Question

the sum of six consecutive even numbers of set-A is 402. what is the sum of another set-B of four consecutive numbers whose lowest number is 15 less than double the lowest number of set-A?

Answer 1

Let x be the lowest number in set A. Since the numbers in set A are consecutive even numbers, we know that the next number in the set is x+2, the next is x+4, and so on. We are told that the sum of the six numbers in set A is 402, so we can set up the equation:

x + (x+2) + (x+4) + (x+6) + (x+8) + (x+10) = 402

We can simplify this equation by combining like terms:

6x + 30 = 402

6x = 372

x = 62

Now, we are told that the lowest number in set B is 15 less than double the lowest number in set A. Double the lowest number in set A is 2*62 = 124, so the lowest number in set B is 124-15 = 109. Since the numbers in set B are consecutive, we know that the next number in the set is 109+1 = 110, the next is 110+1 = 111, and the final number is 111+1 = 112.

The sum of the four numbers in set B can be found by adding them together:

109 + 110 + 111 + 112 = 442