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

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.

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

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

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

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

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

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

Solve for $b.$

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 $\frac{1}{5}$ of the tank.
In $x$ hours, it can fill $\frac{x}{5}$ of the tank.

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

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

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

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

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

Solve for $x.$

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

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

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

Their total cost was \$1000: . $50R + 30C + F \:=\:1000$ .[2]

Subtract [1] from [2]: . $49R + 29C \:=\:900 \quad\Rightarrow\quad R \:=\:\frac{900-29C}{49}$

Since $R$ is a whole number, $900-29C$ must be divisible by 49.

We can try values of $C$ until we find one that works
. . . . . or we can try a little algebra.
We can rewrite the equation: . $R \;=\;16 + \frac{116-29C}{49} \;=\;16 + \frac{29(4 - C)}{40}$

And we see that $C = 4$ gives us: . $\boxed{R = 16}$

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# a merchant has three items on sale

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