Question

Feb 17 A zoo feeds its monkeys a mixture of fruits and biscuits. The diagram shows the ratio of fruit n mass to biscuit mass. Ratios with tape diagrams The table shows the mass, in grams, of fruit and biscuits that the zoo fed two of the monkeys. Monkey A Fruit mass Based on the ratio, complete the missing values in the table. B Biscuit mass Fruit mass (g) Biscuit mass (g) 28 45

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Answer 1

Fruit mass = 63.2 g

Biscuit mass = 94.8 g

Explanation:

Since the ratio of fruit mass to biscuit mass is 2:3, the fraction of fruit mass to the total of fruit and biscuit mass is 2/5 and the fraction of biscuit mass to the total of fruit and biscuit mass is 3/5.

For Monkey A, the total of fruit and biscuit mass is 73 g, which means that the fraction of fruit mass is 2/5 × 73 g = 29.2 g and the fraction of biscuit mass is 3/5 × 73 g = 43.8 g. Therefore, the values for fruit mass and biscuit mass are:

Fruit mass = 29.2 g

Biscuit mass = 43.8 g

For Monkey B, the total of fruit and biscuit mass is 73 + 85 = 158 g, which means that the fraction of fruit mass is 2/5 × 158 g = 63.2 g and the fraction of biscuit mass is 3/5 × 158 g = 94.8 g. Therefore, the values for fruit mass and biscuit mass are:

Fruit mass = 63.2 g

Biscuit mass = 94.8 g

Answer 2

In Monkey A's case, the missing biscuit mass is 20 grams, calculated using the given fruit-to-biscuit ratio of 7:5. For Monkey B, the missing fruit mass is 63 grams, applying the same ratio to the provided biscuit mass of 45 grams.

Let's break down the solution step by step:

Understanding the Ratio

According to the figure, the ratio of fruit mass to biscuit mass is 7:5.

Completing the Table for Monkey A

In the first part, the monkey A's fruit mass is given as 28, and the biscuit mass is missing. Using the ratio 7:5, we can set up the equation:

`\[\frac{\text{Fruit Mass}}{\text{Biscuit Mass}} = (7)/(5)\]`

Substitute the given values:

`\[\frac{28}{\text{Biscuit Mass}} = (7)/(5)\]`

Solving for Biscuit Mass:

`\[\text{Biscuit Mass} = (28 * 5)/(7) = 20\]`

Completing the Table for Monkey B

In the second part, monkey B's biscuit mass is given as 45, and the fruit mass is missing. Use the ratio 7:5 to set up the equation:

`\[\frac{\text{Fruit Mass}}{45} = (7)/(5)\]`

Solving for Fruit Mass:

`\[\text{Fruit Mass} = (45 * 7)/(5) = 63\]`

The missing value for monkey A's biscuit mass is 20, and for monkey B's fruit mass is 63, based on the given ratio of 7:5.

The complete question is:

A zoo feeds its monkeys a mixture of fruits and biscuits. The diagram shows the ratio of fruit mass to biscuit mass.

The table below shows the mass, in grams, of fruit and biscuits that the zoo fed two of the monkeys.

Based on the ratio, complete the missing value in the tables.