Skip to main content

Updates ...

Visit the website 123iitjee.manishverma.site for latest posts, courses, admission & more.

For guest/sponsored article(s), please check this link.

A 5.25 m by 3.78 m rectangular courtyard ...

A 5.25 m by 3.78 m rectangular courtyard is to be paved with square tiles of the same size such that only whole tiles are used. What is the largest possible size of such a tile? Also, find the number of tiles required.

Solution


Let a be the edge of the square tile.

Then, $a\times m=5.25$ and $a\times n=3.78$

Dividing, $\frac {m}{n}=\frac {525}{378}$

= $\frac {175}{126}$ [dividing by 3]

= $\frac {25}{18}$ [dividing by 7]

Smallest value of m = 25 and smallest value of n = 18 (for the square tile to be of largest size, m & n should be the least)

Number of tiles = m.n = 25 x 18 = 450

PS: The solution in file format is embedded below.

Popular posts from this blog

$f(x)=x^6+2x^4+x^3+2x+3 $

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$

$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R $. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $ \Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$