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

Answer: Option A

Video Explanation

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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).

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.

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).

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).

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).

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.

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).

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). 

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).

Answer: Option B

Video Explanation

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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^nd cycle: Work done in 4 minutes = 2 × 9 = 18 units.

Now 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).