R Practice Questions

First, let us rephrase the question. Since we need to find the fraction that is at least twice greater than \(\frac{11}{50}\), we are looking for a fraction that is equal to or greater than \(\frac{22}{50}\). Further, to facilitate our analysis, note that we can come up with an easy benchmark value for this fraction by doubling both the numerator and the denominator and thus expressing it as a percent: \(\frac{22}{50} = \frac{44}{100} = 44\%\). Thus, we can rephrase the question: "Which of the following is greater than or equal to \(44\%\)?"

Now, let's analyze each of the fractions in the answer choices using benchmark values:

\(\frac{2}{5}\): This fraction can be represented as \(40\%\), which is less than \(44\%\).

\(\frac{11}{34}\): This value is slightly less than \(\frac{11}{33}\) or \(\frac{1}{3}\). Therefore, it is smaller than \(44\%\).

\(\frac{43}{99}\): Note that the fraction \(\frac{43}{99}\) is smaller than \(\frac{44}{100}\), since fractions get smaller if the same number (in this case integer \(1\)) is subtracted from both the numerator and the denominator.

\(\frac{8}{21}\): We know that \(\frac{8}{21}\) is a little less than \(\frac{8}{20}\) or \(\frac{2}{5}\). Thus, \(\frac{8}{21}\) is less than \(44\%\).

\(\frac{9}{20}\): Finally, note that by multiplying the numerator and the denominator by \(5\), we can represent this fraction as \(\frac{45}{100}\), thus concluding that this fraction is greater than \(44\%\).

The correct answer is E.

Let's denote the number of juniors and seniors at the beginning of the year as \(j\) and \(s\), respectively.

At the beginning of the year, the ratio of juniors to seniors was \(3\) to \(4\): \(\frac{j}{s} = \frac{3}{4}\). Therefore, \(j = 0.75s\)

At the end of the year, there were \((j - 10)\) juniors and \((s - 20)\) seniors. Additionally, we know that the ratio of juniors to seniors at the end of the year was \(4\) to \(5\). Therefore, we can create the following equation:

\(\frac{j-10}{s-20} = \frac{4}{5}\)

Let's solve this equation by substituting \(j = 0.75s\):

\((j - 10) = 0.8(s - 20)\)

\((0.75s - 10) = 0.8s - 16\)

\(0.8s - 0.75s = 16 - 10\)

\(0.05s = 6\)

\(s = 120\)

Thus, there were \(120\) seniors at the beginning of the year.

The correct answer is E.

The ratio of boys to girls in Class A is 3 to 4. We can represent this as an equation: b/g = \( \frac{3}{4} \). We can isolate the boys:
4b = 3g b = (\( \frac{3}{4} \))g
Let's call the number of boys in Class B x, and the number of girls in Class B y. We know that the number of boys in Class B is one less than the number of boys in Class A. Therefore, x = b – 1. We also know that the number of girls in
Class B is two less than the number of girls in Class A. Therefore, y = g – 2.
We can substitute these in the combined class equation:
The combined class has a boy/girl ratio of 17 to 22: (b + x)/(g + y) = \( \frac{17}{22} \). (b + b – 1)/(g + g – 2) = \( \frac{17}{22} \)
(2b – 1)/(2g – 2) = \( \frac{17}{22} \)

Cross-multiplying yields:
44b – 22 = 34g – 34
Since we know that b = (\( \frac{3}{4} \))g, we can replace the b:
44(\( \frac{3}{4} \))g – 22 = 34g – 34
33g – 22 = 34g – 34
12 = g
Alternatively, because the numbers in the ratios and the answer choices are so low, we can try some real numbers. The ratio of boys to girls in Class A is 3:4, so here are some possible numbers of boys and girls in Class A:
B:G
3:4
6:8
9:12

The ratio of boys to girls in Class B is 4:5, so here are some possible numbers of boys and girls in Class A:
B:G
4:5
8:10
12:15
We were told that there is one more boy in Class A than Class B, and two more girls in Class A than Class B. If we look at our possibilities above, we see that this information matches the case when we have 9 boys and 12 girls in Class A and 8 boys and 10 girls in Class B. Further, we see we would have 9 + 8 = 17 boys and 12
+ 10 = 22 girls in a combined class, so we have the correct 17:22 ratio for a combined class. We know now there are 12 girls in Class A.

The correct answer is E.