# CAT 2018 Set-1 | Question: 86

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A $\text{CAT}$ aspirant appears for a certain number of tests. His average score increases by $1$ if the first $10$ tests are not considered, and decreases by $1$ if the last $10$ tests are not considered. If his average scores for the first $10$ and the last $10$ tests are $20$ and $30$, respectively, then the total number of tests taken by him is ________

Let $n\text{’}$ be the total number of tests taken by a $\text{CAT}$ aspirant, and his average score be $x\text{’}.$

His average score increase by $1,$ if the first $10$ tests are not considered, and his average score for the first $10$ tests is $20.$

So$, (n-10) (x+1) + 10 \times 20 = n \times x$

$\Rightarrow nx + n – 10x – 10 + 200 = nx$

$\Rightarrow -10x + n + 190 = 0$

$\Rightarrow 10x - n - 190 = 0 \quad \longrightarrow (1)$

His average score decreased by $1$ if the last tests are not considered, and his average score for the last $10$ tests is $30.$

So, $(n-10)(x-1) + 10 \times 30 = n \times x$

$\Rightarrow nx – n – 10x + 10 + 300 = nx$

$\Rightarrow -10x – n + 310 = 0$

$\Rightarrow 10x + n – 310 = 0 \quad \longrightarrow (2)$

Adding the equation $(1)$ and $(2),$ we get

$\begin{array} {cc}10x - n - 190 = 0 \\ 10x + n – 310 = 0 \\\hline 20x – 500 = 0 \end{array}$

$\Rightarrow 20x = 500$

$\Rightarrow x = \frac{500}{20}$

$\Rightarrow \boxed{x = 25}$

From, the equation $(2),$ we get

$10x + n – 310 = 0$

$\Rightarrow 10(25) + n – 310 = 0$

$\Rightarrow 250 + n – 310 = 0$

$\Rightarrow \boxed {n = 60}$

$\therefore$ The number of tests taken by $\text{CAT}$ aspirant is $60.$

Correct Answer $: 60$

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