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Sets and Relations question

2022 · 26 Jun · Shift 1 · Q40
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  5. /2022 · 26 Jun · Shift 1 · Q40

Sets and Relations question

2022 · 26 Jun · Shift 1 · Q40

JEE MainMathematicsSets and RelationsNumerical+4 / −1
Let A=∑i=110∑j=110min⁡ {i,j}A = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\min \,\{ i,j\} } }A=i=1∑10​j=1∑10​min{i,j} and B=∑i=110∑j=110max⁡ {i,j}B = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\max \,\{ i,j\} } }B=i=1∑10​j=1∑10​max{i,j}. Then A + B is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 1100

  1. We use the identity
min⁡(i,j)+max⁡(i,j)=i+j\min(i,j)+\max(i,j)=i+jmin(i,j)+max(i,j)=i+j

for every pair (i,j)(i,j)(i,j).

  1. Therefore,
A+B=∑i=110∑j=110(min⁡{i,j}+max⁡{i,j})=∑i=110∑j=110(i+j).A+B=\sum_{i=1}^{10}\sum_{j=1}^{10}\big(\min\{i,j\}+\max\{i,j\}\big) =\sum_{i=1}^{10}\sum_{j=1}^{10}(i+j).A+B=i=1∑10​j=1∑10​(min{i,j}+max{i,j})=i=1∑10​j=1∑10​(i+j).
  1. Split the double sum:
A+B=∑i=110∑j=110i+∑i=110∑j=110j.A+B=\sum_{i=1}^{10}\sum_{j=1}^{10} i + \sum_{i=1}^{10}\sum_{j=1}^{10} j.A+B=i=1∑10​j=1∑10​i+i=1∑10​j=1∑10​j.
  1. Compute each part.

For the first part, for each fixed iii, the term iii appears for all 101010 values of jjj:

∑i=110∑j=110i=∑i=11010i=10∑i=110i.\sum_{i=1}^{10}\sum_{j=1}^{10} i = \sum_{i=1}^{10} 10i = 10\sum_{i=1}^{10} i.i=1∑10​j=1∑10​i=i=1∑10​10i=10i=1∑10​i.

Similarly,

∑i=110∑j=110j=10∑j=110j.\sum_{i=1}^{10}\sum_{j=1}^{10} j = 10\sum_{j=1}^{10} j.i=1∑10​j=1∑10​j=10j=1∑10​j.

So,

A+B=10∑i=110i+10∑j=110j=20∑k=110k.A+B=10\sum_{i=1}^{10} i + 10\sum_{j=1}^{10} j =20\sum_{k=1}^{10} k.A+B=10i=1∑10​i+10j=1∑10​j=20k=1∑10​k.
  1. Now,
∑k=110k=10⋅112=55.\sum_{k=1}^{10} k = \frac{10\cdot 11}{2}=55.k=1∑10​k=210⋅11​=55.

Hence,

A+B=20×55=1100.A+B=20\times 55=1100.A+B=20×55=1100.
  1. Final answer:
1100\boxed{1100}1100​
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