The average of four positive integers is 72.5. The highest integer is 117 and the lowest integer is 15. The difference between the remaining two integers is 12. Which is the higher of these two remaining integers?
A
70
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B
73
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C
85
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D
80
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Solution
The correct option is B 85 Let the higher integer be x. Then the other integer = x - 12 ∴15+(x−12)+x+1174=72.5 ⇒120+2x=72.5×4=290 ⇒2x=170 ⇒x=85