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Quantitative Aptitude

CHAIN RULE MCQs

Total Questions : 326 | Page 3 of 33 pages
Question 21.

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
 Discuss Question
Answer: Option D. -> 12

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)


 


Pumps
4
:
3
3 Pumps, Working 8 Hours A Day, Can Empty A Tank In 2 Days. ...
:: 8 : x
Days
1
:
2

So, 4 x 1 x x = 3 x 2 x 8


x = \(\frac{(3\times2\times8)}{(4)}\)


x = 12.

Question 22.

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.    \(Rs.\left(\frac{xy}{d}\right)\)
  2.    Rs. (xd)
  3.    Rs. (yd)
  4.    \(Rs.\left(\frac{yd}{x}\right)\)
 Discuss Question
Answer: Option D. -> \(Rs.\left(\frac{yd}{x}\right)\)

Cost of x metres = Rs. d.


Cost of 1 metre = Rs. \(\left(\frac{d}{x}\right)\)


Cost of y metres = Rs.  \(\left(\frac{d}{x}.y\right)\)   = Rs.  \(\left(\frac{yd}{x}\right)\)

Question 23.

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
 Discuss Question
Answer: Option B. -> 1800

Let the required number of bottles be x.


More machines, More bottles (Direct Proportion)


More minutes, More bottles (Direct Proportion)


 


Machines
6
:
10
Running At The Same Constant Rate, 6 Identical Machines Can ...
:: 270 : x
Time (in minutes)
1
:
4

So,  6 x 1 x x = 10 x 4 x 270


x = \(\frac{(10\times4\times270)}{(6)}\) 


x= 1800.

Question 24.

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:

  1.    \(29\frac{1}{5}\)
  2.    \(37\frac{1}{4}\)
  3.    42
  4.    54
 Discuss Question
Answer: Option B. -> \(37\frac{1}{4}\)

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)


So,  125 : 150 :: 35 : x   \(\Leftrightarrow\)     125 x x = 150 x 35


x = \(\frac{150\times35}{125}\)


x = 42.

Question 25.

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
 Discuss Question
Answer: Option B. -> 13

Let the required number of days be x.


Less persons, More days (Indirect Proportion)


More working hours per day, Less days (Indirect Proportion)


 


Persons
30
:
39
39 Persons Can Repair A Road In 12 Days, Working 5 Hours A D...
:: 12 : x
Working hours/day
6
:
5

 


39 Persons Can Repair A Road In 12 Days, Working 5 Hours A D... 30 x 6 x x = 39 x 5 x 12


x = \(\frac{(39\times5\times12)}{(30\times6)}\)


x =13.

Question 26.

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?

  1.    5
  2.    6
  3.    7
  4.    \(7\frac{1}{2}\)
 Discuss Question
Answer: Option B. -> 6

Work done = \(\frac{5}{8}\)


Balance work = \(\left(1-\frac{5}{8}\right)=\frac{3}{8}\)


Let the required number of days be x.


Then, \( \frac{5}{8}:\frac{3}{8}=::10:x \Leftrightarrow \frac{5}{8}\times x =\frac{3}{8}\times10\)


x = \(\left(\frac{3}{8}\times10\times\frac{5}{8}\right) \)


x= 6.

Question 27.

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
 Discuss Question
Answer: Option A. -> 48 paise

Let the required weight be x kg.


Less weight, Less cost (Direct Proportion)


So,  250 : 200 :: 60 : x   \(\Leftrightarrow \)    250 x x = (200 x 60)


x = \(\frac{(200\times60)}{250}\)


x = 48.

Question 28.

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.    \(\frac{1}{40}\)
  3.    40
  4.    80
 Discuss Question
Answer: Option C. -> 40

Let the required number of days be x.


Less cows, More days (Indirect Proportion)


Less bags, Less days (Direct Proportion)


 


Cows
1
:
40
In A Dairy Farm, 40 Cows Eat 40 Bags Of Husk In 40 Days. In ...
:: 40 : x
Bags
40
:
1

 


In A Dairy Farm, 40 Cows Eat 40 Bags Of Husk In 40 Days. In ... 1 x 40 x x = 40 x 1 x 40


In A Dairy Farm, 40 Cows Eat 40 Bags Of Husk In 40 Days. In ... x = 40.

Question 29.

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:

  1.    4
  2.    9
  3.    12
  4.    49
 Discuss Question
Answer: Option B. -> 9

Let the required number of revolutions made by larger wheel be x.


Then, More cogs, Less revolutions (Indirect Proportion)


So,  14 : 6 :: 21 : x   \( \Leftrightarrow\)   14 x x = 6 x 21


x =  \(\frac{6\times21}{14}\)


x  = 9.

Question 30.

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

  1.    1
  2.    \(\frac{7}{2}\)
  3.    7
  4.    49
 Discuss Question
Answer: Option C. -> 7

Let the required number days be x.


Less spiders, More days (Indirect Proportion)


Less webs, Less days (Direct Proportion)


 


Spiders
1
:
7
If 7 Spiders Make 7 Webs In 7 Days, Then 1 Spider Will Make ...
:: 7 : x
Webs
7
:
1

 


If 7 Spiders Make 7 Webs In 7 Days, Then 1 Spider Will Make ... 1 x 7 x x = 7 x 1 x 7


If 7 Spiders Make 7 Webs In 7 Days, Then 1 Spider Will Make ... x = 7.

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