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2000 AMC 10 Problem 4

2000 AMC 10 Problem 4AMC 10 Step by Step

Name: Date: September 13, 2026

Linear Equations & Word Problems · difficulty 1 of 5 · about 2 min · position P1-10

Chandra pays an on-line service provider a fixed monthly fee plus an hourly charge for connect time. Her December bill was $\$12.48$ , but in January her bill was $\$17.54$ because she used twice as much connect time as in December. What is the fixed monthly fee?

A)
$ 2.53
B)
$ 5.06
C)
$ 6.24
D)
$ 7.42
E)
$ 8.77

Answer

Key insight

The January bill differs from December only by one extra unit of connect-time cost, so the difference of the bills is the variable part.

Solution

Let $F$ be the fixed fee and $C$ the cost of December's connect time. December: $F + C = 12.48$ . January used twice the time, so $F + 2C = 17.54$ .

Subtracting the first equation from the second gives $C = 17.54 - 12.48 = 5.06$ . Then $F = 12.48 - 5.06 = 7.42$ .

The fixed monthly fee is $\boxed{\textbf{(D)}\ \$7.42}$ .

Why this works

A bill with a fixed part and a proportional part is a linear function of usage. Comparing two bills whose usage differs by a known factor isolates the proportional part; the fixed part is what remains. Whenever a word problem describes "fixed plus per-unit," write $F + kC$ and compare.

Alternative approach

Check the choices: if the fee were $\$7.42$ , December's time cost $\$5.06$ and January's would be $\$10.12$ , giving $\$17.54$ . Only (D) makes both bills work.

The trap

Solving for the connect-time charge (5.06) and reporting it instead of the fixed fee.

Common mistakes

Solving for the connect-time charge (5.06) and reporting it instead of the fixed fee; 5.06 appears as choice (B) precisely to catch this.
Setting January as "double the whole bill" instead of double the connect time only.

Techniques

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