Shamus rode his bicycle 3 times as fast as Grace walked.

Q1: What time was it when Shamus caught up with Grace?

Q2: If the school is 800m from home, did Shamus reach Grace before she arrived at school?

*Use the following to guide you...*

*What information do you know from the problem?**What else do you need to know to solve the problem?**Pick a reasonable number for the information you need.*

Q1)

ReplyDeleteShamus's speed is thrice of Grace, so whatever Grace may cover in 3 minutes, Shamus can cover in 1. Grace also had a start of 10 minutes earlier than Shamus, so we can count their highest common factor from there.

Grace: 10, 11, 12, 13, 14, 15.

Shamus: 0, 3, 6, 9, 12, 15.

15 minutes after 6.30 am would be 6.45 am, hence Shamus caught up with Grace then.

Q2)

800m ÷ 15 minutes ≈ 53.3m/min

If Grace walks faster than that speed, Shamus wouldn't be able to catch her. If Grace walks slower, or equal to her speed, he would be able to catch up with her before she reaches school.

1)

ReplyDeleteAs Shamus's speed is three times that of Grace's, whatever distance Grace is able to travel in 3 minutes, he can cover the distance in 1 minute. But since Grace started 10 minutes earlier than Shamus, we must start taking their HCF from there.

Grace - 10, 11, 12, 13, 14, 15

Shamus - 0, 3, 6, 9, 12, 15

15 minutes after 6.30 am woud be 6.45 am. Therefore Shamus caught up with Grace at 6.45 am.

2)

800m ÷ 15 minutes ≈ 53.3 m/min

Assuming Grace walks at a constant speed, and Shamus cycles at a constant speed, Shamus would be able to catch up with her.

Q1)

ReplyDeleteAs Shamus's speed is thrice that of Grace, the distance that Grace is able to cover in 3 minutes, Shamus would cover that distance in 1 minute. Grace started 10 minutes earlier than Shamus, so we start by taking their Highest Common Factor from there.

Grace: 10, 11, 12, 13, 14, 15

Shamus: 0, 3, 6, 9, 12, 15

Their Highest Common Factor is 15 minutes, so the time that Shamus caught up with Grace would be 15 minutes after 6.30a.m., which is 6.45a.m. So Shamus caught up with Grace at 6.45a.m.

Q2)

800m ÷ 15 min ≈ 53.3 m/min

53.3 m/min is the speed that Shamus travelled at. If Grace travels at a speed faster than 53.3 m/min, Shamus would catch up with Grace only after she reached school. If Grace travels at a speed slower than or equal to 53.3 m/min, Shamus would be able to catch up with Grace before she reaches the school.

Q1) Whatever distance Grace travelled, Shamus could cover it 3 times as fast as he rode his bike 3 times as fast as Grace walked. So, whatever distance Grace travelled in 3 minutes, Shamus was able to cover it in 1 minute. But as Grace had started 10 minutes before Shamus, the Lowest Common Multiple must be calculated from the 10th minute for Grace and the 0th minute for Shamus in order to find out when Shamus caught up with Grace.

ReplyDeleteGrace: 10, 11, 12, 13, 14, 15

Shamus: 0, 3, 6, 9, 12, 15

As Shamus caught up with Grace at the 15th minute, 15 minutes from 6.30 a.m. is 6.45 a.m. and thus, Shamus caught up with Grace at 6.45 a.m.

Q2) 800m ÷ 15mins = 53.3m/min

53.3m/min × 3 = 159.9m/min

If Shamus cycles at a constant speed of 159.9m/min and Grace walks faster than 53.3m/min constantly, Shamus will not be able to catch up to Grace before she reaches the school. But if he cycles at a constant speed of 159.9m/min and Grace walks at a constant speed 53.3m/min or less, Shamus will be able to reach Grace before she reaches the school.

1. Since Shamus's speed is thrice of Grace, the ratio of Shamus's speed and Grace's speed is 3:1 respectively. Also, Grace started 10 minutes earlier than Shamus, so we can start by taking their LCM or Lowest Common Multiple, assuming that Grace takes 1 minute each time.

ReplyDeleteGrace: 10, 11, 12, 13, 14, 15

Shamus: 0, 3, 6, 9, 12, 15

Their Lowest Common Multiple is 15 minutes. So the time taken for Shamus to catch up with Grace would be 15 minutes after 6.30 a.m., which is 6.45 a.m. So, Shamus caught up with Grace at 6.45 a.m.

2. 800m ÷ 15 min ≈ 53.3 m/min

If Grace travels faster than that speed, Shamus would catch up with Grace after she reaches school. If Grace travels slower than that speed, Shamus would be able to catch up with her before she reaches school.

1.)As Grace left the house 10 minutes before Shamus,she already has a 10 minute head start and the fact that Shamus can cover thrice the distance distance walked by Grace with his bicycle means that he can cover whatever Grace covered in three minutes in a minute.So we start finding the lowest common multiple from there.

ReplyDeleteGrace-10,11,12,13,14,15

Shamus-0,3,6,9,12,15

Their lowest common multiple would be 15 so 15 minutes after 6:30a.m is 6:45 a.m.So Shamus caught up with Grace at 6:45 a.m.

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2.)800m÷15min≈53.3m/min

If Grace was able to travel to school faster than this speed,Shamus would not be able to catch up with up her as she would already be at school by the time he got there.If she was traveling at a slower speed than that Shamus should be able to catch up with her before Grace reaches school.

Q1. Since Shamus rode 3 times as fast as Grace walked, if Grace walked for 3 minutes, Shamus could cover the same distance in 1 minute. However, Grace had a head start of 10 minutes before Shamus, so we have to find the lowest common multiple from there.

ReplyDeleteGrace- 10,11,12,13,14,15

Shamus- 0,3,6,9,12,15

This means that Shamus would catch up with Grace 15 minutes after 6.30am (the time Grace left home), which would be 6.45am.

Q2.

Maximum speed Grace should walk at = 800m ÷ 15 minutes ≈ 53.3 m/min

If Grace walked to school faster than this speed, Shamus would not be able to catch up with her. However, if she the speed at which she walked was slower or equal to this speed, Shamus would be able to catch up with Grace before she could reach school.

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ReplyDeleteQ1:

ReplyDeleteShamus: 3x Grace's speed (In other words, Shamus takes a a third of however long Grace takes to reach a certain place.) To get the time Shamus caught up with Grace, LCM is to be used. Grace has a ten-minute head start, so start her LCM at 10.

Working:

Grace: 10, 11, 12, 13, 14, 15

Shamus: 0, 3, 6, 9, 12, 15

LCM: 15.

6.30am + 15mins = 6.45am.

Shamus caught up with Grace at 6.45am.

Q2:

It depends on the speed Grace walks and if her speed is constant. The following statements are if Grace walks at a constant speed. If she walks faster than 53.3m/min*, the answer would be no. If she walks slower than the speed mentioned earlier, the answer would be yes. If her speed is exactly 53.3m/min*, there is a slight chance that Shamus catches up to her right as she reaches school.

Working:

800m ÷ 15mins = 53.3m/min*

*the last three is supposed to have a dot on top.

Question 1:

ReplyDeleteSince Shamus rode his bicycle 3 times as fast as Grace walked,means that whatever distance Grace cover in a certain time,Shamus can cover thrice of her distance she covered within the same time.So if Grace had a 10 minutes head start,so hence we have to find the LCM from 10.

Grace-10,11,12,13,14,15

Shamus-0,3,6,9,12,15

The LCM is then 15.

Hence Shamus will catch up with Grace in 15 minutes.

15 minutes after 6.30am is 6.45am

Shamus hence caught up with Grace at 6.45am.

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Q2: If the school is 800m from home, did Shamus reach Grace before she arrived at school?

800m ÷ 15mins = 53.3m/min

If she continue walking at a constant speed of 53.3m/mins or slower,shamus would be able to catch up with her,but if she walk faster than 53.3m/mins,then shamus would not be able to catch up with her before she reaches school.

Q1. Shamus's speed is thrice of Grace's. If Grace had travelled 1min he would have travelled 3min -worth of Grace's distance in that one minute- in the same time given.

ReplyDeleteThus,

Row: 1 2 3 4 5 6 7 8 9 10

Grace: 11, 12, 13, 14, 15, 16, 17, 18, 19, 20...

Shamus: 3, 6, 9, 12, 15.

Shamus would have took 5 minutes to catch up with Grace after he discovered her wallet 10 minutes after she left.

And if Grace left the house at 6.30am...

5min + 10min = 15min

0630 + 0015 = 0645

Shamus caught up with Grace at 6.45am.

Q2.

800m ÷ 15 minutes ≈ 53.3m/min

As long as Grace does not exceed this speed, Shamus would be able to catch up with her.

Q1.

ReplyDeleteShamus's speed is three times that of Grace's speed. Hence, when Grace had travelled for a minute, Shamus would have travelled 3 minutes of the speed of Grace's.

Grace (min): 10, 11, 12, 13, 14, 15

Shamus (min): 0, 3, 6, 9, 12, 15

LCM is 15. Thus, after 15 minutes that Grace had left the house, which was 6.30pm. 10 minutes after realising Grace forgot her wallet, Shamus caught up with Grace within 5 minutes by cycling. 15 minutes after 6.30pm is 6.45pm

Q2.

800m ÷ 15 minutes = 53.3m/min (rounded off to 3 significant figures)

If Grace walks faster than this speed, Shamus would not be able to catch up with her before she reaches school. Otherwise, if Grace goes at this speed or under this speed, Shamus would then be able to catch up with her.

Q1.

ReplyDeleteAs Shamus rode 3 times as fast as Grace walked and if Grace walked for 3 minutes, Shamus could cover the same distance in 1 minute but Grace started walking 10 minutes before Shamus got onto his bicycle to catch-up with Grace.

Grace(min) : 10 11 12 13 14 15 (1)

Shamus(min) : 0 3 6 9 12 15 (3)

6.30 am ----(15 minutes)--->6.45 am

Shamus caught up with Grace at 6.45 am.

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Q2.

800m ÷ 15 minutes ≈ 53.3 m/min

If Grace walks faster than 53.3m/min, Shamus would not be able to catch-up with her. If she walks slower than 53.3m/min, Shamus would be able to catch-up with her. So as long as she does not exceed 53.3m/min, Shamus would be able to catch-up with her.

1) For every 1 unit Grace walks, Shamus would have rode 3 units. Since Grace started 10 minutes earlier, she would have walked for 10 units.

ReplyDeleteGrace: 10, 11, 12, 13, 14, 15

Shamus, 3, 6, 9, 12, 15,18

15 minutes past 6.30am would be 6.45am. So, Shamus caught up with Grace at 6.45am.

2. 800m ÷ 15 minutes ≈ 53.3m/min

Should Grace walk at a constant speed of 53.3m/min, Shamus would be able to catch up with her. However, if Grace increases her speed and walks at a speed more than 53.3m/min, Shamus would not be able to catch up with her before she reaches school. So as long as Grace walks at a constant speed of 53.3m/min or less, Shamus would be able to catch up with her.

Q1) Shamus needs to make up for the 10 mins 'head start' of Grace. if he walks 3 times faster than grace, it means that he walks 2 units more than Grace. assuming Grace walks at 1 unit per minute,

ReplyDelete10 / 2 = 5 mins

10 + 5 = 15 mins

15 mins after 6.30am is 6.45am

Ans : 6.45 am

Q2) assuming that Grace walked at a reasonable pace of... 70m/min

800/70= 11.42mins

Shamus will not catch up to her and Grace will go hungry for lunch

1) Shamus's speed is thrice of Grace's. So, the distance Grace travels in 3 minutes would be equivalent to The distance Shamus can cover in 1 minute. Grace also had a head start of 10 minutes. We have to find the LCM.

ReplyDeleteGrace: 10, 11, 12, 13, 14, 15

Shamus:3, 6, 9, 12, 15

15 mins

6.30--------->6.45

Ans: 6.45 am

2)800m ÷ 15 minutes ≈ 53.3m/min

If Grace walks faster than 53.3m/min, Shamus would not be able to catch up with her, but if she walks at a constant speed or slower than 53.3m/min, shamus would be able to catch up with her before she reaches the school. Therefore, the answer is no and yes respectively for the explanations.

Q1

ReplyDeleteShamus left 10 minutes later than Grace. However, he travelled 3 times faster than Grace.

So, we assume that Grace travels at a rate of 1 unit per minute while Shamus travels at a rate of 1 unit per minute. From there, we can calculate the Lowest common multiple of the number of units Grace and Shamus travelled.

Grace: 10, 11, 12, 13, 14, 15 Shamus: 0, 3, 6, 9, 12, 15

Shamus caught up with Grace 15 minutes after Grace left home, so Shamus caught up with Grace at 6.45 pm.

Shamus caught up with Grace at 6.45pm.

Q2

800m ÷ 15 minutes ≈ 53.3 m/min

53.3m/min × 3 = 159.9m/min

If Shamus cycles at a constant speed of 159.9m/min and Grace walks at a constant speed 53.3m/min or less, Shamus will be able to arrive at the school before Grace.