1. ## simple word problems

1. A and B working together can finish painting the house in 6 days. A, working alone can finish it in 5 days less than B. How long will it take each of them to finish the work alone?

2. A 400mm diameter pipe can fill the tank alone in 5 hours and another 600mm diameter pipe can fill the tank in 4 hours. A drain pipe 300 mm in diameter can empty the tank in 20 hours. With all the pipes open, how long will it take to fill the tank?

3. A merchant has 3 items on sale: namely, a radio for $50, a clock for$30, and a flashlight for $1. At the end of the day, she has sold a total of 100 of the 3 items and has taken exactly$1000 on the total sales. How many radios did he sell?

I don't have equations to actually get started. Help please!

2. #1:

Both workers can paint 1/6th of the house in one day.

Therefore, $\displaystyle \frac{1}{t}+\frac{1}{t+5}=\frac{1}{6}$

Solve for t.

#2:

The 400mm pipe can fill 1/5th of the tank in one hour and the 600mm can fill 1/4th of it in one hour. But another empties 1/20th of it in one hour.

$\displaystyle \frac{1}{4}+\frac{1}{5}-\frac{1}{20}=\frac{1}{t}$

Solve for t.

For #3, did you forget a portion of the statement?. I may be mistaken, but it appears there's not enough info. The way it is you have two equations and 3 unknowns.

Obviously, setting up the equations is your main task . . .

1. A and B working together can finish painting the house in 6 days.
A, working alone can finish it in 5 days less than B.
How long will it take each of them to finish the work alone?

Let $\displaystyle b$ = number of days for B to do the job alone.
In one day, he can do $\displaystyle \frac{1}{b}$ of the job.
In $\displaystyle 6$ days, he can do $\displaystyle \frac{6}{b}$ of the job.

Then $\displaystyle b-5$ = number of days for A to do the job alone.
In one, he can do $\displaystyle \frac{1}{b-5}$ of the job.
In $\displaystyle 6$ days, he can do $\displaystyle \frac{6}{b-5}$ of the job.

Together, in 6 days, they can do: .$\displaystyle \frac{6}{b} + \frac{6}{b-5}$ of the job.

But we are told that, in 6 days, they can complete the job (1 job).

There is our equation! . . . . $\displaystyle \frac{6}{b} + \frac{6}{b-5} \;=\;1$

Solve for $\displaystyle b.$

2. One pipe can fill the tank alone in 5 hours
and another pipe can fill the tank in 4 hours.
A drain pipe empty the tank in 20 hours.
With all the pipes open, how long will it take to fill the tank?

The first pipe takes 5 hours to fill the tank.
In one hour, it can fill $\displaystyle \frac{1}{5}$ of the tank.
In $\displaystyle x$ hours, it can fill $\displaystyle \frac{x}{5}$ of the tank.

The second pipe takes 4 hours to fill the tank.
In one hour, it can fill $\displaystyle \frac{1}{4}$ of the tank.
In $\displaystyle x$ hours, it can fill $\displaystyle \frac{x}{4}$ of the tank.

The drain pipe takes 20 hours to drain the tank.
In one hour, it empties $\displaystyle \frac{1}{20}$ of the tank.
In $\displaystyle x$ hours, it empties $\displaystyle \frac{x}{20}$ of the tank.

Together, in $\displaystyle x$ hours, they fill: .$\displaystyle \frac{x}{5} + \frac{x}{4} - \frac{x}{20}$ of the tank.

But in $\displaystyle x$ hours, we expect them to fill the entire tank (1 tank).

There is our equation!. . . . $\displaystyle \frac{x}{5} + \frac{x}{4} - \frac{x}{20} \;=\;1$

Solve for $\displaystyle x.$

3. A merchant has 3 items on sale: namely, a radio for $50, a clock for$30,
and a flashlight for $1. At the end of the day, she has sold a total of 100 of the 3 items and has taken exactly$1000 on the total sales.
How many radios did she sell?

Let: .$\displaystyle \begin{array}{ccc}R & = & \text{number of radios sold} \\ C & = & \text{number of clocks sold} \\F & = & \text{number of flashlights sold} \end{array}$

100 items were sold: .$\displaystyle R + C + F \:=\:100$ .[1]

$\displaystyle R$ radios cost $\displaystyle 50R$ dollars.
$\displaystyle C$ clocks cost $\displaystyle 30C$ dollars.
$\displaystyle F$ flashlights cost $\displaystyle 1F$ dollars.

Their total cost was $1000: .$\displaystyle 50R + 30C + F \:=\:1000$.[2] Subtract [1] from [2]: .$\displaystyle 49R + 29C \:=\:900 \quad\Rightarrow\quad R \:=\:\frac{900-29C}{49}$Since$\displaystyle R$is a whole number,$\displaystyle 900-29C$must be divisible by 49. We can try values of$\displaystyle C$until we find one that works . . . . . or we can try a little algebra. We can rewrite the equation: .$\displaystyle R \;=\;16 + \frac{116-29C}{49} \;=\;16 + \frac{29(4 - C)}{40} $And we see that$\displaystyle C = 4$gives us: .$\displaystyle \boxed{R = 16}\$

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# a merchant has three items on sale

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