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Bismuth83
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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
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SatvikVedala
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


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.
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Solomon808
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
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.
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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:

Time0612182124303642
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.
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Solomon808
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
yes I agree
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SatvikVedala
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

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!
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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.
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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!
tomloveless
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.
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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
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Bunuel

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.



TimeLaunches
3
6
7
8
42
56

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"
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xyz12345678
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
Official Solution:
Bunuel

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.



TimeLaunches
3
6
7
8
42
56

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"

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.
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