How to Solve Mixed Number Multiplication in Fractions?

[logicandtruth]

TL;DR Summary: Question about fractions

Hi All, I am currently going through my old maths school book and just finished fractions and decided to google some further questions online, but I am stuck on the following question.

I include the image below with the provided answer from the question I found online. My thought process to find the correct answer was as follows:

• 6 2/3 x 3 1/7 = 20/3 x 22/7 = 440/21. According to the correct answer in the image above, I am correct up to this point - how did they calculate 5 cans are required?
• As each can of paint covers 5m2, to find how many cans of paint required I would then do 440/21 / 5 = 88/21
• 88/21 x £7.50 = £31.42

Could you please explain where my works are incorrect, thank you.

 

[andrewkirk]

You assumed Mrs Holland could buy 88/21 cans, which is 4 4/21 cans. Nobody will sell her 4/21 of a can, so she has to buy 5 cans to have enough paint. She will have 17/21 of a can of paint left over.

 

[phinds]

You assumed Mrs Holland could buy 88/21 cans, which is 4 4/21 cans. Nobody will sell her 4/21 of a can, so she has to buy 5 cans to have enough paint. She will have 17/21 of a can of paint left over.

Sometimes math questions assume you live in the real world (this one does) and sometimes they don't.

 

[anuttarasammyak]

TL;DR Summary: Question about fractions

6 2/3 x 3 1/7 = 20/3 x 22/7 = 440/21. According to the correct answer in the image above, I am correct up to this point - how did they calculate 5 cans are required?

5*4=20 < \frac{440}{21} < 21<25=5*5
4 cans are not enough and 5 cans are enough but with more than 80% of a can not used.

 

[PeroK]

View attachment 354754

Sometimes math questions assume you live in the real world (this one does) and sometimes they don't.

In this case, it depends whether you count the remaining paint as an investment and not part of the cost of painting the wall.

Note that one dimension of the wall ($3\frac 1 7$ metres) is approximately $\pi$ metres.

 

[Mark44]

Sometimes math questions assume you live in the real world (this one does) and sometimes they don't.

This problem falls squarely into the latter category, IMO. How in the world would anyone come up with lengths of 3 1/7 m. or 6 2/3 m.? Every metric length measuring device I've ever seen has subdivisions in tenths, hundredths, etc.

The author of the problem has endeavored to make this problem "relevant" but just made one that is stupid.

 

[phinds]

This problem falls squarely into the latter category, IMO. How in the world would anyone come up with lengths of 3 1/7 m. or 6 2/3 m.? Every metric length measuring device I've ever seen has subdivisions in tenths, hundredths, etc.

Ha. Good catch. I slid right past that one. Yes, it is silly.

 

[anuttarasammyak]

This problem falls squarely into the latter category, IMO. How in the world would anyone come up with lengths of 3 1/7 m. or 6 2/3 m.? Every metric length measuring device I've ever seen has subdivisions in tenths, hundredths, etc.

Two line segments of 22 m and 7 m given, together with 1 m unit line segment , we can get 22/7 m line segment by geometry.

In physics 3 m does not mean 3.00000000… m. Similarly may we say that 22/7 m does not mean 3.1428571…… m ?

[EDIT]
AB = 7m AC = 22 m D is on AB, AD=1m. E is on AC such that BC and DE are parallel. AE= 22/7 m. Is AE inappropriate to be a side of painted area ?

Say fraction is silly, is irrational number also ? E.g. “AB=AC=10m. BAC is rectangle. BC is a side of painted area,etc.”, “how many cans are necessary to paint 10 m radius circle ?“

 

[phinds]

Two line segments of 22 m and 7 m given, together with 1 m unit line segment , we can get 22/7 m line segment by geometry.

(1) I have no idea what you are saying
(2) WhatEVER you are saying, how does it apply to the real world and particularly this problem?

 

[Mark44]

Two line segments of 22 m and 7 m given, together with 1 m unit line segment , we can get 22/7 m line segment by geometry.

But this has nothing to do with the problem as given, in which the dimensions of the wall are 6 2/3 m. by 3 1/7 m. My problem with these dimensions is that I'm not aware of any metric measuring devices with subdivisions in thirds or sevenths.

 

[logicandtruth]

Thank you everyone for your respones with my queries, much appreciated.