Find three conescutive whole numbers whose sum is more than 45 but less
than 54.
Answers
Let the three consecutive whole numbers be x, ( x + 1), ( x +2 ).
According to the question,
➡️ 54 > ( x) + ( x +1 ) + ( x +2 ) > 45
➡️ 54 > 3x + 3 > 45
Solving 54 > 3x +3 separately,
➡️ 54 - 3 > 3x + 3 - 3 [ Subtracting both sides by 3 ]
➡️ 51 > 3x
➡️ 51/3 > 3x/3 [ Dividing both sides by 3 ]
➡️ 17 > x --> ( i )
Now, Solving 3x + 3 > 45
➡️ 3x + 3 - 3 > 45 - 3 [ Subtracting both sides by 3 ]
➡️ 3x > 42
➡️ 3x /3 > 42/3 [ Dividing both sides by 3 ]
➡️ x > 14 --> ( ii )
Now, from equation ( i ) and ( ii ),
17 > x > 14
Three consecutive whole numbers are 14, 15, 16.
Answer:
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Step-by-step explanation:
Let the three conescutive whole number be x, x + 1, x + 2.
From the given information,
45 < x + (x+1) + (x+2) and
x + (x+1) + (x+2) < 54
∴ 45 < 3x + 3 and 3x + 3 < 54
∴ 42 < 3x and 3x < 51
∴ 14 < x and x < 17
∴ x = 15,16
If x = 15,
∴ The other two numbers are 16 and 17.
If x = 16,
∴ The other two numbers are 17 and 18.
Hence, three conescutive whole numbers are 15, 16, 17 or 16, 17, 18.