Problems on Chain Rule

21. A rope can make 70 rounds of the circumference of a cylinder whose radius of the base is 14cm. how many times can it go round a cylinder having radius 20 cm?
A. 49 rounds	B. 42 rounds
C. 54 rounds	D. 52 rounds

answer with explanation

Answer: Option A
Explanation:
Let the required number of rounds be $x$

More radius, less rounds(Indirect proportion)

Hence we can write as
(radius) 14 : 20 :: $x$ : 70

$\Rightarrow 14\times 70=20x\Rightarrow 14\times 7=2x\Rightarrow x=7\times 7=49$

22. 8 persons can build a wall 140m long in 42 days. In how many days can 30 persons complete a similar wall 100 m long?
A. 12	B. 10
C. 8	D. 6

answer with explanation

Answer: Option C
Explanation:
Solution 1 (Chain Rule)

More persons, less days(indirect proportion)
More length of wall, more days(direct proportion)

Hence we can write as

$\begin{array}{cc}\text{(persons)}& 8:30\\ \text{(length of wall)}& 100:140\end{array}\}::x:42$

$\Rightarrow 8\times 100\times 42=30\times 140\times x\Rightarrow x=\frac{8\times 100\times 42}{30\times 140}=\frac{8\times 100\times 14}{10\times 140}=\frac{8\times 100}{10\times 10}=8$

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Solution 2 (Using Time and Work)

Work done by 8 persons working 42 days = 140

Work done by 1 person working 42 days $=\frac{140}{8}$

Work done by 1 person working 1 day $=\frac{140}{8\times 42}$

Work done by 30 persons working 1 day $=\frac{30\times 140}{8\times 42}=\frac{100}{8}$

Assume that 30 persons working $x$ days complete a similar wall 100 m
=> Work done by 30 persons working $x$ days = 100

Hence $x=\frac{100}{\left(\frac{100}{8}\right)}=8$

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Solution 3 (Using Time and Work)

If $M_1$ men can do $W_1$ work in $D_1$ days working $H_1$ hours per day and $M_2$ men can do$W_2$ work in $D_2$ days working $H_2$ hours per day where all men work at the same rate, then

$\frac{M_1{D}_1{H}_1}{W_1}=\frac{M_2{D}_2{H}_2}{W_2}$

M₁ = 8
D₁ = 42
W₁= 140

M₂ = 30
Let D₂ $=x$
W₂= 100

Here H₁ = H₂

$\frac{M_1{D}_1}{W_1}=\frac{M_2{D}_2}{W_2}\Rightarrow \frac{8\times 42}{140}=\frac{30\times x}{100}\Rightarrow \frac{8\times 42}{14}=\frac{30\times x}{10}\Rightarrow 8\times 3=3\times x\Rightarrow x=8$

23. A certain number of persons can finish a piece of work in 100 days. If there were 10 persons less, it would take 10 more days finish the work. How many persons were there originally?
A. 90	B. 100
C. 110	D. 120

answer with explanation

Answer: Option C
Explanation:
Solution 1 (Chain Rule)

Assume that $x$ persons can finish a piece of work in 100 days
Also it is given that $(x-10)$ persons can finish a piece of work in 110 days (∵ 100 + 10 = 110)

More persons, less days(indirect proportion)

Hence we can write as
(persons) $x$ : $(x-10)$ :: 110 : 100

$\Rightarrow 100x=110(x-10)\Rightarrow 100x=110x-1100\Rightarrow 10x=1100\Rightarrow x=\frac{1100}{10}=110$

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Solution 2 (Using Time and Work)

Assume that $x$ persons can finish the work in 100 days
Work done by 1 person in 1 day $=\frac{1}{100x}$ ...(Equation 1)

Also it is given that $(x-10)$ persons can finish the work in 110 days (∵ 100 + 10 = 110)
Work done by 1 person in 1 day $=\frac{1}{110(x-10)}$ ...(Equation 2)

But (Equation 1) = (Equation 2)
$\Rightarrow \frac{1}{100x}=\frac{1}{110(x-10)}\Rightarrow 110(x-10)=100x\Rightarrow 110x-1100=100x\Rightarrow 10x=1100x=\frac{1100}{10}=110$

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Solution 3 (Using Time and Work)

If $M_1$ men can do $W_1$ work in $D_1$ days working $H_1$ hours per day and $M_2$ men can do$W_2$ work in $D_2$ days working $H_2$ hours per day where all men work at the same rate, then

$\frac{M_1{D}_1{H}_1}{W_1}=\frac{M_2{D}_2{H}_2}{W_2}$

Assume that $x$ persons can finish a piece of work in 100 days
Also it is given that $(x-10)$ persons can finish a piece of work in 110 days (∵ 100 + 10 = 110)

M₁ $=x$
D₁ = 100

M₂ $=(x-10)$
D₂ = 110

Here H₁ = H₂ and W₁ = W₂

Hence $M_1{D}_1=M_2{D}_2$

$\Rightarrow 100x=110(x-10)\Rightarrow 100x=110x-1100\Rightarrow 10x=1100\Rightarrow x=\frac{1100}{10}=110$

24. 9 examiners can examine a certain number of answer books in 12 days by working 5 hours a day. How many hours in a day should 4 examiners work to examine twice the number of answer books in 30 days?
A. 9	B. 10
C. 11	D. 12

answer with explanation

Answer: Option A
Explanation:
Solution 1 (Chain Rule)

Let required number of hours be $x$

More examiners, less hours (indirect proportion)
More days, less hours (indirect proportion)
More answer books, more hours (direct proportion)

Hence we can write as

$\begin{array}{cc}\text{(examiners)}& 9:4\\ \text{(days)}& 12:30\\ \text{(answer books)}& 2:1\end{array}\}::x:5$

$\Rightarrow 9\times 12\times 2\times 5=4\times 30\times 1\times x\Rightarrow x=\frac{9\times 12\times 2\times 5}{4\times 30\times 1}=\frac{9\times 3\times 2\times 5}{30}=\frac{9\times 2\times 5}{10}=9$

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Solution 2 (Using Time and Work)

Given that work done by 9 examiners in 12 days by working 5 hours a day = 1
=> Work done by 1 examiner in 1 day in 1 hour $=\frac{1}{9\times 12\times 5}$ ... (Equation 1)

Work needs to be done by 4 examiners in 30 days working $x$ hours a day = 2 (∵ twice work to be completed)
=> Work needs to be done by 1 examiner in 1 day working $x$ hours a day $=\frac{2}{4\times 30}$ ... (Equation 2)

From (Equation 1) and (Equation 2),
$x=\frac{\left(\frac{2}{4\times 30}\right)}{\left(\frac{1}{9\times 12\times 5}\right)}=\frac{2\times 9\times 12\times 5}{4\times 30}=\frac{9\times 12\times 5}{2\times 30}=\frac{9\times 6\times 5}{30}=9$

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Solution 3 (Using Time and Work)

If $M_1$ men can do $W_1$ work in $D_1$ days working $H_1$ hours per day and $M_2$ men can do$W_2$ work in $D_2$ days working $H_2$ hours per day where all men work at the same rate, then

$\frac{M_1{D}_1{H}_1}{W_1}=\frac{M_2{D}_2{H}_2}{W_2}$

M₁ = 9
D₁ = 12
H₁ = 5
W₁ = 1

M₂ = 4
D₂ = 30
H₂ = $x$
W₂ = 2

$\frac{M_1{D}_1{H}_1}{W_1}=\frac{M_2{D}_2{H}_2}{W_2}\Rightarrow \frac{9\times 12\times 5}{1}=\frac{4\times 30\times x}{2}\Rightarrow 9\times 12\times 5=2\times 30\times x\Rightarrow 3\times 12\times 5=2\times 10\times x\Rightarrow 3\times 6\times 5=10\times x\Rightarrow 3\times 3\times 5=5\times x\Rightarrow x=3\times 3=9$

25. 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 with explanation

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 $\frac{1}{3}$ unit which is its the rate of consumption.
If 4 engines of second type consume 1 unit, then 1 engine will consume $\frac{1}{4}$ 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

$\begin{array}{cc}\text{(engines)}& 9:8\\ \text{(consumption rate)}& \frac{1}{3}:\frac{1}{4}\\ \text{(hours)}& 8:13\end{array}\}::24:x$

$\Rightarrow 9\times \frac{1}{3}\times 8\times x=8\times \frac{1}{4}\times 13\times 24\Rightarrow 3\times 8\times x=8\times 6\times 13\Rightarrow 3\times x=6\times 13\Rightarrow x=2\times 13=26$

26. A garrison had provisions for a certain number of days. After 10 days, $\frac{1}{5}$ of the men desert and it is found that the provisions will now last just as long as before. How long was that?
A. 50 days	B. 30 days
C. 40 days	D. 60 days

answer with explanation

Answer: Option A
Explanation:
Solution 1

Assume that initially garrison had provisions for $x$ men for $y$ days.
So, after 10 days, garrison had provisions for $x$ men for $(y-10)$ days

Also, after 10 days, garrison had provisions for $\frac{4x}{5}$ men for $y$ days $(\because x-\frac{x}{5}=\frac{4x}{5})$

More men, Less days (Indirect Proportion)
(men) $x$ : $\frac{4x}{5}$ :: $y$ : $(y-10)$

$\Rightarrow x(y-10)=\frac{4xy}{5}\Rightarrow (y-10)=\frac{4y}{5}\Rightarrow 5(y-10)=4y\Rightarrow 5y-50=4y\Rightarrow y=50$

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Solution 2

Assume that amount of food consumed by 1 man in 1 day $=x$
total men = $y$
and the garrison had provisions for $z$ days

Then, total quantity of food $=xyz$

amount of food consumed by $y$ men in 10 days $=10xy$
Remaining food $=xyz-10xy=xy(z-10)$ ... (Equation 1)

After 10 days, total men $=\frac{4y}{5}$ $(\because y-\frac{y}{5}=\frac{4y}{5})$
food consumed by $\frac{4y}{5}$ men in 1 day = $\frac{4xy}{5}$ ... (Equation 2)

From Equations 1 and 2,
Time taken for $\frac{4y}{5}$ men to complete $xy(z-10)$ food

$=\frac{xy(z-10)}{\left(\frac{4xy}{5}\right)}=\frac{5(z-10)}{4}$

Given that number of days remain the same
$\Rightarrow \frac{5(z-10)}{4}=z\Rightarrow 5z-50=4z\Rightarrow z=50$

27. A garrison of 500 persons had provisions for 27 days. After 3 days a reinforcement of 300 persons arrived. For how many more days will the remaining food last now?
A. 12 days	B. 16 days
C. 14 days	D. 15 days

answer with explanation

Answer: Option D
Explanation:
Solution 1

Given that fort had provision for 500 persons for 27 days
Hence, after 3 days, the remaining food is sufficient for 500 persons for 24 days

Remaining persons after 3 days = 500 + 300 = 800
Assume that after 3 days,the remaining food is sufficient for 800 persons for $x$ days

More men, Less days (Indirect Proportion)
(men) 500 : 800 :: $x$ : 24

$\Rightarrow 500\times 24=800x\Rightarrow 5\times 24=8x\Rightarrow x=5\times 3=15$

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Solution 2

Assume that amount of food consumed by 1 man in 1 day $=x$
Given that the garrison had provisions for 500 persons for 27 days
=> Total quantity of food $=500\times 27\times x=13500x$

Amount of food consumed by 500 persons in 3 days $=500\times 3\times x=1500x$
Remaining food $=13500x-1500x=12000x$ ... (Equation 1)

After 3 days, total persons = 500 + 300 = 800
Food consumed by 800 persons 1 day $=800x$ ... (Equation 2)

From Equations 1 and 2,
Time taken for 800 persons to consume $12000x$ food
$=\frac{12000x}{800x}=\frac{120}{8}=15$

28. A hostel had provisions for 250 men for 40 days. If 50 men left the hostel, how long will the food last at the same rate?
A. 48 days	B. 50 days
C. 45 days	D. 60 days

answer with explanation

Answer: Option B
Explanation:
Solution 1

A hostel had provisions for 250 men for 40 days

If 50 men leaves the hostel, remaining men = 250 - 50 = 200
We need to find out how long the food will last for these 200 men.

Let the required number of days = $x$ days

More men, Less days (Indirect Proportion)
(men) 250 : 200 :: $x$ : 40

$\Rightarrow 250\times 40=200x\Rightarrow 5\times 40=4x\Rightarrow x=5\times 10=50$

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Solution 2

Assume that amount of food consumed by 1 man in 1 day $=x$

Given that the hostel had provisions for 250 men for 40 days
=> Total quantity of food $=250\times 40\times x=10000x$ ...(Equation 1)

If 50 men leave the hostel, remaining men = 250 - 50 = 200
Food consumed by 200 men 1 day $=200x$ ... (Equation 2)

From Equations 1 and 2,
Time taken for 200 men to consume $10000x$ food
$=\frac{10000x}{200x}=\frac{100}{2}=50$

29. 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 with explanation

Answer: Option A
Explanation:
Solution 1

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

$\Rightarrow 2000\times 39=(2000+x)20\Rightarrow 100\times 39=(2000+x)\Rightarrow 3900=2000+x\Rightarrow x=3900-2000=1900$

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Solution 2

Assume that amount of food consumed by 1 person in 1 day $=k$

Given that food was sufficient for 2000 people for 54 days
Hence, total quantity of food $=2000\times 54\times k=108000k$

Amount of food consumed by 2000 persons in 15 days $=2000\times 15\times k=30000k$

Remaining food $=108000k-30000k=78000k$ ...(Equation 1)

Let $x$ number of persons came after 15 days.
Then, total number of people after 15 days $=(2000+x)$
Food consumed by $(2000+x)$ persons 1 day $=(2000+x)k$ ...(Equation 2)

From Equations 1 and 2,
Time taken for $(2000+x)$ persons to consume $78000k$ food
$=\frac{78000k}{(2000+x)k}=\frac{78000}{(2000+x)}$

Given that food lasted only for 20 more days
$\Rightarrow \frac{78000}{(2000+x)}=20\Rightarrow 78000=20(2000+x)\Rightarrow 3900=2000+x\Rightarrow x=3900-2000=1900$

30. If 40 men can make 30 boxes in 20 days, How many more men are needed to make 60 boxes in 25 days?
A. 28	B. 24
C. 22	D. 26

answer with explanation

Answer: Option B
Explanation:
Solution 1 (Chain Rule)

Given that 40 men can make 30 boxes in 20 days

Let $x$ more men are needed to make 60 boxes in 25 days
Then $(40+x)$ men can make 60 boxes in 25 days

More boxes, more men(direct proportion)
More days, less men(indirect proportion)

Hence we can write as

$\begin{array}{cc}\text{(boxes)}& 30:60\\ \text{(days)}& 25:20\end{array}\}::40:(40+x)$

$\Rightarrow 30\times 25\times (40+x)=60\times 20\times 40\Rightarrow 25\times (40+x)=2\times 20\times 40\Rightarrow 5\times (40+x)=2\times 4\times 40\Rightarrow (40+x)=2\times 4\times 8=64\Rightarrow x=64-40=24$

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Solution 2 (Using Time and Work)

Let $x$ more men are needed to make 60 boxes in 25 days
Then $(40+x)$ men can make 60 boxes in 25 days

Work done by 40 men in 20 days = 30
Work done by 1 man in 1 day $=\frac{30}{40\times 20}$
Work done by $(40+x)$ men in 25 days $=\frac{30(40+x)\times 25}{40\times 20}$ ... (Equation 1)

If $(40+x)$ men can make 60 boxes in 25 days,
Work done by $(40+x)$ men in 25 days = 60 ... (Equation 2)

Hence, from equation 1 and equation 2,
$\frac{30(40+x)\times 25}{40\times 20}=60\Rightarrow 30(40+x)\times 25=60\times 40\times 20\Rightarrow (40+x)\times 25=2\times 40\times 20\Rightarrow (40+x)\times 5=2\times 8\times 20\Rightarrow (40+x)=2\times 8\times 4=64\Rightarrow x=64-40=24$

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Solution 3 (Using Time and Work)

If $M_1$ men can do $W_1$ work in $D_1$ days working $H_1$ hours per day and $M_2$ men can do$W_2$ work in $D_2$ days working $H_2$ hours per day where all men work at the same rate, then

$\frac{M_1{D}_1{H}_1}{W_1}=\frac{M_2{D}_2{H}_2}{W_2}$

M₁ = 40
D₁ = 20
W₁= 30

Let M₂ $=x$
Let D₂ = 25
W₂= 60

Here H₁ = H₂

Hence,
$\frac{M_1{D}_1}{W_1}=\frac{M_2{D}_2}{W_2}\Rightarrow \frac{40\times 20}{30}=\frac{x\times 25}{60}\Rightarrow 2(40\times 20)=x\times 25\Rightarrow 2(8\times 20)=x\times 5\Rightarrow x=2(8\times 4)=64$

Hence, additional men required = 64 - 40 = 2
4