Revenues were recorded for Store A and Store B over a period of three months. In the first month, Store A's revenues were $40,000 higher than Store B's revenues. In the second month, Store A's revenues were $8,000 higher than Store B's revenues. If Store A's average (arithmetic mean) monthly revenue for the three months was $2,000 greater than Store B's average monthly revenue, then Store B's revenue in the third month was how much greater than Store A's revenue?
(A) $14,000
(B) $15,000
(C) $42,000
(D) $46,000
(E) $50,000
OA is C.
Source: GMAT Hacks
Revenues: A and B
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In 1st month, revenue of Store A = $40,000 + Store B's revenuesikaplan wrote:Revenues were recorded for Store A and Store B over a period of three months. In the first month, Store A's revenues were $40,000 higher than Store B's revenues. In the second month, Store A's revenues were $8,000 higher than Store B's revenues. If Store A's average (arithmetic mean) monthly revenue for the three months was $2,000 greater than Store B's average monthly revenue, then Store B's revenue in the third month was how much greater than Store A's revenue?
(A) $14,000
(B) $15,000
(C) $42,000
(D) $46,000
(E) $50,000
OA is C.
Source: GMAT Hacks
In 2nd month, revenue of Store A = $8,000 + Store B's revenues
Average revenue of Store A for 3 months = $2,000 + Store B's average monthly revenue
Let in the 3rd month, revenue of store A = X
Then, (40,000 + 8,000 + X)/3 = 2,000
40,000 + 8,000 + X = 6,000
X = -42,000, which implies that Store B's revenue in the third month was $42,000 greater than Store A's revenue.
The correct answer is C.
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Month A B
1 x+40 x
2 y+8 y
3 z+t z
Avg B+2 B
B = (x+y+z)/3
A = B+2 = (x+y+z+48+t)/3 => (48+t)/3 = 2=> t = -42
Since all figures are in thousands' and -42 is in A's side. B's revenue was 42000 more than A.
1 x+40 x
2 y+8 y
3 z+t z
Avg B+2 B
B = (x+y+z)/3
A = B+2 = (x+y+z+48+t)/3 => (48+t)/3 = 2=> t = -42
Since all figures are in thousands' and -42 is in A's side. B's revenue was 42000 more than A.
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