Chain-Rule

1.3 pumps, working 8 hours a day, can empty a tank in 2 days. How many hours a day must 4 pumps work to empty the tank in 1 day?

1. 9
2. 10
3. 11
4. 12

Answer:Option 1

Explanation:

Solution:

Let the required number of working hours per day be x.
More pumps, Less working hours per day (Indirect Proportion)
Less days, More working hours per day (Indirect Proportion)

\begin{matrix} Pumpu  4 : 3 \\ Days 1 : 2 \end{matrix}} :: 8 : x

$\left. \begin{gathered} {\text{Pumpu 4}}:3 \hfill \\ {\text{Days}}\,\,\,\,\,\,\,\,{\text{1}}:2 \hfill \\ \end{gathered} \right\}::8:x$

\begin{aligned} ∴ 4 \times 1 \times x = 3 \times 2 \times 8 \\ \Rightarrow x = \frac{(3 \times 2 \times 8)}{(4)} \\ \Rightarrow x = 12 \end{aligned}

$\eqalign{ & \therefore 4 \times 1 \times x = 3 \times 2 \times 8 \cr & \Rightarrow x = \frac{{\left( {3 \times 2 \times 8} \right)}}{{\left( 4 \right)}} \cr & \Rightarrow x = 12 \cr}$

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2.If the cost of x metres of wire is d rupees, then what is the cost of y metres of wire at the same rate?

1. Rs. $(\frac{xy}{d})$ $\left( {\frac{{{\text{xy}}}}{{\text{d}}}} \right)$
2. Rs. (xd)
3. Rs. (yd)
4. Rs. $(\frac{yd}{x})$ $\left( {\frac{{{\text{yd}}}}{{\text{x}}}} \right)$

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Cost of x metres \\ = Rs . d \\ Coast of 1 metre \\ = Rs . (\frac{d}{x}) \\ Cost of y metres \\ = Rs . (\frac{d}{x} \times y) \\ = Rs . (\frac{y d}{x}) \end{aligned}

$\eqalign{ & {\text{Cost of }}x{\text{ metres}} \cr & = {\text{Rs}}{\text{. }}d \cr & {\text{Coast of 1 metre}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{d}{x}} \right) \cr & {\text{Cost of }}y{\text{ metres}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{d}{x} \times y} \right) \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{yd}}{x}} \right) \cr}$

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3.Running at the same constant rate, 6 identical machines can produce a total of 270 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4 minutes?

1. 648
2. 1800
3. 2700
4. 10800

Answer:Option 1

Explanation:

Solution:

Let the required number of bottles be x.
More machines, More bottles (Direct Proportion)
More minutes, More bottles (Direct Proportion)

\begin{matrix} Machines 4 : 3 \\ Time(in min .) 1 : 2 \end{matrix}} :: 270 : x

$\left. \begin{gathered} {\text{Machines}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{4}}:3 \hfill \\ {\text{Time(in min}}{\text{.)}}\,\,{\text{1}}:2 \hfill \\ \end{gathered} \right\}::270:x$

\begin{aligned} ∴ 6 \times 1 \times x = 10 \times 4 \times 270 \\ \Rightarrow x = \frac{(10 \times 4 \times 270)}{(6)} \\ \Rightarrow x = 1800 \end{aligned}

$\eqalign{ & \therefore 6 \times 1 \times x = 10 \times 4 \times 270 \cr & \Rightarrow x = \frac{{\left( {10 \times 4 \times 270} \right)}}{{\left( 6 \right)}} \cr & \Rightarrow x = 1800 \cr}$

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4.39 persons can repair a road in 12 days, working 5 hours a day. In how many days will 30 persons, working 6 hours a day, complete the work?

1. 10
2. 13
3. 14
4. 15

Answer:Option 1

Explanation:

Solution:

Let the required number of days bex.
Less persons, More days (Indirect Proportion)
More working hours per day, Less days (Indirect Proportion)

\begin{matrix} Persons 30 : 39 \\ Working hour/day 6 : 5 \end{matrix}} :: 12 : x

$\left. \begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Persons }}\,30:39 \hfill \\ {\text{Working hour/day }}\,{\text{6}}:{\text{5}} \hfill \\ \end{gathered} \right\}::12:x$

\begin{aligned} ∴ 30 \times 6 \times x = 39 \times 5 \times 12 \\ \Rightarrow x = \frac{(39 \times 5 \times 12)}{(30 \times 6)} \\ \Rightarrow x = 13 \end{aligned}

$\eqalign{ & \therefore 30 \times 6 \times x = 39 \times 5 \times 12 \cr & \Rightarrow x = \frac{{\left( {39 \times 5 \times 12} \right)}}{{\left( {30 \times 6} \right)}} \cr & \Rightarrow x = 13 \cr}$

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5.A man completes ⁵/₈ of a job in 10 days. At this rate, how many more days will it takes him to finish the job?

1. 5
2. 6
3. 7
4. $7 \frac{1}{2}$ $7\frac{1}{2}$

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Work done = \frac{5}{8} \\ Balance work = (1 - \frac{5}{8}) = \frac{3}{8} \\ Let the required number of days be x \\ Then, \frac{5}{8} : \frac{3}{8} =:: 10 : x \Leftrightarrow \frac{5}{8} \times x = \frac{3}{8} \times 10 \\ \Rightarrow x = (\frac{3}{8} \times 10 \times \frac{8}{5}) \\ \Rightarrow x = 6 \end{aligned}

$\eqalign{ & {\text{Work}}\,{\text{done}} = \frac{5}{8} \cr & {\text{Balance}}\,{\text{work}} = \left( {1 - \frac{5}{8}} \right) = \frac{3}{8} \cr & {\text{Let}}\,{\text{the}}\,{\text{required}}\,{\text{number}}\,{\text{of}}\,{\text{days}}\,{\text{be}}\,x \cr & {\text{Then}},\,\frac{5}{8}:\frac{3}{8} = ::10:x\,\,\,\, \Leftrightarrow \,\,\,\,\frac{5}{8} \times x = \frac{3}{8} \times 10 \cr & \Rightarrow x = \left( {\frac{3}{8} \times 10 \times \frac{8}{5}} \right) \cr & \Rightarrow x = 6 \cr}$

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6.If a quarter kg of potato costs 60 paise, how many paise will 200 gm cost?

1. 48 paise
2. 54 paise
3. 56 paise
4. 72 paise

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Let the required weight be x kg . \\ Less weight, less cost (Direct Proportion) \\ ∴ 250 : 200 :: 60 : x \Leftrightarrow 250 \times x = (200 \times 60) \\ \Rightarrow x = \frac{(200 \times 60)}{250} \\ \Rightarrow x = 48 \end{aligned}

$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{required}}\,{\text{weight}}\,{\text{be}}\,x\,{\text{kg}}. \cr & {\text{Less}}\,{\text{weight,}}\,{\text{less}}\,{\text{cost}}\,\left( {{\text{Direct}}\,{\text{Proportion}}} \right) \cr & \therefore 250:200::60:x\,\,\, \Leftrightarrow \,\,\,250 \times x = \left( {200 \times 60} \right) \cr & \Rightarrow x = \frac{{\left( {200 \times 60} \right)}}{{250}} \cr & \Rightarrow x = 48 \cr}$

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7.In a dairy farm, 40 cows eat 40 bags of husk in 40 days. In how many days one cow will eat one bag of husk?

1. 1
2. $\frac{1}{40}$ $\frac{1}{{40}}$
3. 40
4. 80

Answer:Option 1

Explanation:

Solution:

Let the required number of days be x.
Less cows, More days (Indirect Proportion)
Less bags, Less days (Direct Proportion)

\begin{matrix} Cows 1 : 40 \\ Bags 40 : 1 \end{matrix}} :: 40 : x

$\left. \begin{gathered} {\text{Cows}}\,\,\,\,\,1:40 \hfill \\ {\text{Bags}}\,\,\,\,\,40:1 \hfill \\ \end{gathered} \right\}::40:x$

\begin{aligned} ∴ 1 \times 40 \times x = 40 \times 1 \times 40 \\ \Rightarrow x = 40 \end{aligned}

$\eqalign{ & \therefore 1 \times 40 \times x = 40 \times 1 \times 40 \cr & \Rightarrow x = 40 \cr}$

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8.If 7 spiders make 7 webs in 7 days, then 1 spider will make 1 web in how many days?

1. 1
2. $\frac{7}{2}$ $\frac{7}{2}$
3. 7
4. 49

Answer:Option 1

Explanation:

Solution:

Let the required number days be x.
Less spiders, More days (Indirect Proportion)
Less webs, Less days (Direct Proportion)

\begin{matrix} Spiders 1 : 7 \\ Webs 7 : 1 \end{matrix}} :: 7 : x

$\left. \begin{gathered} {\text{Spiders}}\,\,\,\,\,1:7 \hfill \\ \,\,\,{\text{Webs}}\,\,\,\,\,7:1 \hfill \\ \end{gathered} \right\}::7:x$

\begin{aligned} ∴ 1 \times 7 \times x = 7 \times 1 \times 7 \\ \Rightarrow x = 7 \end{aligned}

$\eqalign{ & \therefore 1 \times 7 \times x = 7 \times 1 \times 7 \cr & \Rightarrow x = 7 \cr}$

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9.A flagstaff 17.5 m high casts a shadow of length 40.25 m. The height of the building, which casts a shadow of length 28.75 m under similar conditions will be:

1. 10 m
2. 12.5 m
3. 17.5 m
4. 21.25 m

Answer:Option 1

Explanation:

Solution:

Let the height of the building x metres.
Less lengthy shadow, Less in the height (Direct Proportion)

\begin{aligned} ∴ 40.25 : 28.75 :: 17.5 : x \Leftrightarrow 40.25 \times x = 28.75 \times 17.5 \\ \Rightarrow x = \frac{28.75 \times 17.5}{40.25} \\ \Rightarrow x = 12.5 \end{aligned}

$\eqalign{ & \therefore 40.25:28.75::17.5:x\,\,\, \Leftrightarrow \,\,\,40.25 \times x = 28.75 \times 17.5 \cr & \Rightarrow x = \frac{{28.75 \times 17.5}}{{40.25}} \cr & \Rightarrow x = 12.5 \cr}$

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10.In a camp, there is a meal for 120 men or 200 children. If 150 children have taken the meal, how many men will be catered to with remaining meal?

1. 20
2. 30
3. 40
4. 50

Answer:Option 1

Explanation:

Solution:

There is a meal for 200 children
150 children have taken the meal
Remaining meal is to be catered to 50 children
Now,
200 children = 120 men
∴ 50 children