F5 Maths compound inequalities

a) Write down three consecutive multiples of 4 if the least number is x.

Answer: x, x + 4, x + 8


= = = = =
b) Hence, find three consecutive multiples of 4 whose sum is between 37 and 97.

37 < x + (x + 4) + (x + 8) < 97
37 < 3x + 12 < 97
37 - 12 < 3x < 97 - 12
25 < 3x < 85
25/3 < x < 85/3

Since x is an multiple of 4,
x = 12, 16, 20, 24, 28

The answer is:
12, 16, 20
or 16, 20, 24
or 20, 24, 28
or 24, 28, 32
or 28, 32, 36