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Re: This problem was already discussed before.. Seems to me that [#permalink]
sharadGmat wrote:
In one week a construction worker worked more than 40 hours, and was paid two times his regular hourly rate for every hour worked in excess of 40 hours. If he earned $780, how many hours did he work in that week?

(1) If he had worked 6 more hours that week, his earnings would have been $960.
(2) If he had worked 20 fewer hours that week, his earnings would have been $390 less that week.


i agree with A but with different solution.

40x + n (2x) = 780......................given.

1) 40x + (n+6) 2x = 960
40x + 2nx + 12x = 960
780 + 12x = 960
12x = 180
x = 180/12
x=15

n= 6 hours

so total hours worked = 46 hours...

2. we know that it doenot help anything...

v1rok wrote:
I am getting perfectly (A). I don't know why you want to consider those special cases.

let x be worker's hourse salary. Then we can write 40x+N(2x)=780,
where N is number of hourse worked in excess of 40.

Stem1. 40x+6*2x=960,
or 52x=960
or x=240/13 (or approx $19/hrs)


Substituting x in the original equation we can determine N /SUFF/

Stem2. Not sufficient because we don't know what his regular salary is, so we can not determine the exact salary he would get while working 20 hours less. So, MOT SUFF.

Answer: (A)
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Re: This problem was already discussed before.. Seems to me that [#permalink]
Getting A too........

Let M be the number of extra hours he worked after 40 hrs.
Let x be the regular pay then 2x is the extra time pay.

The equation we get is

40x + M2x = 780
From 1)

We get
40 + (M+6)2x = 960
40 + M2x +12x = 960

Equating both the equations
We get x =15

2)

he worked 20 hrs less.

Since we do not know how many actual hours he worked we cannot derive a valid equation.
As if he worked only 10 hrs extra. Only the regular salary will be used.

And if he worked 21 hrs extra .... regular + the extra time has to be used.
Hence Insuf
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Re: This problem was already discussed before.. Seems to me that [#permalink]
MA wrote:
sharadGmat wrote:
In one week a construction worker worked more than 40 hours, and was paid two times his regular hourly rate for every hour worked in excess of 40 hours. If he earned $780, how many hours did he work in that week?

(1) If he had worked 6 more hours that week, his earnings would have been $960.
(2) If he had worked 20 fewer hours that week, his earnings would have been $390 less that week.


i agree with A but with different solution.

40x + n (2x) = 780......................given.

1) 40x + (n+6) 2x = 960
40x + 2nx + 12x = 960
780 + 12x = 960
12x = 180
x = 180/12
x=15

n= 6 hours

so total hours worked = 46 hours...

2. we know that it doenot help anything...

v1rok wrote:
I am getting perfectly (A). I don't know why you want to consider those special cases.

let x be worker's hourse salary. Then we can write 40x+N(2x)=780,
where N is number of hourse worked in excess of 40.

Stem1. 40x+6*2x=960,
or 52x=960
or x=240/13 (or approx $19/hrs)


Substituting x in the original equation we can determine N /SUFF/

Stem2. Not sufficient because we don't know what his regular salary is, so we can not determine the exact salary he would get while working 20 hours less. So, MOT SUFF.

Answer: (A)


Could be, could be ... I guess it all depends on interpretation of "6 hours more." But the beauty of DS questions is that both of us would have gotten this question right, because there is no need to actually solve the equation...



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