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51. If 9 men working $7 \frac{1}{2}$ hours a day can finish a piece of work in 20 days, then how many days will be taken by 12 men, working 6 hours a day to finish the work ? (It is being given that 2 men of latter type work as much as 3 men of the former type.)

A. $9 \frac{1}{2}$

B. 11

C. $12 \frac{1}{2}$

D. 13

Discuss

Solution:

Let the required number of days be x
2 men of latter type = 3 men of former type
12 men of latter type
=

(\frac{3}{2} \times 12)

= 18 men of former type
More men, Less days (Indirect proportion)
Less working hours, More days (Indirect proportion)

\begin{matrix} Men 18 : 9 \\ Working hrs 6 : \frac{15}{2} \end{matrix}} :: 20 : x

\begin{aligned} ∴ 18 \times 6 \times x = 9 \times \frac{15}{2} \times 20 \\ \Leftrightarrow 108 x = 1350 \\ \Leftrightarrow x = \frac{25}{2} \\ \Leftrightarrow x = 12 \frac{1}{2} \end{aligned}

52. 15 men take 21 days of 8 hours each to do a piece of work. How many days of 6 hours each would 21 women take, 3 women do as much work as 2 men ?

A. 18

B. 20

C. 25

D. 30

Discuss

Solution:

3 women ≡ 2 men
So, 21 women ≡ 14 men
Less men, More days (Indirect proportion)
Less hours per day, More days (Indirect proportion)

\begin{matrix} Men 14 : 15 \\ Hours per day 6 : 8 \end{matrix}} :: 21 : x

\begin{aligned} ∴ (14 \times 6 \times x) = (15 \times 8 \times 21) \\ \Leftrightarrow x = \frac{(15 \times 8 \times 21)}{(14 \times 6)} \\ \Leftrightarrow x = 30 \end{aligned}

∴ Required number of days = 30

53. In a barrack of soldiers there was stock of food for 190 days for 4000 soldiers. After 30 days 800 soldiers left the barrack. For how many days shall the left over food last for the remaining soldiers ?

A. 175 days

B. 200 days

C. 225 days

D. 250 days

Discuss

Solution:

Let the remaining food last for x days
4000 soldiers had provision for 160 days
3200 soldiers had provision for x days
Less men, More days (Indirect proportion)

\begin{aligned} ∴ 3200 : 4000 :: 160 : x \\ \Leftrightarrow 3200 x = 4000 \times 160 \\ \Leftrightarrow x = \frac{(4000 \times 160)}{3200} \\ \Leftrightarrow x = 200 \end{aligned}

54. A garrison of 500 men had provisions for 27 days. After 3 days a reinforcement of 300 men arrived. For how many more days will the remaining food last now ?

A. 15

B. 16

C. $17 \frac{1}{2}$

D. 18

Discuss

Solution:

Let the remaining food last for x days
500 men had provision for = (27 - 3) = 24 days
(500 + 300) men had provision for x days
More men, Less days (Indirect proportion)

\begin{aligned} ∴ 800 : 500 :: 24 : x \\ \Leftrightarrow (800 \times x) = (500 \times 24) \\ \Leftrightarrow x = \frac{(500 \times 24)}{800} \\ \Leftrightarrow x = 15 \end{aligned}

55. A garrison had provision for a certain number of days. After 10 days, $\frac{1}{5}$ of the men desert and it is found that the provisions will now last just as long as before. How long was that ?

A. 15 days

B. 25 days

C. 35 days

D. 50 days

Discuss

Solution:

Initially, Let there be x men having food for y days
After 10 days, x men had food for days (y - 10)
Also,

(x - \frac{x}{5})

men had food for y days

\begin{aligned} ∴ x (y - 10) = \frac{4 x}{5} \times y \\ \Leftrightarrow 5 x y - 50 x = 4 x y \\ \Leftrightarrow x y - 50 x = 0 \\ \Leftrightarrow x (y - 50) = 0 \\ \Leftrightarrow y - 50 = 0 \\ \Leftrightarrow y = 50 \end{aligned}

56. A fort has provisions for 50 days. If after 10 days they are strengthened by 500 men and the food lasts for 35 days longer, the number of men originally in the fort were ?

A. 2500

B. 3000

C. 3500

D. 4000

Discuss

Solution:

Let there be x men originally
So, x men had provisions 40 days whereas (x + 500) men consumed it in 35 days
More men, Less days (Indirect proportion)

\begin{aligned} ∴ (x + 500) : x :: 40 : 35 \\ \Leftrightarrow 35 \times (x + 500) = 40 x \\ \Leftrightarrow 5 x = 35 \times 500 \\ \Leftrightarrow x = (\frac{35 \times 500}{5}) \\ \Leftrightarrow x = 3500 \end{aligned}

57. A team of workers was employed by a contractor who undertook to finish 360 pieces of an article in a certain number of days. Making four more pieces per day than was planned, they could complete the job a day ahead of schedule. How many days did they take to complete the job ?

A. 8 days

B. 9 days

C. 10 days

D. 12 days

Discuss

Solution:

Let the team take x days to finish 360 pieces
Then, number of pieces made each day =

\frac{360}{x}

More number of pieces per day, Less days (Indirect proportion)

\begin{aligned} ∴ (\frac{360}{x} + 4) : \frac{360}{x} :: x : (x - 1) \\ \Leftrightarrow (\frac{360}{x} + 4) (x - 1) = \frac{360}{x} \times x \\ \Leftrightarrow 360 - \frac{360}{x} + 4 x - 4 = 360 \\ \Leftrightarrow 4 x - \frac{360}{x} - 4 = 0 \\ \Leftrightarrow x - \frac{90}{x} - 1 = 0 \\ \Leftrightarrow x^{2} - x - 90 = 0 \\ \Leftrightarrow (x - 10) (x + 9) = 0 \\ \Leftrightarrow x = 10 \end{aligned}

58. The work done by a women in 8 hours is equal to the work done by a man in 6 hours and by a boy in 12 hours. If working 6 hours per day 9 men can complete a work in 6 days, then in how many days can 12 men, 12 women and 12 boys together finish the same work, working 8 hours per day ?

A. $1 \frac{1}{2} days$

B. $3 days$

C. $3 \frac{2}{3} days$

D. $4 \frac{1}{2} days$

Discuss

Solution:

Ratio of time taken by a woman, a man and a boy

\begin{aligned} = 8 : 6 : 12 \\ = 4 : 3 : 6 \end{aligned}

So, 4 women ≡ 3 men ≡ 6 boy
(12 mens + 12 womens + 12 boys)

\begin{aligned} = [12 + (\frac{3}{4} \times 12) + (\frac{3}{6} \times 12)] men \\ = (12 + 9 + 6) men \\ = 27 men \end{aligned}

Let the required number of days be x
More men, Less days (Indirect proportion)
More working hours, Less days (Indirect proportion)

\begin{matrix} Working hours 8 : 6 \\ Men 27 : 9 \end{matrix}} :: 6 : x

\begin{aligned} ∴ 27 \times 8 \times x = 9 \times 6 \times 6 \\ \Leftrightarrow x = \frac{(9 \times 6 \times 6)}{(27 \times 8)} \\ \Leftrightarrow x = \frac{3}{2} \\ \Leftrightarrow x = 1 \frac{1}{2} \end{aligned}

59. 12 men and 18 boys, working $7 \frac{1}{2}$ hours a day, can do a piece of work in 60 days. If a man works equal to 2 boys, then how many boys will be required to help 21 men to do twice the work in 50 days, working 9 hours a day ?

A. 30

B. 42

C. 48

D. 90

Discuss

Solution:

1 man ≡ 2 boys ⇔ (12 men + 18 boys)
≡ (12 × 2 ×18) boys = 42 boys
Let required number of boys = x
⇒ (21 men + x boys) ≡ (21 × 2 × x) boys = (42 + x) boys
Less days, More boys (Indirect proportion)
More hours per day, Less boys (Indirect proportion)
More work, More boys (Direct proportion)

\begin{matrix} Days 50 : 60 \\ Hours per day 9 : \frac{15}{2} \\ Work 1 : 2 \end{matrix}} :: 42 : (42 + x)

∴ [50 \times 9 \times 1 \times (42 + x)] =

(60 \times \frac{15}{2} \times 2 \times 42)

\begin{aligned} \Leftrightarrow (42 + x) = \frac{37800}{450} \\ \Leftrightarrow 42 + x = 84 \\ \Leftrightarrow x = 42 \end{aligned}

60. Large, medium and small ships are used to bring water. 4 large ships carry as much water as 7 small ships, 3 medium ships carry the same amount of water as 2 large ships and 1 small ship. 15 large, 7 medium and 14 small ships, each made 36 journeys and brought a certain quantity of water. In how many journeys would 12 large, 14 medium and 21 small ships bring the same quantity ?

A. 25

B. 29

C. 32

D. 49

Discuss

Solution:

Let, L = large ships, M = medium ships, S = small ships
Now from the question,
4L ≡ 7S

\begin{aligned} 15L = (\frac{7}{4} \times 15) S = \frac{105}{4} S \\ Also, 2L = (\frac{7}{4} \times 2) S = \frac{7}{2} S \\ 3M \equiv 2 L + 1S \equiv (\frac{7}{2} + 1) S = \frac{9}{2} S \\ \Leftrightarrow 7M = (\frac{9}{2} \times \frac{1}{3} \times 7) S = \frac{21}{2} S \\ ∴ (15L + 7M + 14S) ships \\ \equiv (\frac{105}{4} + \frac{21}{2} + 14) S \\ = \frac{203}{4} S \\ (12 large + 14 medium + 2 1 small) ships \\ \equiv [(\frac{7}{4} \times 12) + (\frac{21}{2} \times 2) + 21] S \\ = (21 + 21 + 21) S \\ = 63 S \end{aligned}

Let the required number of journeys be x
More ships, Less journeys (Indirect proportion)

\begin{aligned} ∴ 63 : \frac{203}{4} :: 36 : x \\ \Leftrightarrow 63 x = \frac{203}{4} \times 36 = 1827 \\ \Leftrightarrow x = \frac{1827}{63} \\ \Leftrightarrow x = 29 \end{aligned}

Alternate:
Now from the question,

\begin{aligned} 4 L = 7 S \\ \Rightarrow \frac{L}{S} = \frac{7}{4} \\ \Rightarrow L = 7 x; S = 4 x \\ 3 M = 2 L + S \\ \Rightarrow M = \frac{2 \times 7 x + 4 x}{3} \\ \Rightarrow M = 6 x \end{aligned}

Thus, ratio of large, medium and small = 7 : 6 : 4
So, numbers of journeys:

\begin{aligned} = \frac{(15 \times 7 + 7 \times 6 + 14 \times 4) 36}{12 \times 7 + 14 \times 6 + 21 \times 4} \\ = \frac{7308}{252} \\ = 29 \end{aligned}

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