Question

the sum of three consecutive multiples of 3 is 333. find the multiples.

Answer 1

$let \: the \: first \: multiple \: = 3x \\ the \: other \: two \: consecutive \: multiples = 3(x + 1) \: and \: 3(x + 2) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: acc. \: to \: ques - \\ 3x + 3(x + 1) + 3(x + 2) = 333 \\ = > 3x + 3x + 3 + 3x + 6 = 333 \\ = > 9x + 9 = 333 \\ = > 9x = 333 - 9 \\ = > 9x = 324 \\ = > x = \frac{324}{9} \\ = > x = 36 \\ \\ the \: first \: consecutive \: multiple \: = 3 \times 36 = 108 \\ other \: consecutive \: multiples = 3(36 + 1) = 111 \: and \: 3(36 + 2) = 114$
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Answer 2

Answer:

• The first required number = 108
• The second required number = 111
• The third required number = 114

Given:

• The sum of three consecutive multiples of 3 is 333.

Need to find:

• The first required number = ?
• The second required number = ?
• The third required number = ?

Solution:

Let,

• The numbers = y, (y + 3), (y + 6).

According To The Question,

y + (y + 3) + (y + 6) = 333

➜ 3y + 9 = 333

➜ 3y = 333 - 9

➜ 3y = 324

➜ y = 324/3

➜ y = 108

Therefore:

• The first number (y) is 108.
• The second number (y + 3) is 111.
• The third Number (y + 6) is 114.

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