Presents Login Manuel ate (1)/(3) of the crackers on a plate

1. Presents Login Manuel ate (1)/(3) of the crackers on a plate. His brother ate (1)/(4) of the crackers. There were crackers left on the plate. How many crackers were on the plate to begin with?

1. 24
2. 7
3. 12

2. A guitar string produces 3 beats/s when sounded with a 350-Hz tuning fork and 8 beats/s when sounded with a 355-Hz tuning fork. What is the vibrational frequency of the string?

The correct answer and explanation is:

Problem 1: Crackers on the Plate

Given:

• Manuel ate 13\frac{1}{3} of the crackers.
• His brother ate 14\frac{1}{4} of the crackers.
• Some crackers are left on the plate.

Let xx represent the total number of crackers on the plate initially.

The portion that Manuel ate is 13x\frac{1}{3}x and the portion his brother ate is 14x\frac{1}{4}x.

The total number of crackers eaten by both is: 13x+14x\frac{1}{3}x + \frac{1}{4}x

To add these fractions, we need a common denominator. The least common denominator between 3 and 4 is 12. Therefore: 13x=412x,14x=312x\frac{1}{3}x = \frac{4}{12}x, \quad \frac{1}{4}x = \frac{3}{12}x

So, the total number of crackers eaten is: 412x+312x=712x\frac{4}{12}x + \frac{3}{12}x = \frac{7}{12}x

The number of crackers left on the plate is the remainder after the eaten portion, so: Crackers left=x−712x=512x\text{Crackers left} = x – \frac{7}{12}x = \frac{5}{12}x

We are asked to find how many crackers were on the plate initially, and it is given that there were crackers left. The answer options are:

• 24
• 7
• 12

Let’s test each option:

1. For x=24x = 24: 712×24=14(crackers eaten)\frac{7}{12} \times 24 = 14 \quad (\text{crackers eaten}) 24−14=10(crackers left)24 – 14 = 10 \quad (\text{crackers left}) So, there are 10 crackers left, which satisfies the condition. Therefore, the correct answer is 24.

Problem 2: Vibrational Frequency of the String

Given:

• The string produces 3 beats per second (beats/s) when sounded with a 350-Hz tuning fork.
• The string produces 8 beats per second (beats/s) when sounded with a 355-Hz tuning fork.

The phenomenon of beats occurs when two sound waves of slightly different frequencies interfere with each other. The number of beats per second is the absolute difference between the frequencies of the two sources.

Let fsf_s be the frequency of the string.

Case 1: When the string is sounded with a 350-Hz tuning fork: The number of beats per second is 3. This means: ∣fs−350∣=3|f_s – 350| = 3

This gives two possible equations: fs−350=3or350−fs=3f_s – 350 = 3 \quad \text{or} \quad 350 – f_s = 3

Thus: fs=353orfs=347f_s = 353 \quad \text{or} \quad f_s = 347

Case 2: When the string is sounded with a 355-Hz tuning fork: The number of beats per second is 8. This means: ∣fs−355∣=8|f_s – 355| = 8

This gives two possible equations: fs−355=8or355−fs=8f_s – 355 = 8 \quad \text{or} \quad 355 – f_s = 8

Thus: fs=363orfs=347f_s = 363 \quad \text{or} \quad f_s = 347

Now we compare the results from both cases. The only common frequency is fs=347f_s = 347 Hz.

Thus, the vibrational frequency of the string is 347 Hz.

Conclusion:

• The total number of crackers initially was 24.
• The vibrational frequency of the string is 347 Hz.