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
let Day 1: 30 *2 =60 dollar
So remaining money =180-60= 120
Day 2: don't get confuse in "If Greg assembled
at least 30 units on each of two days and was paid a total of $180.00 "
That dose not mean that on day2 or day 1 he will do only 30 unit work. its at least means minimum 30 unit. Max anything he can work.
So remain 120 dollar we need to calculate
for 1st 40 unit 2 dollar = 40*2 =80 dollar
Now remaining 120-80 =40 dollar ( Which is extra work)
so 40/2.50 = 16 unit extra work
So either on Day 1 or Day 2 he has work : 40+16= 56 units of work
Now you can ask me why not 60,64 being the answer.
Ok if I will go with 60 or 64:
Lets calculate for 60:
Day 1 = 30 (minimum he has to work) = 30 * 2 =60
Remaining = 180 - 60 = 120
if he has work 60* 2 =120 ( Not possible as for 1st 40 he is getting 2 dollar)
So 40*2 =80
remaining if we will do 20* 2.50 dollar = 50 dollar so total 60+80+50 = 190 ( WRONG)
he has received only 180 dollar , We can't go beyond this. So by looking at your ans choice we can eliminate D,E( Going beyond 180)
In A, B ( it will go less then 180)
So 56 is the correct answer.
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Thanks,
Bijaya
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