AMC 10 Step by Step
Log inSign up
Printable

2000 AMC 10 Problem 8

Problem onlyBack to the problem
2000 AMC 10 Problem 8AMC 10 Step by Step
Name:                             Date: September 13, 2026
Ratios, Percents & Averages · difficulty 1 of 5 · about 1 min · position P1-10

At Olympic High School, 25\frac{2}{5} of the freshmen and 45\frac{4}{5} of the sophomores took the AMC-10. Given that the number of freshmen and sophomore contestants was the same, which of the following must be true?

  1. A)

    Therearefivetimesasmanysophomoresasfreshmen.There are five times as many sophomores as freshmen.

  2. B)

    Therearetwiceasmanysophomoresasfreshmen.There are twice as many sophomores as freshmen.

  3. C)

    Thereareasmanyfreshmenassophomores.There are as many freshmen as sophomores.

  4. D)

    Therearetwiceasmanyfreshmenassophomores.There are twice as many freshmen as sophomores.

  5. E)

    Therearefivetimesasmanyfreshmenassophomores.There are five times as many freshmen as sophomores.

Answer

D

Key insight

Equal contestant counts give (2/5)f = (4/5)s, so f = 2s: the class with the smaller participation fraction must be twice as big.

Solution

Let ff be the number of freshmen and ss the number of sophomores. The contestant counts are 25f\tfrac25 f and 45s\tfrac45 s, and these are equal:
25f=45s⟹2f=4s⟹f=2s.\tfrac25 f = \tfrac45 s \quad\Longrightarrow\quad 2f = 4s \quad\Longrightarrow\quad f = 2s.

So there are twice as many freshmen as sophomores. Sanity check with numbers: 100100 freshmen send 4040 contestants, and 5050 sophomores also send 4040.

The answer is (D) There are twice as many freshmen as sophomores.\boxed{\textbf{(D)}\ \text{There are twice as many freshmen as sophomores.}}

Why this works

If a smaller fraction of one group produces the same count as a larger fraction of another, the first group must be larger, in inverse proportion to the fractions. Writing the equation removes any doubt about direction; then read f=2sf = 2s aloud as "freshmen are double the sophomores."

Alternative approach

Pick a convenient contestant count, say 4040 from each class. Freshmen: 25f=40\tfrac25 f = 40 gives f=100f = 100. Sophomores: 45s=40\tfrac45 s = 40 gives s=50s = 50. Twice as many freshmen.

The trap

Reversing the ratio and answering twice as many sophomores (B), because 4/5 is twice 2/5.

Common mistakes

  • Reversing the ratio and answering twice as many sophomores (B), because 4/5 is twice 2/5.
  • Reading "the number of contestants was the same" as "the classes are the same size" and choosing (C).

Techniques

Set up the equation/formula and compute; no special trick needed