Arithmetic aptitude :: Time and Work :: General Questions

• Time and Work - General Questions
• Time and Work - Data Sufficiency 1
• Time and Work - Data Sufficiency 2
• Time and Work - Data Sufficiency 3
• Time and Work - Important Formulas and Facts

If $A$ can do a part of work in $n$ days, then $A$'s $1$ day work $=\frac{1}{n}$.

If $A$'s $1$ day's work $=\frac{1}{n}$, then $A$ can complete the work in $n$ days.

If $A$ is twice as good a worker as $B$, then:

Ratio of the work done by $A$ and $B$ $=2:1$.

Ratio of times taken by $A$ and $B$ to finish a work $=1:2$.

A, B and C together can complete a piece of work in 10 days. All the three started working at it together and after 4 days A left. Then B and C together completed the work in 10 more days. A alone could complete the work in :

Answer:

Explanation:

Set a = the rate of A in (piece of work)/day

b = the rate of B in (piece of work)/day

c = the rate of C in (piece of work)/day

Below is the equation for all three working together.

($10$ days )$\times$ $(a+b+c)$ = 1(piece of work)

Next we have the equation were all work together for 4 days,

then only B and C continue for $10$ more days.

($4$ days) $\times$ $(a+b+c )$ + ($10$ days)$( b + c)$ = 1(piece of work)

Since each left hand side is equal to 1 piece of work,

we have
($10$ days) $\times$ $( a + b + c )$ = ($4$ days) $\times$ $(a+b+c)$ + ($10$ days)$(b+c)$

and

$10(a+b+c )$ $= 4(a+b+c )$ + $10(b+c)$

$10a + 10b + 10c$ $= 4a + 4b + 4c + 10b + 10c$

combining terms

$10a+10b+10c$ $=4a+14b+14c$

Notice that we can rewrite the above as

$10a + 10(b + c)$ $=4a+4(b+c)$

if we subtract $10(b + c)$ from each side

$10a$ $=4a+4(b+c)$

subtract $4a$ from each side

$6a$ $= 4(b + c)$

divide each side by $4$

$\frac{6a}{4}$ $=(b+c)$

Recall

($10$ days) $\times$ $(a+b+c)$ = 1(piece of work) or

$10(a+b+c)$ $= 1$

So

$10a+10(b+c)$ $=1$

Now substitute $\frac{6a}{4}$ for $(b+c)$ and we have

$10a + 10\left(\frac{6a}{4}\right)$ $= 1$

$10a+15a$ $= 1$

$25a=1$

This means that A at rate a will take 25 days to complete the work.

Two men undertake to do a piece of work for Rs. 240. One could do it alone in 5 days and the other in 8 days. With the help of a boy and a girl they finished it in 3 days. If the work done by the boy be twice that of the girl, what is the share of the girl ?

Answer:

Explanation:

Let name, two men $=A,B$

One boy + girl $=C$

Amount of work $A$ can do in $1$ day $=\frac{1}{5}$

Amount of work $B$ can do in $1$ day $= \frac{1}{8}$

Amount of work $A+B$ can do in $1$ day $=\frac{1}{5}+\frac{1}{8}$ $=\frac{13}{40}$

Amount of work $A+B+C$ can do $=\frac{1}{3}$

Amount of work $C$ can do in $1$ day $= \frac{1}{3}-\frac{13}{40}$ $=\frac{1}{120}$

Work $A$ can do in $1$ day: Work $B$ can do in $1$ day: Work $C$ can do in $1$ day

$=\frac{1}{5} :\frac{1}{8} :\frac{1}{120}$

From the question, one boy work $=2\times$ Girl

So, C = 2Girl + Girl $=\frac{1}{120}$

$\Leftrightarrow$ Girl's one day work $=\frac{1}{360}$

Amount to be paid to Girl $=240\times\frac{1}{120}$ = Rs.2.

Sixteen men can complete a work in twelve days. Twenty-four children can complete the same work in eighteen days. Twelve men and eight children started working and after eight days three more children joined them. How many days will they now take to complete the remaining work ?

Answer:

Explanation:

$1$ man's $1$ day work $=\frac{1}{192}$

child's $1$ day's work = $\frac{1}{432}$.

Work done in $8$ days $=8\left(\frac{12}{192}+\frac{8}{432}\right)$ $=8\left(\frac{1}{16}+\frac{1}{54}\right)$ $=\frac{35}{54}$.

Remaining work $=1-\frac{35}{54}$ $=\frac{19}{54}$.

($12$ men $+11$ children)'s $1$ day work $=\frac{12}{192}+\frac{11}{432}$ $=\frac{19}{216}$.

Now, $\frac{19}{216}$ work is done by them in $1$ day.

$\therefore\frac{19}{54}$ work will be done by them in $\left(\frac{216}{19}\times\frac{19}{54}\right)$ = 4 days.

A and B together can complete a work in 3 days. They start together but after 2 days, B left the work. If the work is completed after 2 more days, B alone can do the work in

Answer:

Explanation:

$(A+B)$'s 2 days work = $\left(\frac{1}{3}\times 2\right)$ $=\frac{2}{3}$.

Remaining work $=\left(1-\frac{2}{3}\right)$ $=\frac{1}{3}$.

Now, $\frac{1}{3}$ work is done by $A$ in 2 days.

Whole work will be done by $A$ in $(2\times 3)$ = 6 days.

$1$ man, $3$ women and $4$ boys can do a piece of work in $96$ hours, $2$ men and $8$ boys can do it in $80$ hours, $2$ men and $3$ women can do it in $120$ hours. $5$ men and $12$ boys can do it in :

Answer:

Explanation:

Let $1$ man's hour's work $=x$, $1$ woman's $1$ hour's work $=y$ and $1$ boy's $1$ hour's work $=z$.

Then, $x+3y+4z=\frac{1}{96}$ ---(1)

$2x+8z=\frac{1}{80}$ ---(2)

$2x+3y=\frac{1}{120}$ ---(3)

Adding (2) and (3) and subtracting (1) from it, we get : $3x+4z=\frac{1}{96}$---(4)

From (2) and (4), we get $x=\frac{1}{480}$.

Substituting, we get : $y=\frac{1}{720}$, $z=\frac{1}{960}$.

($5$ men + $12$ boy's) $1$ hour's work $=\left(\frac{5}{480}+\frac{12}{960}\right)$ $=\left(\frac{1}{96}+\frac{1}{80}\right)$ $=\frac{11}{480}$.

$\therefore$ $5$ men and $12$ boys can do the work in $\frac{480}{11}$

i.e, $43\frac{7}{11}$ hours.

• Time and Work - General Questions
• Time and Work - Data Sufficiency 1
• Time and Work - Data Sufficiency 2
• Time and Work - Data Sufficiency 3