AMC 10 · 2012 · #4
Easy mode Grade 4A bag of marbles is 53 blue, and the rest are red. You double the number of red marbles but leave the blue marbles alone. After that, what fraction of the marbles are red?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A bag of marbles is $\frac{3}{5}$ blue and the rest red. The red marbles are then doubled while the number of blue marbles stays the same. Find what fraction of the marbles are red afterward.
Givens: $\frac{3}{5}$ of the marbles are blue; The rest of the marbles are red; The number of red marbles is doubled; the number of blue marbles is unchanged; Answer choices: (A) $\frac{2}{5}$, (B) $\frac{3}{7}$, (C) $\frac{4}{7}$, (D) $\frac{3}{5}$, (E) $\frac{4}{5}$
Unknowns: The fraction of all the marbles that are red after the red marbles are doubled
Understand
Restated: A bag of marbles is $\frac{3}{5}$ blue and the rest red. The red marbles are then doubled while the number of blue marbles stays the same. Find what fraction of the marbles are red afterward.
Givens: $\frac{3}{5}$ of the marbles are blue; The rest of the marbles are red; The number of red marbles is doubled; the number of blue marbles is unchanged; Answer choices: (A) $\frac{2}{5}$, (B) $\frac{3}{7}$, (C) $\frac{4}{7}$, (D) $\frac{3}{5}$, (E) $\frac{4}{5}$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #16 Change Focus / Count the Complement, #3 Eliminate Possibilities
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) $\frac{2}{5}$ and (D) $\frac{3}{5}$, which are just the original red and blue shares.
Execute — Answer: C
4.NF.B.3 Step 1 Find the starting red fraction
- Blue and red are the only colors, so the reds are everything left after the blues.
- Take the whole bag as $1 = \frac{5}{5}$ and subtract the blue share: $\frac{5}{5} - \frac{3}{5} = \frac{2}{5}$.
- So $\frac{2}{5}$ of the marbles start out red.
💡 Two colors that fill the whole bag must have shares that add up to one whole.
3.NF.A.1 Step 2 Pick a friendly total of 5 marbles
- No real count is given, and doubling keeps the same ratios, so choose the easiest total that makes the fifths whole: $5$ marbles.
- Then $\frac{3}{5}$ of $5$ is $3$ blue, and $\frac{2}{5}$ of $5$ is $2$ red.
- Now the problem is about counting marbles instead of juggling fractions.
💡 Choosing a total that makes each fifth a whole marble turns fractions into things you can count on your fingers.
3.OA.C.7 Step 3 Double the red marbles
- Doubling the reds means $2 \times 2 = 4$ red marbles.
- The blues are untouched, so there are still $3$ blue.
- The bag now holds $4 + 3 = 7$ marbles in all.
💡 Doubling only the red pile grows the total by exactly the reds you added, not the blues.
3.NF.A.1 Step 4 Write the new red fraction
- The red fraction is reds over the new total: $\frac{4}{7}$.
- Check the traps: $\frac{2}{5}$ (A) is the old red share before doubling, and $\frac{3}{5}$ (D) is the blue share, so neither is the new red share.
- The value $\frac{4}{7}$ matches choice (C).
💡 A fraction of a group is just the part you want written over the whole new group.
4.NF.B.3 Blue and red are the only colors, so the reds are everything left after the blue 3.NF.A.1 No real count is given, and doubling keeps the same ratios, so choose the easies 3.OA.C.7 Doubling the reds means $2 \times 2 = 4$ red marbles. The blues are untouched, s 3.NF.A.1 The red fraction is reds over the new total: $\frac{4}{7}$. Check the traps: $\f Review
Reasonableness: Check the direction: doubling the reds should push the red share above its old $\frac{2}{5} = 0.4$, but the blues do not vanish, so red must land between $0.4$ and $1$. $\frac{4}{7} \approx 0.57$ sits there, with blue $\frac{3}{7} \approx 0.43$, and $\frac{4}{7} + \frac{3}{7} = 1$ as it must. That rules out (A) $\frac{2}{5}$ (no change) and (E) $\frac{4}{5}$ (too big), confirming (C).
Alternative: Use Tool #4 (Introduce a Variable) instead of a fixed count. Let the bag hold $5n$ marbles, so $3n$ are blue and $2n$ are red. Doubling the reds gives $4n$ red and still $3n$ blue, a total of $7n$. The red fraction is $\frac{4n}{7n} = \frac{4}{7}$, and the $n$ cancels, showing the real number of marbles never mattered, choice (C).
CCSS standards used (min grade 4)
4.NF.B.3Understand a fraction with numerator greater than one as sum of unit fractions (Subtracting the blue share from one whole, $\frac{5}{5} - \frac{3}{5} = \frac{2}{5}$, to get the starting red fraction.)3.NF.A.1Understand a fraction as quantity formed by parts of a whole (Reading $\frac{3}{5}$ and $\frac{2}{5}$ of $5$ marbles as $3$ and $2$, and writing the final red share as part over whole.)3.OA.C.7Fluently multiply and divide within 100 (Doubling the red count, $2 \times 2 = 4$, and totaling the new bag, $4 + 3 = 7$.)
⭐ 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: $\frac{4}{7}$.
⭐ 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: $\frac{4}{7}$.
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