Normal Distribution - Compare 2 proportions of a class' scores

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Jane scored in the 68th percentile on a test, John scored in the 32nd percentile. Compare quantity A with quantity B (i.e. Quantity A is greater, less or equal to quantity B) given below:

Quantity A: The proportion of the class that received a score less than John’s score.

Quantity B: The proportion of the class that scored equal to or greater than Jane’s score.

My answer was quantity B is greater than quantity A, because B has not only "greater than" but also "equal to", when A only has "less than".

However the correct answer is quantity A is equal to quantity B:

Percentiles define the proportion of a group that scores below a particular benchmark. Since John scored in the 32nd percentile, by definition, 32% of the class scored worse than John. Quantity A is equal to 32%. Jane scored in the 68th percentile, so 68% of the class scored worse than she did. Since 100 – 68 = 32, 32% of the class scored equal to or greater than Jane. Quantity B is also equal to 32%.

Thus am I wrong? Please help me understand it, thank you.

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These test scores are assumed to (approximately) follow a continuous distribution, so $P(T=t)=0$ for some test score $t$. In particular, this means $P(T>t_{0.68})=P(T\ge t_{0.68})=1-0.68=0.32$.