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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.

asked in Numerical Ability by (57 points)
edited by | 158 views
+1
Yes, I also get the answer as $30$

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My take on the question - 

------------------------------------------------------------------------------------

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 $S$.

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

answered by Boss (14.6k points)
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