Consider two rectangular sheets, Sheet M and Sheet N of dimensions 6 cm x 4 cm each.
Folding operation 1: The sheet is folded into half by joining the short edges of the current shape.
Folding operation 2: The sheet is folded into half by joining the long edges of the current shape.
Folding operation 1 is carried out on Sheet M three times.
Folding operation 2 is carried out on Sheet N three times.
The ratio of perimeters of the final folded shape of Sheet N to the final folded shape of Sheet M is ___________.
A. 13:7
B. 3:2
C. 7:5
D. 5:13
GATE 2021 · General Aptitude · Paper Folding · medium
Answer: The ratio of perimeters (Sheet N : Sheet M) = 13 : 7
Fold Sheet M three times with Operation 1: Start: 6x4. After fold 1 (short edge = 4 cm, join -> new long = 6/2 = 3): 3x4. After fold 2 (short edge = 3 cm, join -> new long = 4/2 = 2): 3x2. After fold 3 (short edge = 2 cm, join -> new long = 3/2 = 1.5): 1.5x2.
Fold Sheet N three times with Operation 2: Start: 6x4. After fold 1 (long edge = 6 cm, join -> new short = 4/2 = 2): 6x2. After fold 2 (long edge = 6 cm, join -> new short = 2/2 = 1): 6x1. After fold 3 (long edge = 6 cm, join -> new short = 1/2 = 0.5): 6x0.5.
Compute perimeters of final shapes: Sheet M final shape 1.5 x 2: P_M = 2(1.5 + 2) = 2 x 3.5 = 7 cm. Sheet N final shape 6 x 0.5: P_N = 2(6 + 0.5) = 2 x 6.5 = 13 cm.
Compute the ratio P_N : P_M: P_N : P_M = 13 : 7. This is already in lowest terms since gcd(13, 7) = 1.