Let A-n-n is a 3-digit numberB-9k-2-k- and C-9k-l-k- for some l 0-l-9If the sum of all the elements of the set A - B- C is 274- 400- then l is equal to

3 digit numbers of the form 9k+2 are {101, 109,…,992} ⇒ Sum equals to 1002(1093)=S_1=54650 Now, 274× 400=S_1+S_2 ⇒ S_2=109600 54650 ∴ S_2=54950 S_2=54950=1002[ ...

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Standard VIII

Mathematics

Intersection of Sets

Let A=n∈ℕ:n i...

Question

Let $A = {n \in N : n$ is a $3$ -digit number $}$
$B = {9 k + 2 : k \in N}$ and
$C = {9 k + l : k \in N}$ for some $l (0 < l < 9)$
If the sum of all the elements of the set $A \cap (B \cup C)$ is $274 \times 400$ , then $l$ is equal to

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Solution

$3$ digit numbers of the form $9 k + 2$ are ${101, 109, \dots, 992}$
$\Rightarrow$ Sum equals to $\frac{100}{2} (1093) = S_{1} = 54650$

Now, $274 \times 400 = S_{1} + S_{2}$
$\Rightarrow S_{2} = 109600 - 54650$
$∴ S_{2} = 54950$
$S_{2} = 54950 = \frac{100}{2} [(99 + l) + (990 + l)]$
$\Rightarrow 2 l + 1089 = 1099$
$\Rightarrow l = 5$

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Let A={n∈N:n is a 3-digit number} B={9k+2:k∈N} and C={9k+l:k∈N} for some l (0<l<9) If the sum of all the elements of the set A∩(B∪C) is 274×400, then l is equal to

3 digit numbers of the form 9k+2 are {101,109,…,992} ⇒ Sum equals to 1002(1093)=S1=54650 Now, 274×400=S1+S2 ⇒S2=109600−54650 ∴S2=54950 S2=54950=1002[(99+l)+(990+l)] ⇒2l+1089=1099 ⇒l=5

Let $A = {n \in N : n$ is a $3$ -digit number $}$
$B = {9 k + 2 : k \in N}$ and
$C = {9 k + l : k \in N}$ for some $l (0 < l < 9)$
If the sum of all the elements of the set $A \cap (B \cup C)$ is $274 \times 400$ , then $l$ is equal to