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(#M40010520) TCS question mensuration Keep an EYE

Tbe dimensions of an open box are 50 cm, 40 cm and 23 cm. Its thickness is 2 cm. If 1 cubic cm of metal used in the box weighs 0.5 gms, find the weight of the box.

Asked In TCS harmohit (12 years ago)

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sainath  (12 years ago)

external volume of box is 50*40*23 cm^3
remember the box is open, so if we consider inner part of box,
length will be reduced to 46, because each side 2cm thickness we have so 50-4.
similarly width also reduced to 36 (40-4).
but coming to height top is open, so height will be reduced only from bottom to 21 (23-2).
so internal measurements are 46,36,21
internal volume of box is 46*36*21
so metal volume is external volume-internal volume
ie., (50*40*23)-(46*36*21)
=(2*25*2*20*23)-(2*23*2*18*21)
=(4*25*20*23)-(4*23*18*21)
=4*23*(25*20-18*21)
=4*23*(500-378)
=4*23*122 cm^3

each cm^3 it's of 1/2 gm weight, so total weight of the box is
=4*23*122/2=2*23*122=46*122=5612gms...

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sainath  (12 years ago)

if you want it's simply
volume of metal is external volume of cube-internal volume of cube=(50*40*23)-(46*36*21)=46000-34776
=11224 cm^3
weight of the metal=11224 cm^3 * 0.5 gm/cm^3 =5612 gm or 5.612 kg

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Deepak Kanike  (12 years ago)

External volume of box=(l*b*h)=(50*40*23)=46000cc
box is 2cm thick and open from top
hence, new l=(50-2-2)=46cm
new b=(40-2-2)=36cm
new h=(23-2) =21cm
Internal volume of box=(46*36*21)=34776cc

Metal volume of box=46000-34776=11224cc

Given Density of metal=0.5g/cc
hence, weight of box =(0.5g/cc)*(11224cc)= 5612g = 5.612kg

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pavan pavvi  (12 years ago)

metal is used on the outerpart(circumference of box) oly.. so we have to consider the volume of the outer layer of box
outer volume-inner voluue
=50*40*23-46*36*21 (21 coz one side of box is open )
=11224
11224*0.5=5612 g is ans

Is this solution Helpfull? Yes (4) | No   

varma  (12 years ago)

external volume of box is 50*40*23 cm^3
remember the box is open, so if we consider inner part of box,
length will be reduced to 46, because each side 2cm thickness we have so 50-4.
similarly width also reduced to 36 (40-4).
but coming to height top is open, so height will be reduced only from bottom to 21 (23-2).
so internal measurements are 46,36,21
internal volume of box is 46*36*21
so metal volume is external volume-internal volume
ie., (50*40*23)-(46*36*21)
=(2*25*2*20*23)-(2*23*2*18*21)
=(4*25*20*23)-(4*23*18*21)
=4*23*(25*20-18*21)
=4*23*(500-378)
=4*23*122 cm^3

each cm^3 it's of 1/2 gm weight, so total weight of the box is
=4*23*122/2=2*23*122=46*122=5612gms...

Is this solution Helpfull? Yes (4) | No   

piyush   (12 years ago)

volume of box (50*40*23)-(46*36*19)=14536
no of cube required = 14536/1*1*1=14536
weight of box=14536*0.5=7268 gm= 7.2 kg

Is this solution Helpfull? Yes (3) | No (4)  

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