# PE 5 - Time & Work | Arithmetic - Time & Work

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**PE 5 - Time & Work | Arithmetic - Time & Work**

An overhead water tank in a locality has certain amount of water in it. Additionally it receives some amount of water from municipal corporation on a daily basis. Each resident in the locality consumes a fixed amount of water every day. If there were 120 residents, the water would last for 50 days. Instead if there were 125 residents, the water would last for only 40 days. How long will the water last if there were 110 residents.

- (a)
110 days

- (b)
100 days

- (c)
90 days

- (d)
85 days

Answer: Option B

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**Explanation** :

Let the

amount of water initially in the tank = V liters,

amount supplied by municipal corporation every day = d liters/day

consumption rate of each resident = c liters/day

If there were 120 residents, the water would last for 50 days

⇒ V + 50 × d = c × 120 × 50 …(1)

If there were 125 residents, the water would last for only 40 days

⇒ V + 40 × d = c × 125 × 40 …(2)

Solving (1) and (2), we get

⇒ 6000c – 50d = 5000c – 40d

⇒ 1000c = 10d

⇒ d = 100c

From (1) ⇒ V + 5000c = 6000c

⇒ V = 1000c

Let the water last for x days when there are 110 residents

⇒ V + x × d = c × 110 × x

⇒ 1000c + 100cx = 110cx

⇒ 1000c = 10cx

⇒ x = 100 days.

Hence, option (b).

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**PE 5 - Time & Work | Arithmetic - Time & Work**

A work is done by 30 workers, not all of whom are equally efficient. Every day exactly 2 workers come for work, with no pair of workers working together twice. Even after all possible pairs have worked once, the work isn’t complete. Thus, all the 30 workers work together for three more days to finish the work. Find the number of days in which the whole work would finish, if all the workers worked together.

- (a)
31

- (b)
30

- (c)
29

- (d)
32

Answer: Option D

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**Explanation** :

Here let us focus on work done by each of these 30 workers.

Let the efficiency of first worked by w_{1}.

He would work in pairs with each of the other 29 workers, hence he would work for 29 days. Additionally he would also work for another 3 days when everyone works.

∴ Worker 1 works for a total of 29 + 3 = 32 days

Hence, work done by worker 1 = 32 × w_{1}.

Similarly work done by nth worker would be 32 × w_{n}.

⇒ Total work done by them = 32w_{1} + 32w_{2} + 32w_{3} + … + 32w_{30} = 32 × (sum of their efficiencies) …(1)

Now when all of them work together let’s say they take d days to complete the work.

∴ Work done by all of them together in d days = d × (sum of their efficiencies) …(2)

Now (1) = (2)

⇒ d × (sum of their efficiencies) = 32 × (sum of their efficiencies)

⇒ d = 32 days.

Hence, option (d).

Workspace:

**PE 5 - Time & Work | Arithmetic - Time & Work**

2 men & 3 women complete a task in 70 days. Each woman is at least twice as efficient as a man but not more than thrice that of a man. If 2 men and 6 women take ‘d’ days to complete the same task, find the range of values that ‘d’ can take.

- (a)
38.5 - 40

- (b)
35 – 40

- (c)
32.5 – 34

- (d)
38 – 40.5

Answer: Option A

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**Explanation** :

Let the efficiency of each man and woman be ‘m’ and ‘w’ units/day.

∴ Work done by 2 men and 3 women in 70 days = (2m + 3w) × 70 …(1)

Work done by a men and 6 women in d days = (2m + 6w) × d …(2)

(1) = (2)

⇒ (2m + 3w) × 70 = (2m + 6w) × d

⇒ d = $\frac{2\mathrm{m}+3\mathrm{w}}{2\mathrm{m}+6\mathrm{w}}$ × 70 = $\frac{2\mathrm{m}+3\mathrm{w}}{\mathrm{m}+3\mathrm{w}}$ × 35 = $\left(\frac{\mathrm{m}}{\mathrm{m}+3\mathrm{w}}+\frac{\mathrm{m}+3\mathrm{w}}{\mathrm{m}+3\mathrm{w}}\right)$ × 35

⇒ d = $\left(\frac{1}{1+3{\displaystyle \frac{\mathrm{w}}{\mathrm{m}}}}+1\right)$ × 35

Now, d will be maximum when w/m is minimum i.e., when a woman is only twice as efficient as a man.

⇒ d_{max} = $\left(\frac{1}{1+3\times 2}+1\right)$ × 35 = 40 days

Now, d will be minimum when w/m is maximum i.e., when a woman is thrice as efficient as a man.

⇒ d_{min} = $\left(\frac{1}{1+3\times 3}+1\right)$ × 35 = days

∴ 2 men and 6 women can complete the task in 38.5 - 40 days.

Hence, option (a).

Workspace:

**PE 5 - Time & Work | Arithmetic - Time & Work**

A woman can do as much work in one day as a man can do in 3 days. A child does half the work in a day as a woman. An estate-owner hires 30 people with men, women and children in the ratio 4 : 6 : 5. He pays them ₹ 1120 everyday. Find the daily wages of a child.

- (a)
Rs. 15

- (b)
Rs. 10

- (c)
Rs. 20

- (d)
Rs. 30

Answer: Option D

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**Explanation** :

Number of hired

men = 4/15 × 30 = 8

women = 6/15 × 30 = 12

children = 5/15 × 30 = 10

Let the efficiency of each man, woman and child be m, w and c units/day respectively.

Given, w = 3m and w = 2c

⇒ m : w : c = 2 : 6 : 3

Let’s assume m = 2, w = 6 and c = 2 units/day.

Total work done by

men = 2 × 8 = 16 units/day

women = 6 × 12 = 72 units/day

children = 3 × 10 = 30 units/day

∴ Total work done in a day = 16 + 72 + 30 = 112 units/day

⇒ Fraction of work done by a child = $\frac{3}{112}$ ^{th}

∴ Payment received by each child/day = $\frac{3}{112}$ × 1120 = Rs. 30.

Hence, option (d).

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