MHB Solving a Maths Question: Carpet Tiling for 3.1m x 2.9m Floor

  • Thread starter Thread starter casino281
  • Start date Start date
AI Thread Summary
To determine how many 50cm x 50cm carpet tiles fit a 3.1m x 2.9m floor, first convert the dimensions to centimeters, resulting in 310cm x 290cm, which equals 89,900 cm². Dividing this area by the area of one tile (2,500 cm²) gives 35.96 tiles. Since only whole tiles can be used, 35 tiles would fit, but for complete coverage, 36 tiles are recommended, leaving a small area of 0.01 m² uncovered. The discussion emphasizes the importance of rounding down to whole tiles for practical applications.
casino281
Messages
1
Reaction score
0
I am studying for BTEC level 2 maths.

I need some guidance on solving a maths question.

How many carpet tiles measuring 50cm x 50 cm would fit a floor 3.1meters by 2.9 meters

I thought to convert meters to cm = 310cm x 290 cm = 89900 cm squared

then divide by 50cm x 50cm = 2500 sq cm = 35.96

Is this correct please

thanks(Smile)
 
Mathematics news on Phys.org
Hi casino281,

Welcome to MHB. I'm a little confused about your thread title. Is your name Clive?

I'm glad you found us :)

Jameson
 
casino281 said:
I am studying for BTEC level 2 maths.

I need some guidance on solving a maths question.

How many carpet tiles measuring 50cm x 50 cm would fit a floor 3.1meters by 2.9 meters

I thought to convert meters to cm = 310cm x 290 cm = 89900 cm squared

then divide by 50cm x 50cm = 2500 sq cm = 35.96

Is this correct please

thanks(Smile)

Hey, welcome to the forums (Talking), glad you found us.

Your working and arithmetic are fine, typically this sort of question demands a whole number as an answer so I'd go with 35 since we only deal with whole numbers of tiles (in reality of course you'd use 36 and cut off 0.04 but they don't like that (Rofl))

edit: Is quarter to two really the best time to be doing maths problems? (Wink)
 
SuperSonic4 said:
Hey, welcome to the forums (Talking), glad you found us.

Your working and arithmetic are fine, typically this sort of question demands a whole number as an answer so I'd go with 35 since we only deal with whole numbers of tiles (in reality of course you'd use 36 and cut off 0.04 but they don't like that (Rofl))

edit: Is quarter to two really the best time to be doing maths problems? (Wink)

If you are not allowed to cut the tiles 35 is the largest number of whole tiles you can use, If you have to tile the whole area 36 are sufficient with a piece of area 0.01 m^2 left over.

CB
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Back
Top