Question

A can do a piece of work in 14 days while B can do it in 21 days . they begin together. but 3 days before the completion of the work ,A leaves off .find the total numbers of daysbtaken to complete the work​

Answer 1

${\large\bf{\mid{\overline{\underline{Given:-}}}\mid}}$

• A can do a work in = 14 days .
• B can do it in = 21 days .
• A left before 3 days of the work .

$\Large\underline\mathfrak{Question}$

• we have to Find in how many days the work got completed ?

$\Large\underline{\underline{\sf{Solution}:}}$

we can solve it by either taking x , or we can save time by solving LCM method . ( Thats my favourite ) .

Lets solve this now ,,,

LCM of 14 and 21 = 42

→ so, Let Total unit of work that would be done is = 42 units .

Now, since A completed it alone in 14 days ,

→ His efficiency of per day work = 42/14 = 3 unit/day .

________________

Now, since B complete it alone in 21 days .

→ His efficiency of per day work = 42/21 = 2 unit/day .

___________________________

Again now, we can solve it by taking let they work together for X days and than A left .

But i dont wanna solve by any Equation ,

so i will Try to solve it without X now ....

______________________

Since it is Given that A left before 3 days of the completion of the work ,

That means we can say that, only B can work for that last three days ... ( Hope You understand what i m saying) .

Now, we know that , B work = 2 unit /day .

→ so, in last 3 days he did = 2×3 = 6 units of work...

___________________

Now, total work is 42 units , and 6 units of work done by B alone in last three days, that means rest work was done by both of them .......

$\red{\textbf{ plz think once}}$

______________________

→ Total unit of work was = 42 units,

→ Done by B in three days = 6 units .

→ Left work to be done = 42 -6 = 36 units ..

[ i Hope its clear now, this was done by both of them now, , ]

__________________

Now ,

→ A and B per day work = 2+3 = 5 units .

$\green{time \: for \: them \: to \: complete \: rest \:} \\ \green{36 \: units \: of \: work}\: = \frac{36}{5} = \pink{ 7 \times \frac{1}{5} days}.$

_______________________

$\textbf{</strong><strong>So</strong><strong>,</strong><strong> </strong><strong>Total</strong><strong> </strong><strong>time</strong><strong> </strong><strong>taken</strong><strong> </strong><strong>to</strong><strong> </strong><strong>complete</strong><strong> </strong><strong>whole</strong><strong> </strong><strong>work</strong><strong>}$

= (A+B) Together days + B alone 3 days .

$\red{\boxed\implies} \: \frac{36}{5} + 3 \\ \\ \red{\boxed\implies} \: \frac{36 + 15}{5} \\ \\ \red{\boxed\implies} \: \frac{51}{5} \\ \\ \red{\boxed\implies} \: \red{\large\boxed{\bold{10 \times \frac{1}{5} \: days}}}$

_______________

$\textbf{so the whole work will} \\ \textbf{be competed in } \: \red{10 \times \frac{1}{5} \: \: days..}$

$\large\underline\textbf{Hope it Helps You.}$