AMC 10 · 2012 · #4

Grade 4 rate-ratio
ratio-proportionfraction-arithmetic easier-related-problem ↑ Prerequisites: fraction-arithmetic
📏 Medium solution 💡 2 insights
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Problem
One colour is doubled while the other stays put. Find the new share of that colour.

Pick an answer.

(A)
$\frac{2}{5}$
(B)
$\frac{3}{7}$
(C)
$\frac{4}{7}$
(D)
$\frac{3}{5}$
(E)
$\frac{4}{5}$
How to solve
Strategy Solve an Easier Related Problem

The fractions are abstract and no real count is given, so Tool #9 (Solve an Easier Related Problem) replaces them with a concrete total of 5 marbles, turning fractions into objects you can count. Tool #16 (Count the Complement) gets the starting red share as everything that is not blue. Tool #3 (Eliminate Possibilities) guards the traps (A) 2/5 and (D) 3/5, which are just the original red and blue shares.

1STEP 1

Find the starting red fraction

The other colour starts at 2/5.

1 - 3/5 = 5/5 - 3/5 = 2/5
2STEP 2

Pick a friendly total of 5 marbles

A friendly total makes both counts whole.

5 marbles → 3 blue, 2 red
3STEP 3

Double the red marbles

Doubling changes the total too.

2 × 2 = 4 red, 3 blue, 4 + 3 = 7 total
4STEP 4

Write the new red fraction

The new share is 4/7, choice (C).

4/(4+3) = 4/7 → (C)
Answer
4/7
Check the direction: doubling the reds should push the red share above its old 2/5 = 0.4, but the blues do not vanish, so red must land between 0.4 and 1. 4/7 ≈ 0.57 sits there, with blue 3/7 ≈ 0.43, and 4/7 + 3/7 = 1 as it must. That rules out (A) 2/5 (no change) and (E) 4/5 (too big), confirming (C).
💡Key takeaway

When a problem gives only fractions, pick a small total that makes them whole — here 5 marbles — count what happens, then read off the new fraction: 4/7.

  • Find the starting red fraction
  • Pick a friendly total of 5 marbles
  • Double the red marbles
  • Write the new red fraction