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In a cricket tournament 16 school teams participated. A sum of $₹ 8000$ is to be awarded among themselves as prize money. If the last placed team is awarded ₹ 275 in prize money and the award increases by the same amount for successive finishing places, how much amount will the first place team receive?
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Verified Answer
Let the first place team got ₹ $a$.
Since, award money increased by the same amount for successive finishing places. Therefore series becomes an AP
Let the fixed amount be $d$.
Here, $l=275, n=16$ and $S_{16}=8000$
$$
\begin{aligned}
&\begin{array}{l}
\therefore l=a+(n-1) d \\
\Rightarrow l=a+(16-1)(-d) \\
\qquad \text { [we take common difference }(-\mathrm{ve}) \\
\text { because series is decreasing] }
\end{array} \\
&\Rightarrow 275=a-15 d \\
&\text { and } S_{16}=\frac{16}{2}[2 a+(n-1) \cdot(-d)] \\
&\Rightarrow 8000=8[2 a+(16-1)(-d)] \\
&\Rightarrow 8000=8[2 a-15 d] \\
&\Rightarrow 1000=2 a-15 d \\
&\text { On subtracting Eq. (i) from Eq. (ii), we get } \\
&\begin{array}{l}
\text { (2a } \\
\Rightarrow 15 d)-(a-15 d)=1000-275
\end{array} \\
&\Rightarrow 2 a-15 d-a+15 d=725 \Rightarrow a=725
\end{aligned}
$$
Hence, first place team receive $₹ 725$.
Since, award money increased by the same amount for successive finishing places. Therefore series becomes an AP
Let the fixed amount be $d$.
Here, $l=275, n=16$ and $S_{16}=8000$
$$
\begin{aligned}
&\begin{array}{l}
\therefore l=a+(n-1) d \\
\Rightarrow l=a+(16-1)(-d) \\
\qquad \text { [we take common difference }(-\mathrm{ve}) \\
\text { because series is decreasing] }
\end{array} \\
&\Rightarrow 275=a-15 d \\
&\text { and } S_{16}=\frac{16}{2}[2 a+(n-1) \cdot(-d)] \\
&\Rightarrow 8000=8[2 a+(16-1)(-d)] \\
&\Rightarrow 8000=8[2 a-15 d] \\
&\Rightarrow 1000=2 a-15 d \\
&\text { On subtracting Eq. (i) from Eq. (ii), we get } \\
&\begin{array}{l}
\text { (2a } \\
\Rightarrow 15 d)-(a-15 d)=1000-275
\end{array} \\
&\Rightarrow 2 a-15 d-a+15 d=725 \Rightarrow a=725
\end{aligned}
$$
Hence, first place team receive $₹ 725$.
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