Question

The sum of three consecutive multiples of 5 is 345. Find the multiples.

Answer 1

Question: The sum of three consecutive multiples of $5$ is $345$. Find the multiples.

Solution:

Take the three consecutive multiples of $5$ which give the sum of 345 are $x$, $x + 5$ and $x + 10$.

So, the equation is $x + x + 5 + x + 10 = 345$

$x + x + 5 + x + 10 = 345$

$3x + 15 = 345$

$3x = 345 - 15$

$3x = 330$

$x = \frac{330}{3}$

$x = 110$

The multiples would be:

$x = 110$

$x +5 = 110+5 = 115$

$x + 10 = 110+10 = 120$

Therefore, the three consecutive multiples of $5$ which give the sum of $345$ are $110$, $115$ and $120$.

Answer 2

Answer:

110,115,120

Step-by-step explanation:

Here, it was asked for consecutive multiples of 5.

Any multiple of 5 can be written as 5x.

The next multiple will be when x is increased by 1.

Thus, the consecutive multiples are 5x, 5(x + 1), 5(x + 2).

According to the given condition,

⇒ 5x + 5(x + 1) + 5(x + 2) = 345

⇒ 5x + 5x + 5 + 5x + 10 = 345

⇒ 15x + 15 = 345

⇒ 15x = 330

⇒ x = 22

Then:

5x = 110

5(x + 1) = 115

5(x + 2) = 120

Therefore, the multiples are 110,115 and 120.

Hope it helps!