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An infinite two-dimensional pattern is indicated below.

The smallest closed figure made by the lines is called a unit triangle. Within every unit triangle, there is a mouse.

At every vertex there is a laddoo. What is the average number of laddoos per mouse?

1. $\quad 3$
2. $\quad 2$
3. $\quad 1$
4. $\left(\dfrac{1}{2}\right)$
5. $\left(\dfrac{1}{3}\right)$

edited | 460 views

Let the number of lines per direction be $1$ as shown bellow:

Here $x,y,z$ depict the directions of the line.

$\eta_{laddoo} = 1$

$\eta_{mouse}= 0$

Add one more parallel line to each dimension $x,y,z$ as shown bellow:

Encircled points represent laddoo  $\Rightarrow \eta_{laddoo} = 3$

and triangle enclosed by them represent mouse $\implies \eta_{mouse} = 1$

Similarly for $3$ lines in each direction

$\eta_{laddoo} = 6\quad (1+2+3)$

$\eta_{mouse} = 4 \quad(2^{2})$

As we continue we get a series which depends upon the no. of lines per direction $($let say $l)$

So, $\eta_{laddoo} = \frac{l\left ( l+1 \right )}{2}$

$\eta_{mouse} =\left ( l-1 \right )^{2}$

$\lim _{l\to \infty} \frac{ \eta_{laddoo}}{\eta_{mouse}} =1/2.$

So, $D:$ $1/2$ is the correct answer.

by (235 points)
edited
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this should be chosen as best answer!! (y)
+3

Consider the middle laddoo marked with blue color. This blue laddoo will be equally shared by 6 mice surrounding it. So each mouse will be having 1/6th share of this laddoo.

Share on one laddoo -> 1/6

Share on 3 laddoo-s -> 3*1/6 = 1/2

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In 3rd figure, there are Only 2 lines in z-direction
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Share on 3 laddoo-s -> 3*1/6 = 1/2

@MiNiPanda Can you please explain this?

On what basis you said this. Didn't able to get it.

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@techbd123

Can you please check this commented answer, if it's mathematically valid or not?

Ans will be 1/2.Every Laddoo shared by 2 mouse

by Veteran (119k points)
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can you please clarify, I am getting 1
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every laddoo is shared by 6 mouse ....there are 6 triangles surrounding a laddoo
This question can also answered by using little bit of graph theory.

Consider graph $G=(V,E)$ where $V$ is infinite set of all laddoos and all mouses. $E$ is edge set where it contains an edge between mouse $M_i \in V$ to laddoo $L_j \in V$ if and only if $M_i$ can legally eat laddoo $L_j$.

Now, total number of edges be $E$ = $6L$ = $3M$ (Because every edge is only between some mouse and some laddoo. No other case is possible.)

$\therefore \frac{L}{M} = \frac{3}{6} = 1/2$

And that remains correct even if $L, M \rightarrow \infty.$
by (321 points)