Three Variables with Close Ratios Numbers x, y, z satisfy xy = 96050 and xz = 95625, with y exceeding z by one (y = z + 1). Find the value of z.

Introduction / Context:This problem blends ratio reasoning with products of unknowns. The key is to divide the two product equations to relate y and z directly; then use the given difference condition to pin down exact values.

Given Data / Assumptions:

• xy = 96050.
• xz = 95625.
• y = z + 1.

Concept / Approach:Divide the equations: (xy)/(xz) = y/z = 96050/95625. Reduce the fraction to simplest terms to express y : z. Then insert y = z + 1 and solve for both as integer multiples of the simplified ratio.

Step-by-Step Solution:y/z = 96050/95625.Reduce 96050/95625 by gcd 425 → 226/225.Thus y : z = 226 : 225.But y = z + 1 fits perfectly with 226 and 225 → let z = 225 and y = 226.Hence z = 225.

Verification / Alternative check:With z = 225 and y = 226, the ratio y/z = 226/225 matches the reduced fraction. Existence of some x is guaranteed because xy and xz are given; x = (xz)/z = 95625/225 is consistent.

Why Other Options Are Wrong:425, 220, 525, and 221 do not satisfy both the ratio 226:225 and the “y = z + 1” constraint simultaneously.

Common Pitfalls:Failing to reduce the fraction fully; assuming y − z is greater than 1; mixing y and z when applying the ratio.

Final Answer:225