A shopkeeper displays bangles in a perfect square. He has 38 bangles left over. If he increases the side of the square by 1, he falls short by 25 bangles. Find the actual number of bangles he has.

Introduction:
This is a classic perfect-square boundary problem involving leftovers and shortfalls when moving between consecutive squares k^2 and (k + 1)^2.

Given Data / Assumptions:

• Total bangles = N
• N = k^2 + 38 (left over after forming k by k square)
• N = (k + 1)^2 − 25 (25 short of the next square)

Concept / Approach:
Equate the two expressions for N and solve for k; then substitute back to get N.

Step-by-Step Solution:
k^2 + 38 = (k + 1)^2 − 25k^2 + 38 = k^2 + 2k + 1 − 2538 = 2k − 24 ⇒ 2k = 62 ⇒ k = 31N = k^2 + 38 = 31^2 + 38 = 961 + 38 = 999

Verification / Alternative check:
(k + 1)^2 − N = 32^2 − 999 = 1024 − 999 = 25, consistent with the shortfall.

Why Other Options Are Wrong:
1690, 538, 1024: Do not satisfy both square-boundary conditions.Cannot be determined: The equations determine N uniquely.

Common Pitfalls:
Arithmetic errors when expanding (k + 1)^2; mixing up leftover vs shortfall signs.

Final Answer:
999