[Math Revolution GMAT math practice question]
The average (arithmetic mean) of the five numbers in a data set is 72. The average (arithmetic mean) of the first three numbers is 98. What is the average (arithmetic mean) of the last two numbers?
A. 31
B. 32
C. 33
D. 34
E. 35
The average (arithmetic mean) of the five numbers in a data
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- Max@Math Revolution
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Let a, b, c, d and e be the five numbers in the set.Max@Math Revolution wrote: The average (arithmetic mean) of the five numbers in a data set is 72. The average (arithmetic mean) of the first three numbers is 98. What is the average (arithmetic mean) of the last two numbers?
A. 31
B. 32
C. 33
D. 34
E. 35
The average (arithmetic mean) of the five numbers in a data set is 72.
So, (a + b + c + d + e)/5 = 72
Multiply both sides of the equation by 5 to get: a + b + c + d + e = 360
The average (arithmetic mean) of the first three numbers is 98.
So, (a + b + c)/3 = 98
Multiply both sides of the equation by 3 to get: a + b + c = 294
What is the average (arithmetic mean) of the last two numbers?
In other words, what is the average of d and e?
Take a + b + c + d + e = 360 and replace a + b + c with 294
We get: 294 + d + e = 360
Subtract 294 from both sides to get: d + e = 66
So, the average of d and e = (d + e)/2
= 66/2
= 33
Answer: C
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Sum = (quantity)(average).Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
The average (arithmetic mean) of the five numbers in a data set is 72. The average (arithmetic mean) of the first three numbers is 98. What is the average (arithmetic mean) of the last two numbers?
A. 31
B. 32
C. 33
D. 34
E. 35
The average (arithmetic mean) of the five numbers in a data set is 72.
Sum = 5*72 = integer with a UNITS DIGIT OF 0.
The average (arithmetic mean) of the first three numbers is 98.
Sum = 3*98 = integer with a UNITS DIGIT OF 4.
For the sum of all 5 numbers to have a units digit of 0, the sum of the remaining 2 numbers must have a UNITS DIGIT OF 6, since INTEGER WITH A UNITS DIGIT OF 6 + INTEGER WITH A UNITS DIGIT OF 4 = INTEGER WITH A UNITS DIGIT OF 0.
Only C will yield for the other 2 numbers a sum with a units digit of 6:
2*33 = 66.
The correct answer is C.
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=>
Let a, b, c, d and e be the five numbers.
Now, the average of these numbers is ( a + b + c + d + e ) / 5 = 72, so a + b + c + d + e = 72 * 5 = 360.
The average of the first three numbers is ( a + b + c ) / 3 = 98, so a + b + c = 98 * 3 = 294.
Thus, the sum of the last two numbers is d + e = ( a + b + c + d + e ) - ( a + b + c ) = 360 - 294 = 66.
Their average is ( d + e ) / 2 = 66 / 2 = 33.
Therefore, the answer is C.
Answer: C
Let a, b, c, d and e be the five numbers.
Now, the average of these numbers is ( a + b + c + d + e ) / 5 = 72, so a + b + c + d + e = 72 * 5 = 360.
The average of the first three numbers is ( a + b + c ) / 3 = 98, so a + b + c = 98 * 3 = 294.
Thus, the sum of the last two numbers is d + e = ( a + b + c + d + e ) - ( a + b + c ) = 360 - 294 = 66.
Their average is ( d + e ) / 2 = 66 / 2 = 33.
Therefore, the answer is C.
Answer: C
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