basic but needs brains

**adhyeta** · April 30th 2009, 01:09 AM

please help(as i said, it needs brains

)

1. a worker suffers a 20 percent cut in wages. he regains his original pay buy obtaining a rise of?

(its not 20...!)

2. m men can do a job in d days, then the number of days in which m+r men would do the same?

3. a boy walks from his home to school at 6kmph . walks back at 2 kmph. avg. speed=?

**earboth** · April 30th 2009, 01:40 AM

Originally Posted by adhyeta

please help(as i said, it needs brains

)

1. a worker suffers a 20 percent cut in wages. he regains his original pay buy obtaining a rise of?

(its not 20...!)

After the cut he earns 80% of his original income. To rise these 80% to a 100% again he has to add 20%. The percentage is $p=\dfrac{20}{80} \cdot 100\% = 25\%$

...

3. a boy walks from his home to school at 6kmph . walks back at 2 kmph. avg. speed=?

Let s denote the distance from home to school. Then the first elapsed time is:

$t_1 = \dfrac s{6\ \frac{km}{h}}$ and

$t_2 = \dfrac s{2\ \frac{km}{h}}$

The boy walks the distance twice. Then the average speed is:

$v=\dfrac{2s}{t_1+t_2}=\dfrac{2s}{\dfrac s{6\ \frac{km}{h}} + \dfrac s{2\ \frac{km}{h}}} = \dfrac{2s}{\dfrac23 s \ \frac{km}{h}} = 3 \ \frac{km}{h}$

**adhyeta** · April 30th 2009, 01:58 AM

thnx. just one more.........

**HallsofIvy** · April 30th 2009, 02:46 AM

Well? We are waiting impatiently!

**earboth** · April 30th 2009, 11:41 PM

Originally Posted by HallsofIvy

Well? We are waiting impatiently!

Probably adhyeta is refering to the question #2.

But I can't answer this question because I don't know what these variables m, r mean ...

**Prove It** · May 1st 2009, 12:05 AM

Originally Posted by earboth

After the cut he earns 80% of his original income. To rise these 80% to a 100% again he has to add 20%. The percentage is $p=\dfrac{20}{80} \cdot 100\% = 25\%$

...

Let s denote the distance from home to school. Then the first elapsed time is:

$t_1 = \dfrac s{6\ \frac{km}{h}}$ and

$t_2 = \dfrac s{2\ \frac{km}{h}}$

The boy walks the distance twice. Then the average speed is:

$v=\dfrac{2s}{t_1+t_2}=\dfrac{2s}{\dfrac s{6\ \frac{km}{h}} + \dfrac s{2\ \frac{km}{h}}} = \dfrac{2s}{\dfrac23 s \ \frac{km}{h}} = 3 \ \frac{km}{h}$

I'd do question 1 differently.

Let $p_{o}$ represent his original salary and $p_n$ represent the new.

From our information, he has received a 20% pay cut, so he is receiving 80% of what he was originally getting.

$p_n = 80\%\textrm{ of }p_o$

$= \frac{80}{100}\times p_o$

$= \frac{4}{5}p_o$ .

Therefore to get back to the original salary

$p_o = \frac{5}{4} p_n$

$= \frac{125}{100} \times p_n$

$= 125\%\textrm{ of }p_n$ .

So he would need a 25% pay increase to get back to his original salary.

**adhyeta** · May 1st 2009, 01:04 AM

Originally Posted by earboth

Probably adhyeta is refering to the question #2.

But I can't answer this question because I don't know what these variables m, r mean ...

a typo.
the questions said-if m men can do a job in d days, then the days reqd for m+r men=?

**earboth** · May 1st 2009, 07:04 AM

Originally Posted by adhyeta

a typo.
the questions said-if m men can do a job in d days, then the days reqd for m+r men=?

$\begin{array}{lcr}m\ men&\rightarrow\ need \ \rightarrow& \ d\ days \\ 1\ man&\rightarrow\ needs \ \rightarrow& \ m \cdot d\ days \\ m+r\ men&\rightarrow\ need \ \rightarrow &\ \dfrac{m\cdot d}{m+r}\ days \end{array}$

**adhyeta** · May 1st 2009, 08:36 AM

lol. thnx earboth. i was doing m men-d days, then 1 man-d/m days.

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