AMC 8 · 1999 · #3

Grade 7 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

fraction-arithmeticsystematic-enumerationmental-arithmetic systematic-enumeration ↑ Prerequisites: fraction-arithmeticmulti-digit-arithmetic

📏 Medium solution 💡 2 insights

Problem

Which triplet of numbers has a sum NOT equal to 1?

Pick an answer.

(A)

(1/2,1/3,1/6)

(B)

(2,-2,1)

(C)

(0.1,0.3,0.6)

(D)

(1.1,-2.1,1.0)

(E)

(-3/2,-5/2,5)

AMC 8 1999 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Five triplets of numbers are listed. Four of them have a sum of $1$. Find the one triplet whose sum is NOT equal to $1$.

Givens: Triplet (A): $\left(\tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{6}\right)$; Triplet (B): $(2, -2, 1)$; Triplet (C): $(0.1, 0.3, 0.6)$; Triplet (D): $(1.1, -2.1, 1.0)$; Triplet (E): $\left(-\tfrac{3}{2}, -\tfrac{5}{2}, 5\right)$

Unknowns: Which one of the five triplets has a sum that is NOT $1$

Understand

Restated: Five triplets of numbers are listed. Four of them have a sum of $1$. Find the one triplet whose sum is NOT equal to $1$.

Plan

Primary tool: #3 Eliminate Possibilities

The question hands you five labeled candidates and asks which one fails a single, easy-to-check property (sum equals $1$). That is exactly Tool #3 (Eliminate Possibilities): walk through the candidates, compute the sum of each, and cross off any whose sum is $1$. The one that survives the elimination is the answer. No algebra or pattern-hunting is needed — every triplet is just three numbers to add.

Execute — Answer: D

#3 Eliminate Possibilities 5.NF.A.1 Step 1

Test (A).
Add the fractions using a common denominator of $6$.

$\tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{6} = \tfrac{3}{6} + \tfrac{2}{6} + \tfrac{1}{6} = \tfrac{6}{6} = 1$. Sum $= 1$, so (A) is eliminated.

💡 Adding fractions with unlike denominators by rewriting with a common denominator is the Grade 5 standard.

#3 Eliminate Possibilities 7.NS.A.1 Step 2

Test (B).
Add the signed integers left to right.

$2 + (-2) + 1 = 0 + 1 = 1$. Sum $= 1$, so (B) is eliminated.

💡 Adding a positive and its opposite gives $0$ — the Grade 7 additive-inverse idea.

#3 Eliminate Possibilities 5.NBT.B.7 Step 3

Test (C).
Add the decimals (each is a tenth).

$0.1 + 0.3 + 0.6 = 1.0$. Sum $= 1$, so (C) is eliminated.

💡 Adding decimals to the hundredths is the Grade 5 standard; here all three numbers already line up at the tenths place.

#3 Eliminate Possibilities 7.NS.A.1 Step 4

Test (D).
Add the signed decimals.

$1.1 + (-2.1) + 1.0 = -1.0 + 1.0 = 0$. Sum $= 0$, NOT $1$. (D) survives the elimination.

💡 $1.1$ and $-2.1$ differ in size by exactly $1.0$, so adding them gives $-1.0$, which then cancels the last $+1.0$ to leave $0$.

#3 Eliminate Possibilities 7.NS.A.1 Step 5

Check (E) as a sanity pass, since the answer should be the unique survivor.

$-\tfrac{3}{2} + \left(-\tfrac{5}{2}\right) + 5 = -\tfrac{8}{2} + 5 = -4 + 5 = 1$. Sum $= 1$, so (E) is eliminated. The only triplet left is $\textbf{(D)}$.

💡 Both fractions share denominator $2$, so they combine directly to $-4$, and $-4 + 5 = 1$.

[1] #3 5.NF.A.1 Test (A). Add the fractions using a common denominator of $6$.

[2] #3 7.NS.A.1 Test (B). Add the signed integers left to right.

[3] #3 5.NBT.B.7 Test (C). Add the decimals (each is a tenth).

[4] #3 7.NS.A.1 Test (D). Add the signed decimals.

[5] #3 7.NS.A.1 Check (E) as a sanity pass, since the answer should be the unique survivor.

Review

Reasonableness: Four sums came out to exactly $1$ and one came out to $0$. The problem promises exactly one triplet fails the test, so finding exactly one failure (and four passes) matches the structure perfectly. A quick second look at (D): $1.1 + 1.0 = 2.1$, and $2.1 + (-2.1) = 0$ — same answer either order. The unique mismatch is (D).

Alternative: Tool #16 (Change Focus): instead of computing each sum from scratch, look at each triplet for cancellation. (B) has $2$ and $-2$ that cancel; (E) has $-\tfrac{3}{2} - \tfrac{5}{2} = -4$ that cancels with $+5$ down to $1$. For (D), $1.1$ and $-2.1$ leave $-1.0$, but the remaining number is $+1.0$, which cancels to $0$ — not $1$. So (D) is the only one whose leftover after cancellation is wrong.

CCSS standards used (min grade 7)

5.NF.A.1 Add and subtract fractions with unlike denominators (Computing $\tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{6}$ in triplet (A) by rewriting all three fractions with denominator $6$.)
5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths (Computing $0.1 + 0.3 + 0.6$ in triplet (C) as straight decimal addition at the tenths place.)
7.NS.A.1 Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers (Adding signed numbers in triplets (B), (D), and (E), including the negative decimal $-2.1$ and the negative fractions $-\tfrac{3}{2}, -\tfrac{5}{2}$.)

⭐ When the question is "which one is different?", just test each candidate and cross off the ones that pass. The single triplet that fails is your answer — here, (D) sums to $0$, not $1$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 7)

More like this