Example 1
Food Item |
Serving size |
Servings per container |
Calories per serving |
PDV Dietary Fiber per serving |
PDV Saturated Fat per serving |
Peanut Butter |
2 Tbsp |
14 |
190 |
7% |
12% |
Soup |
1 cup |
2 |
250 |
4% |
30% |
The table shows two certain food items, peanut butter and soup. It shows their serving sizes, servings per container, calories per serving, percent of daily value (PDV) of dietary fiber per serving, and percent of daily value (PDV) of saturated fat per serving.
Erica loves peanut butter in her soup. At lunch, she adds one tablespoon of peanut butter to the entire can of soup as she is warming it. Assuming that she will eat the entire can of soup, by what percent has she increased the caloric value of her lunch soup?
Solution:
We first determine that she will consume the entire can of soup, which contains two servings. Each serving has 250 calories, so the soup alone yields 500 calories.
Now we determine the number of calories in one tablespoon of peanut butter. A tablespoon of peanut butter is half a serving, so she will add $\frac{190}{2}=95$ calories to the soup. Thus, her final concoction will contain (500 + 95) = 595 calories.
We need to calculate the percent increase of her calories by adding the peanut butter.
$\begin{array}{c}\Rightarrow \%\text{change =}\left(\frac{Final\text{\hspace{0.17em}}Value-Initial\text{\hspace{0.17em}}Value}{Initial\text{\hspace{0.17em}}Value}\right)\times 100\\ \Rightarrow \%\text{change}=\frac{595-500}{500}\times 100\%\\ \Rightarrow \%\text{change}=\frac{95}{500}\times 100\%\\ \Rightarrow \%\text{change}=19\%\end{array}$
This is a positive value, so the percent change is an increase of 19%.
Example 2
A particular triangle has a height that is twice its base. If the area of the triangle is 100, what is the length of the base?
Since the height of the triangle is twice its base, let the base be x and the height be 2x. To solve for x, we can put these variables into the area formula:
$\Rightarrow \mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}=\frac{\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e}\times \mathrm{h}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}{2}\to 100=\frac{\left(\mathrm{x}\right)\left(2\mathrm{x}\right)}{2}\to 200=2{\mathrm{x}}^{2}\to 100={\mathrm{x}}^{2}\to \mathrm{x}=10=\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e}$
Example 3
In a fishing box with red, blue, and green worms, the ratio of red worms to blue worms is 5 to 6. The ratio of red worms to green worms is 4 to 3. How many blue worms could be in the box?
Indicate all such answers.
We can let R = the number of red worms, B = the number of blue worms, and G = the number of green worms.
We know that the ratio of R to B is 5: 6 and that R to G is 4: 3.
The ratios do not share the same numbers, but they share a common variable, R. Because R is present in both ratios, we can find the LCM of the given values of R. In the first ratio, R = 5, but in the second ratio, R = 4. The LCM of 5 and 4 is 20. Multiplying the first ratio by 4 and the second ratio by 5 yields
$\Rightarrow $ R:B = 20:24 $\Rightarrow $ R:G = 20:15
Finally, we can put together the multi-part ratio, which is R : B : G = 20 : 24 : 15
We know that the number of blue worms must be a multiple of 24.
Thus the correct answer consists of Choices C (24), E (48), and G (72).
Example 4
If 24^{4 } × 16^{x+3} = 18^{2}, then x is equal to which of the following?
To determine the value of x, we must get like bases. We can prime factorize each base. Once the bases are the same, we can set the exponents equal to each other.
⇒ 24^{4 } × 16^{x+3} = 18^{2}
⇒ (3×2^{3})^{4 } × (2^{4})^{x+3} = (3^{2} × 2^{1})^{2}
⇒ 3^{4 } × 2^{12} × 2^{4x+12} = 3^{4 } × 2^{2}
The 3^{4} on the left hand side of the equation and on the right hand side of the equation cancel out and we are left with:
⇒ 2^{12} × 2^{4x+12} = 2^{2}
⇒ 2^{4x+24} = 2^{2}
Since the bases are equal we can drop the bases and determine the value of x.
⇒ 4x + 24 = 2
⇒ 4x = -22
⇒ x = $-\frac{22}{4}=-\frac{11}{2}=-5.5$
Example 5
2x + 2y = – 4 and $\frac{\mathrm{z}}{2}=\mathrm{x}+\mathrm{y}$
Quantity A:
z
Quantity B:
z^{2}
We must determine the value of z to solve for both Quantity A and Quantity B. We are given two equations: 2x + 2y = –4 and $\frac{\mathrm{z}}{2}=\mathrm{x}+\mathrm{y}$.
With these two equations we can produce a unique value for z. The first equation can be reduced to
x + y = –2. We can substitute –2 for x + y in equation two to yield $\frac{\mathrm{z}}{2}=-2\to \mathrm{z}=-4$. Since z is negative, z^{2 }is greater than z.
Thus, Quantity B is greater than Quantity A.