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# At a local appliance manufacturing facility, the workers received a 20

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At a local appliance manufacturing facility, the workers received a 20  [#permalink]

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25 Jun 2015, 03:44
00:00

Difficulty:

35% (medium)

Question Stats:

73% (01:46) correct 27% (01:53) wrong based on 186 sessions

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At a local appliance manufacturing facility, the workers received a 20% hourly pay raise due to extraordinary performance. If one worker decided to reduce the number of hours that he worked so that his overall pay would remain unchanged, by approximately what percent would he reduce the number of hours that he worked?

A) 83%
B) 80%
C) 20%
D) 17%
E) 12%

Source: Platinum GMAT
Kudos for a correct solution.

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Re: At a local appliance manufacturing facility, the workers received a 20  [#permalink]

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25 Jun 2015, 04:42
Bunuel wrote:
At a local appliance manufacturing facility, the workers received a 20% hourly pay raise due to extraordinary performance. If one worker decided to reduce the number of hours that he worked so that his overall pay would remain unchanged, by approximately what percent would he reduce the number of hours that he worked?

A) 83%
B) 80%
C) 20%
D) 17%
E) 12%

Source: Platinum GMAT
Kudos for a correct solution.

Let, Initial No of Hours worked = x
Initial Pay per hour = y

Therefore, New Hourly Pay = x+20% of x = 1.2x
New No. of Hours worked = (y-a) [If the hours worked are reduced by a hours]
Total pay = 1.2x * (y-a)

Pay Remains Unchanged

i.e. x*y = 1.2x*(y-a)
i.e. y = 1.2y - 1.2a
i.e. o.2y = 1.2a
i.e. y = 6a

i.e. Hours worked changed from 6a to 5a
i.e. %Change = [(a)/(6a)]*100 = 16.66%

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Re: At a local appliance manufacturing facility, the workers received a 20  [#permalink]

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25 Jun 2015, 06:07
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Let's say he works usually 10 hours and earns 100 per hour.

10 * 100 = 1000
10 * 120 = 1200 (this are the new earnings after the raise)

To figure out how much he needs to work with the new salary in order to earn the original 1000:

1000/120 = 8.3333

So he can reduce his work by 1.6666 hours. Which is >15%.

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Re: At a local appliance manufacturing facility, the workers received a 20  [#permalink]

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25 Jun 2015, 08:51
1
Bunuel wrote:
At a local appliance manufacturing facility, the workers received a 20% hourly pay raise due to extraordinary performance. If one worker decided to reduce the number of hours that he worked so that his overall pay would remain unchanged, by approximately what percent would he reduce the number of hours that he worked?

A) 83%
B) 80%
C) 20%
D) 17%
E) 12%

Source: Platinum GMAT
Kudos for a correct solution.

Let us assume, Hourly Rate=\$100
No of hours worked=6
Salary earned=\$600
Now, the worker wants to keep the pay unchanged at \$600
New Hourly Rate=120% of \$100=\$120
New no of hours worked=\$600/\$120=5 hours
Required %=6-5/6=1/6=16.66%=17%
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Re: At a local appliance manufacturing facility, the workers received a 20  [#permalink]

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25 Jun 2015, 09:06
Option D. 16.66 %. Closest answer 17%

Bunuel wrote:
At a local appliance manufacturing facility, the workers received a 20% hourly pay raise due to extraordinary performance. If one worker decided to reduce the number of hours that he worked so that his overall pay would remain unchanged, by approximately what percent would he reduce the number of hours that he worked?

A) 83%
B) 80%
C) 20%
D) 17%
E) 12%

Source: Platinum GMAT
Kudos for a correct solution.

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Re: At a local appliance manufacturing facility, the workers received a 20  [#permalink]

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25 Jun 2015, 10:23
Assume hourly pay for the worker =x and number of hours worked =y
Therefore as per question we can write
Xy =1.2x *(1-k)y
1.2(1-k)=1 => K =1/6 =.166666
Therefore by approximately reducing 17% of hours

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Re: At a local appliance manufacturing facility, the workers received a 20  [#permalink]

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25 Jun 2015, 17:28
1
Hi All,

Many Test Takers would deal with this question by using decimals, but if you're comfortable using fractions (and you know your fraction-to-decimal conversions), then you can save some time on the 'math work' that this question requires.

We're told that an employee receives a 20% pay raise per hour and plans to reduce the number of hours so that his total overall pay remains the same. We're asked for the percentage DECREASE in the number of hours that he plans to work.

P = Pay Rate/Hour
H = Number of hours

Total Pay before raise = (P)(H)

Pay increases by 20% --> (6/5)(P)

To "counter" the 6/5 increase, we would need to multiply the number of hours by 5/6 to get back to the original Total Pay...

(6/5)(P)(5/6)(H) = (P)(H)

Thus, we're reducing the total number of hours by 1/6 of the original number of hours...

1/6 = .16666666 = about 17%

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Posts: 50544
Re: At a local appliance manufacturing facility, the workers received a 20  [#permalink]

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29 Jun 2015, 04:09
Bunuel wrote:
At a local appliance manufacturing facility, the workers received a 20% hourly pay raise due to extraordinary performance. If one worker decided to reduce the number of hours that he worked so that his overall pay would remain unchanged, by approximately what percent would he reduce the number of hours that he worked?

A) 83%
B) 80%
C) 20%
D) 17%
E) 12%

Source: Platinum GMAT
Kudos for a correct solution.

Platinum GMAT Official Solution:

This question can be solved using either real numbers or algebra.

Method 1: Use Real Numbers.

Assume that the worker in question, call him John, earned \$50 by working 10 hours for \$5 per hour.

After the pay-raise, the total amount earned will be the same at \$50, but the hourly rate will increase by 20% to \$5(1.2) = \$6 per hour.

Using the intuition that salary=(hours)(wage), you can find that \$50=(hours)(\$6) and therefore the number of hours worked after the change in pay will equal (50/6) = 8.33 hours

Since John worked 10 hours before the change in wage and he now works 8.33 hours, John reduced the number of hours he worked by (10-8.33)/10 = .1666666% = 17%

Note: It does not matter what numbers you use, as long as the original pay equals the post wage-change pay and the wage is increased by 20%.

Method 2: Use Algebra.

Assign variables to pieces of the problem:
Let p = the percentage of old hours worked after the raise
if p = 95%, the worker labors 95% of the old hours after the raise (i.e., 5% fewer hours after the raise)
Let r = hourly rate that employees were paid before the raise
Let R = hourly rate that employees were paid after the raise
Let n = number of hours worked before pay-raise
Let N = number of hours worked after pay-raise

We know that the total pay before and after the pay-raise equal each other. The pay before was based on the rate r with hours n while the pay afterwards was based on rate R and hours N:
rn = RN

The rate after the raise was 20% more than the rate before the raise:
R = 1.20r
Plug this value into the equation for R:
rn = RN
rn = 1.20rN

We are looking for the percentage of hours worked after the raise, so substitute N in terms of n and percent p.
N = pn; the new number of hours equals the old number of hours multiplied by the percent of old hours worked after the raise
rn = (1.2r)(pn).

Divide both sides by r and n:
1 = 1.2p

Divide by 1.2 to solve for p:
p = 1/1.2 = 10/12 = 5/6 = .8333 = 83.33%.

Be careful: the question asks what percent the worker would reduce his hours, so subtract from 100% to yield 16.66%, or 17%.

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Re: At a local appliance manufacturing facility, the workers received a 20  [#permalink]

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08 Aug 2016, 02:28
Think that previously the hourly rate was \$100. Since hourly rate has been increased by 20% then the new rate is 120 dollar per hour. If a worker wants to earn \$100 then he will work 100/120=83% of his usual time. Hence he can reduce 17% of his working time.
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Re: At a local appliance manufacturing facility, the workers received a 20  [#permalink]

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11 Aug 2016, 23:16
Here is my approach =>
Let the wage be \$x
and the number of hours he worked initially be y
Now wage => \$xy
New wage => x(1+20/100)=6x/5
let the new hours => y'
new wage => 6xy'/5
New wage = old wage => y/y'=5/6
now percentage change =>p
y(1+p/100)=y'
=> p=16.67%
Smash that E
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Re: At a local appliance manufacturing facility, the workers received a 20  [#permalink]

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16 Sep 2017, 08:04
Formula tip = -100*r/100+r %

Here r takes the respective sign for fall or rise.
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Re: At a local appliance manufacturing facility, the workers received a 20  [#permalink]

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13 May 2018, 17:01
Bunuel wrote:
At a local appliance manufacturing facility, the workers received a 20% hourly pay raise due to extraordinary performance. If one worker decided to reduce the number of hours that he worked so that his overall pay would remain unchanged, by approximately what percent would he reduce the number of hours that he worked?

A) 83%
B) 80%
C) 20%
D) 17%
E) 12%

We can let the current wage = w, the current number of hours = h, and the percent reduction = n and create the equation:

wh = 1.2w x h(1 - n/100)

1 = 1.2(100 - n)/100

100 = 120 - 1.2n

1.2n = 20

n = 20/1.2 = 200/12 = 50/3 ≈ 17

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Re: At a local appliance manufacturing facility, the workers received a 20 &nbs [#permalink] 13 May 2018, 17:01
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