Divisor–Dividend Relation with Given Remainder In a division problem, the divisor is two-thirds (2/3) of the dividend and also equals two times the remainder. If the remainder is 5, determine the value of the dividend.

Introduction / Context:
This problem tests understanding of the standard division identity and how to interpret verbal relationships among dividend, divisor, quotient, and remainder. It is a classic number system application that rewards setting up clean equations before computing.

Given Data / Assumptions:

• Remainder R = 5.
• Divisor d is two times the remainder, so d = 2 * R.
• Divisor d is also two-thirds of the dividend D, so d = (2/3) * D.
• Division identity: D = d * q + R for some integer quotient q.

Concept / Approach:
Use the two given equalities involving the divisor to solve for D directly. Then verify with the division identity that an integer quotient exists and the remainder matches 5.

Step-by-Step Solution:
From d = 2 * R and R = 5 → d = 10.Also d = (2/3) * D → 10 = (2/3) * D → D = 10 * 3 / 2 = 15.Check the division identity: 15 = 10 * q + 5 → q = 1 (an integer), valid.

Verification / Alternative check:
Confirm consistency: divisor equals two-thirds of the dividend (10 = 2/3 of 15) and also equals twice the remainder (10 = 2 * 5). Both conditions hold simultaneously.

Why Other Options Are Wrong:
25, 18, 24, and 30 fail at least one of the relationships d = (2/3)D or d = 2R with R = 5, or fail the identity D = d * q + R with integer q.

Common Pitfalls:
Confusing “two-thirds of the dividend” with “dividend is two-thirds of divisor,” or forgetting to check the division identity after computing D.

Final Answer:
15