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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?

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)
$\begin{array}{c}\text{Pumpu 4}:3\\ \text{Days}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{1}:2\end{array}\right\}::8:x$

$\begin{array}{rl}& \therefore 4×1×x=3×2×8\\ & ⇒x=\frac{\left(3×2×8\right)}{\left(4\right)}\\ & ⇒x=12\end{array}$

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?

Explanation:

Solution:

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?

Explanation:

Solution:
Let the required number of bottles be x.
More machines, More bottles (Direct Proportion)
More minutes, More bottles (Direct Proportion)
$\begin{array}{c}\text{Machines}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{4}:3\\ \text{Time(in min}\text{.)}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{1}:2\end{array}\right\}::270:x$

$\begin{array}{rl}& \therefore 6×1×x=10×4×270\\ & ⇒x=\frac{\left(10×4×270\right)}{\left(6\right)}\\ & ⇒x=1800\end{array}$

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?

Explanation:

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

$\begin{array}{rl}& \therefore 30×6×x=39×5×12\\ & ⇒x=\frac{\left(39×5×12\right)}{\left(30×6\right)}\\ & ⇒x=13\end{array}$

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?

Explanation:

Solution:
$\begin{array}{rl}& \text{Work}\phantom{\rule{thinmathspace}{0ex}}\text{done}=\frac{5}{8}\\ & \text{Balance}\phantom{\rule{thinmathspace}{0ex}}\text{work}=\left(1-\frac{5}{8}\right)=\frac{3}{8}\\ & \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{required}\phantom{\rule{thinmathspace}{0ex}}\text{number}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{days}\phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}x\\ & \text{Then},\phantom{\rule{thinmathspace}{0ex}}\frac{5}{8}:\frac{3}{8}=::10:x\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}⇔\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{5}{8}×x=\frac{3}{8}×10\\ & ⇒x=\left(\frac{3}{8}×10×\frac{8}{5}\right)\\ & ⇒x=6\end{array}$

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

Explanation:

Solution:
$\begin{array}{rl}& \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{required}\phantom{\rule{thinmathspace}{0ex}}\text{weight}\phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}x\phantom{\rule{thinmathspace}{0ex}}\text{kg}.\\ & \text{Less}\phantom{\rule{thinmathspace}{0ex}}\text{weight,}\phantom{\rule{thinmathspace}{0ex}}\text{less}\phantom{\rule{thinmathspace}{0ex}}\text{cost}\phantom{\rule{thinmathspace}{0ex}}\left(\text{Direct}\phantom{\rule{thinmathspace}{0ex}}\text{Proportion}\right)\\ & \therefore 250:200::60:x\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}⇔\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}250×x=\left(200×60\right)\\ & ⇒x=\frac{\left(200×60\right)}{250}\\ & ⇒x=48\end{array}$

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?

Explanation:

Solution:
Let the required number of days be x.
Less cows, More days (Indirect Proportion)
Less bags, Less days (Direct Proportion)
$\begin{array}{c}\text{Cows}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1:40\\ \text{Bags}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}40:1\end{array}\right\}::40:x$

$\begin{array}{rl}& \therefore 1×40×x=40×1×40\\ & ⇒x=40\end{array}$

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

Explanation:

Solution:
Let the required number days be x.
Less spiders, More days (Indirect Proportion)
Less webs, Less days (Direct Proportion)
$\begin{array}{c}\text{Spiders}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1:7\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{Webs}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}7:1\end{array}\right\}::7:x$

$\begin{array}{rl}& \therefore 1×7×x=7×1×7\\ & ⇒x=7\end{array}$

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:

Explanation:

Solution:
Let the height of the building x metres.
Less lengthy shadow, Less in the height (Direct Proportion)
$\begin{array}{rl}& \therefore 40.25:28.75::17.5:x\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}⇔\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}40.25×x=28.75×17.5\\ & ⇒x=\frac{28.75×17.5}{40.25}\\ & ⇒x=12.5\end{array}$

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?

$\begin{array}{rl}& =\left(\frac{120}{200}×50\right)\text{men}\\ & =\text{30 men}\end{array}$