, 2015

The total number of students at School A is the sum of the number of students who received a ride and the number of students who did not, or $330+420=750$.

Since 330 of 750 students received a ride, the percentage of students at School A who received a ride is $\frac{330}{750}\times 100\%=0.44\times 100\%=44\%$.
Solution 1

At school B, 30% of 240 students or $\frac{30}{100}\times240=\frac{7200}{100}=72$ students received a ride.

If 50% or $\frac{1}{2}$ of the students in School B were to receive a ride, then $\frac{1}{2}\times240=120$ students would get a ride.

Therefore, $120-72=48$ more students needed to receive a ride so that 50% of the students in School B got a ride.

Solution 2

As a percent, the difference between 50% of students receiving a ride and the 30% of students who did receive a ride is 20%.

Therefore, 20% of 240 students or $\frac{20}{100}\times240=\frac{4800}{100}=48$ additional students would need to receive a ride so that 50% of students in School B got a ride.
Solution 1

At school C, 45% of 200 students or $\frac{45}{100}\times200=\frac{9000}{100}=90$ students received a ride.

At school D, $x\%$ of 300 students or $\frac{x}{100}\times300=\frac{300x}{100}=3x$ students received a ride.

The total number of students at School C and School D is $200+300=500$.

The total number of students receiving a ride at School C and School D is $90+3x$.

Since 57.6% of the combined group of students from the two schools received a ride, then $\frac{90+3x}{500}=\frac{57.6}{100}$.

Multiplying both sides of this equation by 500, we get $90+3x=57.6\times5$ or $90+3x=288$ or $3x=198$ and so $x=66$.

Solution 2

At school C, 45% of 200 students or $\frac{45}{100}\times200=\frac{9000}{100}=90$ students received a ride.

The total number of students at School C and School D is $200+300=500$.

Since 57.6% of the combined group of students from the two schools received a ride, then $\frac{57.6}{100}\times500=\frac{28800}{100}=288$ students received a ride.

Out of the 288 students who received a ride, 90 students were from School C and so the remaining $288-90=198$ students were from School D.

Since there are 300 students at School D, the percentage of students receiving a ride is $\frac{198}{300}\times 100\%=0.66\times 100\%=66\%$.

Therefore, the value of $x$ is 66.
At school E, $n\%$ of 200 students or $\frac{n}{100}\times200=\frac{200n}{100}=2n$ students received a ride.

At school F, $2n\%$ of 250 students or $\frac{2n}{100}\times250=\frac{500n}{100}=5n$ students received a ride.

The total number of students at School E and School F is $200+250=450$.

The total number of students receiving a ride at School E and School F is $2n+5n=7n$.

Between 55% and 60% of the 450 students from the two schools received a ride.

Since 55% of 450 is 247.5 and 60% of 450 is 270, then $7n>247.5$ and $7n<270$.

Solving $7n>247.5$ we get $n>35.35$, after rounding, and $7n<270$ gives $n<38.57$, after rounding.

Since $n$ is a positive integer and $n>35.35$ and $n<38.57$, then the possible values of $n$ are $36,37$ and 38.