can any one solve this sum?
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The dimensions of an open box are 50cm, 40cm and 23 cm. Its thickness is 3 cm. If 1 cubic cmof metal used in the box weight 0.5 gms, find the weight of the box.
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can any one solve this sum?
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The dimensions of an open box are 50cm, 40cm and 23 cm. Its thickness is 3 cm. If 1 cubic cmof metal used in the box weight 0.5 gms, find the weight of the box.
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Outer Dimensions:
Given, length of box = 50 cm.
Given, breadth of box = 40 cm.
Given, height of box = 23 cm.
We know that Volume of box = l * b * h.
⇒ 50 * 40 * 23
⇒ 46000 cm^3.
Inner Dimensions:
Given thickness of the box = 3 cm.
Length of the box = Outer length - Twice the width
= 50 - (2 * 3)
= 44 cm.
Breadth of the box = Outer Breadth - Twice the width
= 40 - (2 * 3)
= 34.
Height of the box = (23 - 3) {It is an open box}
= 20 cm.
We know that Volume of the box = L * b * h
⇒ 44 * 34 * 20
⇒ 29920.
Hence, Volume of the metal used in the box = Outer volume - Inner volume
⇒ 46000 - 29920
⇒ 16080 cm^3.
Now,
Weight of the metal = (16080 * 0.5)/1000
⇒ 8.04 kg.
Therefore, weight of the box = 8.04 kg.
Hope it helps!