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Let $T$ be the set of integers $\{3,11,19,27, ..... , 451, 459, 467\}$ and $S$ be a subset of $T$ such that the sum of no two elements of $S$ is $470$. The maximum possible number of elements in $S$ is ?

1. $31$
2. $28$
3. $29$
4. $30$

Answer is given as c. 29. But as per my calculations the answer has to be d.30. Let me know if my answer is correct.

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+1
Yes, I also get the answer as $30$

+1 vote

My take on the question -

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First thing, we've to notice few things-

1. the elements of $T$ are in Arithmetic Progression or AP.
2. the sum of the first term & the last term is $470$
3. Similarly, the sum of the $2^{nd}$ term & the $2^{nd}$ last term is $470$ and the pattern continues ......

Now, as we know,

the formula to find $n^{th}$ term of an AP series

$\qquad a+(n-1)d = t_n$

$\qquad where, a = \text{first term,}$

$\qquad \text{n=no. of terms,}$

$\qquad \text{d=common difference,}$

$\qquad t_n = n^{th} term$

∴ from this formula, we can determine the no. of elements in the set $T$.

in set $T$ -

$\qquad \text{first term (a) = 3}$

$\qquad \text{common difference between elements = (d) = 8}$

$\qquad n^{th} \text{term = 467}$

$\qquad \text{no. of terms(n) = ?}$

∴ $467 = 3+(n-1)8$

$\Rightarrow 467 = 3+8n-8$

$\Rightarrow 467 = 8n-5$

$\Rightarrow 467+5 = 8n$

$\Rightarrow n = \dfrac{472}{8} = 59$

∴ In set $T$, there are a total of $59$ elements.

∴ In set $T$, there will be $\left (\dfrac{59}{2} = 29 \right ) pairs$ (if we add the elements in each pair, then we'll get $470$) , & only $1$ element left i.e. $\left (\dfrac{3+467}{2} = \dfrac{470}{2} = 235\right)$.

From set $T$, we've to form a subset $S$ where the sum of no two elements will be $470$.

∴ We can take $1$ element from $29$ pairs i.e. $29$ elements & place them to set $S$.

And we can also place the only element $( i.e. 235)$ which has no pair, to $S$.

∴ $\color{violet}{\text{Set S can have maximum}}$ $\color{maroon}{30}$ $\color{violet}{\text{elements in it.}}$

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