On July 4th, a town holds an annual fireworks show. The shows finale

Question

On July 4th, a town holds an annual fireworks show. The show’s finale features two types of fireworks—Firework A and Firework B—launched at regular intervals to create a synchronized effect in the sky.

• Firework A is launched every 6 seconds.

• Firework B is launched every 21 seconds.

They are first launched together at the exact start of the finale, and the time required to launch a firework is negligible.

Based on the above information, select for  Time  the number of seconds after the start of the finale both Firework A and Firework B next launch simultaneously and select for  Launches  the number of times Firework A will have been launched before that moment.

Bunuel · Accepted Answer

Official Solution:

Firework A launches every 6 seconds: 0, 6, 12, 18, 24, 30, 36, 42, ...

Firework B launches every 21 seconds: 0, 21, 42, ...
 
The first time both launch together again after time 0 is at 42 seconds.
 
By then, Firework A has launched at: 0, 6, 12, 18, 24, 30, 36 — 7 times before 42.

Correct answer:

Time  "42"

Launches  "7"

Solomon808 · Answer

For Time column, you have to find the least common multiple for 6 and 21 which is 42. This means that both fireworks can be launched together again after 42 seconds.

For the Launch column, the question asked for the number of times it will take Firework A to reach 42 before both fireworks start again. That can be done by dividing 42 by 6 which is 7.

Thank you

SatvikVedala · Answer

Hi For the second column (Launch), Since in the 42nd minute Both A & B are launched simultaneously
Shouldn't the answer be  6 ?
Since the Question asks "Launches the number of times Firework A will be launched by the time they launch together again".

Thank You
Satvik

Solomon808 · Answer

Question Stem
Launches column, the number of times Firework A will be launched by the time they launch together again.

Analysis 
From the Times column, we know that after the initial launch, both fireworks (A & B) can be launched together again after 42 seconds. 
The question asked for the number of times firework A will catch up with firework B to launch together again. 
Since firework A is launched after 6 seconds, It follows that, firework A has to be launched 7 times before it can be launched together with firework B again after their initial launch together.

Marius94 · Answer

For the lunch column, is it possible to find our figure with a ratio, 6 to 21 is equivalent with 1: 3,5. Even if your method appears way simplier.

Bismuth83 · Answer

1. The question asks us to  find the next point fireworks A and B launch together and how many launches did Firework A do before then .

2. Let’s try to visualize all the launches over time:

Time 
 0 
 6 
 12 
 18 
 21 
 24 
 30 
 36 
 42 
 
  Firework A Launches 
  #  
 # 
 # 
 # 
  
 # 
 # 
 # 
  #  
 
  Firework B Launches 
  #

#

3. We see that the first time fireworks A and B launch together after the beginning is 42 seconds later. In addition, the number of Firework A launches before then is 7.

4. So, our answer is:   Time - 42 and Launches - 7  .

Sparsh2604 · Answer

yes I agree

coolsies · Answer

The series for launch of A will be: 6*0, 6*1, 6*2,.....,6*6 seconds before launch instance. So basically, we missed the 0th instance and that's why got it wrong.

Hope this helps!

tomloveless · Answer

If the initial launch is at zero then the answer is 8 and not 7.

A = 0...6...12...18...24...30...36...42 = 8 There has to be an initial launch to start the interval. 
B = 0..................21...................42 = 3

Question is worded weird.

Bismuth83 · Answer

You are correct with your time diagram and that we should count the initial launch.

However, the last launch shouldn't be counted. This is because the question states that we're looking for "the number of times Firework A will be launched    by the time    they launch together again." So, it's only 7 launches that should be counted.

I hope that helped!

xyz12345678 · Answer

But doesnt it ask number of times A launched  before  it launched simultaneously with B.In that case shoudnt the answer for 2nd part be 6 since the 7th time was when it launched simultaneously with B

Bunuel · Answer

No, the answer is not 6. Firework A launches at 0, 6, 12, 18, 24, 30, and 36. All of these happen before 42 seconds, which is when it launches with B again. That makes 7 launches before the simultaneous one at 42. The 42-second launch is the eighth, so it is not included. The correct count before that moment is 7.

Tinytrotter · Answer

I agree with your analysis.

But when A launches at 0 seconds, doesn't B also launch at the same time? Next A has to launch 6 times (6, 12, 18, 24, 30, 36) before launching for the 7th time which will again coincide with B.

My reading of the question and again I can be wrong was how many launches does A have before both A & B are launched together again. Therefore, I selected the answer as 6 as A will have have to be launched 6 times in between successive joint launches. "

Please let me know if I read the question wrong.

Bunuel · Answer

Yes, the issue is in the reading of “before that moment.”

“That moment” is 42 seconds. Firework A launches at 0, 6, 12, 18, 24, 30, and 36 seconds before that moment. The launch at 0 is included because it also occurs before 42 seconds.

On July 4th, a town holds an annual fireworks show. The shows finale

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Time	0	6	12	18	21	24	30	36	42
Firework A Launches	#	#	#	#		#	#	#	#
Firework B Launches	#				#				#

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On July 4th, a town holds an annual fireworks show. The shows finale

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