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

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.

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).

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).

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).

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.

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.

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).

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 - x²
⇒ 14 - 2x = 15 + 2x - x²
⇒ 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).

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 T₁ in 56 minutes, hence T₁’s volume = 1 × 56 = 56 units.
Volume of T₂ = 2 × 56 = 112 units.

Time taken by P, Q and R to fill T₂ = $\frac{112}{2+8+18}$ = 4 minutes.

Hence, option (c).