Out of four numbers a, b, c, d, the average of first three is 16 and that of the last three is 17. If d = 19, a is:

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Solution

The correct option is
C

16
Average of first three is 16, thus ( a + b + c ) / 3 = 16 or a + b + c = 48
Average of last three is 17, thus ( b + c + d ) / 3 = 17 or b + c + d = 51
From the difference of two equations, we get
d - a = 3
given d = 19, thus a = 19 - 3 = 16
$∴ a = 16$

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Out of four numbers a, b, c, d, the average of first three is 16 and that of the last three is 17. If d = 19, a is:

The correct option is C 16Average of first three is 16, thus ( a + b + c ) / 3 = 16 or a + b + c = 48Average of last three is 17, thus ( b + c + d ) / 3 = 17 or b + c + d = 51From the difference of two equations, we get d - a = 3given d = 19, thus a = 19 - 3 = 16∴a=16

The correct option is
C

16
Average of first three is 16, thus ( a + b + c ) / 3 = 16 or a + b + c = 48
Average of last three is 17, thus ( b + c + d ) / 3 = 17 or b + c + d = 51
From the difference of two equations, we get
d - a = 3
given d = 19, thus a = 19 - 3 = 16
$∴ a = 16$