Question

A work was completed by three persons of equal ability, first one doing m hours for m days, second one doing n hours for n days (m and n being integers) and third one doing 16 hours for 16 days. The work could have been completed in 29 days by third person alone with his respective working hours. If all of them do the work together with their respective working hours, then they can complete it in about

Answer 1

Answer:

solved

Explanation:

first person = m hours

second person = n hours

third person = 45 hours

then they can complete it in about 45 days /3 = 15 days

Answer 2

They can complete it in about "13 days".

Explanation:

We have,

$(m^{2} +n^{2} +16^{2} )k=1$      ...(1)

and

16 × 29k = 1

∴ k = $\dfrac{1}{16\times 29}$

Putting the value of k in equation (1), we get

$m^{2} +n^{2} +16^{2} =16\times 29$

⇒$m^{2} +n^{2} +16^{2} =16(29-16)=16\times 13=208$

Now, the last digits of m, n cannot be (0, 8), (1, 7), (2, 6), (3, 5).

Hence, it can only be (4, 4) or (9, 9).

On checking,

We find,

$m^{2} +n^{2}=12^{2} +8^{2}$

∴ They can together do the work in

= $\dfrac{1}{(m+n+16)k} =\dfrac{16\times 29}{(12+8+16)}$

$=\dfrac{16\times 29}{36}=12.88$

≈ 13 days

Hence, they can complete it in about 13 days.