Work and time

Question

A can do piece of work in 30 days while Bonku alone can do it in 40 days. If they work together, how long will it take to complete the work?

Comments

A ---> 30 days

B --->40 days

let total 120 unit(e.i. LCM of 30 and 40) of work they have to do 

A ---> 120/30 = 4 unit/day

B ---> 120/40 = 3 unit/day

A + B together in one day ---> 3+ 4 = 7 units of work

so, to do 120 units of work  ---> 120/7 days

Answer 1

A ---> 30 days

B --->40 days

let total 120 unit(e.i. LCM of 30 and 40) of work they have to do 

A ---> 120/30 = 4 unit/day

B ---> 120/40 = 3 unit/day

A + B together in one day ---> 3+ 4 = 7 units of work

so, to do 120 units of work  ---> 120/7 days

Answer 2

THIS CAN MADE SIMPLE TOO MUCH BY SOLVING USING PERCENTAGE METHODS:

let suppose that total work be 100 %

work done by Monku (A)  in one day  =100/33=3.33% of work done in one day

work done by Bonku (B) in one day  =100/40=2.25% of work done in one day 

total Work done by both of them in one day =3.33+2.25 =5.83% of work at one Day 

than the times  required for 100% work done by both of them =100/5.83 =17 days 

Answer 3

Let us done by Below Method it will help you slove every problem easily and understand your concept 

Men             A                          B                    A+B

Rate            4                           3                   3+4 = 7

Time           30                         40                    120/7 (Answer)

---

Work          120                      120                  120

Work will be lcm of both A and B time i.e 120. Now we have to calculate each Rate(Efficiency) R1= W/T = 120/30 = 4 same for B i.e 3

Comments

$\text{Total work} =\text{ No of days} \times \text{Efficiency}$

Answer 4

$\text{Total work} =\text{Efficiency (or) One day work}\times \text{Total time}$ 

$\textbf{W = DMTE}$ 

D = Number of days

M = Number of men

T = Number of hours per day

E = Efficiency

W = Amount of work

Now, $D_{1}M_{1} = D_{2}M_{2}$

$\implies 30\times A = 40\times B$

$\implies 3A =4B\implies B = \dfrac{3A}{4}$

Again, $D_{1}M_{1} = D_{2}M_{2}$

$\implies 30 \times A = x \times (A+B)$

$\implies 30A = x \times \bigg(A+\dfrac{3A}{4}\bigg)$

$\implies 30A = x \times \dfrac{7A}{4}$

$\therefore x = \dfrac{120}{7}$  days