The outer dimensions of a cubical room are 8m*6m*4m and inner dimension are 7.8m*5.8m*4m (roof is no

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16 Apr 2015 Number of bricks

The outer dimensions of a cubical room are 8m*6m*4m and inner dimension are 7.8m*5.8m*4m (roof is not included in the dimension). If the size of brick used (including cement) are 20cm*10cm*8cm, then number of bricks used to construct wall will be

OPtion

1) 7100
2) 6800
3) 6900
4) 6400
5) 6700
6) 5700
7) 6200
8) 6500
9) 7300
10)7500

Solution

Total volume = 8*6*4 = 192 m^3
Total inner volume = 7.8*5.8*4 = 180.96 m^3
So volume of bricks = 192-180.96 m^3
= 11.04 m^3
Number of bricks
11.04 / .2*.1*.08
= 110400/16 = 6900
Option 3)

Correct Option: 3 Best Solution (14)

SAURABH KUMAR SINHA   9 years AGO

Empty Volume=800*600*400-780*580*400=11040000

no of bricks=11040000÷(20*10*8)=6900 ANS..

Like? Yes (7) | No (3)  

Supriya Mondal   9 years AGO

Volume of the room respect to outer dimension is=8*6*4=192 m^3
And volume of the room respect to inner dimension is=7.8*5.8*4=180.96 m^3
So, difference between two volumes=192-180.96=11.04 m^3
Volume of each brick is=0.2*0.1*0.08=1.6*10^-3 m^3
So, numbers of bricks needed=(11.04/1.6)*10^3=6900
So, option 3)6900 is the right answer

Like? Yes (3) | No   

MRITUNJAY   9 years AGO

volume of cubical room=External -internal
=8*6*4-7.8*5.8*4=11.04m^3=11040000cm^3
volume of 1 brick=20*10*8=1600 cm^3
number of bricks=11040000/1600=6900

Like? Yes (2) | No   

Yathiraj   9 years AGO

Height of the Brick = 8cm
Height of the walls = 4m => 400cm
Hence, no. of lines of bricks required in all the 4 walls = 400/8 => 50 lines

If the inner dimension between the walls is less by 0.2m then brick thickness on each wall will be 0.1m => 10cm
Hence, the bricks will be layed down along 20cm for each wall.

2 walls of 6m => 600cm
No. of bricks per line => 600/20 => 30
Hence, no. of bricks on one wall = 30x50 => 1500. For 2 such walls => 1500*2 => 3000 ---- 1

2 other walls 7.8m (less brick on either sides) => 780m
no. of bricks per line => 780/20 => 39
Hence, no. of bricks on one wall = 39x50 => 1950. For 2 such walls => 1950*2 => 3900 ----- 2

1 + 2 => 3000 + 3900 => 6900

Option 3)

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sompon temtheerakij   9 years AGO

2*(40+29)*50 = 6900

Like? Yes (1) | No   

syed kashif ullah   9 years AGO

volume of outer - volume of inner= volume of bricks
volume of bricks = 8*6*4- 7.8*5.8*4
number of brucks = volume of bricks/volume of one brick=6900...0option 3.)

Like? Yes (1) | No   

Nandha kumar   9 years AGO

The volume of outer dimension is 192000000.
The volume of inner dimension is 180960000.
The difference between them is 11040000.
The volume of brick is 1600.
Dividing 11040000 by 1600 the ans is 6900.
Thus the no.of bricks used to construct the wall is 6900.
The answer is option 3)6900.

Like? Yes (1) | No   

Mohanraj K   9 years AGO

outer volume of cube=8*6*4=192 cum
inner volume of cube =7.8*5.8*4=180.96 cum

The volume of cube occupied by brick=192-180.96=11.04 cum
volume of single brick=0.2*0.1*0.08=0.0016 cum
So,number of bricks =0.0016*X=11.04cum
so X=6900(option 3)

Like? Yes (1) | No   

kranthi kanchanapally   9 years AGO

The outer volume is=8*6*4=192m^3
inner volume is=7.8*5.8*4=180.96m^3
then the volume is=11.04m^3
the size of the bricks is=20*10*8=1600
THen the total number of bricks used in a wall is=
volume of the room/volume of the brick
11040000/1600=6900

Like? Yes (1) | No   

Paresh   9 years AGO

Volume of outer dimensions=800*600*400
Volume of inner dimensions=780*580*400
net volume=outer-inner
no of bricks=(800*600*400)-(780*580*400)/20*10*8=6900

Like? Yes (1) | No   

vivek   9 years AGO

(outer volume- inner volume)/ volume of brick= 11040000/1600=6900

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sanjeet   9 years AGO

Total volume of WAll (in cm)=(800*600*400) - (780*580*400)
=11040000 cm^3
total volume of bricks= 20*10*8=1600 cm^3
so total number of bricks require for building=11040000/1600=6900

Like? Yes (2) | No (2)  

aman aditya   9 years AGO

11040000/(20*10*8)= 6900

Like? Yes (1) | No (1)  

sangrambadi sahoo   9 years AGO

800*600*400)-780*580*400)/1600=6900

Like? Yes (1) | No (1)