In a street there are some houses to be painted using a fixed set of colours. If 4 houses are painted with each colour, exactly 1 house remains unpainted. If instead 5 houses are painted with each colour, exactly 1 colour remains unused. How many houses are there in the street?

Introduction / Context:
This is a combinational arithmetic reasoning puzzle involving houses and colours. We are told that the same set of colours is used in two different ways. In one scheme, houses are grouped by colour leaving one house unpainted. In the other scheme, houses are grouped differently, leaving one colour unused. By interpreting these conditions algebraically, we can determine how many houses are on the street.

Given Data / Assumptions:
- Let H be the total number of houses.
- Let C be the total number of colours available.
- Scenario 1: If 4 houses are painted with each colour, one house is left unpainted.
- Scenario 2: If 5 houses are painted with each colour, exactly one colour is not used at all.
- Each house receives exactly one colour and each used colour is applied to the stated number of houses in that scenario.

Concept / Approach:
We translate each scenario into an equation relating H and C. In the first scenario, painting 4 houses per colour for all C colours covers 4C houses and leaves 1 unpainted, so H = 4C + 1. In the second scenario, one colour stays unused, so only C - 1 colours are used with 5 houses each, covering 5(C - 1) houses; this must equal the total H. Equating these two expressions for H gives a linear equation in C, from which we can find C and then H. Finally, we match H with the correct option.

Step-by-Step Solution:
Step 1: From scenario 1, 4 houses per colour and one house left unpainted imply H = 4C + 1. Step 2: From scenario 2, 5 houses per colour with one unused colour imply that only C - 1 colours are used, so H = 5(C - 1). Step 3: Equate the two expressions for H: 4C + 1 = 5(C - 1). Step 4: Expand the right-hand side: 5(C - 1) = 5C - 5, so we have 4C + 1 = 5C - 5. Step 5: Rearrange: 4C + 1 - 5C + 5 = 0 leads to -C + 6 = 0, so C = 6. Step 6: Substitute C = 6 into H = 4C + 1 to obtain H = 4 × 6 + 1 = 24 + 1 = 25 houses.

Verification / Alternative check:
Check scenario 1 with H = 25 and C = 6: 4 houses per colour uses 4 × 6 = 24 houses, leaving exactly 1 house unpainted, as required. Check scenario 2: using 5 houses per colour but only C - 1 = 5 colours gives 5 × 5 = 25 houses, so one colour is unused and all houses are painted. Both conditions are satisfied, confirming that 25 houses is correct.

Why Other Options Are Wrong:
Option 6 and 7: These are far too small to accommodate the described painting schemes; they do not fit both equations simultaneously.
Option 32: If H = 32, then from the first scenario C would be (32 - 1) / 4 = 31 / 4, not an integer, contradicting the notion of an integer number of colours.
Option None of these: Not correct, because 25 houses perfectly satisfy both scenarios.

Common Pitfalls:
Learners often confuse houses left unpainted with colours left unused, or misread the second condition as leaving one house unpainted again instead of leaving a colour unused. Another mistake is assuming that the same number of colours is used in both cases without adjusting for the unused colour. Writing clear equations for each scenario before solving helps avoid such confusion.

Final Answer:
There are 25 houses in the street.