10. A large cube is formed from the material obtained by melting three smaller cubes
with sides 3, 4 and 5 cm respectively. What is the ratio of the total surface areas of the
smaller cubes and the large cube?
a. 2:1
b. 25:18
c. 3:2
d. 27:20
Cube Surface Area Ratio 3 4 5 cm Melted Large Cube
The correct answer is b. 25:18. Three smaller cubes with sides 3 cm, 4 cm, and 5 cm are melted to form a large cube, and the ratio of their total surface areas (smaller to large) is calculated using volume conservation.
Step-by-Step Solution
Volumes remain constant when melted: \(3^3 = 27\) cm³, \(4^3 = 64\) cm³, \(5^3 = 125\) cm³. Total volume equals \(27 + 64 + 125 = 216\) cm³, so the large cube’s side is \(\sqrt[3]{216} = 6\) cm.
Surface area of a cube is \(6a^2\). Smaller cubes: \(6 \times 3^2 = 54\) cm², \(6 \times 4^2 = 96\) cm², \(6 \times 5^2 = 150\) cm². Total: \(54 + 96 + 150 = 300\) cm². Large cube: \(6 \times 6^2 = 216\) cm².
Ratio \(300:216\) simplifies by dividing by 12 to \(25:18\).
Option Analysis
- a. 2:1 (300:150): Too high; assumes large surface area of 150 cm², ignoring correct 216 cm² volume-derived side.
- b. 25:18 (300:216): Matches exact calculation after simplification.
- c. 3:2 (300:200): Close but underestimates large surface area by 16 cm².
- d. 27:20 (270:200): Wrong totals; confuses volumes (27+64+125≠270) with areas.
In competitive exams like CSIR NET, the cube surface area ratio 3 4 5 cm problem tests volume conservation and geometry basics. Three smaller cubes (sides 3 cm, 4 cm, 5 cm) melt into one large cube—find the ratio of total smaller surface areas to the large cube’s surface area.
Quick Calculation Guide
Total volume: \(27 + 64 + 125 = 216\) cm³ → large side 6 cm.
Smaller areas sum to 300 cm²; large area 216 cm² → \(25:18\).
This appears in aptitude tests; always verify volume first, as surface area decreases with consolidation.
