Question

2. In a hockey tournament , a total of 153 matches were played. If each team played one match with every other team, find the total number of teams that participated in the tournament​

Answer 1

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$\it\huge\mathfrak\red{Answer:-}$]

There are 153 matches to be played with let;s assume n teams.

Therefore the number of matches that can be played with these n teams is:

( 1 + 2 + 3 + . . . + n - 1 )

Σ ( n - 1 ) = 153

Σ ( n ) - Σ ( 1 ) = 153

( n ( n- 1 ) ) / 2 - n = 153

n ^ 2 - n = 153 * 2

n ^ 2 - n - 306

n ^ 2 - 18 * n + 17 * n - 306 = 0

n( n - 18 ) + 17( n - 18) = 0

(n - 18)( n + 17 ) = 0

n = 18 because there cannot be -ve teams.

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$\it\huge\mathfrak\red{Thanks}$]

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Answer 2

ANSWER

Here, let the number of teams played the matches be n.

Now,

If n teams play matches with every other team then the formula will be =>

$(n\times\dfrac{n-1}{2})$

So according to the question we have=>

$(n\times\dfrac{n-1}{2}$)=153

=>$n^2$-n = 306

=>$n^2$-n -306 =0

Since this is a quadratic equation we shall use the middle term split method to solve it=>

=>$n^2$-18n +17n -306 =0

=>n(n-18) + 17(n-18) =0

=>(n-18)(n+17) =0

Hence,

either n= 18

or

n = (-17)

{neglected value as the number of teams cannot be negative)

Hence the team played in the matches were 18.

Remember

1)An = a+(n-1)d

where An = last term of an AP.

2)Sn = a+$\dfrac{n}{2}${2a + (n-1)d}

where Sn= sum till the nth term of an AP.