AMC 8 Daily Practice - Solid Figures

Complete problem set with solutions and individual problem pages

Problem 8 Medium

There is a sufficiently deep water tank with a rectangular base of length 16cm and width 12cm. Originally, the tank contains water with a depth of 6cm and oil (above the water) with a depth of 6cm. An iron block with length 8cm, width 8cm, and height 12cm is placed into the tank. What is the height of the oil layer at this time, in centimeters?

  • A.

    6

  • B.

    6.5

  • C.

    7

  • D.

    8

  • E.

    9

Answer:C

After placing the iron block, the base area of the water layer becomes: 16 \times 12 - 8 \times 8 = 128 \text{ square centimeters}.

The height of the water layer changes to: \frac{16 \times 12 \times 6}{128} = 9 \text{ centimeters}.

This indicates that a 9-centimeter height of the iron block is submerged in water, and a 3-centimeter height of the iron block is immersed in oil.

The volume of the oil layer increases by: 3 \times 8 \times 8 = 192 \text{ cubic centimeters}.

The increased height of the oil layer is: \frac{192}{16 \times 12} = 1 \text{ centimeter}.

Thus, the height of the oil layer at this time is: 6 + 1 = 7 \text{ centimeters}.

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