Osmosys
Company
Numerical Ability
Age Problem
In order to construct a wall of length of 10m,height 6m and thickness 22cm. Brick dimensions are long:22cm, Breadth:10cm, Thickness:6cm. In B/w each brick a cement and sand is mixture is placed with 0.5cm and
In b/w each row mixture of cement and sand with 1cm is placed. Cement to sand ratio is 2:1 and each row requires 1.5kg of mixture Condition: Bricks are placed in horizantal way only. Find total number of Bricks, Weight of Cement and Sand used?
Read Solution (Total 2)
-
- hi it is very simple .. calculate area of wall and area of brick area of wall 1000cm*600cm*22cm=1000*600*22 cm^3 area of brick=22*10*6cm^3 now divide area of wall /area of brick then we will get no of bricks 1000*600*22/22*6*10= 10000 bricks they asked for gap so . in order to find gap bw bricks length of wall/brick =1000/22=45.45(no of bricks in row) 45.45*0.5=22.5cm extra cm so subtract 1000-22.5=977.25 similarly height 572.23 now then calculate area 977.25*572.25*22/1320 that eqals 9328 bricks i think u may got idea
- 10 years agoHelpfull: Yes(10) No(1)
- Any other problems bro Plzz help.
- 8 years agoHelpfull: Yes(1) No(0)
Osmosys Other Question