This module will teach you the basics of direct and indirect proportions. These concepts will further help you in time and work questions.
Important Formulas – chain rule
 Direct Proportion
 Indirect Proportion (inverse proportion)
Two quantities are said to be indirectly proportional (inversely proportional) if on the increase of the one, the other decreases to the same extent and viceversa.
Examples

 Number of days needed to complete a work is indirectly proportional (inversely proportional) with the number of persons who does the work (More Persons, Less Days needed)
 The time taken to travel a distance is indirectly proportional (inversely proportional) with the speed in which one is travelling (More Speed, Less Time)
Solved Examples
Level 1
1. 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. (xd/y)  B. Rs. x/d 
C. Rs. (yd/x)  D. Rs. y/d 
Answer : Option C
Explanation :
cost of x metres of wire = Rs. d
cost of 1 metre of wire = Rs.(d/x)
cost of y metre of wire = Rs.(y×d/x)=Rs. (yd/x)
2. 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?  
A. 50  B. 30 
C. 40  D. 10 
Answer : Option B
Explanation :
Meal for 200 children = Meal for 120 men
Meal for 1 child = Meal for 120/200 men
Meal for 150 children = Meal for (120×150)/200 men=Meal for 90 men
Total mean available = Meal for 120 men
Renaming meal = Meal for 120 men – Meal for 90 men = Meal for 30 men
3. 36 men can complete a piece of work in 18 days. In how many days will 27 men complete the same work?  
A. 26  B. 22 
C. 12  D. 24 
Answer : Option D
Explanation :
Let the required number of days be x
More men, less days (indirect proportion)
Hence we can write as
Men36:27}::x:18 ⇒36×18=27×x ⇒12×18=9×x
⇒12×2=x
⇒x=24
4. A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. If the smaller wheel has made 21 revolutions, what will be the number of revolutions made by the larger wheel?  
A. 15  B. 12 
C. 21  D. 9 
Answer : Option D
Explanation :
Let the number of revolutions made by the larger wheel be x
More cogs, less revolutions (Indirect proportion)
Hence we can write as
Cogs 6:14}: x: 21⇒6×21=14×x ⇒6×3=2×x ⇒3×3=x ⇒x=9
5. 3 pumps, working 8 hours a day, can empty a tank in 2 days. How many hours a day should 4 pumps work in order to empty the tank in 1 day?  
A. 10  B. 12 
C. 8  D. 15 
Answer : Option B
Explanation :
Let the required hours needed be x
More pumps, less hours (Indirect proportion)
More Days, less hours (Indirect proportion)
Hence we can write as
Pumps 3:4
::x:8
Days 2:1
⇒3×2×8=4×1×x
⇒3×2×2=x
⇒x=12
6. 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. 9  B. 12 
C. 10  D. 13 
Answer : Option D
Explanation :
Let the required number of days be x
More persons, less days (indirect proportion)
More hours, less days (indirect proportion)
Hence we can write as
Persons 39:30
::x:12
Hours 5:6
⇒39×5×12=30×6×x ⇒39×5×2=30×x ⇒39=3×x ⇒x=13
7. A certain industrial loom weaves 0.128 meters of cloth every second. Approximately how many seconds will it take for the loom to weave 25 meters of cloth?  
A. 205  B. 200 
C. 180  D. 195 
Answer : Option D
Explanation :
Let the required number of seconds be x
More cloth, More time, (direct proportion)
Hence we can write as
Cloth 0.128:25} :: 1:x
⇒0.128x=25 ⇒x=25/0.128 ⇒25000/128=3125/16≈195
8. 21 goats eat as much as 15 cows. How many goats each as much as 35 cows?  
A. 49  B. 32 
C. 36  D. 41 
Answer : Option A
Explanation :
15 cows ≡ 21 goats
1 cow ≡21/15 goats
35 cows ≡ (21×35)/15 goats≡(21×7)/3 goats≡7×7 goats ≡ 49 goats
Level 2
1. 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. 40 
C. 20  D. 26 
Answer : Option B
Explanation :
Assume that in x days, one cow will eat one bag of husk.
More cows, less days (Indirect proportion)
More bags, more days (direct proportion)
Hence we can write as
Cows 40:1 ::x:40
Bags 1:40
⇒40×1×40=1×40×x ⇒x=40
2. If a quarter kg of potato costs 60 paise, how many paise does 200 gm cost?  
A. 65 paise  B. 70 paise 
C. 52 paise  D. 48 paise 
Answer : Option D
Explanation :
Let 200 gm potato costs x paise
Cost of ¼ Kg potato = 60 Paise
=> Cost of 250 gm potato = 60 Paise (∵ 1 Kg = 1000 gm => ¼ Kg = ^{1000}/_{4} gm = 250 gm)
More quantity, More Paise (direct proportion)
Hence we can write as
Quantity 200:250} :: x:60
⇒200×60=250×x ⇒4×60=5×x ⇒4×12=x ⇒x=48
3. A contract is to be completed in 56 days if 104 persons work, each working at 8 hours a day. After 30 days, ^{2}/_{5} of the work is completed. How many additional persons should be deployed so that the work will be completed in the scheduled time, each person’s now working 9 hours a day.  
A. 160  B. 150 
C. 24  D. 56 
Answer : Option D
Explanation :
Persons worked = 104
Number of hours each person worked per day = 8
Number of days they worked = 30
Work completed = ^{2}/_{5}
Remaining days = 56 – 30 = 26
Remaining Work to be completed = 1 – ^{2}/_{5} = ^{3}/_{5}
Let the total number of persons who do the remaining work = x
Number of hours each person needs to be work per day = 9
More days, less persons(indirect proportion) More hours, less persons(indirect proportion)
More work, more persons(direct proportion)
Hence we can write as
Days 30:26
Hours 8:9 ::x:104
Work 35:25
⇒30×8×3/5×104=26×9×2/5×x
⇒x=(30×8×3/5×104)/(26×9×2/5)=(30×8×3×104)/(26×9×2)
=(30×8×104)/(26×3×2)=(30×8×4)/(3×2)=5×8×4=160
Number of additional persons required = 160 – 104 = 56
4. x men working x hours per day can do x units of a work in x days. How much work can be completed by y men working y hours per day in y days?  
A. x^{2}/y^{2} units  B. y^{3}/x^{2} units 
C. x^{3}/y^{2} units  D. y^{2}/x^{2} units 
Answer : Option B
Explanation :
Let amount of work completed by y men working y hours per in y days = w units
More men, more work(direct proportion)
More hours, more work(direct proportion)
More days, more work(direct proportion)
Hence we can write as
Men x:y
Hours x:y ::x:w
Days x:y
⇒x^{3}w=y^{3}x ⇒w=y^{3}x/x^{3}=y^{3}/x^{2}
5. A flagstaff 17.5 m high casts a shadow of length 40.25 m. What will be the height of a building, which casts a shadow of length 28.75 m under similar conditions?  
A. 12.5 m  B. 10.5 m 
C. 14  D. 12 
Answer : Option A
Explanation :
Let the required height of the building be x meter
More shadow length, More height (direct proportion)
Hence we can write as
Shadow length 40.25:28.75}:: 17.5:x
⇒40.25×x=28.75×17.5 ⇒x=(28.75×17.5)/40.25=(2875×175)/40250
= (2875×7)/1610=2875/230=575/46=12.5
6. If the price of 357 apples is Rs.1517.25, what will be the approximate price of 49 dozens of such apples?  
A. Rs. 2500  B. Rs. 2300 
C. Rs. 2200  D. Rs. 1400 
Answer : Option A
Explanation :
Let the required price be x
More apples, More price (direct proportion)
Hence we can write as
Apples 357:(49×12)} :: 1517.25:x
⇒357x = (49×12)×1517.25⇒x = (49×12×1517.25)/357=(7×12×1517.25)/51
= (7×4×1517.25)/17
=7×4×89.25≈2500
7. 9 engines consume 24 metric tonnes of coal, when each is working 8 hours day. How much coal is required for 8 engines, each running 13 hours a day, if 3 engines of former type consume as much as 4 engines of latter type?  
A. 20 metric tonnes  B. 22 metric tonnes 
C. 24 metric tonnes  D. 26 metric tonnes 
Answer : Option D
Explanation :
Let required amount of coal be x metric tonnes
More engines, more amount of coal (direct proportion)
If 3 engines of first type consume 1 unit, then 1 engine will consume 1/3 unit which is its the rate of consumption.
If 4 engines of second type consume 1 unit, then 1 engine will consume 1/4 unit which is its the rate of consumption
More rate of consumption, more amount of coal (direct proportion)
More hours, more amount of coal(direct proportion)
Hence we can write as
Engines 9:8
rate of consumption 13:14 ::24:x
hours 8:13
⇒9×1/3×8×x=8×1/4×13×24 ⇒3×8×x=8×6×13 ⇒3xX=6×13⇒x=2×13=26
8. in a camp, food was was sufficient for 2000 people for 54 days. After 15 days, more people came and the food last only for 20 more days. How many people came?  
A. 1900  B. 1800 
C. 1940  D. 2000 
Answer : Option A
Explanation :
Given that food was sufficient for 2000 people for 54 days
Hence, after 15 days, the remaining food was sufficient for 2000 people for 39 days (∵ 54 – 15 =39)
Let x number of people came after 15 days.
Then, total number of people after 15 days = (2000 + x)
Then, the remaining food was sufficient for (2000 + x) people for 20 days
More men, Less days (Indirect Proportion)⇒Men 2000:(2000+x)} :: 20:39
⇒2000×39=(2000+x)20⇒100×39=(2000+x)⇒3900=2000+x⇒x=3900−2000=1900