Question

katie has a puppy that weighs 7 pounds. katies brother, ryan, has a puppy that is $\frac{8}{7}$ times the weight of katies puppy. which statement about the weight of ryans puppy is true?

a. the weight of ryans puppy is greater than 7 pounds because $\frac{8}{7}$ is less than 1.

b. the weight of ryans puppy is less than 7 pounds because $\frac{8}{7}$ is greater than 1.

c. the weight of ryans puppy less than 7 pounds because $\frac{8}{7}$ is less than 1.

d. the weight of ryans puppy is greater than 7 pounds because $\frac{8}{7}$ is greater than 1.

Explanation:

To determine the weight of Ryan's puppy, we first analyze the multiplier \(\frac{8}{7}\). A fraction greater than 1 (when the numerator is greater than the denominator, like \(\frac{8}{7}\) since \(8>7\)) means multiplying a number by it will result in a larger number. Katie's puppy weighs 7 pounds, so Ryan's puppy's weight is \(7\times\frac{8}{7} = 8\) pounds, which is greater than 7 pounds. Now we check the options:

• Option A: Claims \(\frac{8}{7}\) is less than 1, which is false (since \(8>7\), \(\frac{8}{7}>1\)).
• Option B: Claims the weight is less than 7 and \(\frac{8}{7}>1\), but a number multiplied by a factor greater than 1 should be larger, so this is false.
• Option C: Claims the weight is less than 7 and \(\frac{8}{7}<1\), both false.
• Option D: Correctly states the weight is greater than 7 because \(\frac{8}{7}>1\).

Answer:

D. The weight of Ryan’s puppy is greater than 7 pounds because \(\frac{8}{7}\) is greater than 1.