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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

 

by Level 11 User (81.5k points)
edited by
The values listed on the final line don't work! The average of 22 + 16 is 19,
which would make a = 38.

Let one of the numbers be x and y another be . The average of these two numbers is x+y/2, so twice the average of these two numbers is x+y. So the sum of the three numbers is:
2x+2y=64
x + y = 32...(1)
x - y = 6...(2)
is also true since the choice of or for the larger of the two is arbitrary.

Solve the 2X2 system for x and y, then calculate


Solving Math Word Problems
 

by Level 8 User (30.1k points)

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