The sum of three numbers is 64. One of the numbers is twice the average of the other two. The difference between the other two is 6. What are the three numbers?
Let the three numbers be a, b, c.
We are given,
The sum of three numbers is 64, i.e.
a + b + c = 64 ------------------ (1)
One of the numbers (call it a) is twice the average of the other two ( b and c), i.e.
a = 2*{(b + c)/2} --------------- (2)
The difference between the other two is 6 (let b be the larger of the two numbers), i.e.
b – c = 6 ------------------------ (3)
From (3), b = 6 + c. Substitute for b into (1) and (2), giving
a + (6 + c) + c = 64
a = (6 + c) + c
Expanding these two eqns gives us,
a + 2c = 58
a – 2c = 6
Adding together the above two eqns,
2a = 64
a = 32
Giving also, c = 13, b = 19
Sorry, copied over the final results wrongly. Corrected values below.
Answer: a = 32, b = 19, c = 13