True discount for half the time (same sum and rate) $20 is the true discount on $260 due after a certain time. What will be the true discount on the same amount due after half of that time, at the same rate?

Introduction / Context:
True discount TD for amount S at simple interest satisfies TD = S * x / (1 + x), where x = r*t. Knowing TD for an original time lets us infer x, then recompute TD for half the time (i.e., x/2), while keeping S and r the same.

Given Data / Assumptions:

• S = $260.
• TD₁ = $20 for time t at rate r.
• Find TD₂ for time t/2 (i.e., x/2).

Concept / Approach:
From 20 = 260 * x / (1 + x), solve for x. Then set x₂ = x/2 and compute TD₂ = 260 * x₂ / (1 + x₂).

Step-by-Step Solution:
20(1 + x) = 260x ⇒ 20 + 20x = 260x ⇒ 20 = 240x ⇒ x = 1/12.Half time ⇒ x₂ = (1/12)/2 = 1/24.TD₂ = 260 * (1/24) / (1 + 1/24) = 260 * (1/24) * (24/25) = 260/25 = $10.

Verification / Alternative check:
Present worth: PW₁ = 260 − 20 = 240; growing by factor 1 + 1/12 = 13/12 gives 240 * 13/12 = 260. For half time: PW₂ = 260 − 10 = 250; factor 1 + 1/24 = 25/24 gives 250 * 25/24 = 260, confirming correctness.

Why Other Options Are Wrong:
$10.40, $15.20, and $13 stem from proportionality errors that ignore the 1 + x denominator effect.

Common Pitfalls:
Assuming TD is directly proportional to time without the corrective denominator. Always use TD = S * x / (1 + x).

Final Answer:
$ 10