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he is paid $2.00 per unit for the first 40

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he is paid $2.00 per unit for the first 40

by BTGModeratorVI » Wed Jul 29, 2020 2:52 pm

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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

Answer: C
Source: Official guide
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Re: he is paid $2.00 per unit for the first 40

by Brent@GMATPrepNow » Thu Jul 30, 2020 2:20 pm
BTGModeratorVI wrote: ↑
Wed Jul 29, 2020 2:52 pm
 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

Answer: C
Source: Official guide
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: he is paid $2.00 per unit for the first 40

by Scott@TargetTestPrep » Sat Aug 01, 2020 6:39 am
BTGModeratorVI wrote: ↑
Wed Jul 29, 2020 2:52 pm
 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

Answer: C
Source: Official guide
Solution:

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 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: he is paid $2.00 per unit for the first 40

by Ignite » Sun Aug 02, 2020 12:06 am
Greg assembled at least 30 units on each of two days.

So, consider on Day 1 he assembled 30 units.
So money made = 30 * 2 = $60

Now Day 2 if he assembles 40 units, money he makes is 40 * 2 = $80
Now he obviously made more than that to receive $180.

$60 + $80 = $140
i.e he made $40 more in Day 2

Number of additional units to earn $40 = $40/2.5 = 16

Thus he made at max = 40 + 16 = 56 units.


Answer C

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Ignite
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