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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?

A. 9

B. 10

C. 11

D. 12

Discuss

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} Pumps  4 : 3 \\ Days 1 : 2 \end{matrix}} :: 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}

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?

A. Rs. $\frac{xy}{d}$

B. $Rs . x d$

C. $Rs . y d$

D. Rs. $\frac{yd}{x}$

Discuss

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}

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?

A. 648

B. 1800

C. 2700

D. 10800

Discuss

Solution:

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

\begin{matrix} Machines 6 : 10 \\ Time(in min .) 1 : 4 \end{matrix}} :: 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}

4. A fort had provision of food for 150 men for 45 days. After 10 days, 25 men left the fort. The number of days for which the remaining food will last, is:

A. $29 \frac{1}{5}$

B. $37 \frac{1}{4}$

C. 42

D. 54

Discuss

Solution:

After 10 days : 150 men had food for 35 days.
Suppose 125 men had food for x days.
Now, Less men, More days (Indirect Proportion)

\begin{aligned} ∴ 125 : 150 :: 35 : x \\ \Rightarrow 125 \times x = 150 \times 35 \\ \Rightarrow x = \frac{150 \times 35}{125} \\ \Rightarrow x = 42 \end{aligned}

5. 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?

A. 10

B. 13

C. 14

D. 15

Discuss

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

\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}

6. A man completes $\frac{5}{8}$ of a job in 10 days. At this rate, how many more days will it takes him to finish the job?

A. 5

B. 6

C. 7

D. $7 \frac{1}{2}$

Discuss

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 \\ \Rightarrow \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}

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

A. 48 paise

B. 54 paise

C. 56 paise

D. 72 paise

Discuss

Solution:

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

8. 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?

A. 1

B. $\frac{1}{40}$

C. 40

D. 80

Discuss

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

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

9. A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. When the smaller wheel has made 21 revolutions, then the number of revolutions mad by the larger wheel is:

A. 4

B. 9

C. 12

D. 49

Discuss

Solution:

Let the required number of revolutions made by larger wheel be x.
Then, More cogs, Less revolutions (Indirect Proportion)

\begin{aligned} ∴ 14 : 6 :: 21 : x \\ \Rightarrow 14 \times x = 6 \times 21 \\ \Rightarrow x = \frac{6 \times 21}{14} \\ \Rightarrow x = 9 \end{aligned}

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

A. 1

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

C. 7

D. 49

Discuss

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

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

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