Question

The sum of three consecutive multiples of 3 is 90. What is the largest of these numbers

Answer 1

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let the first multiple be X
so
next two multiples are
x+3
&
x+6

according to question
x + x +3 + x +6=90
3x+9=90
3x=90-9=81

x=81/3=27

therefore the multiples are
27
30
33

Answer 2

Answer: 33

Step-by-step explanation:

Given,

• Three consecutive multiples of 3
• Their sum is 90

To Find: The largest of the three consecutive multiples of 3 whose sum is 90.

Explanation:

• If a number is a multiple of 3, then vice-versa 3 will also be a factor of that number.
• Let us assume three consecutive multiples of 3, say,

      3n, 3(n + 1) & 3(n + 2).

• As per the condition mentioned in the question-statement,

       $[3n + 3(n + 1) + 3(n + 2)] = 90\\or, 3n+3n+3+3n+6=90\\or, (3n+3n+3n)+(3+9)=90\\or, 9n+9=90\\or, 9n=(90-9)\\or, 9n=81\\or, n=\frac{81}{9}\\ or, n=9$

• Therefore, 3n = (3 * 9) = 27
• 3(n + 1) = 3(9 + 1) = (3 * 10) = 30
• 3(n + 2) = 3(9 + 2) = (3 * 11) = 33
• Among 27, 30 and 33, the largest multiple of 3 is 33.
• Thus, our required answer is 33.

The sum of three consecutive multiples of 9 is 378. Then find the three multiples.

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If three consecutive multiples of 11 are multiplied by 2, 3 and 4 respectively and then added, the sum is 814, then find each of the multiples.

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