Ratio Practice Questions

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 B:
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.

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.

We know that the student to teacher ratio at the school is 16 to 1, and the total number of people is 510. Therefore:
Number of students = (\( \frac{16}{17} \))(510) = 480
Number of teachers = (\( \frac{1}{17} \))(510) = 30

Kindergarten students make up \( \frac{1}{5} \) of the student population, so:
Number of kindergarten students = (\( \frac{1}{5} \))(480) = 96

Fifth and sixth graders account for \( \frac{1}{3} \) of the remainder (after kindergarten students are subtracted from the total), therefore:
Number of 5th and 6th grade students = (\( \frac{1}{3} \))(480 – 96) = (\( \frac{1}{3} \))(384) = 128

Students in first and second grades account for \( \frac{1}{4} \) of all the students, so:
Number of 1st and 2nd grade students = (\( \frac{1}{4} \))(480) = 120
So far, we have accounted for every grade but the 3rd and 4th grades, so they must consist of the students left over:
Number of 3rd and 4th grade students = Total students – students in other grades
Number of 3rd and 4th grade students = 480 – 96 – 128 – 120 = 136

If there are an equal number of students in the third and fourth grades, then:
Number of 3rd grade students = \( \frac{136}{2} \) = 68

The number of students in third grade is 68, which is fewer than 96, the number of students in kindergarten. The number of students in 3rd grade is thus 96 – 68 = 28 fewer than the number of kindergarten students.

The correct answer is C.