Question

A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is k, then k – 20 = ?

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Answer 1

Explanation:

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Question:-

A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is k, then k – 20 = ?

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Given :-

• Pack contains n cards numbered from 1 to n. .
• Two consecutive numbered cards are removed from the pack .
• The sum of the numbers on the remaining cards is 1224.
• The smaller of the numbers on the removed cards is k

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To Find :-

• then k – 20 = ?

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According to the given task :-

n(n +1)/2-k(-k+1) =1224

On solving

square are cut

k = n(n+1)-2448

n( n+1)>2448

n =48 in satisfying :-

then k = 24 and k = k – 20

putting values :-

24-20 =4

Answer is K = 4

Answer 2

The sum of numbers on all $\sf{n}$ cards is,

$\longrightarrow\sf{1+2+3+\,\dots\,+n=\dfrac{n(n+1)}{2}}$

 

If smaller of the numbers on the two removed cards is $\sf{k,}$ then the number on the other card is $\sf{k+1}$ because two consecutively numbered cards are removed.

Given that the sum of numbers on the remaining cards is 1224. Thus we have,

$\longrightarrow\sf{\dfrac{n(n+1)}{2}-(k+k+1)=1224}$

 

$\longrightarrow\sf{\dfrac{n(n+1)}{2}-2k-1=1224}$

$\longrightarrow\sf{\dfrac{n(n+1)}{2}-1225=2k}$

$\longrightarrow\sf{n(n+1)-2450=4k}$

$\longrightarrow\sf{n(n+1)-49\times50=4k\quad\quad\dots(1)}$

Since $\sf{k\ \textgreater\ 0,}$

$\longrightarrow\sf{4k\ \textgreater\ 0}$

From (1),

$\longrightarrow\sf{n(n+1)-49\times50\ \textgreater\ 0}$

$\longrightarrow\sf{n^2+n-49\times50\ \textgreater\ 0}$

$\longrightarrow\sf{n^2+50n-49n-49\times50\ \textgreater\ 0}$

$\longrightarrow\sf{(n-49)(n+50)\ \textgreater\ 0}$

$\Longrightarrow\sf{n\ \textgreater\ 49\quad or\quad n\ \textless\ -50}$

But since $\sf{n\geq1,}$

$\longrightarrow\sf{n\ \textgreater\ 49}$

We also see that,

$\longrightarrow\sf{k\ \textless\ n}$

$\longrightarrow\sf{4k\ \textless\ 4n}$

From (1),

$\longrightarrow\sf{n(n+1)-2450\ \textless\ 4n}$

$\longrightarrow\sf{n^2+n-2450\ \textless\ 4n}$

$\longrightarrow\sf{n^2-3n-2450\ \textless\ 0}$

$\Longrightarrow\sf{n\ \textless\ \dfrac{3+\sqrt{3^2-(4\times1\times-2450)}}{2}\quad and\quad n\ \textgreater\ \dfrac{3-\sqrt{3^2-(4\times1\times-2450)}}{2}}$

$\longrightarrow\sf{n\ \textless\ \dfrac{3+\sqrt{9-(-9800)}}{2}\quad and\quad n\ \textgreater\ \dfrac{3-\sqrt{9-(-9800)}}{2}}$

$\longrightarrow\sf{n\ \textless\ \dfrac{3+\sqrt{9809}}{2}\quad and\quad n\ \textgreater\ \dfrac{3-\sqrt{9809}}{2}}$

Since $\sf{\sqrt{9809}\approx99,}$ we simply take the condition as,

$\longrightarrow\sf{n\ \textless\ \dfrac{3+99}{2}\quad and\quad n\ \textgreater\ \dfrac{3-99}{2}}$

$\longrightarrow\sf{n\ \textless\ 51\quad and\quad n\ \textgreater\ -48}$

$\longrightarrow\sf{n\in(-48,\ 51)}$

Since $\sf{n\ \textgreater\ 49,}$

$\longrightarrow\sf{n\in(49,\ 51)}$

Since $\sf{n\in\mathbb{N},}$

$\longrightarrow\sf{n=50}$

[Note: $\sf{n=51}$ is also a possibility if the approximation $\sf{\sqrt{9809}\approx99}$ is not considered. But it does not give an integer value for $\sf{k.}$]

Hence (1) becomes,

$\longrightarrow\sf{50\times51-49\times50=4k}$

$\longrightarrow\sf{50(51-49)=4k}$

$\longrightarrow\sf{50\times2=4k}$

$\longrightarrow\sf{4k=100}$

$\longrightarrow\sf{k=25}$

$\longrightarrow\underline{\underline{\sf{k-20=5}}}$

Hence $\bf{5}$ is the answer.