Domino

Aman verma wrote:Q: A Domino was build using match-boxes.All the matchboxes were arranged in the form of an equilateral triangle i.e one matchbox in the front row , 2 matchboxes in the second row,3 matchboxes in the third row, 4 matchboxes in the fourth row and so on . If 669 more matchboxes are added in such a way that all the matchboxes are now in the form of a square and each of the sides now contain 8 matchboxes less than each side of the equilateral triangle, then ,initially ,how many matchboxes were there:

Say if there were n match boxes on the base for the equilateral "triangle" ; total no of match boxes = n(n+1)/2

Square has (n-8)*(n-8) = (n-8)^2 match boxes

n(n+1) = 2*(n-8)^2 -2*669
=> n^2 -n = 2n^2 -32n + 128 -2*669
=>n^2-33n-1210=0

1210= 2*5*11*11
(n-55) (n+22) =0
n = 55

[spoiler]No of match boxes = 55*56/2 =1540[/spoiler]

Aman verma wrote:Q: A Domino was build using match-boxes.All the matchboxes were arranged in the form of an equilateral triangle i.e one matchbox in the front row , 2 matchboxes in the second row,3 matchboxes in the third row, 4 matchboxes in the fourth row and so on . If 669 more matchboxes are added in such a way that all the matchboxes are now in the form of a square and each of the sides now contain 8 matchboxes less than each side of the equilateral triangle, then ,initially ,how many matchboxes were there:

a) 2056

b)1540

c)1400

d)1220

e)1115



Note: I guess everybody knows what a domino is .

Let there be x matchboxes put in n such rows that made an equilateral triangle, so that
n/2 (n + 1) = x, the longest row here would be containing the maximum, n matchboxes to it, if 1 unit is the length of each matchbox (may I assume that the matchboxes are put along its lengths and also that these are square?), then n is the side of the equilateral triangle so formed. Now we have to imagine a square of side (n – 8), such that

(n – 8)^2 = x + 669, and we’re to find x.

Can we do n/2 (n + 1) = (n – 8)^2 – 669, first?

OK, so n^2 – 33 n – 1210 = 0

Or (n + 22) (n – 55) = 0

Or n = 55, and hence x = n/2 (n + 1) = 55 × 28 = [spoiler]1540[/spoiler].

[spoiler]B[/spoiler]

If that's the OA then the question really challanges our general awareness about all sorts of matchboxes in the world, too!

Ans.[spoiler]b)1540[/spoiler]. I couldn't have possibly solved this algebraically. My approach was backsolving.

And the question is concerned with the number of matcheboxes, not with the dimension of matchboxes!