kind of wine = 589 gallons

To get the highest capacity of the container we will find the H.C.F. of all amounts of the wine.

H.C.F. of 403, 527 and 589 = 31

Now the number of full casks required = $\frac{\text{Total amount of wine}}{\text{H.C.F. of amounts}}$

= $\frac{403+527+589}{31}$

= $\frac{1519}{31}$

= 49

Therefore, number of casks required will be 49.

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