MHB Two elementary algebra problems

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To hold 1,000,000 inflated balloons, approximately 49 classrooms are needed, calculated by dividing the total balloon volume by the volume of a typical classroom, which is 20,250 ft³. For the second problem, a stack of 1 million $1 bills would be about 119 yards high, determined by dividing the total number of bills by the number of bills per inch and converting inches to yards. The discussion emphasizes the need for clear problem-solving methods and encourages users to avoid duplicate posts. Overall, the thread provides solutions to both algebra problems while reinforcing proper forum etiquette.
jangoom
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#1. How many classrooms would be necessary to hold 1,000,000 inflated balloons? (Assume one balloon is about 1 ft3 and a typical classroom is about 30 ft × 45 ft × 15 ft. Round your answer to the nearest number of classrooms.)

#2. Approximately how high would a stack of 1 million \$1 bills be? (Assume there are 233 new \$1 bills per inch. Round your answer to the nearest yard.

Its been years! I need an answer and how you got there please.. much appreciated.
 
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I've re-titled this thread and moved it into the appropriate forum. Also, I've deleted another thread of yours as it duplicates a question found here. In future, please do not post duplicates, post in the appropriate forum and give your threads a descriptive title. Thanks! :)

Using the method we have used in your other thread, can you now attempt these questions?
 
Well I'm still lost. Is it the same method? Can you show me again how you do it.. I need a learn it...
 
jangoom said:
1. How many classrooms would be necessary to hold 1,000,000 inflated balloons?
Assume one balloon is about 1 ft³ and a typical classroom is about 30 ft x 45 ft x 15 ft.
Round your answer to the nearest number of classrooms.
A classroom holds: 30 \times 45\times 15 \:=\:20,250 \text{ ft}^3

To hold a million balloons requires: \frac{1,000,000}{20.250} \:=\:49.3827...\;\approx\; 49\text{ classrooms.}


#2. Approximately how high would a stack of 1 million \$1 bills be?
Assume there are 233 new \$1 bills per inch. Round your answer to the nearest yard.
\frac{1,000,000}{233} \:=\:4291.8454... \text{ inches}

\text{Divide by 36; }\;119.2179... \; \approx\; 119\text{ yards}



 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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