Time and Work - 04

Explanation:-

$$\eqalign{ & {\text{Let}}\,{\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = x\,{\text{and}} \cr & {\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = y \cr & {\text{Then}},\,x + y = \frac{1}{{30}} \cr & 16x + 44y = 1 \cr & {\text{Solving}}\,{\text{these}}\,{\text{two}}\,{\text{equations,}}\,{\text{we}}\,{\text{get}} \cr & x = \frac{1}{{60}}\,{\text{and}}\,y = \frac{1}{{60}} \cr & \therefore {\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{60}} \cr & {\text{Hence,}}\,{\text{B}}\,{\text{alone}}\,{\text{shall}}\,{\text{finish}}\,{\text{the}} \cr & {\text{whole}}\,{\text{work}}\,{\text{in}}\,{\text{60}}\,{\text{days}} \cr}$$

Explanation:-

$$\eqalign{ & \left( {{\text{B + C}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = \left( {\frac{1}{9} + \frac{1}{{12}}} \right) = \frac{7}{{36}} \cr & {\text{Work}}\,{\text{done}}\,{\text{by}}\,{\text{B}}\,{\text{and}}\,{\text{C}}\,{\text{in}}\,{\text{3}}\,{\text{days}} \cr & = \left( {\frac{7}{{36}} \times 3} \right) = \frac{7}{{12}} \cr & {\text{Remaining}}\,{\text{work}} \cr & = \left( {1 - \frac{7}{{12}}} \right) = \frac{5}{{12}} \cr & {\text{Now}},\,\frac{1}{{24}}\,{\text{work}}\,{\text{is}}\,{\text{done}}\,{\text{by}}\,{\text{A}}\,{\text{in}}\,{\text{1}}\,{\text{day}} \cr & {\text{So}},\,\frac{5}{{12}}\,{\text{work}}\,{\text{is}}\,{\text{done}}\,{\text{by}}\,{\text{A}}\,{\text{in}} \cr & \left( {24 \times \frac{5}{{12}}} \right) = 10\,{\text{days}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{2(A + B + C)'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = \left( {\frac{1}{{30}} + \frac{1}{{24}} + \frac{1}{{20}}} \right) \cr & = \frac{{15}}{{120}} = \frac{1}{8} \cr & \therefore \left( {{\text{A + B + C}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = \frac{1}{{2 \times 8}} = \frac{1}{{16}} \cr & {\text{Work}}\,{\text{done}}\,{\text{by}}\,{\text{A,}}\,{\text{B,}}\,{\text{C}}\,{\text{in}}\,{\text{10}}\,{\text{days}} \cr & = \frac{{10}}{{16}} = \frac{5}{8} \cr & {\text{Remaining}}\,{\text{work}} \cr & = \left( {1 - \frac{5}{8}} \right) = \frac{3}{8} \cr & {\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = \left( {\frac{1}{{16}} - \frac{1}{{24}}} \right) = \frac{1}{{48}} \cr & {\text{Now}},\,\frac{1}{{48}}\,{\text{work}}\,{\text{isdone}}\,{\text{by}}\,{\text{A}}\,{\text{in}}\,{\text{1}}\,{\text{day}} \cr & {\text{So}},\,\frac{3}{8}\,{\text{work}}\,{\text{will}}\,{\text{be}}\,{\text{done}}\,{\text{by}}\,{\text{A}}\,{\text{in}} \cr & \left( {48 \times \frac{3}{8}} \right) = 18\,{\text{days}} \cr}$$

Explanation:-

Number of page copied by (Subhash + Prakash) in 1 hour = $$\frac{{300}}{{40}}$$ = 7.5 pages;
Subhash copied pages in one hour = $$\frac{{50}}{{10}}$$ = 5 pages
Hence, Prakash copied pages in one hour = 7.5 - 5 = 2.5
Thus,
Prakash can copied 30 pages in = $$\frac{{30}}{{2.5}}$$ = 12 hour

Explanation:-

Man = 3 boy
Daily wages for them = $$\frac{{800}}{5}$$ = 160
4 boy (1 man + 1 boy) = 160
4 boy = 160
Or, boy = Rs. 40

Explanation:-

Leak emptied the tank per minute = $$\frac{{100}}{{10}}$$ = 10% of water per hour
Quantity of water emptied per hour = 4 × 60 = 240
Thus, 10% = 240 liter
Hence, capacity of water,
100% = $$\frac{{240 \times 100}}{{10}}$$   = 2400 liter.

Explanation:-

Water filled by (boy + girl) in one minute,
$$\eqalign{ & \frac{4}{3} + \frac{3}{4} = \frac{{16 + 9}}{{12}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{25}}{{12}}\,{\text{liter}} \cr}$$
Hence, time taken to fill 100 liter,
$${\text{ = 100}} \times \frac{{12}}{{25}} = 48\,{\text{minutes}}$$

Explanation:-

Let the pipes fill the tank in x minutes.
Part of tank filled by pipes A and B in one minute: $$\frac{1}{{36}}$$
Part of the tank filled by pipe A in 1 minute = $$\frac{1}{{36}} - \frac{1}{x}$$

According to the question,

$$\eqalign{ & 30 \times \frac{1}{{\text{x}}} + \left( {\frac{1}{{36}} - \frac{1}{{\text{x}}}} \right) = 1 \cr & \Rightarrow \frac{{30}}{{\text{x}}} + \frac{{10}}{9} - \frac{{40}}{{\text{x}}} = 1 \cr & \Rightarrow \frac{{10}}{{\text{x}}} = \frac{1}{9} \cr}$$
Hence, x = 90 minutes.

Alternatively,

% of tank filled by (A+B) in 1 minute = $$\frac{{100}}{{36}}$$ = 2.78%
% of tank filled in 30 minutes = 2.78 × 30 = 83%
That means 17% of tank can be filled by A in last 10 minutes
Then, % tank filled by A in 1 minute = $$\frac{{17}}{{10}}$$ = 1.7%
% of tank filled by B in 1 minute = 2.78 - 1.7 = 1.1%
So, B alone can fill 100% tank in,
= $$\frac{{100}}{{1.1}}$$ = 90 minutes (approx.)

Explanation:-

% Cistern is filled by 1^st pipe in one minute = $$\frac{{100}}{{20}}$$ = 5%;
% Cistern is filled by 2^nd pipe in one minute = $$\frac{{100}}{{30}}$$ = 3.33%;
% cistern filled by 1^st and 2^nd pipes in one minute = 8.33%;

According to question,
Cistern is totally filled by 2^nd pipe in last 10 minute.
That means 2^nd pipe filled 33.3% of the cistern in last 10 minute and 66.66% of cistern is filled by 1^st and 2^nd pipe together in = $$\frac{{66.66}}{{8.33}}$$   = 8 minutes;
Thus, after 8 minute, 1^st pipe must be turned off.

Explanation:-

P can fill cistern in 12 minutes
P fills cistern in 1 minute = $$\frac{1}{{12}}$$
Q can fill cistern in 15 minutes
Q fills cistern in 1 minute = $$\frac{1}{{15}}$$ part
P and Q together can fill cistern in 1 minute,
$$= \frac{1}{{12}} + \frac{1}{{15}} \Rightarrow \frac{9}{{60}}$$    part
So, they can together fill cistern in 3 minute,
$$= 3 \times \frac{9}{{60}} \Rightarrow \frac{9}{{20}}$$    part
Rest Cistern = $$1 - \frac{9}{{20}}$$   = $$\frac{{11}}{{20}}$$   part
$$\frac{{11}}{{20}}$$ part cistern must be filled by Q in
$$\frac{{\frac{{11}}{{20}}}}{{\frac{1}{{15}}}} = 8\frac{1}{4}$$   minutes

Alternatively,
Pipe P can fill cistern in one minute = $$\frac{{100}}{{12}}$$ = 8.33%
Pipe Q can fill cistern in one minute = $$\frac{{100}}{{15}}$$ = 6.66%
(P + Q) together can fill the cistern in one minute = 15%;
In 3 minutes cistern is filled = 45%
P become off, then rest of cistern will be filled by Pipe Q in = $$\frac{{55}}{{6.66}}$$ = $$8\frac{1}{4}$$ minutes.