ASVAB/PICAT Arithmetic Reasoning Test: Tank Capacity Example

ASVAB/PICAT Arithmetic Reasoning Test: Maximum Water Capacity in Tanks

Problem Overview

  • We have two tanks, each partially filled with water.
  • Tank A: 3/4 full.
  • Tank B: 1/2 full.
  • Each tank currently has 60 gallons of water.

Objective

  • Determine the maximum amount of water that can be held in both tanks together.

Analyzing the Problem

  • Full Capacity Calculation for Each Tank:
    • If Tank A is 3/4 full with 60 gallons, then the total capacity of Tank A can be calculated as follows:
    • ext{Capacity of Tank A} = rac{60}{ rac{3}{4}} = 60 imes rac{4}{3} = 80 ext{ gallons}
    • For Tank B, which is 1/2 full with 60 gallons:
    • ext{Capacity of Tank B} = rac{60}{ rac{1}{2}} = 60 imes 2 = 120 ext{ gallons}

Total Capacity Calculation

  • Total Maximum Capacity of Both Tanks:
    • Combine the capacity of both tanks:
    • extTotalCapacity=extCapacityofTankA+extCapacityofTankBext{Total Capacity} = ext{Capacity of Tank A} + ext{Capacity of Tank B}
    • extTotalCapacity=80+120=200extgallonsext{Total Capacity} = 80 + 120 = 200 ext{ gallons}

Answer Choices

  • A) 180 gallons
  • B) 200 gallons
  • C) 210 gallons
  • D) 220 gallons

Conclusion

  • Based on the calculations, the maximum amount of water that can be held in both tanks together is:
    • B) 200 gallons.

Additional Calculations (Correlating Data)

  • The values presented on the page appear to represent a set of calculations or logical reasoning to derive values, but the specific numerical relationships are unclear.
  • Variables such as "x" could represent unknowns related to tank volumes or other operations, although they are not further explained in the transcript. Clarifications may be needed for further contextual understanding.