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

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Greg assembles units of a certain product at a factory. Each day he is  [#permalink]

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New post 25 Jun 2018, 06:18
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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


NEW question from GMAT® Official Guide 2019


(PS02102)

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Re: Greg assembles units of a certain product at a factory. Each day he is  [#permalink]

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New post 25 Jun 2018, 06:27
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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


NEW question from GMAT® Official Guide 2019


(PS02102)


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
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Re: Greg assembles units of a certain product at a factory. Each day he is  [#permalink]

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New post 25 Jun 2018, 06:38
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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 $180
In 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 unit
So, 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

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Brent
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Re: Greg assembles units of a certain product at a factory. Each day he is  [#permalink]

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New post 25 Jun 2018, 07:42

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
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Re: Greg assembles units of a certain product at a factory. Each day he is  [#permalink]

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New post 25 Jun 2018, 09:25
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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


NEW question from GMAT® Official Guide 2019


(PS02102)


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
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Re: Greg assembles units of a certain product at a factory. Each day he is  [#permalink]

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New post 26 Jun 2018, 17:59
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1
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



Since we want to determine the greatest possible number of units he could have assembled in one day, we can assume he assembled only 30 units (the least number of units) in one of the two days. Thus, the number of units he will will have assembled on the other day must be maximum. If he assembled 30 units in one day, he will have earned 2(30) = $60 on that day and $120 on the other day. We can let x = number of units he assembled on the other day, and so we have:

2(40) + 2.5(x - 40) = 120

80 + 2.5x - 100 = 120

2.5x = 140

x = 56

Answer: C
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Re: Greg assembles units of a certain product at a factory. Each day he is  [#permalink]

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New post 29 Oct 2018, 03:19
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
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Re: Greg assembles units of a certain product at a factory. Each day he is   [#permalink] 29 Oct 2018, 03:19
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