Find three consecutive whole numbers whose sum is more than 45 but less than 54.
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Answered by
4
15,16,17 or 16,17,18
Step-by-step explanation:
Define x:
Let x be the smallest number
The other 2 numbers are (x + 1) and (x + 2)
Solve x:
The sum is more than 45 but less that 54
45 < x + (x + 1) + (x + 2) < 54
45 < x + x + 1 + x + 2 < 54
45 < 3x + 3 < 54
42 < 3x < 51
42 < 3x < 51
14 < x < 17
So the smallest number must be between 14 and 17
⇒ The possible numbers are 15 and 16
If x = 15
x + 1 = 15 + 1 = 16
x + 2 = 15 + 2 = 17
Total sum = 15 + 16 + 17 = 48
If x = 16
x + 1 = 16 + 1 = 17
x + 2 = 16 + 2 = 18
Total sum = 16 + 17 + 18 = 51
Answer: The two possible set of numbers are 15 16 and 17 or 16, 17 and 18
Answered by
0
the three consecutive numbers are 13,17,19
theory:here we have given in the question that the sum is greter than 45 and lessthan 54in whole numbers we have 2pairssatiafying the given ondition they are 15;16;17and 16;17;18
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