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In 2007, the arithmetic mean of the annual incomes of Jack a
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11 Apr 2019, 16:17
11 Apr 2019, 16:17
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In 2007, the arithmetic mean of the annual incomes of Jack and Jill was $3800. The arithmetic mean of the annual incomes of Jill and Jess was $4800, and the arithmetic mean of the annual incomes of Jess and Jack was $5800. What is the arithmetic mean of the incomes of the three?
(A) $4000
(B) $4200
(C) $4400
(D) $4800
(E) $5000
(A) $4000
(B) $4200
(C) $4400
(D) $4800
(E) $5000
Re: In 2007, the arithmetic mean of the annual incomes of Jack a
[#permalink]
01 Jul 2020, 23:18
01 Jul 2020, 23:18
1
Assuming that the annual incomes are equal to:
Jess=x
Jack=y
Jill=z
The problem is asking for the arithmetic mean of the three, this is equal to (x+y+z)/3
We know that:
The arithmetic mean of Jack and Jill incomes can be expressed as=
(y+z)/2 =3800...............(1)
The arithmetic mean of Jill and Jess incomes can be expressed as=
(z+x)/2 =4800...............(2)
The arithmetic mean of Jess and Jack incomes can be expressed as=
(x+y)/2 =5800...............(3)
The result of the sum of the left terms of (1),(2) and (3) is equal to = 2(x+y+z)/2
Reducing we obtain
x+y+z
So the sum of the right terms should be divided by 3 to get the answer
Result= 14400/3 = 4800
Answer=D
Jess=x
Jack=y
Jill=z
The problem is asking for the arithmetic mean of the three, this is equal to (x+y+z)/3
We know that:
The arithmetic mean of Jack and Jill incomes can be expressed as=
(y+z)/2 =3800...............(1)
The arithmetic mean of Jill and Jess incomes can be expressed as=
(z+x)/2 =4800...............(2)
The arithmetic mean of Jess and Jack incomes can be expressed as=
(x+y)/2 =5800...............(3)
The result of the sum of the left terms of (1),(2) and (3) is equal to = 2(x+y+z)/2
Reducing we obtain
x+y+z
So the sum of the right terms should be divided by 3 to get the answer
Result= 14400/3 = 4800
Answer=D
gmatclubot
Re: In 2007, the arithmetic mean of the annual incomes of Jack a [#permalink]
01 Jul 2020, 23:18
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