Ratio 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 \(44\%\)?"<br><br>

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

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

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

\(\frac{43}{99}\): 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.<br><br>

\(\frac{8}{21}\): 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 \(44\%\).<br><br>

\(\frac{9}{20}\): 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 \(44\%\).<br><br>

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.<br><br>

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

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:<br><br>

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

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

\((j - 10) = 0.8(s - 20)\)<br><br>

\((0.75s - 10) = 0.8s - 16\)<br><br>

\(0.8s - 0.75s = 16 - 10\)<br><br>

\(0.05s = 6\)<br><br>

\(s = 120\)<br><br>

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

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:<br>4b = 3g<br>b = \( \frac{3}{4} \)g<br><br>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.<br><br>We can substitute these in the combined class equation:<br>The combined class has a boy/girl ratio of 17 to 22: (b + x)/(g + y) = \( \frac{17}{22} \)<br>(b + b – 1)/(g + g – 2) = \( \frac{17}{22} \)<br>(2b – 1)/(2g – 2) = \( \frac{17}{22} \)<br><br>Cross-multiplying yields:<br>44b – 22 = 34g – 34<br><br>Since we know that b = \( \frac{3}{4} \)g, we can replace the b:<br>44(\( \frac{3}{4} \))g – 22 = 34g – 34<br>33g – 22 = 34g – 34<br>12 = g<br><br>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:<br>B:G<br>3:4<br>6:8<br>9:12<br><br>The ratio of boys to girls in Class B is 4:5, so here are some possible numbers of boys and girls in Class B:<br>B:G<br>4:5<br>8:10<br>12:15<br><br>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.<br><br>The correct answer is E.