Question

Find the volume of iron required to construct an open box whose external dimensions are 36cm,25cm,and 16.5cm,provided that the thickness of the metal is 1.5cm.If1cm cube of iron weight 7.5g find the weight of the box.

Answer 1

outside volume
= 36(25)(16.5) cm^3
= 14850 cm^2

volume of inner part (the empty space)
= 33(22)(13.5) cm^2
= 9801 cm^3

volume of iron = 14850 - 9801
= 5049 cm^3

1 cm^3 weighs 7.5 kg
so weight of box = .....

Answer 2

External volume of box= (36*25*16.5)cm cube
= 14850 cm cube
Again,
Internal volume of box where,
Length= (36 - (1.5*2)) = (36-3) = 33 cm
Breadth= (25-(1.5*2)) = (25 - 3)= 22 cm
Height = (16.5 - (1.5 *2)) =(16.5 - 3) = 13.5 cm
then inner volume = (33*22*13.5) = 9801cm cube
Volume of iron = (14850 - 9801) = 5049cm cube
1cm cube weight 7.5 g ,then weight of 5049cm cube = 37867.5 g.-ANS.

Answer 3

Math but iam a student of class8 in manav bharti national school - Shubham Raj from Gaya, Thursday, January 5, 2017 at 9:56pm
External volume of box= (36*25*16.5)cm cube
= 14850 cm cube
Again,
Internal volume of box where,
Length= (36 - (1.5*2)) = (36-3) = 33 cm
Breadth= (25-(1.5*2)) = (25 - 3)= 22 cm
Height= (16.5 - 1.5) = 15 cm
Inner volume = ( 33*22*15) = 10890cm cube
Volume of iron = (14850 - 10890) = 3960 cm cube.
Weight of box = (3960 * 7.5)g = 29700g or (29700/1000) kg.