# Time and work. Arithmetic.

$A$, $B$ and $C$ working together completed a job in $10$ days. However, $C$ only worked for the first three days when $\dfrac{37}{100}$ of the job was done. Also, the work done by $A$ in $5$ days equals the work done by $B$ in $4$ days. How many days will the fastest worker take to complete the given work alone?

Is my approach right?

My attempt:

Let the rate of work done by $A,B,C$ be $a \,\text{units/day}, b \,\text{units/day}, c \,\text{units/day}$ respectively.

Since, $C$ works only on the first $3$ days (along with $A$ and $B$), and $B$ and $A$ works together for the rest of the time,

$$\text{Total work}=(a+b+c)\cdot3+(a+b)\cdot7=10(a+b)+3c$$

Also, $$3c=\frac{37}{100}[10(a+b)+3c]$$ so $$189c=370(a+b) \tag{1}$$

and $$5a=4b \tag{2}$$

Where is the third equation that would solve the problem?

• $7a+7b=63/100$ (job completed in 10 days) – Vasya Dec 13 '17 at 19:01
• Do you have the answer to the problem? – Michael Rybkin Dec 13 '17 at 20:50
• You should have $3(a + b + c) = \frac{37}{100}[10(a + b) + 3c]$ since the total amount of work done in the first three days was $37\%$ of the total work done. – N. F. Taussig Dec 13 '17 at 21:30
• "Where is the third equation that would solve the problem?" What do you mean? You have three equations: 1) $\text{Total work}=(a+b+c)\cdot3+(a+b)\cdot7=10(a+b)+3c$ 2) $3c=\frac{37}{100}[10(a+b)+3c]$ And 3) $5a = 4b$, htat's three equations. So what do you mean "where is the third equation"? – fleablood Dec 14 '17 at 1:00
• What do you consider a "unit" when you say A does a units per day? If instead you say A does $a$ amount of the job in a day the $1 = (a+b+c)*3 + (a+b)*7$ and $3c = .37$. In other words, 1 unit = total work. Or a unit can be any proportion you want. Totall work = 100 units. Or a 1000 units. It doesn't matter as it is proportional. – fleablood Dec 14 '17 at 1:09

Let $A$, $B$ and $C$ be the amounts of work that can be done in a day by each worker. What this means is that $C$, for example, represents the numerical value of the amount of work worker $C$ can accomplish in a day. In one day, they together can do this much work: $A + B + C$. In two days, $2(A+B+C)$. In three days, $3(A + B + C)=\frac{37}{100}$. We also know that during the last $7$ days only workers $A$ and $B$ were working. And we also know the amount of work that they had done: $1-\frac{37}{100} = \frac{100}{100}-\frac{37}{100}=\frac{67}{100}$. This can be expressed like this: $7(A + B)=\frac{63}{100}$. And we also know that the amount of work done in $5$ days by worker $A$ is the same as the amount of work done by worker $B$ in $4$ days (by the way, we already know that worker $B$ works faster than worker $A$). So, here's what we have so far:

$$3(C + A + B)=\frac{37}{100}\\ 7(A + B)=\frac{63}{100}\\ 5A=4B$$

Now let's solve all that. The amount of work $A$ can do per day:

$$5A=4B\implies B=\frac{5}{4}A\\ 7(A + B)=\frac{63}{100}\implies 7\left(A + \frac{5}{4}A\right)=\frac{63}{100}\implies A=0.04\\$$ Now we can find the amount of work $B$ can do per day: $$B=\frac{5}{4}A\implies B=0.05\\$$ The amount of work C can do per day: $$3(C + A + B)=\frac{37}{100}\implies C=\frac{1}{30}$$

$0.05 > 0.04 > \frac{1}{30}≈0.03$: the fastest worker is worker $B$ because he can accomplish a larger amount of work per day than the other two workers. The whole job when finished is represented by the quantity $\frac{100}{100}$ which is equivalent to $1$. Now, remember that $0.05$ means that it takes worker $B$ one day to do accomplish this much work: $0.05$. How many days will it take him to complete the entire job ($\frac{100}{100}=1$) with that rate? If we let $x$ be the number of days, then it will take worker $B$ to complete the given work alone:

$$x \cdot 0.05 = \frac{100}{100}\\ x \cdot 0.05 = 1\\ x = \frac{1}{0.05}\\ x = 20 \text{ days}$$

I think I interpreted the problem a bit differently, thinking of $a,b$, and $c$ as the proportion of the total job $A$, $B$, and $C$ complete per day, respectively.
In three days, with $A, B$, and $C$ working, $0.37$ of the work gets done: $$3(a+b+c)=\dfrac{37}{100}$$
The job takes $10$ days to complete, where $A, B$ work all ten days, but $C$ only works $3$: $$10(a+b) + 3c = 1$$
Finally, $A$ does $5$ times the work of $B$: $$5a=4b$$
Solving this system yields $a = \dfrac{4}{100}$, $b=\dfrac{5}{100}$, and $c = \dfrac{1}{30}$, so $B$ is the fastest worker. So, it follows that the number of days it takes $B$ to complete the job solo is given by $$\frac{5}{100}t = 1 \iff t = 20 \, \text{days}.$$
• but $C$ is not the fastest worker.. – Surb Dec 13 '17 at 21:51