AMC 10 · 2007 · #3
Easy mode Grade 5A glass tank has a rectangular bottom that is 100 cm long and 40 cm wide, and it stands 50 cm tall. Water fills it to a depth of 40 cm. A solid brick shaped like a box, 40 cm long, 20 cm wide, and 10 cm tall, is set on the bottom of the tank. By how many centimeters does the water level rise?
Pick an answer.
AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: An aquarium with a rectangular base $100$ cm by $40$ cm and height $50$ cm holds water $40$ cm deep. A solid brick measuring $40$ cm by $20$ cm by $10$ cm is set on the bottom. Find how many centimeters the water surface rises.
Givens: The tank's base is $100$ cm by $40$ cm and it is $50$ cm tall; The water starts $40$ cm deep; The brick is a solid box $40$ cm by $20$ cm by $10$ cm; The brick is placed on the bottom of the tank; Answer choices: (A) $0.5$, (B) $1$, (C) $1.5$, (D) $2$, (E) $2.5$
Unknowns: The number of centimeters by which the water level rises after the brick is placed in
Understand
Restated: An aquarium with a rectangular base $100$ cm by $40$ cm and height $50$ cm holds water $40$ cm deep. A solid brick measuring $40$ cm by $20$ cm by $10$ cm is set on the bottom. Find how many centimeters the water surface rises.
Givens: The tank's base is $100$ cm by $40$ cm and it is $50$ cm tall; The water starts $40$ cm deep; The brick is a solid box $40$ cm by $20$ cm by $10$ cm; The brick is placed on the bottom of the tank; Answer choices: (A) $0.5$, (B) $1$, (C) $1.5$, (D) $2$, (E) $2.5$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #8 Analyze the Units
A quick cross-section drawing (Tool #1, Draw a Diagram) shows the key fact: the brick sits fully under the water and shoves aside a chunk of water equal to its own volume, and that water can only go up, forming a thin slab across the entire base. Once that picture is clear, the number splits into two clean subproblems (Tool #7): first the brick's volume, then the height of a slab with that volume spread over the tank's base. Tool #8 (Analyze the Units) confirms the last step, since a volume in cm$^3$ divided by a base area in cm$^2$ must give a height in cm.
Execute — Answer: D
5.MD.C.3 Step 1 See what the brick does to the water
- Sketch the tank from the side.
- The water is $40$ cm deep, and the brick is only $10$ cm tall, so once it rests on the bottom the whole brick is under water.
- A solid object under water takes up space that water used to fill, so it pushes that water out of the way.
- The pushed water has nowhere to go but up, and it spreads out across the full base of the tank, lifting the surface everywhere by the same small amount.
💡 A solid sunk in water shoves aside exactly its own volume of water.
5.MD.C.5 Step 2 Find the brick's volume
- The brick is a rectangular box, so its volume is length times width times height.
- Multiply the three side lengths: $40\times20\times10=8000$.
- So the brick takes up $8000$ cubic centimeters, and that is exactly how much water gets pushed upward.
💡 The volume of a box is just how many unit cubes fill it: length by width by height.
5.NBT.B.6 Step 3 Spread that volume across the base
- The lifted water forms a thin slab that covers the whole base of the tank.
- The base area is $100\times40=4000$ square centimeters.
- This slab has volume $8000$ cm$^3$, so its height $h$ satisfies (base area) times height $=$ volume, that is $4000\times h=8000$.
- Dividing, $h=8000\div4000=2$.
- The water rises $2$ cm, which is choice (D).
💡 Volume divided by the base area gives the height, since cm$^3\div$ cm$^2=$ cm.
5.MD.C.3 Sketch the tank from the side. The water is $40$ cm deep, and the brick is only 5.MD.C.5 The brick is a rectangular box, so its volume is length times width times height 5.NBT.B.6 The lifted water forms a thin slab that covers the whole base of the tank. The b Review
Reasonableness: The answer is small and positive, which fits — a modest brick should nudge a wide tank's water up only a little. Check the size: the brick's footprint is $40\times20=800$ cm$^2$, one fifth of the tank's $4000$ cm$^2$ base, and the brick is $10$ cm tall, so spreading that $10$ cm of solid over a base five times wider gives about $10\times\tfrac{800}{4000}=2$ cm — matching. Also the tank does not overflow: the level goes from $40$ cm to $42$ cm, still under the $50$ cm rim, so the whole brick really does stay submerged and the reasoning holds. Answer $2$ cm is (D).
Alternative: Use a variable and conservation of water. The amount of water never changes: it starts at $4000\times40=160000$ cm$^3$. After the brick goes in, the water fills the tank up to height $40+h$ everywhere except the space taken by the brick, so $4000\times(40+h)-8000=160000$. Expanding gives $160000+4000h-8000=160000$, so $4000h=8000$ and $h=2$, the same answer.
CCSS standards used (min grade 5)
5.MD.C.3Recognize volume as an attribute of solid figures (Understanding that the submerged brick occupies volume and displaces an equal volume of water upward.)5.MD.C.5Relate volume to the operations of multiplication and addition (Computing the brick's volume as $40\times20\times10=8000$ cm$^3$.)5.NBT.B.6Find whole-number quotients with up to four-digit dividends and two-digit divisors (Dividing the displaced volume by the base area to get the rise: $8000\div4000=2$.)
⭐ A sunk object pushes up its own volume of water, so spread that volume over the tank's base to find how high the water climbs.
⭐ A sunk object pushes up its own volume of water, so spread that volume over the tank's base to find how high the water climbs.
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