Machine A produces pencils at a constant rate of 9000

BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

This topic has expert replies

Post new topic Post Reply

Previous Topic Next Topic

VJesus12: Legendary Member; Posts: 2276; Joined: Sat Oct 14, 2017 6:10 am; Followed by:3 members

Machine A produces pencils at a constant rate of 9000

by VJesus12 » Mon May 21, 2018 7:35 am

Timer

00:00

Your Answer

Global Stats

Machine A produces pencils at a constant rate of 9000 pencils per hour, and machine B produces pencils at a constant rate of 7000 pencils per hour. If the two machines together must produce 100,000 pencils and if each machine can operate for at most 8 hours, what is the least amount of time, in hours, that machine B must operate?

A. 4
B. 4 2/3
C. 5 1/3
D. 6
E. 6 1/4

The OA is the option A .

What is the equation that I should set here in order to get the correct answer? Could anyone help me? Please.

Quote

Source: Beat The GMAT — Problem Solving | Mon May 21, 2018 7:35 am

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Mon May 21, 2018 7:49 am

VJesus12 wrote:Machine A produces pencils at a constant rate of 9000 pencils per hour, and machine B produces pencils at a constant rate of 7000 pencils per hour. If the two machines together must produce 100,000 pencils and if each machine can operate for at most 8 hours, what is the least amount of time, in hours, that machine B must operate?

A. 4
B. 4 2/3
C. 5 1/3
D. 6
E. 6 1/4

To minimize B's time, we must maximize A's time.
Since each machine can work for at most 8 hours, let A's time = 8 hours.
Since A's rate = 9000 pencils per hour, the number of pencils produced by A in 8 hours = rt = 9000*8 = 72,000.
Remaining pencils = 100,000 - 72,000, = 28,000.
Since B's rate = 7000 pencils per hour, the time for B to produce the remaining 28,000 pencils = w/r = 28,000/7000 = 4 hours.

The correct answer is A.

Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

Quote

Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pm; Thanked: 6 times; Followed by:5 members

by Vincen » Mon May 21, 2018 11:51 pm

VJesus12 wrote:Machine A produces pencils at a constant rate of 9000 pencils per hour, and machine B produces pencils at a constant rate of 7000 pencils per hour. If the two machines together must produce 100,000 pencils and if each machine can operate for at most 8 hours, what is the least amount of time, in hours, that machine B must operate?

A. 4
B. 4 2/3
C. 5 1/3
D. 6
E. 6 1/4

The OA is the option A .

What is the equation that I should set here in order to get the correct answer? Could anyone help me? Please.

Hello Vjesus12.

I will solve it as follows:

1- Machine A produces 9,000 pencils per hour.
2- Machine B produces 7,000 pencils per hour.
3- Both machines have to produce 100,000 pencils.
4- Each machine can operate for at most 8 hours.

(?) What is the least amount of time, in hours, that machine B must operate?

Since machine A produces more pencils per hour than machine B, then in order to find the least amount of time that machine B must operate, machine A must operate all 8 hours.

In 8 hours, machine A will produce $$9,000\cdot8=72,000.$$ Hence, machine B must produce 100,000-72,000 = 28,000 pencils.

Since the rate of the machine B is 7,000 per hour, then producing 28,000 pencils will take machine B $$\frac{28,000}{7,000}=4\ hours .$$ Hence, the correct answer is the option A .

I hope it is clear enough.

Quote

swerve: Legendary Member; Posts: 2499; Joined: Sun Oct 29, 2017 2:04 pm; Followed by:6 members

by swerve » Tue May 22, 2018 10:01 am

9,000 x 8hours + 7,000 x Xhours = 100,000

72,000 + 7,000 x X hours = 100,000

X hours = 28,000 / 7,000 = 4. Option A.

Regards!

Quote

GMAT/MBA Expert

Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Thu May 24, 2018 12:32 pm

VJesus12 wrote:Machine A produces pencils at a constant rate of 9000 pencils per hour, and machine B produces pencils at a constant rate of 7000 pencils per hour. If the two machines together must produce 100,000 pencils and if each machine can operate for at most 8 hours, what is the least amount of time, in hours, that machine B must operate?

A. 4
B. 4 2/3
C. 5 1/3
D. 6
E. 6 1/4

If machine B's operational time is to be minimized, we must maximize the time for machine A to operate. Since each machine can operate for at most 8 hours, we can let machine A operate for 8 hours. Since the rate of machine A is 9000 pencils per hour, machine A produces 8 x 9000 = 72,000 pencils, and thus 28,000 pencils are left to be produced.

Thus, it will take machine B 28,000/7,000 = 4 hours to produce the remaining pencils.

Answer: A

Scott Woodbury-Stewart
Founder and CEO
[email protected]