SOLUTION: Divide 105 peaches into three bags such that bag B shall have four times the peaches in bag A, and bag C will be twice the sum of bag A and bag B. How many peaches in bag C? A)

Question 252791: Divide 105 peaches into three bags such that bag B shall have four times the
peaches in bag A, and bag C will be twice the sum of bag A and bag B. How many peaches in bag C?
A) 7 B) 28 C) 60 D) 70 E) 80
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
a + b + c = 105

a,b,c represent the amount of peaches in bags A, B, and C.

b = 4*a; (bag B has 4 times the peaches in bag A).

c = 2 * (a + b); (bag C contains 2 times the amount in bags A and B).

since b = 4*a, then:

c = 2 * (a + b) becomes:

c = 2 * (a + 4*a).

remove parentheses to get:

c = 2*a + 8*a = 10*a

we have:

b = 4*a
c = 10*a

a + b + c = 105 becomes:

a + 4*a + 10*a = 105 which becomes:

15*a = 105

solve for a to get:

a = 105/15 = 7

b = 4*a = 4 * 7 = 28

c = 10*a = 10*7 = 70

70 + 28 + 7 = 105 (this is good).

b = 2 * (a + b) = 2 * (7 + 28) = 2 * 35 = 70 (this is also good).

You have confirmed that:

bag A contains 7 peaches
bag B contains 28 peaches
bag C contains 70 peaches

Your answer is bag C contains 70 peaches which is selection D.

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