• # question_answer A tank internally measuring $150\,\,cm\times 120\text{ }cm\times 100\text{ }cm$ has $1281600\text{ }c{{m}^{3}}$water in it Porous bricks are placed in the water until the tank is full up to its brim. Each brick absorbs one tenth of its volume of water. How many bricks, of $20\text{ }cm\times 6\text{ }cm\times 4\text{ }cm$, can be put in the tank without spilling over the water? A)  1150                           B)  1175         C)  1200    D)  1250

(c): Let n bricks can be put in the tank without spilling over the water. According to the question the volume of the tank should be totally occupied by the available water and the bricks. Also, since the question tells us that each brick absorbs 10% of its own volume of water, the additional volume added to the current water by each brick added to the tank would only by 90% of the brick's own volume. Let the number of bricks required by n. Then, $150\times 120\times 100=n\times 20\times 6\times 4$ $\left( 1-\frac{10}{100} \right)+1281600$ $150\times 120\times 100-1801600=n\times 20\times 6\times 4\times 0.9$ $n=\frac{518400}{20\times 6\times 4\times 0.9}=1200$ Note: This problem requires mental thinking on the part of student as only 90% of volume of bricks is equal to available volume since 10% of water has gone into pores of the bricks.