(a) Let x be the price of each large flag in dollars. Then the price of each small flag is (2/3)x dollars. Frida bought 5 flags for a total of 5x dollars. We know that 5x = 24 because she spent $24. Solving for x, we get x = 4.80. Therefore, the price of each small flag is (2/3)($4.80) = $3.20.
(b) The flag is 4 feet tall, which is 48 inches. Each stripe takes up 1/13 of the total height, so each stripe is (1/13)(48) = 3.69 inches tall. This can be written as a mixed number as 3 11/13 inches tall.
(c) Janine's height and the distance between her and the flagpole form a right triangle, where the flagpole is the vertical side and Janine's shadow is the horizontal side. The ratio of the lengths of the vertical side to the horizontal side is the same as the ratio of Janine's height to the length of her shadow. In other words:
20 / x = 5 / (x - 6'3" or 75 inches)
Cross-multiplying and solving for x, we get x = 24 feet. Therefore, the length of the flagpole's shadow is 24 - 6'3" or 18'9" or 225 inches.
(a) A stand at a local parade is selling flags for observers to wave. The small flags each cost 2/3 the price of each large flag, and each flag costs a whole number of dollars. If Frida bought 5 flags and spent $24.00, how much do each of the small flags cost in dollars?
(b) The American flag has 13 horizontal stripes. If the flag outside of Victor’s house is 4 feet tall and 6 feet wide, how many inches tall is each of the 13 stripes? Express your answer as a mixed number.
(c) On a bright, sunny day, Janine is standing 6 feet 3 inches away from the base of a 20-foot-tall flagpole. She notices that her shadow just touches the base of the flagpole. If Janine is exactly 5 feet tall, how long is the shadow of the flagpole at this same time?
…………
3 answers
(a) wrong, since each flag costs a whole number of dollars.
If each flag costs a whole number of dollars, we need to find two whole numbers that satisfy the conditions in the problem. Let x be the price of each large flag in dollars. Then the price of each small flag is (2/3)x dollars. Since each flag costs a whole number of dollars, we know that x must be a multiple of 3. We also know that 5 flags cost $24.
Let's try some values of x that are multiples of 3:
- If x = 3, then the price of each small flag is (2/3)($3) = $2, and the total cost of 5 flags is $13, which is too low.
- If x = 6, then the price of each small flag is (2/3)($6) = $4, and the total cost of 5 flags is $26, which is too high.
- If x = 9, then the price of each small flag is (2/3)($9) = $6, and the total cost of 5 flags is $30, which is too high.
- If x = 12, then the price of each small flag is (2/3)($12) = $8, and the total cost of 5 flags is $40, which is too high.
We can continue trying larger values of x, or we can notice that if x = 15, then the price of each small flag is (2/3)($15) = $10, and the total cost of 5 flags is $50, which is too high. Therefore, the only possibility is that x = 18.
If x = 18, then the price of each small flag is (2/3)($18) = $12, and the total cost of 5 flags is $60, which is too high. Therefore, x cannot be 18.
We can try x = 21, but it's clear that this will be too high since the total cost of 5 flags will be at least $70. Therefore, we conclude that there is no solution to this problem if each flag must cost a whole number of dollars.
Let's try some values of x that are multiples of 3:
- If x = 3, then the price of each small flag is (2/3)($3) = $2, and the total cost of 5 flags is $13, which is too low.
- If x = 6, then the price of each small flag is (2/3)($6) = $4, and the total cost of 5 flags is $26, which is too high.
- If x = 9, then the price of each small flag is (2/3)($9) = $6, and the total cost of 5 flags is $30, which is too high.
- If x = 12, then the price of each small flag is (2/3)($12) = $8, and the total cost of 5 flags is $40, which is too high.
We can continue trying larger values of x, or we can notice that if x = 15, then the price of each small flag is (2/3)($15) = $10, and the total cost of 5 flags is $50, which is too high. Therefore, the only possibility is that x = 18.
If x = 18, then the price of each small flag is (2/3)($18) = $12, and the total cost of 5 flags is $60, which is too high. Therefore, x cannot be 18.
We can try x = 21, but it's clear that this will be too high since the total cost of 5 flags will be at least $70. Therefore, we conclude that there is no solution to this problem if each flag must cost a whole number of dollars.
1 answer