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+1 vote

Let $A=\{10,11,12,13, \dots ,99\}$. How many pairs of numbers $x$ and $y$ are possible so that $x+y\geq 100$ and $x$ and $y$ belong to $A$?

- $2405$
- $2455$
- $1200$
- $1230$

+1 vote

Answer - **Option (B)**

Dividing the set into two sets for convenience without loss of generality.

$S_1 = \{ 10, 11, 12, 13, ..., 49 \}$ and $S_2 = \{50,51,52,53, ..., 99 \}$

Here, No. of elements in $S_1 = 40$ and No. of elements in $S_2 = 50$

Now, to make $x + y \geq 100$

we can choose either any 2 numbers from $S_2$ [Part A] or one number from $S_1$ and one number from $S_2$ [Part B].

Since ordered pairs are not asked, pairs $(a,b)$ and $(b,a)$ will be same, hence

[Part A]

**Total no. of choices** = ways of selecting $2$ distinct numbers from set $S_2$= $^{50} C_2 = \frac{50 \times 49}{2}$

but the choices like $(50,50)$, $51,51$ are missing above. So, no. of choices such that $x$ and $y$ are equal are $= 50$

[Part B]

Now this part is tricky. We have to choose $1$ number from $S_1$ and $1$ number from $S_2$.

For numbers $\{90, 91, 92, ..., 99 \}$ we can choose any number from $S_1$ and have $x + y \geq 100$

So, for these $10$ numbers, we have 40 choices.

So,** total no. of choices ** $ = 10 \times 40 = 400$

Now for $89$, we can choose everything from $S_1$ except number $10$ since that would fail the requirement of $x+y \geq 100$.

So,

for $89$ we have $40-1 = 39$ choices

Similarly, for $88$ we have $40 -2 = 38$ choices since $\{10, 11 \}$ from $S_1$ will fail our criteria of sum greater than 100.

Similarly, for $87$, we have $40-3 = 37$ choices

...

Similarly, for $51$, we have $1$ choice (which is $49$)

Similarly, for $50$, we have $0$ choices.

Hence **total no. of choices** $ = 0 + 1 + 2+ 3+ 4+ ... + 39 = \frac{(39)(40)}{2} = 780$

Hence, Answer $= 1225 +50+780 + 400 = 2455$

0

@srestha is this pair valid ? $(50,50)$ i.e. can I use same element twice ? I have excluded those. If I can include these, then the answer will be $2455$.

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