# PE 3 - Time & Work | Arithmetic - Time & Work

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**PE 3 - Time & Work | Arithmetic - Time & Work**

Two candles of the same length are lighted at the same time. The first is consumed in 6 hours and the second in 4 hours. Assuming that each candle burns at a constant rate, in how many hours after being lighted was the first candle twice the length of the second?

- (a)
3 hours

- (b)
2.5 hours

- (c)
1.5 hours

- (d)
2 hours

- (e)
3.5 hours

Answer: Option A

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**Explanation** :

Let the length of each of the candles be 1 unit.

Let the first candle be twice the length of second candle after x hours of burning.

Amount of candle 1 burnt in x hours = x/6

Amount of candle 2 burnt in x hours = x/4

After x hours of burning, length remaining of

**Candle 1**: (1 – x/6)

**Candle 2**: (1 – x/4)

∴ (1 – x/6) = 2 × (1 – x/4)

⇒ 1 – x/6 = 2 – x/2

⇒ x = 3

∴ After 3 hours the first candle is twice the length of second.

Hence, option (a).

Workspace:

**PE 3 - Time & Work | Arithmetic - Time & Work**

A man can do a piece of work in 80 days. He started doing the work and on every successive day, he does twice the amount of work done on the previous day. If he goes on at the same rate, in how many days will he be able to complete the work?

Answer: 7

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**Explanation** :

Assuming he can do 1 unit of work every day.

∴ Total work done in 80 days = 80 × 1 = 80 units.

Now, when his efficiency doubles every day.

Work done till day 1 = 1 unit.

Work done till day 2 = 1 + 2 = 3 units.

Work done till day 3 = 3 + 4 = 7 units.

Work done till day 4 = 7 + 8 = 15 units.

Work done till day 5 = 15 + 16 = 31 units.

Work done till day 6 = 31 + 32 = 63 units.

Work done till day 7 = 63 + 64 = 127 units.

Since only 80 units of work is to be complete, it would get completed on 7^{th} day.

Hence, 7.

Workspace:

**PE 3 - Time & Work | Arithmetic - Time & Work**

A can do a piece of work 3 hours faster than B. They worked together for 1 hour and then the remaining part of the work was done by B alone in 3 hours. How many hours does B take to do the work if he works alone?

- (a)
9 hours

- (b)
6 hours

- (c)
8 hours

- (d)
10 hours

Answer: Option B

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**Explanation** :

Let B take x hours, A takes (x - 3) hours

∴ $\left(\frac{1}{x-3}+\frac{1}{x}\right)\times 1+\frac{1}{x}\times 3$ = 1

⇒ x = 6

∴ A takes 3 hours and B takes 6 hours.

Hence, option (b).

Workspace:

**PE 3 - Time & Work | Arithmetic - Time & Work**

A can do a piece of work in 12 days, B in 18 days. They work together for 6 days. The remaining work was finished by C in 4 days. If C received Rs. 45 for the part of the work he did, find how much B received for the work done by him.

- (a)
Rs. 120

- (b)
Rs. 100

- (c)
Rs. 150

- (d)
Rs. 90

Answer: Option D

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**Explanation** :

Let the total work to be done = LCM(12, 18) = 36 units.

∴ A’s 1 day work = 36/12 = 3 units/day

∴ B’s 1 day work = 36/18 = 2 units/day

In 6 days,

Word done by A = 3 × 6 = 18 units, and

Word done by B = 2 × 6 = 12 units

∴ C completed the remaining work = 36 – 18 – 12 = 6 units.

For 6 units of work done by C, payment = Rs. 45.

∴ For 12 units of work done by B, payment = 45 × 2 = Rs. 90.

Hence, option (d).

Workspace:

**PE 3 - Time & Work | Arithmetic - Time & Work**

Pipes A and B can fill a tank in 30 minutes and 40 minutes respectively and C can empty it in 60 minutes. A is opened for a minute and then closed. B is then opened for a minute and then closed. C is then opened for a minute and then closed. This process is repeated until the tank is filled. Find the time taken to fill the tank (in minutes).

- (a)
71

- (b)
72

- (c)
70 minutes and 20 seconds

- (d)
None of these

Answer: Option C

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**Explanation** :

Let the total capacity of tank = LCM(30, 40, 60) = 120 units

∴ Efficiency of pipe A = 120/30 = 4 units/minute

∴ Efficiency of pipe B = 120/40 = 3 units/minute

∴ Efficiency of pipe C = 120/60 = - 2 units/minute (-ve indicates that pipe C empties the tank)

Now work done in 1 cycle (i.e., 3 minutes) = 4 + 3 – 2 = 5 units

⇒ Number of cycles required to complete the work = 120/5 = 24 cycles.

Since C is emptying the tank, let’s break the last cycle.

Work done till 23 cycles (i.e., in 23 × 3 = 69 minutes) = 23 × 5 = 115 units.

Work left for 24th cycle = 120 – 115 = 5 units.

Now, pipe A will fill 4 units in 70th minute.

Work left now = 5 – 4 = 1 unit

Hence, pipe B will completely fill the tank in 1/3 minutes = 20 seconds.

Hence, the tank gets completely filled in 70 minutes and 20 seconds.

Since the tank is full now, we will not be operating the pipes anymore.

Hence, option (c).

Workspace:

**PE 3 - Time & Work | Arithmetic - Time & Work**

Pipes A and B can fill a tank in 20 minutes and 30 minutes respectively and C can empty it in 15 minutes. A is opened for a minute and then closed. B is then opened for a minute and then closed. C is then opened for a minute and then closed. This process is repeated until the tank is filled. Find the time taken to fill the tank. Enter your answer as the nearest possible integer in minutes.

Answer: 167

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**Explanation** :

Let the total capacity of tank = LCM(20, 30, 15) = 60 units

∴ Efficiency of pipe A = 60/20 = 3 units/minute

∴ Efficiency of pipe B = 60/30 = 2 units/minute

∴ Efficiency of pipe C = 60/15 = - 4 units/minute (-ve indicates that pipe C empties the tank)

Now work done in 1 cycle (i.e., 3 minutes) = 3 + 2 – 4 = 1 units

Now this becomes a little tricky situation because C is emptying a large amount of water which is filled by A and B.

⇒ Number of cycles required to complete the work = 60/1 = 60 cycles.

Since C is emptying the tank, let’s break the last few cycles.

Work done till 53 cycles (i.e., in 53 × 3 = 159 minutes) = 53 × 1 = 53 units.

**In 54 ^{th} cycle**,

In next 2 minutes A and B will fill 3 + 2 = 5 units. Hence, the total work done = 58 units till 161 minutes.

In next 1 minute C will empty 4 units. Hence, the total work done = 54 units till 162 minutes.

**In 55 ^{th} cycle**,

In next 2 minutes A and B will fill 3 + 2 = 5 units. Hence, the total work done = 59 units till 164 minutes.

In next 1 minute C will empty 4 units. Hence, the total work done = 55 units till 165 minutes.

**In 56 ^{th} cycle**,

In next 2 minutes A and B will fill 3 + 2 = 5 units. Hence, the total work done = 60 units till 167 minutes.

Now, since the work is completed, pipes will no longer be operated.

∴ It takes 167 minutes for the tank to be completely filled.

Hence, 167.

Workspace:

**PE 3 - Time & Work | Arithmetic - Time & Work**

A cistern is provided with two supply pipes A and B and a waste pipe C. A and B fill it in 12 and 20 minutes respectively and C discharges 20 gallons of water per minute. If all the pipes are opened at once, the empty cistern is filled in 10 minutes. Find the capacity of the cistern?

- (a)
600 gallons

- (b)
550 gallons

- (c)
500 gallons

- (d)
None of these

Answer: Option A

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**Explanation** :

Let the total volume of the cistern be V gallons.

When all three work together it takes 10 hours for the cistern to be filled.

∴ $\frac{V}{12}+\frac{V}{20}-20=\frac{V}{10}$

⇒ $\frac{8V}{60}-\frac{V}{10}=20$

⇒ V = 600 gallons.

Hence, option (a).

Workspace:

**PE 3 - Time & Work | Arithmetic - Time & Work**

A contractor undertook a certain work to be done in 80 days and employed 66 men to do it. After 25 days he found that 1/3^{rd} of the work has already been finished. How many men should he dismiss so that the work may be finished on the date agreed upon?

- (a)
18

- (b)
12

- (c)
6

- (d)
4

- (e)
None of these

Answer: Option C

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**Explanation** :

66 men can finish 1/3^{rd} of the work in 25 days.

⇒ They can complete the entire work in 75 days only.

∴ They can complete the remaining work in 50 days.

But question requires them to complete the work in another (80 – 25 =) 55 days.

Let x men need to be dismissed to finish the remaining work in 55 days.

The remaing work which 66 men could complete in 50 more days needs to be completed in 55 days by (66 - x) men.

∴ 50 × 66 = 55 × (66 - x)

⇒ 66 - x = 60

⇒ x = 6

So, to complete the work in another 55 days, the contractor has to dismiss 6 men.

Hence, option (c).

Workspace:

**PE 3 - Time & Work | Arithmetic - Time & Work**

A can do a piece of work in 18 days. If B joins him and they work on alternate days starting with A, the work will be completed in 24 days. In how many days will B complete the same work alone?

- (a)
12 days

- (b)
18 days

- (c)
24 days

- (d)
30 days

- (e)
36 days

Answer: Option E

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**Explanation** :

A and B work alternately for 24 days, hence A would have worked for 12 days and B would have worked for 12 days.

Work done by A in 12 days = $\frac{12}{18}$ = $\frac{2}{3}$ ^{rd}

Work done by B in 12 days = 1 - $\frac{2}{3}$ = $\frac{1}{3}$ ^{rd}

∴ B completes 1/3^{rd} of the work in 12 days.

Hence he can complete the entire work in 12 × 3 = 36 days.

**Alternately**,

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

∴ Efficiency of A = 72/18 = 4 units/day

∴ Efficiency of B = b units/day

In a cycle of 2 days, A works on 1st day and B on 2nd day. The work done in 1 cycle = (4 + b) units.

Since the work is completed in 24 days i.e., 12 cycles.

⇒ 12 × (4 + b) = 72

⇒ b = 2 units/day

∴ Time taken by B alone = 72/2 = 36 days.

Hence, option (e).

Workspace:

**PE 3 - Time & Work | Arithmetic - Time & Work**

Two pipes can separately fill a cistern in 4 and 5 minutes respectively. If the taps are opened turn by turn, each for a minute, starting with first pipe. Find the time taken by them to fill the cistern?

- (a)
4 mins 30 seconds

- (b)
4 mins 24 seconds

- (c)
4 mins 48 seconds

- (d)
None of these

Answer: Option B

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**Explanation** :

Let the work to be done = LCM(4, 5) = 20 units.

∴ Efficiency of pipe A = 20/4 = 5 units/minute

∴ Efficiency of pipe B = 20/5 = 4 units/minute

They work alternately.

Work done in each cycle = 5 + 4 = 9 units.

**Till 1 ^{st} cycle**: Work done in 2 minutes = 1 × 9 = 9 units.

**Till 2**: Work done in 4 minutes = 2 × 9 = 18 units.

^{nd}cycleNow the remaining 2 units of work will be done by A in 5^{th} minute.

∴ Time taken by A = 2/5 = 0.4 minute or 24 seconds.

∴ total time taken = 4 minutes and 24 seconds.

Hence, option (b).

Workspace:

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