Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

he is paid $2.00 per unit for the first 40

Expert replies
by BTGModeratorVI » Wed Jul 29, 2020 2:52 pm
Timer
00:00
Your answer
Difficulty–

right wrong

Pick an answer to see how members did

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
Join the discussion
Source: — Problem Solving |

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

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion

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

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Join the discussion

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

Thanks,
Ignite
Join the discussion