# Time and Work Questions

**FACTS AND FORMULAE FOR TIME AND WORK QUESTIONS**

**1. **If A can do a piece of work in n days, then A's 1 day's work =$\frac{1}{n}$

**2. **If A’s 1 day's work =$\frac{1}{n}$, then A can finish the work in n days.

**3. **A is thrice as good a workman as B, then:

Ratio of work done by A and B = 3 : 1.

Ratio of times taken by A and B to finish a work = 1 : 3.

NOTE :

$Efficiency\propto \frac{1}{Nooftimeunits}$

$\therefore Efficiency\times Time=Cons\mathrm{tan}tWork$

Hence, $Requiredtime=\frac{Work}{Efficiency}$

Whole work is always considered as 1, in terms of fraction and 100% , in terms of percentage.

In general, number of day's or hours = $\frac{100}{Efficiency}$

A) 11 days | B) 12 days |

C) 13 days | D) 14 days |

Explanation:

One day's work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10.

C leaves 4 days before completion of the work, which means only A and B work during the last 4 days.

Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10.

Remaining Work = 7/10, which was done by A,B and C in the initial number of days.

Number of days required for this initial work = 7 days.

Thus, the total numbers of days required = 4 + 7 = 11 days.

A) 200, 250, 300 | B) 300, 200, 250 |

C) 200, 300, 400 | D) None of these |

Explanation:

A's 5 days work = 50%

B's 5 days work = 33.33%

C's 2 days work = 16.66% [100- (50+33.33)]

Ratio of contribution of work of A, B and C = $50:33\frac{1}{3}:16\frac{2}{3}$ = 3 : 2 : 1

A's total share = Rs. 1500

B's total share = Rs. 1000

C's total share = Rs. 500

A's one day's earning = Rs.300

B's one day's earning = Rs.200

C's one day's earning = Rs.250

A) 60/11 | B) 61/11 |

C) 71/11 | D) 72/11 |

Explanation:

P can complete the work in (12 x 8) hrs = 96 hrs

Q can complete the work in (8 x 10) hrs=80 hrs

Therefore, P's 1 hour work=1/96 and Q's 1 hour work= 1/80

(P+Q)'s 1 hour's work =(1/96) + (1/80) = 11/480. So both P and Q will finish the work in 480/11 hrs

Therefore, Number of days of 8 hours each = (480/11) x (1/8) = 60/11

A) 8 | B) 12 |

C) 16 | D) 24 |

Explanation:

1 man's 1 day work =1/96 ; 1 woman's 1 day work = 1/192

Work done in 6 days=$6\left(\frac{8}{96}+\frac{8}{192}\right)=6\times \frac{1}{8}=\frac{3}{4}$

Remaining work = 1/4

(8 men +8 women)'s 1 day work = $1\left(\frac{8}{96}+\frac{8}{192}\right)$ =1/8

Remaining work =1/4 - 1/8 = 1/8

1/96 work is done in 1 day by 1 man

Therefore, 1/8 work will be done in 1 day by 96 x (1/8) =12 men

A) 18 minutes | B) 19 minutes |

C) 22 minutes | D) 24 minutes |

Explanation:

Let the two conditioners be A and B

'A' cools at 40min

'B' at 45min

Together = **(a x b)/(a + b)**

= (45 x 40)/(45 + 40)

= 45 x 40/85

= 21.1764

= **22 min** (approx).

Ratio of times taken by A and B = 1 : 2.

The time difference is (2 - 1) 1 day while B take 2 days and A takes 1 day.

If difference of time is 1 day, B takes 2 days.

If difference of time is 30 days, B takes 2 x 30 = 60 days.

So, A takes 30 days to do the work.

A's 1 day's work = 1/30

B's 1 day's work = 1/60

(A + B)'s 1 day's work = 1/30 + 1/60 = 1/20

A and B together can do the work in 20 days.

A) 25 | B) 24 |

C) 23 | D) 21 |

Explanation:

(A+B+C) do 1 work in 10 days.

So (A+B+C)'s 1 day work=1/10 and as they work together for 4 days so workdone by them in 4 days=4/10=2/5

Remaining work=1-2/5=3/5

(B+C) take 10 more days to complete 3/5 work. So( B+C)'s 1 day work=3/50

Now A'S 1 day work=(A+B+C)'s 1 day work - (B+C)'s 1 day work=1/10-3/50=1/25

A does 1/25 work in in 1 day

Therefore 1 work in 25 days.

A) 17 men | B) 14 men |

C) 13 men | D) 16 men |

Explanation:

M x T / W = Constant

where, M= Men (no. of men)

T= Time taken

W= Work load

So, here we apply

M1 x T1/ W1 = M2 x T2 / W2

Given that, M1 = 4 men, T1 = 7 hours ; T2 = 2 hours, we have to find M2 =?

Note that here, W1 = W2 = 1 road, ie. equal work load.

Clearly, substituting in the above equation we get, M2 = 14 men.