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Mathematics

The external dimensions of an open rectangular wooden box are 98 cm by 84 cm by 77 cm. If the wood is 2 cm thick all around, find

(i) the capacity of the box

(ii) the volume of the wood used in making the box, and

(iii) the weight of the box in kilograms correct to one decimal place, given that 1 cm3 of wood weighs 0.8 g.

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Answer

It is given that

External dimensions of open rectangular wooden box = 98 cm, 84 cm and 77 cm

Thickness = 2 cm

So, the internal dimensions of open rectangular wooden box = (98 - 2 × 2) cm, (84 - 2 × 2) cm and (77 - 2) cm

= (98 - 4) cm, (84 - 4) cm, 75 cm

= 94 cm, 80 cm, 75 cm.

(i) We know that,

Capacity of the box = Internal volume of box = 94 cm × 80 cm × 75 cm

= 564000 cm3.

Hence, capacity of the box = 564000 cm3.

(ii) Internal volume of box = 564000 cm3

External volume of box = 98 cm × 84 cm × 77 cm = 633864 cm3.

Volume of wood used in making the box = External volume - Internal volume

= 633864 – 564000 = 69864 cm3.

Hence, volume of box used in making the box = 69864 cm3.

(iii) Weight of 1 cm3 wood = 0.8 gm

So the weight of 69864 cm3 wood = 0.8 × 69864 gm

= 0.8×698641000\dfrac{0.8 \times 69864}{1000} kg

= 55891.21000\dfrac{55891.2}{1000} kg

= 55.89 kg = 55.9 kg.

Hence, weight of box = 55.9 kg.

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