Boundary High School question

[MathsFormula]

This is a similar thread to the questions here: http://www.freemathhelp.com/forum/t...ding-question-Not-sure-my-thinking-is-correct but are worded differently and what I learnt there (taught by Ishudu) is not working for Q1 below.


Q1 Peter transports metal bars in his van.

The van has a safety notice "Maximum Load 1200 kg".

Each metal bar has a label "Weight 60 kg".

For safety reasons Peter assumes that 1200 is rounded correct to 2 significant figures and 60 is rounded correct to 1 significant figure.

Calculate the greatest number of bars that Peter can safely put into his van if his assumptions are correct.



A1. The ZERO’s at the end of 1200 kg are units and tens. So the range is 100

<-------------------- range is 100 kg ------------------>
1950 kg <-------------- 1200 kg ---------------------> 1250 kg



The ZERO at the end of 60 kg are units. So the range is 10

<-------------------- range is 100 kg ------------------>
55 kg <-------------- 60 kg ---------------------> 65 kg

Maximum number of bars that can fit into the truck = 1250/55 = 22.72 = 22 bars. This is the WRONG ANSWER. Answer in book = 19 bars

I thought I should make the maximum weight of the load to be 1249 kg because this can be round DOWN so that it is safe for the truck to carry this load. (1250 means rounding UP). But still the answer is wrong 1249/55 = 22.7 = 22 bars

Answer in book = 19 bars and this answer is revealed by the calculation 1250/65 = 19.2 (this would be the minimum number of bars so I think the book is wrong).


Q2. A cyclist travels 60 m at a speed of 8 m/s.
Both values are measured to correct the nearest whole number. What is the greatest possible time taken.


The ZERO at the end of 60 are units. So the range is 10

<-------------------- boundary is 1 m ------------------>
55.5 m <-------------- 60 m ---------------------> 60.5 m


<-------------------- boundary is 1 m/s ------------------>
7.5 m/s <-------------- 8 m/s ---------------------> 8.5 m/s


Speed = d/t

t = d/speed

t = 60.5/7.5 = 8.06 = 8.1 seconds (2 sf). This is correct according to the BOOK (and is the method taught to be by Ishudu) but just need someone to confirm this because lost my confidence after Q1. I think my book is full of errors and that's messing with my mind.

 

[Deleted member 4993]

This is a similar thread to the questions here: http://www.freemathhelp.com/forum/t...ding-question-Not-sure-my-thinking-is-correct but are worded differently and what I learnt there (taught by Ishudu) is not working for Q1 below.


Q1 Peter transports metal bars in his van.

The van has a safety notice "Maximum Load 1200 kg".

Each metal bar has a label "Weight 60 kg".

For safety reasons Peter assumes that 1200 is rounded correct to 2 significant figures and 60 is rounded correct to 1 significant figure.

Calculate the greatest number of bars that Peter can safely put into his van if his assumptions are correct.



A1. The ZERO’s at the end of 1200 kg are units and tens. So the range is 100

<-------------------- range is 100 kg ------------------>
1950 kg <-------------- 1200 kg ---------------------> 1250 kg



The ZERO at the end of 60 kg are units. So the range is 10

<-------------------- range is 100 kg ------------------>
55 kg <-------------- 60 kg ---------------------> 65 kg

Maximum number of bars that can fit into the truck = 1250/55 = 22.72 = 22 bars. This is the WRONG ANSWER. Answer in book = 19 bars

I thought I should make the maximum weight of the load to be 1249 kg because this can be round DOWN so that it is safe for the truck to carry this load. (1250 means rounding UP). But still the answer is wrong 1249/55 = 22.7 = 22 bars

Answer in book = 19 bars and this answer is revealed by the calculation 1250/65 = 19.2 (this would be the minimum number of bars so I think the book is wrong).[minimum # of bars for safe transport - (not economical transport) is 1]

When we are looking for maximum # of bars that can be carried "safely" ---- [1150/65 = 17.69 =] 17 is the correct #. [edited]

In general, when looking for "maximum" value - we need to round down (each situation should be thought through).

Suppose the carrying capacity of the truck is actually is 1150 kg and each rod is actually 65 kg.

Then carrying 22 bars would be unsafe!!

So to be safe - Peter must not put more than 17 rods in the truck. With 17 rods (or less) - is safe.


Q2. A cyclist travels 60 m at a speed of 8 m/s.
Both values are measured to correct the nearest whole number. What is the greatest possible time taken.


The ZERO at the end of 60 are units. So the range is 10

<-------------------- boundary is 1 m ------------------>
55.5 m <-------------- 60 m ---------------------> 60.5 m


<-------------------- boundary is 1 m/s ------------------>
7.5 m/s <-------------- 8 m/s ---------------------> 8.5 m/s


Speed = d/t

t = d/speed

t = 60.5/7.5 = 8.06 = 8.1 seconds (2 sf). This is correct according to the BOOK (and is the method taught to be by Ishudu) but just need someone to confirm this because lost my confidence after Q1. I think my book is full of errors and that's messing with my mind.

.

 

[Ishuda]

This is a similar thread to the questions here: http://www.freemathhelp.com/forum/t...ding-question-Not-sure-my-thinking-is-correct but are worded differently and what I learnt there (taught by Ishudu) is not working for Q1 below.


Q1 Peter transports metal bars in his van.

The van has a safety notice "Maximum Load 1200 kg".

Each metal bar has a label "Weight 60 kg".

For safety reasons Peter assumes that 1200 is rounded correct to 2 significant figures and 60 is rounded correct to 1 significant figure.

Calculate the greatest number of bars that Peter can safely put into his van if his assumptions are correct.
...

First your numbers are, IMO, correct
Van Load = 1200 \(\displaystyle \pm\) 50 kg
Bar Weight = 60 \(\displaystyle \pm\) 5 kg

Now
Number of bars = Van Load / Bar Weight
but here we have a question of safety so we want the minimum value possible for the number of bars to be carried. So what if the maximum van load was really 1150 kg [at the minimum end of the range] and each bar really weighed 65 kg [at the maximum end of the range]? The number of bars which could be carried in safety would then be
Number of bars = 1150 / 65 ~ 17.69
or 17 bars rounded to an integer. Once again, I disagree with the book which has an answer of 19.

In general when you have something like this:
N = a / b
where both and b have a possible range of values, we have
N min = Minimum value of a / Maximum value of b
N max = Maximum value of a / Minimum value of b

 

[MathsFormula]

A1. The ZERO’s at the end of 1200 kg are units and tens. So the range is 100

<-------------------- range is 100 kg ------------------>
1950 kg <-------------- 1200 kg ---------------------> 1250 kg



The ZERO at the end of 60 kg are units. So the range is 10

<-------------------- range is 100 kg ------------------>
55 kg <-------------- 60 kg ---------------------> 65 kg

I spotted an error above in my first post. I should have used a calculator.

Should have been:

The ZERO’s at the end of 1200 kg are units and tens. So the range is 100

<-------------------- range is 100 kg ------------------>
1150 kg <-------------- 1200 kg ---------------------> 1250 kg



The ZERO at the end of 60 kg are units. So the range is 10

<-------------------- range is 100 kg ------------------>
55 kg <-------------- 60 kg ---------------------> 65 kg

 

[MathsFormula]

I've got it ...... hope my thinking is correct ... so going to blurt it out now while fresh in my mind:



The ZERO’s at the end of 1200 kg are units and tens. So the range is 100

<-------------------- range is 100 kg ------------------>
1150 kg <-------------- 1200 kg ---------------------> 1250 kg



The ZERO at the end of 60 kg are units. So the range is 10

<-------------------- range is 100 kg ------------------>
55 kg <-------------- 60 kg ---------------------> 65 kg


For SAFETY we need to err on the side of caution. So we assume the manufacturers scale that measured the weight of the van is more likely to be correct at 1150 kg (because if we assume 1250 kg then we may pile too many metal bars on the van when the weight tolerance of the van was actually much less). So for SAFETY assume the van can only lift 1150 kg.
OR assume that the van people have ROUNDED UP


For SAFETY we need to err on the side of caution. So we assume the manufacturers scale that measured the weight of the metal bars is more likely to be correct at 65 kg (because if we assume 55 kg then we may pile too many metal bars on the van when the weight of the bars was in reality much more than 55 kg). So for safety assume the bars to be the heaviest they can be which is 65 kg
OR assume that the metal people have ROUNDED DOWN

So 1150/65 = 17.6 = 17 bars maximum for safety

Book answer of 19 is wrong

 

[Ishuda]

I've got it ...... hope my thinking is correct ... so going to blurt it out now while fresh in my mind:



The ZERO’s at the end of 1200 kg are units and tens. So the range is 100

<-------------------- range is 100 kg ------------------>
1150 kg <-------------- 1200 kg ---------------------> 1250 kg



The ZERO at the end of 60 kg are units. So the range is 10

<-------------------- range is 100 kg ------------------>
55 kg <-------------- 60 kg ---------------------> 65 kg


For SAFETY we need to err on the side of caution. So we assume the manufacturers scale that measured the weight of the van is more likely to be correct at 1150 kg (because if we assume 1250 kg then we may pile too many metal bars on the van when the weight tolerance of the van was actually much less). So for SAFETY assume the van can only lift 1150 kg.
OR assume that the van people have ROUNDED UP


For SAFETY we need to err on the side of caution. So we assume the manufacturers scale that measured the weight of the metal bars is more likely to be correct at 65 kg (because if we assume 55 kg then we may pile too many metal bars on the van when the weight of the bars was in reality much more than 55 kg). So for safety assume the bars to be the heaviest they can be which is 65 kg
OR assume that the metal people have ROUNDED DOWN

So 1150/65 = 17.6 = 17 bars maximum for safety

Book answer of 19 is wrong

I agree with that reasoning.

 

[MathsFormula]

^^Thanks