# PE 4 - Time & Work | Arithmetic - Time & Work

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**PE 4 - Time & Work | Arithmetic - Time & Work**

A piece of work can be completed by 15 men in a certain number of days. If there were three more men the work will be completed one day earlier. In how many days can 45 men complete the same work?

Answer: 2

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**Explanation** :

Let 15 men take ‘d’ days to complete the work.

∴ 18 men will take (d – 1) days to complete the same work.

∴ 15 × d = 18 × (d – 1)

⇒ d = 6 days.

∴ Total work = 15 × 6 = 90 units.

Let 45 men take ‘x’ days to complete the same work

∴ 90 = 45 × x

⇒ x = 2 days.

Hence, 2.

Workspace:

**PE 4 - Time & Work | Arithmetic - Time & Work**

Certain number of men can complete a work in 16 days. They start working together. Starting from second day a man leaves at the start of the day. The work now gets completed at the end of the day when the last person works. How many men were there at the start.

Answer: 31

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**Explanation** :

Let the number of men at the start be ‘m’.

Let efficiency of each man be 1 unit/day.

When all of them work it takes 16 days to complete the work.

∴ Total work done = m × 16 units.

Now, if a person leaves at the start of every day.

Work done on 1^{st} day = m × 1 = m units

Work done on 2^{nd} day = (m – 1) × 1 = (m – 1) units

.

.

.

Work done on m^{th} day = 1 × 1 = 1 units

∴ Total work done = m + (m – 1) + … + 1 = m(m + 1)/2 units.

⇒ m(m + 1)/2 = m × 16

⇒ m = 31

Hence, 31.

Workspace:

**PE 4 - Time & Work | Arithmetic - Time & Work**

A and B can do a certain piece of work in 18 days, B and C can do it in 12 days and C and A can do it in 24 days. How long would each take separately to do it (in days resp)?

- (a)
144, 115, 95

- (b)
144/7, 110, 80

- (c)
114/5, 120, 45

- (d)
None of these

Answer: Option D

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**Explanation** :

Let the work to be done = LCM (18, 12, 24) = 72 units

∴ Efficiency of A + B = 72/18 = 4 units/day …(1)

Efficiency of B + C = 72/12 = 6 units/day …(2)

Efficiency of C + A = 72/24 = 3 units/day …(3)

Adding (1), (2) and (3), we get

(A + B + C) = 6.5 …(4)

Now, solving (4), (1), (2) and (3), we get

Efficiency of A = 0.5 ⇒ Time taken by A alone = 72/0.5 = 144

Efficiency of B = 3.5 ⇒ Time taken by B alone = 72/3.5 = 144/7

Efficiency of C = 2.5 ⇒ Time taken by C alone = 72/2.5 = 144/5

Hence, option (d).

Workspace:

**PE 4 - Time & Work | Arithmetic - Time & Work**

A pipe can fill a tank in 10 minutes and another pipe can fill it in 15 minutes. A person opens both the pipes simultaneously. When the tank should have been full, he finds that the drain pipe is open. He then closes the drain pipe and after six minutes, the tank is full. In how much time can the drain pipe empty the tank?

- (a)
5 mins

- (b)
8 mins

- (c)
4 mins

- (d)
3 mins

- (e)
6 mins

Answer: Option E

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**Explanation** :

First pipe can fill the tank in 10 mins while second pipe can fill the tank in 15 mins.

Together, both pipes can fill the tank in $\frac{1}{{\displaystyle \frac{1}{10}}+{\displaystyle \frac{1}{15}}}$ = 6 mins

Now, during the first 6 mins drain pipe was also open which was closed and it takes another 6 mins for the tank to be completely filled.

∴ 6$\left(\frac{1}{10}+\frac{1}{15}-\frac{1}{x}\right)$ + 6$\left(\frac{1}{10}+\frac{1}{15}\right)$ = 1

[Where x = Time taken by drain pipe to empty the tank]

⇒ $\left(\frac{1}{10}+\frac{1}{15}-\frac{1}{x}\right)$ + $\left(\frac{1}{10}+\frac{1}{15}\right)$ = $\frac{1}{6}$

⇒ $\frac{2}{10}$ + $\frac{2}{15}$ - $\frac{1}{x}$ = $\frac{1}{6}$

⇒ $\frac{1}{3}$ - $\frac{1}{x}$ = $\frac{1}{6}$

⇒ x = 6

**Alternately**,

Time taken for both filling pipes to fill the tank is 6 mins.

Due to drain pipe it taken another 6 mins to fill the tank. Hence, during the first 6 mins effectively no work was done.

∴ Work done by filling pipes in first 6 mins = Work done by the draining pipe in first 6 pipes.

In 6 mins filling pipes would’ve completely filled the tank, hence the draining pipe must have completely emptied the tank in first 6 mins.

∴ It takes 6 mins for the draining pipe to completely empty the tank.

Hence, option (e).

Workspace:

**PE 4 - Time & Work | Arithmetic - Time & Work**

To completely fill a half filled tank, tap X is opened and the tank will be fully filled in 30 hours. But after 6 hours, the water supply was cut-off. Before the supply resumes, exactly half of the water in the tank has been used up. How much time would it take now to fill the tank completely?

- (a)
12 hours

- (b)
15 hours

- (c)
42 hours

- (d)
24 hours

- (e)
20 hours

Answer: Option C

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**Explanation** :

Half of the tank is filled by tap X in 30 hours. Hence, the tank is fully filled by tap X in 60 hours.

The tank is initially half full and it would take 30 hours to fill the tank.

Part of the tank filled in 6 hours = $\frac{6}{10}$ = $\frac{1}{10}$ ^{th}

Now, according to the question, amount of water in the tank = $\frac{1}{2}$ + $\frac{1}{10}$ = $\frac{6}{10}$ ^{th}

After this, half of water has been used up during the supply cut.

∴ The remaining water = $\frac{3}{10}$ ^{th}

The part of tank still empty = 1 – $\frac{3}{10}$ = $\frac{7}{10}$ ^{th}

Time taken to fill $\frac{7}{10}$ ^{th} of the tank = $\frac{7}{10}$ × 60 = 42 hours

Hence, option (c).

Workspace:

**PE 4 - Time & Work | Arithmetic - Time & Work**

An inlet tap can fill a tank in 20 hours. This tap was opened at time zero, but after 4 hours (t = 4), a leak developed at the bottom. After 8 hours (t = 8), the cross section of the leak became triple because of which, it took 30 hours to fill the tank. How much time will it take to fill the tank if the cross section of the hole does not change at t = 8?

[Type in your answer as the nearest possible integer in seconds.]

Answer: 81600

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**Explanation** :

Let the hole formed initially can empty the entire tank in x hours.

After the cross section triples, it empties the entire tank in x/3 hours.

The inlet pipe worked for 30 hours. The outlet pipe worked for 4 hours at normal efficiency and then worked for another 22 hours at three times the efficiency.

∴ $\frac{1}{20}$ × 30 - $\frac{1}{x}$ × 4 - $\frac{3}{x}$ × 22 = 1

⇒ $\frac{3}{2}$ - $\frac{70}{x}$ = 1

⇒ $\frac{70}{x}$ = $\frac{1}{2}$

⇒ x = 140 hours.

Now had the cross section not changed let’s say it would’ve taken t hours to fill the tank.

∴ The inlet pipe works for t hours, whereas the outlet pipe works at normal efficiency for (t – 4) hours.

∴ $\frac{1}{20}$ × t - $\frac{1}{140}$ × (t - 4) = 1

⇒ t$\left(\frac{1}{20}-\frac{1}{140}\right)$ = 1 - $\frac{4}{140}$

⇒ t(7 - 1) = 140 - 4

⇒ t = 136/6 = 81600 seconds.

Hence, 81600.

Workspace:

**PE 4 - Time & Work | Arithmetic - Time & Work**

The amount of work done by a man on nth day after starting a piece of work is n times the amount of work that can be done on the first day. If 15 men start the work, it can be completed in 7 days. How many men should work if the work has to be completed in 3 days?

Answer: 70

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**Explanation** :

Let each of the 8 men does x units of work on the first day.

Total work done = 15(x + 2x + 3x + 4x + 5x + 6x + 7x) = 420x

To do the same 420x units work in 15 days, let n men be required.

So, n(x + 2x + 3x) = 420x

⇒ n × 6x = 420x

⇒ n = 70

Hence, 70.

Workspace:

**PE 4 - Time & Work | Arithmetic - Time & Work**

A man started a piece of work on the first day. The second day, one more person joined him. The next day, one more person joined. Everyday, one new person joined until the work got completed. Also every day the efficiency of all men doubled as compared to previous day. If the work was completed in 5 days, how many days would 43 men take to complete the same work, given that they worked regularly and with the same efficiency on all days as on day 1?

- (a)
$3\frac{1}{3}$ days

- (b)
$2\frac{1}{2}$ days

- (c)
3 days

- (d)
2 days

- (e)
5 days

Answer: Option C

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**Explanation** :

Let the initial efficiency of each man = 1 unit/day

Work done on

Day 1 = 1 × 1 = 1 unit

Day 2 = 2 × 2 = 4 units

Day 3 = 3 × 4 = 12 units

Day 4 = 4 × 8 = 32 units

Day 5 = 5 × 16 = 80 units

∴ Total work done in 5 days = 1 + 4 + 12 + 32 + 80 = 129 units.

Now, 43 men need to complete 129 units of work

Combined efficiency of these 43 men = 43 × 1 = 43 units/day

∴ Time taken = 129/43 = 3 days.

Hence, option (c).

Workspace:

**PE 4 - Time & Work | Arithmetic - Time & Work**

Three cooks together have to make 80 samosas. While working together, they can make 20 pieces per minute. The first cook starts the work alone and makes 20 pieces while working for more than three minutes. The remaining part of the work is done by the second cook and the third cook, working together. If they take a total of 8 minutes to prepare 80 samosas, then how many minutes will the first cook take alone to cook 160 samosas?

- (a)
32

- (b)
16

- (c)
24

- (d)
80

- (e)
40

Answer: Option A

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**Explanation** :

Suppose,

the 1st cook makes A samosas per minute

the 2nd cook makes B samosas per minute

the 3rd cook makes C samosas per minute.

Given: A + B + C = 20 … (i)

Let the 1st cook take (3 + x) minutes to make 20 samosas.

∴ (3 + x)A = 20

A = $\frac{20}{3+x}$ … (ii)

So, the 2nd cook and the 3rd cook take (5 - x) minutes to make 60 samosas.

∴ (5 - x)(B + C) = 60

B + C = $\frac{60}{5-x}$ … (iii)

Substituting the values from (ii) and (iii), we get

⇒ $\frac{20}{3+x}$ + $\frac{60}{5-x}$ = 20

⇒ $\frac{1}{3+x}$ + $\frac{3}{5-x}$ = 1

⇒ 5 - x + 9 + 3x = 15 + 2x - x^{2}

⇒ 14 - 2x = 15 + 2x - x^{2}

⇒ 14 = 15 - x2

⇒ x = 1

So, the 1st cook takes 4 minutes to make 20 samosas.

∴ He will take 4 × 8 = 32 minutes to make 160 samosas.

Hence, option (c).

Workspace:

**PE 4 - Time & Work | Arithmetic - Time & Work**

Consider two tanks - T_{1} & T_{2} such that the volume of T2 is twice that of T_{1}. T_{1} is fitted with 1 pipe A of circular cross-section, whereas T_{2} has 3 pipes – P, Q & R of circular cross-sections. The ratio of diameters of P, Q & R is 1 : 2 : 3. P’s cross-sectional area is 2 times A’s cross-sectional area. A takes 56 min to fill T_{1}. Find the time taken to fill T_{2}, if all the 3 pipes are opened.

- (a)
5 minutes

- (b)
4.5 minutes

- (c)
4 minutes

- (d)
None of these

Answer: Option C

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**Explanation** :

Let the efficiency of pipe A be 1 unit/min.

Since P’s cross-sectional area is 2 times A’s cross-sectional area, P’s efficiency will be twice of A i.e., 2 units/min.

Now since, the diameter of P, Q and R are in the ratio 1 : 2 : 3 their efficiencies will be in the ratio 1 : 4 : 9.

∴ Efficiency of P = 2 units/min

Efficiency of Q = 8 units/min

Efficiency of R = 18 units/min

A can fill T_{1} in 56 minutes, hence T_{1}’s volume = 1 × 56 = 56 units.

Volume of T_{2} = 2 × 56 = 112 units.

Time taken by P, Q and R to fill T_{2} = $\frac{112}{2+8+18}$ = 4 minutes.

Hence, option (c).

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