Question

A tank internally measuring 150 cm × 120 cm × 100 cm has 1281600 cm\(^3\) 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 size 20 cm × 6 cm × 4 cm, can be put in the tank without spilling over the water?

• A. 1100
• B. 1200
• C. 1150
• D. 1250
• E. None of the above

Hint:

In such problems, always adjust the effective volume of objects when they absorb or displace some liquid. Here, each brick absorbs 48 cm\(^3\) water, so it effectively occupies only 432 cm\(^3\) inside the tank.

Answer 1

The Correct Option is B

Step 1: Calculate total capacity of the tank.
Tank dimensions: 150 cm × 120 cm × 100 cm.
Therefore, volume of the tank = \(150 \times 120 \times 100 = 18{,}00{,}000 \; \text{cm}^3\).
Step 2: Calculate water already in the tank.
Water volume = 1281600 cm\(^3\).
Step 3: Calculate remaining capacity to be filled by bricks.
Remaining volume = Total tank volume − Water volume.
= \(18{,}00{,}000 - 12{,}81{,}600 = 5{,}18{,}400 \; \text{cm}^3\).
So the combined “effective volume” of bricks that can be put inside must equal 5,18,400 cm\(^3\).
Step 4: Volume of one brick.
Brick dimensions = 20 cm × 6 cm × 4 cm.
So, volume of one brick = \(20 \times 6 \times 4 = 480 \; \text{cm}^3\).
Step 5: Effect of absorption.
Each brick absorbs one tenth of its own volume in water.
So, each brick absorbs = \(480 \div 10 = 48 \; \text{cm}^3\).
Thus, when one brick is put into the tank, it effectively occupies only
= Brick volume − absorbed water
= \(480 - 48 = 432 \; \text{cm}^3\).
Step 6: Number of bricks that can be placed.
Let number of bricks = \(x\).
Then total effective volume occupied = \(432 \times x\).
We require:
\(432x = 5{,}18{,}400\).
So, \(x = 5{,}18{,}400 \div 432 = 1200\).
Step 7: Conclude.
Hence, the maximum number of bricks that can be placed in the tank without spilling water is 1200. \[ \boxed{1200 \; \text{(Option B)}} \]