Chain Rule Questions and Answers

Q: A contractor undertakes to do a piece of work in 40 days. He engages 100 men at the begining and 100 more after 35 days and completes the work in stipulated time. If he had not engaged the additional men, how many days behind schedule would it be finished?

Count the man-days actually worked.\[ 100 \times 35 + 200 \times 5 = 3500 + 1000 = 4500 \]With only the original 100 men:\[ \frac{4500}{100} = 45 \text{ days} \]\[ 45 - 40 = 5 \text{ days behind schedule} \]

Q: 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?

Let the required number of days be x. Less persons, More days (Indirect Proportion) More working hours per day, Less days (Indirect Proportion) Persons30:39Working hours/day6:5⋮⋮ 12:x => 30 * 6 * x = 39 * 5 * 12 => x= 13

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

Let the required number days be x. Less spiders, More days (Indirect Proportion) Less webs, Less days (Direct Proportion) Spiders1:7Webs7:1⋮⋮ 7:x => 1 * 7 * x = 7 * 1 * 7 => x= 7

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11.

A certain number of men can complete a piece of work in 180 days. If there are 30 men less, it will take 20 days more for the work to be completed. How many men were there originally?

A.135
B.165
C.150
D.180
E.300

Answer: E

Let there be x men originally.
They were to complete the work in 180 days but as the number of persons is reduced to x - 30.
∴ Work takes 20 more days. So the equation is 180x = (x - 30)200 ⇒ 20x = 6000 ⇒ x = 300.

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12.

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.12
B.36
C.42
D.54

Answer: C

Count the food in man-days, which does not change when men leave. After 10 days there remain \( 45 - 10 = 35 \) days of provision for 150 men:

\[ \text{food remaining} = 150 \times 35 = 5{,}250 \text{ man-days} \]

With 25 men gone the strength is \( 150 - 25 = 125 \), so

\[ \text{days} = \frac{5{,}250}{125} = 42 \]

Fewer men must make the food last longer, so any answer below 35 is wrong on inspection - which disposes of 12 at once.

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13.

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

A.4
B.6
C.8
D.7⁄2

Answer: B

Find the daily rate first, then apply it to what is left. Completing \( \tfrac{5}{8} \) in 10 days means

\[ \text{rate} = \frac{5/8}{10} = \frac{1}{16} \text{ of the job per day} \]

The remaining fraction is \( 1 - \tfrac{5}{8} = \tfrac{3}{8} \), so

\[ \text{days} = \frac{3/8}{1/16} = \frac{3}{8} \times 16 = 6 \]

The question asks how many more days, so 6 is the answer; 16 would be the total time for the whole job from the start.

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14.

4 mat-weavers can weave 4 mats in 4 days. At the same rate, how many mats would be woven by 8 mat-weaves in 8 days?

A.2
B.4
C.8
D.16

Answer: D

Reduce to a single weaver before scaling up. Four weavers making 4 mats in 4 days means

\[ 1 \text{ weaver} \rightarrow 1 \text{ mat in } 4 \text{ days} \]

Output is proportional to both the number of weavers and the time given, so

\[ \text{mats} = 8 \text{ weavers} \times \frac{8 \text{ days}}{4 \text{ days per mat}} = 8 \times 2 = 16 \]

The intuitive answer of 8 comes from doubling only one of the two factors. Both the workforce and the time double here, so the output goes up fourfold.

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15.

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.20
C.40
D.60

Answer: C

Read the rate per cow rather than for the herd. If 40 cows finish 40 bags in 40 days, then in those same 40 days

\[ \text{each cow eats } \frac{40 \text{ bags}}{40 \text{ cows}} = 1 \text{ bag} \]

so one cow takes 40 days for one bag.

The number of cows never enters the answer: each is eating alongside the others, not sharing a single bag. Dividing 40 by 40 to get 1 day is the classic slip, and it would mean the herd polished off 40 bags in a single day.

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16.

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

Answer: D

Let the required no 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)

Pumps4 : 3Days1 : 2⋮⋮ 8:x

=> (4 * 1 * x) = (3 * 2 * 8)

=> x=12

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17.

A contractor undertakes to do a piece of work in 40 days. He engages 100 men at the begining and 100 more after 35 days and completes the work in stipulated time. If he had not engaged the additional men, how many days behind schedule would it be finished?

Answer: B

Count the man-days actually worked.

\[ 100 \times 35 + 200 \times 5 = 3500 + 1000 = 4500 \]

With only the original 100 men:

\[ \frac{4500}{100} = 45 \text{ days} \]

\[ 45 - 40 = 5 \text{ days behind schedule} \]

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18.

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

Answer: B

Let the required number of days be x.

Less persons, More days (Indirect Proportion)

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

Persons30:39Working hours/day6:5⋮⋮ 12:x

=> 30 * 6 * x = 39 * 5 * 12

=> x= 13

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19.

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

A.1
B.3
C.7
D.14

Answer: C

Let the required number days be x.

Less spiders, More days (Indirect Proportion)

Less webs, Less days (Direct Proportion)

Spiders1:7Webs7:1⋮⋮ 7:x

=> 1 * 7 * x = 7 * 1 * 7

=> x= 7

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20.

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 created to with the remaining meal?

A.10
B.30
C.50
D.70

Answer: B

The meal is one fixed quantity, described two ways, so the two descriptions are equivalent:

\[ 120 \text{ men} \equiv 200 \text{ children} \]

After 150 children have eaten, what is left would feed \( 200 - 150 = 50 \) children. Converting that share to men:

\[ 50 \text{ children} \times \frac{120 \text{ men}}{200 \text{ children}} = 30 \text{ men} \]

The step that matters is doing the subtraction in children and converting only at the end; converting first and then subtracting mixes the two units.

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