Math, asked by anuragdhellrajer, 3 months ago

A set of 30 consecutive natural numbers is divided into two subsets, A and B. In set A there are 15 consecutive natural numbers whereas the rest are in set B. If the average of the numbers in set B is 83, what is the last number in set A?

Answers

Answered by prince5311
0

Answer:

Answer:

(a

2

−b

2

) is divisible by 3

i.e (a−b)(a+b) is divisible by 3.

Case(1) (a+b) is divisible by 3

If a is chosen from (3k+1) set ,then b must be from (3k+2) set and viceversa.

If a is chosen from (3k) set, then b must be from (3k).

So, number of ways = 10*10

Note that when a is from 3k and b is from 3k,

then (a−b) is also divisible by 3.

Case(2) (a−b) but not (a+b) is divisible by 3

If a is chosen from (3k+1) set ,then b must be from (3k+1) set .

Same for (3k+2) set.

So, number of ways = 2* 10*10

Number of ways of choosing such that a,b such that (a

2

−b

2

) is divisible by 3

=30*30

Probability

=

30×30

5×10×10

=

9

5

Step-by-step explanation:

hope u ike

THANKS

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