True discount doubles when time doubles (same sum and rate) $21 is the true discount on $371 for a certain time at a certain rate. If the time is doubled at the same rate, what is the new true discount on $371?

Introduction / Context:
True discount TD relates future amount S, rate r, and time t via TD = S − S/(1 + r*t) = S * (r*t) / (1 + r*t). For a fixed S and r, TD depends on x = r*t in the form TD = S * x / (1 + x). This lets us solve for x from one scenario and then adjust x when time changes.

Given Data / Assumptions:

• S = $371 (future amount).
• TD₁ = $21 at time t.
• Same rate r; doubled time ⇒ x doubles.

Concept / Approach:
Let x = r*t. Then 21 = 371 * x / (1 + x). Solve for x, double it to get 2x, and compute TD₂ = 371 * (2x) / (1 + 2x).

Step-by-Step Solution:
21(1 + x) = 371x ⇒ 21 + 21x = 371x ⇒ 21 = 350x ⇒ x = 21/350 = 0.06.Double time ⇒ 2x = 0.12.TD₂ = 371 * 0.12 / 1.12 = 371 * 0.107142... = $39.75.

Verification / Alternative check:
Present worths: PW₁ = 371 − 21 = 350; growing at rate x = 0.06 gives 350 * 1.06 = 371. For double time, PW₂ = 371 − 39.75 = 331.25; 331.25 * 1.12 = 371, confirming correctness.

Why Other Options Are Wrong:
$44.38 and $33.25 arise from incorrect proportionality or missing the denominator (1 + x). “None of these” is unnecessary since the exact value is computable.

Common Pitfalls:
Assuming true discount doubles linearly with time. It increases but not strictly linearly because of the divisor (1 + x). Use the exact formula to avoid errors.

Final Answer:
$ 39.75