Bunuel wrote:

Greg assembles units of a certain product at a factory. Each day he is paid $2.00 per unit for the first 40 units that he assembles and $2.50 for each additional unit that he assembles that day. If Greg assembled at least 30 units on each of two days and was paid a total of $180.00 for assembling units on the two days, what is the greatest possible number of units that he could have assembled on one of the two days?

A. 48

B. 52

C. 56

D. 60

E. 64

GIVEN: Total pay for 2 days is $180In order to MAXIMIZE the number of units assembled on one of the days, we must MINIMIZE the number of units assembled on the other day

We're told that

Greg assembled at least 30 units each day.

So, the MINIMUM is 30 units

Let's say that Greg assembled 30 units on DAY 1

Payment for Day 1 = (30)($2) = $60

If Greg is paid $60 for Day 1, then he must have received $120 on Day 2.

If Greg was paid $120 on Day 2, how many units did he assemble?

GIVEN: Greg is paid $2.00 per unit for the first 40 units that he assembles and $2.50 for each additional unitSo, for the first

40 units, Greg is paid $80 (since 40 x $2 = 80)

$120 - $80 = $40

So, the remaining $40 was paid to Greg for the additional units he assembled at a rate of $2.50/unit

$40/$2.50 =

16 units So, the TOTAL number of units assembled =

40 +

16 = 56

Answer: C

Cheers,

Brent

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