The Back Half Pays 20.227 Percent, and the Fifth Root Cancels

Because the horizons are ten and five, both sides of the no-arbitrage equation are fifth powers and the root disappears, leaving 1 + f = 1.15 squared over 1.10 = 529/440, so f = 89/440 exactly. Reflecting 10 percent around 15 to get 20 is low by exactly (b - a) squared over (1 + a), a square over a positive number, which is why the reflection can never overshoot for any pair of rates. Under continuous compounding the same problem is linear and 20 percent is exactly right, so the instinct is correct machinery pointed at the wrong convention.

Lambdia
August 20, 2026 · 9 min read

Money locked away for five years earns 10 percent a year. Locked away for ten years it earns 15 percent a year. What must the second five years pay?

Ten is five below fifteen, so the back half should sit five above: 20 percent. That is a good estimate and it is not the answer. The answer is exactly $89/440$ , which is 20.2272 recurring percent, and the extra 0.227 points has an exact closed form worth knowing.

One equation, from a condition rather than a formula

Two ways of holding money for ten years have to end in the same place, or one of them is free money. Lock for ten years directly, or lock for five and then lock again for the rate the second half implies:

1.10^{5}\,(1+f)^{5} = 1.15^{10}

(1)

That is the whole model. The 1.10 and the 1.15 are per-year growth factors under annual compounding, and equation (1) is a definition of $f$ rather than a theory about it.

Implied forward rate

The rate for a future period that makes two holding strategies break even today. It is the market's indifference point, not a forecast: nothing in equation (1) claims the second five years will pay 20.227 percent, only that a curve which quotes 10 and 15 has already committed to that number.

The fifth root disappears

Equation (1) looks like it wants a calculator, and it does not, because the two horizons are ten and five and their ratio is exactly 2:

(1+f)^{5} = \frac{1.15^{10}}{1.10^{5}} = \left(\frac{1.15^{2}}{1.10}\right)^{5} \;\Longrightarrow\; 1+f = \frac{1.15^{2}}{1.10}

(2)

Both sides are fifth powers, so the fifth root is free. What remains is one squaring and one division:

1 + f = \frac{1.3225}{1.10} = \frac{529}{440}, \qquad f = \frac{529}{440} - 1 = \frac{89}{440} = 0.20227\overline{27}

(3)

Exact, in rationals, with no rounding anywhere. Anyone who reaches for a fifth root here has not looked at the exponents, and the collapse is the one thing in this problem worth carrying into the next one.

Fig. 1 — Equation (1) as a picture. The two blocks compound in sequence, and 20 percent leaves the money 0.94 percent short of where ten years at fifteen puts it.

What the reflex gets right

The reflection $2b - a$ is not a blunder. Check where it lands: compounding 10 percent for five years and 20 percent for five years gives

1.10^{5} \times 1.20^{5} = 4.007464 \;<\; 4.045558 = 1.15^{10}

(4)

Short by 0.94 percent of the target, which for a mental estimate under time pressure is a respectable place to be. Calling 20 percent wrong is a little unfair; calling it low is exactly right, and the next section says by how much.

A magnitude check is worth doing before trusting equation (3), since a slip of a decimal place would be invisible in a fraction like $89/440$ . At 20.227 percent a year, five years turn one unit into $(529/440)^5 = 2.51197$ , and multiplying by the first block's 1.61051 gives 4.045558, which is $1.15^{10}$ exactly. So the back half of this curve more than doubles the money, which is what a jump from 10 to 15 percent on the headline rate is quietly asking for.

The error has a closed form

Let the five-year rate be $a$ and the ten-year rate be $b$ . From equation (2) the true forward is $(1+b)^2/(1+a) - 1$ , and subtracting the reflection $2b - a$ gives, after expanding the numerator,

\underbrace{\frac{(1+b)^2}{1+a} - 1}_{\text{true forward}} \;-\; \underbrace{(2b - a)}_{\text{reflection}} \;=\; \frac{(b-a)^2}{1+a}

(5)

A square divided by a positive number. So the reflection is never above the true forward, for any pair of rates above $-100$ percent, and it is equal only when the curve is flat. Equation (5) licenses the word "always", which a numerical check at one pair of rates never could.

Substituting the numbers from the problem,

\frac{(0.15 - 0.10)^2}{1.10} = \frac{0.0025}{1.1} = \frac{1}{440} = 0.227\ \text{percentage points}

(6)

which is precisely the difference between 20 and 20.227. The quadratic in $b - a$ also explains why nobody notices this on a gentle curve. With a five-year rate of 10 percent and a ten-year rate of 11 percent the gap is under a hundredth of a point, and the forward is 12.009 percent against a reflected 12. Steepen the curve to 5 percent and 15 percent and the gap is 0.95 points: the forward is 25.95 percent against a reflected 25.

Fig. 2 — Equation (5) plotted with the five-year rate fixed at 10 percent. The error grows as the square of the steepness, so the reflex is nearly exact on a flat curve and badly low on a steep one.

Why the wrong answer feels so right

Here is the part that makes this problem more than an arithmetic trap. Redo it under continuous compounding, where a rate $r$ over time $T$ grows money by $e^{rT}$ . Equation (1) becomes

e^{0.10 \times 5}\,e^{5f} = e^{0.15 \times 10} \;\Longrightarrow\; 0.5 + 5f = 1.5 \;\Longrightarrow\; f = 0.20

(7)

Exactly 20 percent. Under continuous compounding the exponents add instead of multiplying, so rates are linear in time and the reflection is not an approximation at all. The instinct is correct machinery pointed at the wrong convention.

That is worth stating out loud whenever this answer is given, because it means the question is under-specified until somebody says how the rates compound. Annual compounding gives 20.227. Continuous compounding gives 20. Semi-annual gives something in between. The convention is part of the problem, not part of the solution.

Where the clean answer stops

Both of the nice features above lean on the horizons being 10 and 5. In general, with a rate $a$ to time $T_1$ and $b$ to time $T_2$ ,

1 + f = \left[\frac{(1+b)^{T_2}}{(1+a)^{T_1}}\right]^{\frac{1}{T_2 - T_1}}

(8)

and the root only disappears when $T_2 = 2T_1$ , which is why the puzzle picked those numbers. A three-year rate against a ten-year rate leaves a seventh root that no amount of cleverness removes. Equation (5) is specific to the same structure and should not be quoted for arbitrary horizon pairs.

One more caution about equation (8) that has nothing to do with algebra. It produces a rate that two strategies agree on given today's quotes, and steep curves produce forwards that look implausible as forecasts. A curve rising from 10 to 15 over five years implies a back half above 20, which is arithmetic rather than prophecy.

Sources and further reading

The convention that decides the answer — Compound interest
The condition in equation (1) — Forward rate
Why the reflection is a mean and therefore low — Inequality of arithmetic and geometric means
The alternative convention in equation (7) — Yield curve

The answer was confirmed by exact rational arithmetic, by a 300-step bisection on equation (1) in exact fractions, which landed on $89/440$ with a residual of $9 \times 10^{-92}$ , and by an ordinary floating-point calculation returning 20.227272727272695 percent. Equation (5) was checked exactly at 400 pairs of rates from 1 to 20 percent, inverted curves included, and the reflection was never above the true forward at any of them.

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