Question

valerie needs to buy ( 3\frac{4}{5} ) pounds of nuts. when the clerk places some nuts in a container and weighs it, the scale shows ( 3\frac{1}{5} ) pounds. the container weighs ( \frac{3}{20} ) pound. how many more pounds of nuts should be added to the scale to get the amount that valerie needs? explain how you solved the problem. a. subtract the amount on the scale from the amount she needs. then, subtract the weight of the container. the clerk needs to add (square) more pounds of nuts. b. subtract the weight of the container from the amount on the scale. then, add the amount she needs. the clerk needs to add (square) more pounds of nuts. c. subtract the weight of the container from the amount on the scale. then, subtract this value from the amount she needs. the clerk needs to add (square) more pounds of nuts. d. subtract the weight of the container from the amount she needs. then, subtract the amount on the scale. the clerk needs to add (square) more pounds of nuts.

Explanation:

Step1: Analyze the problem

Valerie needs \( 3\frac{4}{5} \) pounds of nuts. The scale shows \( 3\frac{1}{5} \) pounds, and the container weighs \( \frac{3}{20} \) pound. First, we need to find the weight of the nuts on the scale by subtracting the container's weight from the scale's reading. Then, subtract that from the total nuts needed.

First, find the weight of nuts on the scale: \( 3\frac{1}{5}-\frac{3}{20} \). Convert \( 3\frac{1}{5} \) to improper fraction: \( 3\frac{1}{5}=\frac{16}{5}=\frac{64}{20} \). Then \( \frac{64}{20}-\frac{3}{20}=\frac{61}{20}=3\frac{1}{20} \) pounds (nuts on scale).

Step2: Find the difference needed

Valerie needs \( 3\frac{4}{5} \) pounds, which is \( \frac{19}{5}=\frac{76}{20} \) pounds. Subtract the nuts on scale from needed: \( \frac{76}{20}-\frac{61}{20}=\frac{15}{20}=\frac{3}{4} \) pounds.

Now, let's analyze the options:

Option A: Subtract scale reading from needed (\( 3\frac{4}{5}-3\frac{1}{5}=\frac{3}{5} \)), then subtract container weight? No, that's wrong.

Option B: Subtract container from scale (\( 3\frac{1}{5}-\frac{3}{20}=3\frac{1}{20} \)), then add to needed? No, we need to subtract from needed.

Option C: Subtract container from scale (\( 3\frac{1}{5}-\frac{3}{20}=3\frac{1}{20} \)), then subtract this from needed (\( 3\frac{4}{5}-3\frac{1}{20}=\frac{3}{4} \)). This matches our calculation.

Option D: Subtract container from needed (\( 3\frac{4}{5}-\frac{3}{20}=3\frac{13}{20} \)), then subtract scale? No, that's incorrect.

So the correct option is C. Now, for the first blank in option C: subtract container from scale (\( 3\frac{1}{5}-\frac{3}{20} \)), the result is \( 3\frac{1}{20} \)? Wait, no, the first blank in option C: "Subtract the weight of the container from the amount on the scale. Then, subtract this value from the amount she needs. The clerk needs to add [blank] more pounds of nuts." Wait, no, let's re-express the steps for option C:

1. Subtract container weight from scale reading: \( 3\frac{1}{5}-\frac{3}{20} \)
2. Subtract this result from the needed amount (\( 3\frac{4}{5} \)) to find how much more to add.

First, calculate \( 3\frac{1}{5}-\frac{3}{20} \):

\( 3\frac{1}{5}=\frac{16}{5}=\frac{64}{20} \), \( \frac{64}{20}-\frac{3}{20}=\frac{61}{20}=3\frac{1}{20} \)

Then, \( 3\frac{4}{5}-3\frac{1}{20}=\frac{19}{5}-\frac{61}{20}=\frac{76}{20}-\frac{61}{20}=\frac{15}{20}=\frac{3}{4} \)

So the first blank in option C (the amount to add after first subtraction? Wait, no, the option C says: "Subtract the weight of the container from the amount on the scale. Then, subtract this value from the amount she needs. The clerk needs to add [blank] more pounds of nuts." Wait, no, the structure is:

- Step 1: Subtract container from scale: let's call this \( N \) (nuts on scale)
- Step 2: Subtract \( N \) from needed: \( 3\frac{4}{5}-N \), which is the amount to add.

So the first blank in option C (the "more pounds" after "add") is \( \frac{3}{4} \), but let's check the option structure again. Wait, the option C is: "Subtract the weight of the container from the amount on the scale. Then, subtract this value from the amount she needs. The clerk needs to add [blank] more pounds of nuts." Wait, no, maybe the first blank is the result of subtracting container from scale, but no, the options have two blanks? Wait, the original problem has two blanks in each option? Wait, looking back:

Option C: "Subtract the weight of the container from the amount on the scale. Then, subtract this value from the amount she needs. The clerk needs to add [blank] more pounds of nuts." Wait, no, the user's image shows:

Option C: "Subtract the weight of the container from the amount on the scale. Then, subtract this value from the amount she needs. The clerk needs to add [blank] more pounds of nuts." Wait, maybe the first blank is the result of subtracting container from scale, but no, let's re-express:

Wait, the problem is a multiple-choice with blanks. Let's focus on the correct option first. The correct option is C. Now, let's find the values:

First, subtract container from scale: \( 3\frac{1}{5}-\frac{3}{20} \). Convert \( 3\frac{1}{5} \) to \( \frac{16}{5}=\frac{64}{20} \), so \( \frac{64}{20}-\frac{3}{20}=\frac{61}{20}=3\frac{1}{20} \). Then subtract this from needed: \( 3\frac{4}{5}-3\frac{1}{20} \). Convert \( 3\frac{4}{5} \) to \( \frac{19}{5}=\frac{76}{20} \), so \( \frac{76}{20}-\frac{61}{20}=\frac{15}{20}=\frac{3}{4} \). So the first blank in option C (the value after subtracting container from scale) is \( 3\frac{1}{20} \)? Wait, no, the option C says: "Subtract the weight of the container from the amount on the scale. Then, subtract this value from the amount she needs. The clerk needs to add [blank] more pounds of nuts." Wait, maybe the first blank is the result of subtracting container from scale, and the second blank is the result of subtracting that from needed. But the question is to select the correct option (C) and fill the blanks.

Wait, the correct option is C. Now, for the first blank in option C (the value to subtract from the amount she needs), which is the weight of the nuts on the scale, calculated as \( 3\frac{1}{5}-\frac{3}{20}=3\frac{1}{20} \) pounds. Then, subtracting this from \( 3\frac{4}{5} \) gives \( 3\frac{4}{5}-3\frac{1}{20}=\frac{3}{4} \) pounds, which is the amount to add.

So the correct option is C. The first blank (the value after subtracting container from scale) is \( 3\frac{1}{20} \), and the second blank (the amount to add) is \( \frac{3}{4} \). But let's confirm with the calculation:

1. Weight of nuts on scale: \( 3\frac{1}{5}-\frac{3}{20}=3\frac{4}{20}-\frac{3}{20}=3\frac{1}{20} \) (wait, \( 3\frac{1}{5} \) is \( 3 + \frac{1}{5} = 3 + \frac{4}{20} \), so \( 3\frac{4}{20}-\frac{3}{20}=3\frac{1}{20} \), correct).

2. Amount needed to add: \( 3\frac{4}{5}-3\frac{1}{20}=3\frac{16}{20}-3\frac{1}{20}=\frac{15}{20}=\frac{3}{4} \), correct.

So the correct option is C, with the first blank being \( 3\frac{1}{20} \) and the second blank being \( \frac{3}{4} \).

Answer:

C. Subtract the weight of the container from the amount on the scale. Then, subtract this value (\( 3\frac{1}{20} \)) from the amount she needs. The clerk needs to add \( \frac{3}{4} \) more pounds of nuts.