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A part-time employee whose hourly wage was increased by 25 [#permalink]
22 Oct 2006, 07:25

00:00

A

B

C

D

E

Difficulty:

5% (low)

Question Stats:

100% (04:06) correct
0% (00:00) wrong based on 10 sessions

A part-time employee whose hourly wage was increased by 25 percent decided to reduce the number of hours worked per week so that the employee's total weekly income would remain unchanged. By what percent should the number of hours worked be reduced?

Let original hourly wage be x and let the no of hours worked be y
Total wage will be =x*y
After the increment the wage will be=1.25 x
Now we need to find number of hours worked so that

x*y=1.25x*z i.e z=4/5 y

%decrease = (y-4/5y)/y *100=100/5=20%.
Thus my answer is B.

Re: Number of hours worked [#permalink]
23 Jul 2011, 21:37

enigma123 wrote:

A part time employee whose hourly wage was increased by 25% decided to reduce the number of hours worked per week so that the employee's total weekly income would remain unchanged. By what percent should the number of hours worked be reduced?

Ans: 20%

Can anyone please tell me the approach to solve this question?

Number of Hours he works today = N Hourly Wage today = P

Number of Hours he works today = M Hourly wage after increase = P + 25% = P(1+25/100) = P * (5/4)

To keep the total wage constant : N*P = M * P * (5/4)

M / N = 4/5 (N-M) / N = 1/5 (fraction by which Number of hours can be decreased while keeping total wage same)

To get the percentage , multiply both side by 100

percent of the number of hours worked be reduced ((N-M) *100/ N) = 100/5 = 20% _________________

Re: Number of hours worked [#permalink]
23 Jul 2011, 21:58

A = hours of work BEFORE wage increase B= hours of work AFTER wage increase when answering about increase or decrease, we need to make the denomenator being the old one. Here we know that to get the same total wage, A must be 1.25 times of B; (x is a multiple of #hours, it will get cancel out)

Re: Number of hours worked [#permalink]
04 Aug 2011, 04:28

Let initial salary be X. Then new salary is 1.25X. Let the initial no. of hours = h1 and final no. of hours = h2; given: X(h1) = 1.25X(h2) => h1 = 1.25(h2) (cancelling X on both sides) We need to find the percent change in no. of hours:

Formula: Percent change = (New - Original)/Original * 100 => (h2 - h1)/h1 * 100 Now, We know that h1 = 1.25(h2) => h2/h1 = 1/1.25; Using dividendo; we get (h2 - h1)/h1 = (1-1.25)/1.25 ------(1) We get our answer by multiplying (1) with 100

Re: Number of hours worked [#permalink]
04 Aug 2011, 21:38

Expert's post

enigma123 wrote:

A part time employee whose hourly wage was increased by 25% decided to reduce the number of hours worked per week so that the employee's total weekly income would remain unchanged. By what percent should the number of hours worked be reduced?

Ans: 20%

Can anyone please tell me the approach to solve this question?

Consider this:

Total Income = Hourly Wage * No of hours If we want to keep the total income same,

Hourly Wage * No of hours = New Hourly Wage * New No of hours Hourly Wage * No of hours = 5/4 * Hourly Wage * New No of hours

So New No of hours/No of hours = 4/5 A decrease of 1/5 i.e. 20%

Or think of it this way: Since total income has to remain the same, Hourly wage *No of hours has to remain the same. If Hourly wage gets multiplied by 5/4 (an increase of 25%), no of hours should be multiplied by 4/5 to keep their product same i.e. no of hours should be reduced by 1/5 i.e. 20%.

Re: Number of hours worked [#permalink]
05 Aug 2011, 19:47

hahaha...If you read this question and are a consultant who is paid hourly wages....you will think :

Hey, my current hourly rate is 100 $. Now it has increased by 25%.I can reduce the time I work, because I am now earning more and want to spend time on some of my hobbies/free time :D

So how much % time can I work less !!!

100 $ per hr(1 hr) = 125$ per hr(new time in hrs) ==> new time = 100/125 hrs = 100/25.5 = 20/25 = 80/100 Old time = 1 hr

Time decrease = 1-80/100 = 20/100 or 20%

I am not putting any approach, just a practical way I would have seen this problem...
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