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61. An employee pays Rs. 26 for each day a worker and forfeits Rs. 7 for each day he idle. At the end of 56 days, if the worker got Rs. 829, for how many days did the worker remain idle?

A. 21

B. 15

C. 19

D. 13

Discuss

Solution:

His Per day pay = Rs. 26
Total pay employee got = Rs. 829
Total pay he gets if he did not remain idle a single day,
= 26 × 56 = Rs.1456
He Forfeits or fined = 1456 - 829 = Rs. 627
Per day he Forfeits Rs. 7 Means per idle day he loses = 26 + 7 = Rs. 33
So, Total idle days =

\frac{627}{33}

= 19 days

Alternatively,
Let he works for X days and remain idle for (56 - X)
Now, According to question,
X × 26 - 7 × (56 - X) = 829
26X + 7X - 392 = 829
33X = 829 + 392 = 1221
X = 37 days
Number of days remain idle = 56 -37 =19 days

62. A is 60% more efficient than B. In how many days will A and B working together complete a piece of work which A alone takes 15 days to finish?

A. $\frac{124}{13}$ days

B. $\frac{113}{13}$ days

C. $\frac{108}{13}$ days

D. $\frac{131}{13}$ days

Discuss

Solution:

Given,
A is 60% more efficient of B Means,

\begin{aligned} A = B + 60 % of B \\ A = B + \frac{60 B}{100} \\ A = \frac{100 B + 60 B}{100} \\ A = \frac{160 B}{100} \\ A = \frac{8 B}{5} \end{aligned}

A can complete whole work in 15 days. So,
One day work of A =

\frac{1}{15}

One day work of A =

\frac{8 B}{5}

\frac{1}{15}

One day work of B =

\frac{5}{120}

\frac{1}{24}

One day work, (A + B) =

\frac{1}{15} + \frac{1}{24}

One day work, (A + B) =

\frac{24 + 15}{360}

\frac{39}{360}

So, Time taken to finish the work by A and B together =

\frac{360}{39}

\frac{120}{13}

days

Alternatively
We can solve it through percentage method
Given,
A =

\frac{8 B}{5}

A can complete whole work in 15 days
Work rate of A =

\frac{100}{15}

= 6.66% per day
Work rate of

\frac{8 B}{5}

= 6.66%
Work rate of B = 4.16% per day
Work rate of (A + B) = 6.66 + 4.16 = 10.82% per day
So, A and B can complete 100% work in

\frac{120}{13}

days

63. A pipe can fill a tank in 0.9 hours and another pipe can empty in 0.7 hours. If tank is completely filled and both pipes are opened simultaneously then 450 liters of water is removed from the tank is 2.5 hours. What is the capacity of the tank?

A. 200 liters

B. 350 liters

C. 456 liters

D. 567 liters

Discuss

Solution:

Pipe A can fill the empty tank in = 0.9 hours
So work rate of the Pipe A =

\frac{100}{0.9}

% per hour
Pipe B can empty the tank in = 0.7 hours
Negative Work rate of B =

\frac{100}{0.7}

% per hour. (B is removing water, so, taken as negative work)
Tank fill per hour =

\frac{100}{0.7} - \frac{100}{0.9}

= 31.75% per hour
Time Taken to empty the tank =

\frac{100}{31.75}

≈ 3.15 hours
So, Capacity of the tank = 3.15 × 180 = 567 liters

64. A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is :

A. $\frac{1}{4}$

B. $\frac{1}{10}$

C. $\frac{7}{15}$

D. $\frac{8}{15}$

Discuss

Solution:

\begin{aligned} A's 1 day's work = \frac{1}{15}; \\ B's 1 day's work = \frac{1}{20}; \\ (A + B)'s 1day's work \\ = \frac{1}{15} + \frac{1}{20} = \frac{7}{60} \\ (A + B)'s 4 day's work \\ = \frac{7}{60} \times 4 = \frac{7}{15} \\ ∴ Remaining work = 1 - \frac{7}{15} = \frac{8}{15} \end{aligned}

65. A can lay railway track between two given stations in 16 days and B can do the same job in 12 days. With help of C, they did the job in 4 days only. Then, C alone can do the job in:

A. $9 \frac{1}{5}$ days

B. $9 \frac{2}{7}$ days

C. $9 \frac{3}{5}$ days

D. 10

Discuss

Solution:

\begin{aligned} (A + B + C)'s 1 day's work = \frac{1}{4} \\ A's 1 day's work = \frac{1}{16} \\ B's 1 day's work = \frac{1}{12} \\ ∴ C's 1 day's work \\ = \frac{1}{4} - (\frac{1}{16} + \frac{1}{12}) = \frac{1}{4} - \frac{7}{48} = \frac{5}{48} \\ So, C alone can do the work in \\ \frac{48}{5} = 9 \frac{3}{5} days \end{aligned}

66. A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?

A. 12 days

B. 15 days

C. 16 days

D. 18 days

Discuss

Solution:

\begin{aligned} A's 2 day's work = \frac{1}{20} \times 2 = \frac{1}{10} \\ (A + B + C)'s 1 day's work \\ = \frac{1}{20} + \frac{1}{30} + \frac{1}{60} = \frac{6}{60} = \frac{1}{10} \\ Work done in 3 days = \frac{1}{10} + \frac{1}{10} = \frac{1}{5} \\ Now, \frac{1}{5} work is done in 3 days \\ ∴ Whole work will be done in \\ 3 \times 5 = 15 days \end{aligned}

67. A is thrice as good as workman as B and therefore is able to finish a job in 60 days less than B. Working together, they can do it in:

A. 20 days

B. $22 \frac{1}{2}$ days

C. 25 days

D. 30 days

Discuss

Solution:

\begin{aligned} Ratio of times taken by A and B = 1:3 \\ The time difference is (3 - 1) 2 days \\ while B take 3 days and A takes 1 day . \\ If difference of time is 2 days, B takes 3 days . \\ If difference of time is 60 days, \\ B takes (\frac{3}{2} \times 60) = 90 days \\ So, A takes 30 days to do the work . \\ A's 1 day's work = \frac{1}{30} \\ B's 1 day's work = \frac{1}{90} \\ (A + B)'s 1 day's work \\ = \frac{1}{30} + \frac{1}{90} = \frac{4}{90} = \frac{2}{45} \\ ∴ A and B together can do the work in \\ \frac{45}{2} = 22 \frac{1}{2} days \end{aligned}

68. A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs. 3200. With the help of C, they completed the work in 3 days. How much is to be paid to C?

A. Rs. 375

B. Rs. 400

C. Rs. 600

D. Rs. 800

Discuss

Solution:

\begin{aligned} C's 1 day's work \\ = \frac{1}{3} - (\frac{1}{6} + \frac{1}{8}) \\ = \frac{1}{3} - \frac{7}{24} \\ = \frac{1}{24} \\ A's wages:B's wages:C's wages \\ = \frac{1}{6} : \frac{1}{8} : \frac{1}{24} \\ = 4 : 3 : 1 \\ ∴ C's share (for 3 days) \\ = R s . (3 \times \frac{1}{24} \times 3200) \\ = R s . 400 \end{aligned}

69. If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:

A. 4 days

B. 5 days

C. 6 days

D. 7 days

Discuss

Solution:

\begin{aligned} Let 1 man's 1 day's work = x and \\ 1 boy's 1day's work = y \\ Then, 6 x + 8 y = \frac{1}{10} and \\ 26 x + 48 y = \frac{1}{2} \\ Solving these two equations, we get \\ x = \frac{1}{100} a n d y = \frac{1}{200} \\ (15 men + 20 boy)'s 1 day's work \\ = \frac{15}{100} + \frac{20}{200} = \frac{1}{4} \\ ∴ 15 men and 20 boys can do the work in 4 days \end{aligned}

70. A can do a piece of work in 4 hours; B and C together can do it in 3 hours, while A and C together can do it in 2 hours. How long will B alone take to do it?

A. 8 hours

B. 10 hours

C. 12 hours

D. 24 hours

Discuss

Solution:

\begin{aligned} A's 1 hour's work = \frac{1}{4}; \\ (B + C)'s 1 hour's work = \frac{1}{3}; \\ (A + C)'s 1 hour's work = \frac{1}{2} . \\ (A + B + C)'s 1 hour's work \\ = \frac{1}{4} + \frac{1}{3} = \frac{7}{12} \\ B's 1 hour's work \\ = \frac{7}{12} - \frac{1}{2} = \frac{1}{12} \\ ∴ B along will take 12 hours to do the work \end{aligned}

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