**Suppose A and B together can do a job in 12 days, while B alone can finish a job in 24 days. In how many days can A alone finish the work?**

**Solution: **

Number of days in which A and B together can finish the work = 12 days

Number of days in which B alone can finish the work = 30

^{1}/_{T} = ^{1}/_{m} + ^{1}/_{n}

^{1}/_{12} = ^{1}/_{m} + ^{1}/_{30}

^{1}/_{m} = ^{1}/_{30} – ^{1}/_{12} = ^{5-2}/_{60} = ^{3}/_{60} = ^{1}/_{20}

A can finish the work in 20 days.

Suppose A is twice as good a workman as B and together they can finish a job in 24 days. How many days A alone takes to finish the job?

Solution:

A is twice as good a workman as B

i.e if B can finish a work in t days A can finish it in ^{1}/_{2} days

^{1}/_{T} = ^{1}/_{m} + ^{1}/_{n}

^{1}/_{24} = ^{1}/_{t/2} + ^{1}/_{t} = ^{2}/_{t} + ^{1}/_{t} = ^{3}/_{t}

^{1}/_{24} = ^{3}/_{t}

t = 24 x 3 = 72

i.e, B takes 72days to finish the job

A takes ^{72}/_{2} = 36 days to finish it

- Suppose B is 60% more efficient them A. if A can finish a job in 15 days how many days B needs to finish the same job?

Solution:

A can finish a work in 15 days.

Work done A in 1 day = ^{1}/_{15}

B is 60% more efficient

Work done by B in 1 day

^{1}/_{15} + ^{1}/_{15} x ^{60}/_{100}

= ^{1}/_{15} (1 + ^{60}/_{100})

= ^{1}/_{15} ( ^{8}/_{5})

= ^{8}/_{75}

Number of days in which B alone can finish the work = ^{1}/_{(8/75)} = ^{75}/_{8} = 9^{3}/_{8} days

- Suppose A can do a piece of work in 14 days while B can do it in 21 days. They begin together and worked at it for 6 days. Then A fell ill B had to complete the work alone. In how many days was the work completed?

Solution:

M = 14 days

N = 21 days

Part of work done in 6 days

= (^{1}/_{14} + ^{1}/_{21})^{6}

= 6(^{3+2}/_{42}) = ^{5×6}/_{42} = ^{5}/_{7}

Remaining part of the work = ^{1-5}/_{7} = ^{2}/_{7}

Days taken by B to finish

^{2}/_{7} part of the work = ^{(2/7)}/_{(1/21)} = ^{2}/_{7} x ^{21}/_{7} = 6 days

Total number of days in which the work is completed = 6+6 = 12 days

- Suppose A takes twice as much time as B and thrice as much time as C to complete a work. If all of them work together they can finish the work in 2 days. How much time B and C working together will take to finish it?

Solution:

If A alone takes to t1 days to do the work , B finishes it in t_{1}/2 and C is t_{1}/_{3} days

^{1}/_{T} = ^{1}/_{t1} +^{1}/_{t2} + ^{1}/_{t3}

= ^{1}/_{t1} +^{1}/_{(t1/2)} + ^{1}/_{(t1/3)}

= ^{1}/_{t1} +^{2}/_{t1} + ^{3}/_{t1}

= ^{6}/_{t1}

^{1}/_{T} = ^{1}/_{2}

^{1}/_{2} = ^{6}/_{T1}

i.e. t_{1} = 12 dyas

B takes ^{12}/_{2} = 6days

C takes ^{12}/_{3} = 4 days

Part of work done by B

In one day = ^{1}/_{6}

Part of work done by C in one day = ^{1}/_{4}

If B and C together takes t days to finish the work ^{1}/_{T} = ^{1}/_{6} + ^{1}/_{4} = ^{2+3}/_{12} = ^{5}/_{12}

T = ^{12}/_{5} = 2.4 days

Pingback: IX – Table of Contents | Breath Math