AMC 8 · 2014 · #7

Easy mode Grade 6

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Problem

Ms. Raub's class has $28$ students in total.

The class has $4$ more girls than boys.

What is the ratio of girls to boys in the class?

$\textbf{(A) }3 : 4\qquad\textbf{(B) }4 : 3\qquad\textbf{(C) }3 : 2\qquad\textbf{(D) }7 : 4\qquad \textbf{(E) }2 : 1$

Pick an answer.

(A)

3 : 4

(B)

4 : 3

(C)

3 : 2

(D)

7 : 4

(E)

2 : 1

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Toolkit + CCSS Solution

Understand

Restated: Ms. Raub's class has $28$ students total, and there are exactly $4$ more girls than boys. Find the ratio of girls to boys in lowest terms.

Givens: Total students $= 28$; Girls $=$ boys $+ 4$; Answer choices: (A) $3:4$, (B) $4:3$, (C) $3:2$, (D) $7:4$, (E) $2:1$

Unknowns: The ratio (girls $:$ boys), simplified

Understand

Restated: Ms. Raub's class has $28$ students total, and there are exactly $4$ more girls than boys. Find the ratio of girls to boys in lowest terms.

Givens: Total students $= 28$; Girls $=$ boys $+ 4$; Answer choices: (A) $3:4$, (B) $4:3$, (C) $3:2$, (D) $7:4$, (E) $2:1$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #6 Guess and Check

The class has $4$ extra girls on top of an otherwise equal split. Tool #7 (Identify Subproblems) lets us peel off those $4$ extra girls first, so the remaining $28 - 4 = 24$ students split evenly into $12$ boys and $12$ girls. Adding the $4$ back gives the counts in one clean step. We keep Tool #6 (Guess and Check) on hand because it's the natural way to double-check the answer against the multiple-choice ratios: only $16:12$ satisfies both "sums to $28$" and "differs by $4$". Algebra (Tool #13) would also work, but for two unknowns connected by a sum and a difference, splitting off the extra is faster and more intuitive.

Execute — Answer: B

#7 Identify Subproblems 2.OA.A.1 Step 1

Set aside the $4$ extra girls.
Imagine removing $4$ girls from the class so that boys and girls would be equal.
That leaves $28 - 4 = 24$ students who split evenly.

$$28 - 4 = 24$$

💡 Subtracting the "extra" first is the classic subproblem move: turn an unequal split into a fair one.

#7 Identify Subproblems 3.OA.A.2 Step 2

Split the $24$ evenly between boys and girls.
Half of $24$ is $12$, so there are $12$ boys (and also $12$ girls before we put the extras back).

$$24 \div 2 = 12 \text{ boys}$$

💡 Dividing a total into two equal groups is a Grade 3 "partitive division" idea.

#7 Identify Subproblems 2.OA.A.1 Step 3

Put the $4$ extra girls back to find the actual number of girls.

$$12 + 4 = 16 \text{ girls}$$

💡 Once the equal split is found, the extras only land in the girls' column.

#7 Identify Subproblems 6.RP.A.1 Step 4

Write the ratio girls $:$ boys and simplify by dividing both parts by their greatest common factor, $\gcd(16, 12) = 4$.

$$\text{girls} : \text{boys} = 16 : 12 = \dfrac{16}{4} : \dfrac{12}{4} = 4 : 3 \;\Rightarrow\; \textbf{(B)}$$

💡 A ratio in lowest terms divides both numbers by their greatest common factor, just like reducing a fraction.

[1] #7 2.OA.A.1 Set aside the $4$ extra girls. Imagine removing $4$ girls from the class so that

[2] #7 3.OA.A.2 Split the $24$ evenly between boys and girls. Half of $24$ is $12$, so there are

[3] #7 2.OA.A.1 Put the $4$ extra girls back to find the actual number of girls.

[4] #7 6.RP.A.1 Write the ratio girls $:$ boys and simplify by dividing both parts by their grea

Review

Reasonableness: Check both conditions on $16$ girls and $12$ boys. Total: $16 + 12 = 28$ ✓. Difference: $16 - 12 = 4$ ✓. The ratio $16:12 = 4:3$ also makes sense as a "slightly more girls than boys" answer — not as lopsided as $2:1$ (which would mean $\tfrac{2}{3}$ of the class is girls, i.e. about $19$ girls) and clearly more than $1:1$.

Alternative: Tool #6 (Guess and Check) on the choices: each ratio $g:b$ with $g + b = 28$ gives a candidate split. (A) $3:4$ would mean fewer girls than boys — wrong direction. (B) $4:3$ scales to $16:12$, total $28$, difference $4$ ✓. (C) $3:2$ scales to $\tfrac{3}{5}(28) : \tfrac{2}{5}(28)$, which isn't a whole number. (D) $7:4$ doesn't divide $28$ evenly either. (E) $2:1$ gives $\tfrac{56}{3}$ girls — not a whole number. Only (B) survives.

CCSS standards used (min grade 6)

2.OA.A.1 Use addition and subtraction within $100$ to solve word problems (Subtracting the $4$ extra girls to get $28 - 4 = 24$, and adding them back with $12 + 4 = 16$.)
3.OA.A.2 Interpret whole-number quotients as partitioning into equal shares (Splitting the $24$ remaining students into $2$ equal groups: $24 \div 2 = 12$.)
6.RP.A.1 Understand the concept of a ratio and use ratio language to describe a relationship (Forming the ratio girls $:$ boys $= 16:12$ and simplifying it to lowest terms $4:3$.)

⭐ Once you set the $4$ extra girls aside, the rest of the class splits in half — and the ratio idea you need is right at the Grade 6 level.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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