Greg assembles units of a certain product at a factory. Each day he is

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To get the max on one day, minimize on the second day..
So let on one day, it be only 30, so payment for these =30*2=60$
Payment on second day =180-60=120..

Subtract for first 40@2per piece=40*2=80
So remaining payment for excess pieces = 120-80=40
Number of pieces@2.5=40/2.5=400/25=16

Total =40+16=56

C

Solution

Given:

• Payment per unit = $2 for the first 40 units, $2.50 for each additional unit
• Greg assembled at least 30 units on each of two days, and was paid $180 in total

To find:

• The greatest possible number of units that he could have assembled in one of the two days

Approach and Working:

• As we are trying to maximise the number of units that Greg could have assembled on a given day, we should minimise the number of units assembled on the other day
• As the minimum units assembled in a day is 30 units, the payment for those 30 units = 30 * $2 = $60
• Therefore, the payment for the other day = $180 – $60 = $120

Out of the payment of $120, the payment for the first 40 units will be 40 * 2 = $80

• Hence, the payment for the remaining units = $120 - $80 = $40

As the payment for each of the remaining units is $2.50,

• The number of remaining units = \(\frac{40}{2.50}\) = 16
• Therefore, the maximum possible units assembled = 40 + 16 = 56

Hence, the correct answer is option C.

Answer: C

A minimum of 30 units are assembled in 1 day.

Total earnings = $180 in 2 days.

Day 1: 30 units *2 = $60 (Assumed since we need to find the maximum number of units assembled in a day.)

Day 2: 40 units *2 = $80 Total $140 (Balance $40 (180-140))

For every additional unit, Greg earns $2.5 per unit. That means he earned $40 in the enhanced rate.

Therefore, additional units = 40 / 2.5 = 16 units. (These are assumed to be assembled in Day 2 since we need to find the maximum number of units assembled in a day.)

Therefore, total number of units assembled in Day 2 is 40 + 16 = 56 units. Answer: C

Total earning =$ 180
Minimum earned in two days = 30*2*2 = $60
Remaining = 180-120=60

Day1+Day2 = 60 = 2*10+2.5*x1 +2*10+2.5*x2

Earning on one of the days could be maximum if the other day's earning is minimum hence assuming it was minimum=0

Therefore, 2*10+2.5*x1=60 => x1=16
So total no of hours worked =40+16 = 56
2

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.

Hello,

It is mentioned that he is paid $2 per unit for first 40 units. Should we assume that $2 per unit would be the payment for first 30 units as well?

Rishov

Given: Greg assembles units of a certain product at a factory.
1. 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.
2. Greg assembled at least 30 units on each of two days
3. He was paid a total of $180.00 for assembling units on the two days.

Asked: What is the greatest possible number of units that he could have assembled on one of the two days?

Greg assembled at least 30 units on each of two days
To maximize number of units on a particular day, we have to minimize number of units on other day.
On day 1, he assembles 30 units.
He was paid for 30 units = $60 on day 1

He was paid $180 - $60 = $120 on day 2
$120 = $2 * 40 + $2.5 x
x = 16
He assembled number of units on day 2 (max) = 40 + 16 = 56

IMO C

Least earning of Greg in one of the two days \(30*2=60\); rest amount \(=180-60=120\)

Earning on the other day \(=40*2+2.50x=120\) [x is the number of hours in addition to 40 hours]

\(=2.50x=120-80=40\)

\(x=\frac{40}{2.50}=\frac{40*100}{250}=16\)

Total maximum hours on a single day \(=40+16=56\) hours

The answer is \( C\)

This is a max/min question

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

The min on day 1 (or 2) must be 30 units x $2 each = $60. So, the remainder is 120.

120 = 80 + 2.50 (x - 40)
x = 56

C.

Answer: Option C

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It’s given that $180 was paid for 2 days.
In order to find the greatest possible number of units that he could have assembled on one of the two days, the number of units assembled in other day should be minimum.

Minimum 30 units are assembled in each day; we can say that no of units assembled in day 1 is 30.
Payment made for day 1 =30 * 2 = 60 $
So, we can conclude that payment for day 2 = 180 – 60 = 120. $

For first 40 units, payment is 2 $ per unit and for the remaining units it is 2.5 $ per unit.
40 * 2 + x *2.5 = 120
X= 40/2.5 = 16
So, Greg assembled 40 + 16 = 56 units in day 2 and this is the maximum possible value.

Option C.

Hope this helps,
Clifin J Francis

Shortest Solution -

Minimum per day -> 30

Day 1 -> 30 units => $60

Day 2 :

Need - $120
For first 40 units -> 2*40=$80
$ Left -> 120-80 = 40

$40 = x units * $2.5(per unit)
x = 40/2.5 => 16
Max Units = 16 + 40 = 56 => Answer

For the second part of the question
Consider 2 units at 5 $ (2.5$ each)
So we need to find number of units in 40 dollar

So we have 2*16 = 5*8

Which is 16 units