Fundamentals of Mathematics Part III Solutions

Let’s denote the certain number as $( x )$.

According to the problem, if the sum of $( x )$ and $8$ is divided by $5$, the quotient is $5$. This can be written as:

$$ \frac{x + 8}{5} = 5 $$

To find $( x )$, we solve the equation step by step:

1. Multiply both sides of the equation by $5$ to eliminate the fraction:

$$ x + 8 = 5 \times 5 $$

2. Simplify the right-hand side:

$$ x + 8 = 25 $$

3. Subtract $8$ from both sides to isolate $( x )$:

$$ x = 25 - 8 $$

4. Simplify the right-hand side:

$$ x = 17 $$

Therefore, the number is $( 17 )$.

The correct answer is:

D. $17$

We are asked to find the product of $( 0.4 \times 0.04 \times 0.004 )$.

First, let’s multiply the first two numbers using fraction multiplication strategy.

$$ 0.4 \times 0.04 $$

$$ 0.4 = \frac{4}{10}, \quad 0.04 = \frac{4}{100} $$

$$ \frac{4}{10} \times \frac{4}{100} = \frac{16}{1000} $$

$$ \frac{16}{1000} = 0.016 $$

Now, multiply this result by $0.004$:

$$ 0.016 \times 0.004 $$

$$ 0.016 = \frac{16}{1000}, $$

$$ 0.004 = \frac{4}{1000} $$

$$ \frac{16}{1000} \times \frac{4}{1000} = \frac{64}{1000000} $$

$$ \frac{64}{1000000} = 0.000064 $$

Thus, the product of $( 0.4 \times 0.04 \times 0.004 )$ is $( 0.000064 )$.

The correct answer is:

D. $0.000064$.

To determine how much Mikey should pay for a ride covering $11$ kilometers, we can break down the fare calculation as follows:

1. The first four kilometers cost ₱$7.50$.
2. For the additional kilometers (beyond the first four), the cost is ₱$0.50$ per kilometer.

First, we calculate the number of additional kilometers:

$$ 11 - 4 = 7 \text{km} $$

Next, we calculate the cost for the additional kilometers:

$$ 7 \times 0.50 = P3.50 $$

Now, we can find the total fare:

$$ 7.50 + 3.50 = P11.00 $$

Thus, Mikey should pay ₱$11.00$ for a ride covering $11$ kilometers.

The correct answer is:

B. ₱$11.00$.

Let’s evaluate each expression one by one to determine which is greater than $1$.

Letter A.

$$ \frac{1}{3} - \frac{1}{4} $$

To find a common denominator ($12$):

$$ \frac{1}{3} - \frac{1}{4} = \frac{4}{12} - \frac{3}{12} = \frac{1}{12} $$

$$ \frac{1}{12} < 1 $$

Letter B.

$$ \frac{1}{3} + \frac{1}{4} $$

Using a common denominator ($12$):

$$ \frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12} $$

$$ \frac{7}{12} < 1 $$

Letter C.

$$ \frac{1}{3} \times \frac{1}{4} $$

$$ \frac{1}{3} \times \frac{1}{4} = \frac{1}{12} $$

$$ \frac{1}{12} < 1 $$

Letter D.

$$ \frac{1}{3} \div \frac{1}{4} $$

Dividing by a fraction is equivalent to multiplying by its reciprocal:

$$ \frac{1}{3} \div \frac{1}{4} = \frac{1}{3} \times 4 = \frac{4}{3} $$

$$ \frac{4}{3} > 1 $$

Conclusion: The only expression that is greater than $1$ is D..

The correct answer is:

D. $\dfrac{1}{3} \div \dfrac{1}{4} = \dfrac{4}{3}$

To determine which set contains equivalent fractions, we need to check if the fractions in each set can be simplified to the same value.

Set A: $\dfrac{2}{3}, \dfrac{4}{6}, \dfrac{8}{12}$

• Simplify $\dfrac{4}{6}$:

$$ \frac{4}{6} = \frac{2 \times 2}{2 \times 3} = \frac{2}{3} $$

• Simplify $\dfrac{8}{12}$:

$$ \frac{8}{12} = \frac{2 \times 4}{3 \times 4} = \frac{2}{3} $$

All fractions in Set A simplify to $\dfrac{2}{3}$.

Set B: $\dfrac{2}{3}, \dfrac{4}{6}, \dfrac{7}{10}$

• Simplify $\dfrac{4}{6}$:

$$ \frac{4}{6} = \frac{2 \times 2}{2 \times 3} = \frac{2}{3} $$

• $\dfrac{7}{10}$ cannot be simplified further and is not equivalent to $\dfrac{2}{3}$.

The fractions in Set B are not all equivalent.

Set C and D: $\dfrac{4}{5}, \dfrac{3}{4}, \dfrac{7}{10}$

• $\dfrac{4}{5}$ cannot be simplified further.

• $\dfrac{3}{4}$ cannot be simplified further.

• $\dfrac{7}{10}$ cannot be simplified further.

The fractions in Set C and D are not equivalent.

Therefore, the set of equivalent fractions is:

A. $\dfrac{2}{3}, \dfrac{4}{6}, \dfrac{8}{12}$

To find out how much cake was left after Karen ate $\dfrac{2}{3}$ of the $\dfrac{3}{4}$ pan of cake, we need to calculate the fraction of the cake that Karen ate and then subtract that from the original amount left by Robert.

First, let’s find out how much cake Karen ate:

Karen ate $\dfrac{2}{3}$ of $\dfrac{3}{4}$ of the cake.

To multiply these fractions, we multiply the numerators together and the denominators together:

$$ \frac{2}{3} \times \frac{3}{4} = \frac{2 \times 3}{3 \times 4} = \frac{6}{12} $$

We can simplify $\dfrac{6}{12}$ by dividing both the numerator and the denominator by their greatest common divisor, which is $6$:

$$ \frac{6}{12} = \frac{6 \div 6}{12 \div 6} = \frac{1}{2} $$

So, Karen ate $\dfrac{1}{2}$ of the cake.

Now, let’s find out how much cake was left:

Robert left $\dfrac{3}{4}$ of the cake, and Karen ate $\dfrac{1}{2}$ of the cake. To find the remaining cake, we subtract the fraction Karen ate from the fraction Robert left:

$$ \frac{3}{4} - \frac{1}{2} $$

To subtract these fractions, we need a common denominator. The least common denominator for 4 and 2 is 4. We convert $\dfrac{1}{2}$ to a fraction with a denominator of 4:

$$ \frac{1}{2} = \frac{1 \times 2}{2 \times 2} = \frac{2}{4} $$

Now we can subtract:

$$ \frac{3}{4} - \frac{2}{4} = \frac{3 - 2}{4} = \frac{1}{4} $$

So, the amount of cake left is $\dfrac{1}{4}$.

Therefore, the correct answer is:

B. $\dfrac{1}{4}$