The sum of four consecutive multiples of 5 is 230. What is the greatest of these numbers?

5 n + 5 ( n + 1 ) + 5 ( n + 2 ) + 5 ( n + 3 ) = 230 5 n + 5 n + 5 + 5 n + 10 + 5 n + 15 = 230 20 n + 30 = 230 20 n = 200 / : 20 = 10 ⇒ 5 ( n + 3 ) = 5 ⋅ ( 10 + 3 ) = 5 ⋅ 13 = 65 A n s . t h e g re a t es t i s 65.

Answered by kate200468 | 2024-06-10

To find the greatest of four consecutive multiples of 5 summing to 230, we set up an equation with the first multiple as x. Solving this, we discover that the first multiple is 50, making the greatest multiple 65.

The question asks for the greatest of four consecutive multiples of 5 that add up to 230. To solve this problem, let the first multiple be represented as x. Therefore, the four consecutive multiples are x, x+5, x+10, and x+15. Their sum is given by the equation:
x + (x + 5) + (x + 10) + (x + 15) = 230
Combining like terms, we get:
4x + 30 = 230
Subtracting 30 from both sides, we get:
4x = 200
Dividing both sides by 4, we find:
x = 50
So, the greatest number, which is x+15, is:
50 + 15 = 65
Therefore, the greatest of these four consecutive multiples of 5 is 65.

Answered by SravyaDa | 2024-06-25

The greatest of the four consecutive multiples of 5 that sum to 230 is 65. This was found by setting up an equation with the consecutive multiples and solving for n, which equals 10. The corresponding multiples are 50, 55, 60, and 65, with 65 being the greatest.
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Answered by kate200468 | 2024-09-27