AMC 10 · 2020 · #2

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

mean-median-mode-rangelinear-equations-one-varwork-backwards work-backwardsidentify-subproblems ↑ Prerequisites: mean-median-mode-range

📏 Short solution 💡 2 insights

Problem

The numbers $3, 5, 7, a,$ and $b$ have an average (arithmetic mean) of $15$ . What is the average of $a$ and $b$ ?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: Five numbers — $3$, $5$, $7$, $a$, and $b$ — have an average of $15$. Find the average of just $a$ and $b$.

Givens: The five numbers are $3, 5, 7, a, b$; Their arithmetic mean is $15$; Answer choices: (A) $0$, (B) $15$, (C) $30$, (D) $45$, (E) $60$

Unknowns: The average of $a$ and $b$ alone

Understand

Restated: Five numbers — $3$, $5$, $7$, $a$, and $b$ — have an average of $15$. Find the average of just $a$ and $b$.

Givens: The five numbers are $3, 5, 7, a, b$; Their arithmetic mean is $15$; Answer choices: (A) $0$, (B) $15$, (C) $30$, (D) $45$, (E) $60$

Plan

Primary tool: #11 Work Backwards

Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities

Tool #11 (Work Backwards) is the spine: the given is the *end product* (the average $15$), and we need the *missing pieces* ($a$ and $b$). Undo the average by multiplying back: total sum $= 15 \times 5 = 75$. Tool #7 (Subproblems) splits the work cleanly — first the sum of all five, then the sum of $a + b$, then the average. Tool #3 (Eliminate) is a sanity check: since the known three $(3, 5, 7)$ are all *below* the average of $15$, the missing two must be substantially *above* $15$ — so (A) $0$ and (B) $15$ are impossible.

Execute — Answer: C

#11 Work Backwards 6.SP.A.3 Step 1

Undo the average.
If five numbers have mean $15$, their total sum is $5 \times 15 = 75$.

$$\text{sum of all five} = 5 \times 15 = 75$$

💡 Mean $\times$ count $=$ total — the definition of average run in reverse.

#7 Identify Subproblems 4.OA.A.3 Step 2

Subtract the three known numbers to find $a + b$.
We have $3 + 5 + 7 = 15$, so $a + b = 75 - 15 = 60$.

$$3 + 5 + 7 = 15, \quad a + b = 75 - 15 = 60$$

💡 Split "sum of five" into "sum of three knowns" plus "sum of two unknowns", then subtract.

#11 Work Backwards 6.SP.A.3 Step 3

Now take the average of just $a$ and $b$: divide their sum by $2$.
$(a + b)/2 = 60/2 = 30$.

$$\dfrac{a + b}{2} = \dfrac{60}{2} = 30 \;\Rightarrow\; \textbf{(C)}$$

💡 Average of two numbers $=$ their sum divided by $2$.

#3 Eliminate Possibilities 6.SP.A.3 Step 4

Quick elimination check.
The three known values $3, 5, 7$ sit $12, 10, 8$ below the target average $15$, a total of $30$ below.
The other two numbers must make up that deficit, so they average $30/2 = 15$ above $15$, i.e.
$30$.
This confirms (C) without redoing the arithmetic.

$$(15{-}3) + (15{-}5) + (15{-}7) = 30 \;\Rightarrow\; \text{avg of}\ a, b = 15 + 15 = 30$$

💡 Balancing deviations above and below the mean must total zero.

[1] #11 6.SP.A.3 Undo the average. If five numbers have mean $15$, their total sum is $5 \times 1

[2] #7 4.OA.A.3 Subtract the three known numbers to find $a + b$. We have $3 + 5 + 7 = 15$, so $

[3] #11 6.SP.A.3 Now take the average of just $a$ and $b$: divide their sum by $2$. $(a + b)/2 =

[4] #3 6.SP.A.3 Quick elimination check. The three known values $3, 5, 7$ sit $12, 10, 8$ below

Review

Reasonableness: Sanity: if $a + b = 60$ then the five-number sum is $3 + 5 + 7 + 60 = 75$, and $75 / 5 = 15$. The original condition holds. Magnitude-wise, $30$ is larger than $15$ which makes sense because $a$ and $b$ have to push the low values $3, 5, 7$ up to a mean of $15$.

Alternative: Tool #6 (Guess and Check) with a tidy pair, e.g. $a = b = 30$: then five-number sum $= 3 + 5 + 7 + 30 + 30 = 75$, average $= 15$ $\checkmark$. Same answer (C) $30$. Any symmetric pair around $30$ (like $20$ and $40$) also works because only the sum matters.

CCSS standards used (min grade 6)

4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Splitting the total sum of $5$ numbers into a sum of $3$ knowns plus a sum of $2$ unknowns, then subtracting.)
6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number (Using the mean-as-balance definition both forward (sum $= $ mean $\times$ count) and to compute the average of $a$ and $b$.)

⭐ This AMC 10 problem only needs Grade 6 "the mean is the total split evenly" you already know — multiply back to $75$, take out $3 + 5 + 7 = 15$, and the missing pair sums to $60$, so their average is $30$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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