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NUMBERS

by vaibhav101 » Tue May 22, 2018 11:09 am

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The signboard outside the department store 'Ron and Harry' lights up as described below-
When the switch is turned on, all three words light up and remain lighted for 3 seconds. After that, the first word is switched off for
$$7\frac{5}{6}$$ seconds , the second word is switched off for $$1\frac{1}{3}$$ seconds and the third word is switched off for $$5\frac{2}{3}$$ seconds.
Then each word is again switched on for 3 seconds and switched off for the time duration mentioned. This process continues , repeatedly. After how many seconds of switching on the signboard will the entire board be switched on for the second time for 3 seconds?
A $$40\frac{1}{3}$$
B $$41\frac{2}{3}$$
C $$42\frac{2}{3}$$
D $$43\frac{1}{3}$$
E $$43\frac{2}{3}$$
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by GMATGuruNY » Tue May 22, 2018 2:31 pm
vaibhav101 wrote:The signboard outside the department store 'Ron and Harry' lights up as described below-
When the switch is turned on, all three words light up and remain lighted for 3 seconds. After that, the first word is switched off for
$$7\frac{5}{6}$$ seconds , the second word is switched off for $$1\frac{1}{3}$$ seconds and the third word is switched off for $$5\frac{2}{3}$$ seconds.
Then each word is again switched on for 3 seconds and switched off for the time duration mentioned. This process continues , repeatedly. After how many seconds of switching on the signboard will the entire board be switched on for the second time for 3 seconds?
A $$40\frac{1}{3}$$
B $$41\frac{2}{3}$$
C $$42\frac{2}{3}$$
D $$43\frac{1}{3}$$
E $$43\frac{2}{3}$$
For the 3 words to be switched on simultaneously, each must reach the END OF ITS RESPECTIVE ON-OFF CYCLE.

On-off cycle for RON:
(3 seconds on) + (47/6 seconds off) = 18/6 + 47/6 = 65/6 seconds.
On-off cycle for AND:
(3 seconds on) + (4/3 seconds off) = 18/6 + 8/6 = 26/6 seconds.
On-off cycle for HARRY:
(3 seconds on) + (17/3 seconds off) = 18/6 + 34/6 = 52/6 seconds.

For each word to reach the end of its respective cycle, the total number of seconds must yield an integer when divided by each of the 3 values in blue.
The smallest value that will satisfy this condition = (the LCM of 65, 52 and 26)/6.

65 = 5*13
52 = 2*2*13
26 = 2*13
The LCM of 65, 52 and 13 is equal to the product of the red factors above:
2*2*5*13 = 260.
The reason:
2*2*5*13=260 is the least value divisible by 5*13 (65), by 2*2*13 (52) and by 2*13 (26).

Implication:
260/6 is the smallest value that will yield an integer when divided by each of the 3 blue values above.
(260/6) ÷ (65/6) = 4, implying that RON will complete 4 on-off cycles in 260/6 seconds.
(260/6) ÷ (52/6) = 5, implying that AND will complete 5 on-off cycles in 260/6 seconds.
(260/6) ÷ (26/6) = 10, implying that HARRY will complete 10 on-off cycles in 260/6 seconds.

260/6 = 130/3 = 43â…“ seconds.
Thus, each of the three words will reach the end of its respective cycle and be switched on simultaneously after the passage of 43â…“ seconds.

The correct answer is D.
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