Question

Find 3 consecutive whole numbers whose sum is more than 45 but less than 54.

Answer 1

Answer:

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

Answer 2

Answer:

16,17,18 and 15,16,17

Step-by-step explanation:

Let x be the smallest whole Number

then the other two whole numbers are (x+1) and x(+2)

According to given condition

45< x + (x+1) + (x+2) <54

45<x + x +1 +x+2 < 54

45 < 3x + 3 < 54

Subtracting 3 from all expression

45-3 < 3x +3-3<54-3

42 < 3x < 51

Dividing whole equation by 3

$\frac{42}{3}<\frac{3x}{3}<\frac{51}{3}$

14<x<17

so the smallest whole number is in between 14 and 17

which are 15 and 16

If x= 15 then

x+1 = 16

x+2 = 17

Now according to first condition

x + (x+1) + (x+2) = 15 + 16 + 17

                         =48

which satisfies both conditions

If x= 16 then

x+1 = 17

x+2 = 18

Now according to first condition

x + (x+1) + (x+2) = 16 + 17 + 18

                         =51

which satisfies both conditions

so the required consecutive whole numbers are

16,17,18 and 15,16,17