Ribbon Lengths and Blinking Lights on the Annual-Day Stage
The cultural committee of a Zilla Parishad high school is preparing the stage for the annual day. Three rolls of ribbon, measuring 3 m 60 cm, 4 m 20 cm and 5 m 40 cm, must be cut into pieces of equal length so that every roll is used completely and each piece is as long as possible. Separately, three strings of serial lights are tested: the red string blinks every 6 seconds, the green string every 8 seconds and the blue string every 10 seconds. All three blink together at 6:00:00 p.m., when the programme begins. Before the cutting starts, the secretary writes in the plan: 'Pieces of 30 cm each will use up all three rolls exactly, and the three light strings will blink together again 4 minutes after the start.' The teacher in charge asks the committee to check both statements against the actual numbers before any ribbon is cut or any lights are rewired.
1. Find the length of the longest ribbon piece that cuts all three rolls with no waste, and the total number of pieces obtained.
Read the worked answer
Convert to a common unit first. 3 m 60 cm = 360 cm, 4 m 20 cm = 420 cm, 5 m 40 cm = 540 cm.
Choose the right tool. 'Equal pieces, no waste, longest possible' means the piece length must divide each roll exactly and must be the greatest such number - that is the HCF, not the LCM. (The LCM, 7560 cm, is longer than every roll, so it cannot be a piece length at all.)
360 = 2³ × 3² × 5
420 = 2² × 3 × 5 × 7
540 = 2² × 3³ × 5
HCF = 2² × 3 × 5 = 60 cm
Number of pieces = 360/60 + 420/60 + 540/60 = 6 + 7 + 9 = 22 pieces.
2. Test the secretary's statement that pieces of 30 cm will use up all three rolls exactly. Is the plan acceptable?
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30 cm does divide each roll exactly: 360/30 = 12, 420/30 = 14, 540/30 = 18, so the 'no waste' part of the claim is true.
But the committee's requirement was the longest possible piece. Since 30 is a common factor but not the HCF (60), it fails that condition: it produces 12 + 14 + 18 = 44 pieces instead of 22 - twice as many pieces, each half as long.
Verdict: the statement is arithmetically true about waste but wrong as a plan. The correct piece length is 60 cm. Extra pieces also mean extra joints when the ribbon is stitched along the stage, which is the practical cost of choosing a smaller common factor.
3. Determine when the three light strings next blink together, and judge the secretary's '4 minutes' claim.
Read the worked answer
6 = 2 × 3, 8 = 2³, 10 = 2 × 5, so LCM(6, 8, 10) = 2³ × 3 × 5 = 120 seconds = 2 minutes.
The strings therefore coincide at every multiple of 2 minutes after 6:00:00 p.m. - at 2, 4, 6, 8, 10 minutes and so on.
Judgement: the first repeat is at 2 minutes, not 4. The claim is false if 'again' means the first repeat. It is true only in the weaker sense that 4 minutes is a multiple of 2 minutes, so at 4 minutes the strings do coincide once more (for the second time).
The examination-safe answer states the LCM and writes: 'together again first at 2 minutes, and again at 4 minutes and every 2 minutes thereafter'.