Latest Elitmus Aptitude Question SOLUTION: A square was given. Inside the square there are white tiles and black tiles. Black tiles was among the diagonal of the square and dimensions of both white as well as black tiles is

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(#M40153922) ELITMUS Numerical Ability question No of white tile Keep an EYE

A square was given. Inside the square there are white tiles and black tiles. Black tiles was among the diagonal of the square and dimensions of both white as well as black tiles is 1cmx1cm. If there are 81black tiles in the sqaure. Then find the no of white tiles in it.

Asked In Elitmus amit kumar (8 years ago)

Solved Vikas Jha Read Solution (15)

Is this Puzzle helpful?   (15)   (6) Submit Your Solution Area and Volume

Read Solution (15) :

Sourabh Prasad  (8 years ago)

If n is no. of sides of a square.
Then,
No. of tiles in Diagonal = 2n-1 (odd)
And No. of tiles in Diagonal = 2n (even)

Here given no of tiles in Diagonal =81 which is odd.

So 2n-1=81
2n= 82
n=41

So, total no of tiles = 41*41
= 1681

Is this solution Helpfull? Yes (30) | No (4)  

Vikas Jha  (8 years ago) Best Solution

Total no of white tiles will be 1600..
Total no of tiles will be 1681(including black and white)
In a 3*3 square ,,no of black tiles will be 5.. Similarly in 5*5 square,, no of black tiles will be 4extra around every corner hence 9.
So we get two series..
1). 5,9,13......81
2). 3,5,7.....
From first series total no of term will be 20. Find the 20th term in second series.. It will be 41.. So 41*41 will be the total no of tiles i.e 1681 out of which 81 are black hence 1600 are white..

Is this solution Helpfull? Yes (26) | No (2)  

adi  (8 years ago)

1600...
As total no of black tiles are 81 and also they are on diagonals of square.. so length if diagonal of square= 41*√2
So lengthy of side of square= 41
So a real of square= 41*41
So total area of square= area of black tiles+area of square tiles
41*41= 81+ area of white tiles
Area if white tiles= 41*41-81=1600
So total no - 1600/1=1600

Is this solution Helpfull? Yes (11) | No (6)  

Vikas Jha  (8 years ago)

There's one more easy trick.. Each diagonal will share 1 common cube.. And 80 will be the sum of two diagonals ignoring the common one for a while.. Hence each diagonal will have 40+1(common one).. So the square will be of 41*41. Because no of cube in diagonal will also be equal to no of cube along side.
Note:-We are not talking about length so no of cube in diagonal will not be root two times that of side.. It will be same.

Is this solution Helpfull? Yes (9) | No   

Paras Jain  (8 years ago)

The question was to find the total number of tiles, not to find the total no of white tiles,
let me remind you , You are wrong dude.

Is this solution Helpfull? Yes (4) | No (6)  

vivek  (8 years ago)

As given that total number of black tiles is 81 which are placed in diagonals only.
So, each diagonal contains 41 black marble. i.e 41+41=82 and 1 black marble is common, which is in middle.
Also size of each marble is 1 cm.
Now take ! diagonal one time to count the white marble in upper half triangle which is in series 40,39,38,....,1.
so the sum of series is as, 40*41/2=820.
Also in case of lower half triangle we get same no. of white marbles i.e, 820
so total white marble used in square is 820*2=1640.
But we have also taken 40 black marbles while counting the white marbles as we consider only one diagonal but there is one more diagonal which contributes its 40 marbles in counting of black barbes.
So, total white marble = 1640-40=1600
And total marble= 1600+81=1681

Is this solution Helpfull? Yes (2) | No (4)  

Abhishek Desai  (8 years ago)

Total tiles= 1681
White tiles=1681-81=>1600

Is this solution Helpfull? Yes (2) | No (2)  

Nawal Kishor Sharma  (8 years ago)

saurabh prasad i thik i want to know if n is side of square then how much the diagonal plz explain.....

Is this solution Helpfull? Yes (2) | No   

kuldeep kumar  (8 years ago)

1600 is the answer

Is this solution Helpfull? Yes (1) | No   

shiwangi bhatia  (7 years ago)

Problem

The floor of a square room is tiled with square tiles. Along the two diagonals of room measuring 5x5 tiles there are nine tiles.


If there is 81 tiles along both diagonals, how many tiles are there on the floor?

Solution

The diagonal running from top left to bottom right will have one square in every column. So an n x n square will have n tiles in the diagonal.

The same will be true for the other diagonal,
but if the dimensions of the square are odd, the diagonals will have a common tile at the centre.

n being even: Tiles in diagonal = 2n
n being odd: Tiles in diagonal = 2n - 1 (which itself is odd)

So if there are 81 tiles in the diagonal, n must be odd.
Solving 2n minus 1 = 81 implies 2n = 82 implies n = 41.
Hence, there are 41 times 41 = 1681 tiles in the room.

Is this solution Helpfull? Yes (1) | No   

amit kumar  (8 years ago)

@Adi how u take 41root2.
think of a small case. if we have a 9 block of squre. then in that 6 black nd rest white. then if 81 black then how many white?? xplain the logic clearly buddy

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pawan  (8 years ago)

kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

Is this solution Helpfull? Yes | No (1)  

poonam  (8 years ago)

Diagonal= 41 cm. Side= 41√2 by hypotenuse formula.

Is this solution Helpfull? Yes | No (2)  

KIRAN  (8 years ago)

i think all are wrong,he had given that 81 black tiles are there,then for 5*5 matrix we have 9 tiles as diagonals,so,9 black tile,remaining are white tiles,they are 16,for 16 white tiles-->9black tiles,
81 black tiles-->144 white tiles

Is this solution Helpfull? Yes | No (6)  

shalini kumari  (8 years ago)

P(A U B U C) = P(A) + P(B) + P(C) - P(A∩B) - P(B∩C) - P(C∩A) + P(A∩B∩C)
this gives 36 ( we calculate here P(A∩B) - P(B∩C) - P(C∩A)=)
is this we get the answer for who written exam two and more than two
we want of only two exams written members are
36-3*common
=36-3*5=21

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