AREAS AND VOLUMES 351

Example.- -Find the space required to stow—

(i) 100 rolls paper, length 36 in., diameter 32 in.
(ii) 400 tons sugar in bags, stowage factor 42
(iii) 200 bales, 4 ft. X2 ft. X2 ft. 6 in.
(iv) 1000 cases, stowage factor 60

If the freight is 35s. per shipping ton measurement less 5 per cent,
required the net freight.

/•x T> i «? 22 16X16 36 1

(i) Paper, volume * r* I =~ X — — X y X — g

= 1676 cub. ft.

(ii) Sugar volume 400x42 =16,800

(iii) Bales „ 200x4x2x2-5= 4000
(iv) Cases „ 1000x60 X 60,000

Divide by 40)82,476 cub. ft.
Shipping tons measurement 2061*9 tons

2062 tons, at 35s, .... £3608 10 0

less 5 per cent. - - = 180 8 6

Net freight ..... £3428 1 6

Example. — A compartment measures 10,000 cubic feet and is to be
filled with a total weight of 400 tons made up of bales stowing at 60
cubic feet per ton and pig lead stowing at 10 cubic feet per ton.
Required the maximum tons and cubic capacities of each that could be
stowed in the compartment.

This involves a simultaneous equation. Let x = the bales and y
the lead, then (i) x + y = 400 tons weight

and (ii) 60 x + 10 y = 10,000 tons measurement

multiply (i) by 10 10 x + 10 y = 4000

subtract 50 x = 6000

x =120 tons bales

x + =400

y = 280 tons lead

120 cons bales X 60 = 7200 cubic feet
280 tons lead X 10 = 2800 „