Problems on Chain Rule

Problems on Chain Rule - Solved Examples

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. $\text{Rs.}\left(\frac{xd}{y}\right)$	B. $\text{Rs. xd}$
C. $\text{Rs.}\left(\frac{yd}{x}\right)$	D. $\text{Rs. yd}$

answer with explanation

Answer: Option C
Explanation:
cost of $x$ metres of wire = Rs.$d$

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

cost of $y$ metre of wire = Rs.$(y\times \frac{d}{x})=\text{Rs.}\left(\frac{yd}{x}\right)$

2. 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 with explanation

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

$\begin{array}{cc}\text{(cows)}& 40:1\\ \text{(bags)}& 1:40\end{array}\}::x:40$

$\Rightarrow 40\times 1\times 40=1\times 40\times x\Rightarrow x=40$

3. If 7 spiders make 7 webs in 7 days, then how many days are needed for 1 spider to make 1 web?
A. 1	B. 7
C. 3	D. 14

answer with explanation

Answer: Option B
Explanation:
Let, 1 spider make 1 web in $x$ days.

More spiders, Less days (Indirect proportion)
More webs, more days (Direct proportion)

Hence we can write as

$\begin{array}{cc}\text{(spiders)}& 7:1\\ \text{(webs)}& 1:7\end{array}\}::x:7$

$\Rightarrow 7\times 1\times 7=1\times 7\times x\Rightarrow x=7$

4. 4 mat-weavers can weave 4 mats in 4 days. At the same rate, how many mats would be woven by 8 mat-weavers in 8 days?
A. 4	B. 16
C. 8	D. 1

answer with explanation

Answer: Option B
Explanation:
Let the required number of mats be $x$

More mat-weavers, more mats (direct proportion)
More days, more mats (direct proportion)

Hence we can write as

$\begin{array}{cc}\text{(mat-weavers)}& 4:8\\ \text{(days)}& 4:8\end{array}\}::4:x$

$\Rightarrow 4\times 4\times x=8\times 8\times 4\Rightarrow x=2\times 2\times 4=16$

5. 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 with explanation

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

Let 200 gm potato costs $x$ paise

Cost of $\frac{1}{4}$ Kg potato = 60 Paise
=> Cost of 250 gm potato = 60 Paise $(\because \frac{1}{4}\text{kg}=\frac{1000}{4}\text{gm}=250\text{gm})$

More quantity, More Paise (direct proportion)

Hence we can write as
(quantity) 200 : 250 :: $x$ : 60

$\Rightarrow 200\times 60=250\times x\Rightarrow 4\times 60=5\times x\Rightarrow 4\times 12=x\Rightarrow x=48$

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Solution 2

Cost of $\frac{1}{4}$ kg potato = 60 Paise

=> Cost of 250 gm potato = 60 Paise $(\because \frac{1}{4}\text{kg}=\frac{1000}{4}\text{gm}=250\text{gm})$

=> Cost of 200 gm potato $=\frac{60\times 200}{250}=\frac{60\times 4}{5}=48\text{paise}$

6. 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 with explanation

Answer: Option B
Explanation:
Meal for 200 children = Meal for 120 men

=> Meal for 1 child = Meal for $\frac{120}{200}$ men

=> Meal for 150 children
= Meal for $\frac{120\times 150}{200}\text{men}$ = Meal for 90 men

Total meal available = Meal for 120 men

Remaining meal
= Meal for 120 men - Meal for 90 men
= Meal for 30 men

7. 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 with explanation

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

Let the required number of days be $x$

More men, less days (indirect proportion)

Hence we can write as
(men) 36 : 27 :: $x$ : 18

$\Rightarrow 36\times 18=27\times x\Rightarrow 12\times 18=9\times x\Rightarrow 12\times 2=x\Rightarrow x=24$

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Solution 2 (Using Time and Work)

Amount of work 36 men can do in 1 day $=\frac{1}{18}$

Amount of work 1 man can do in 1 day $=\frac{1}{18\times 36}$

Amount of work 27 men can do in 1 day $=\frac{27}{18\times 36}=\frac{3}{18\times 4}=\frac{1}{24}$

27 men can complete the work in 24 days

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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

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

In this case,

$M_1$ = 36, $M_2$ = 27

$D_1$ = 18, $D_2=x$

$W_1=W_2$

$H_1=H_2$ (∵ We can assume like this as these vales are not explicitly given)

Hence,

$M_1{D}_1=M_2{D}_2\Rightarrow 36\times 18=27x\Rightarrow x=\frac{36\times 18}{27}=\frac{4\times 18}{3}=4\times 6=24$

8. 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 with explanation

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

$\Rightarrow 6\times 21=14\times x\Rightarrow 6\times 3=2\times x\Rightarrow 3\times 3=x\Rightarrow x=9$

9. 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 with explanation

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

$\begin{array}{cc}\text{(pumps)}& 3:4\\ \text{(days)}& 2:1\end{array}\}::x:8$

$\Rightarrow 3\times 2\times 8=4\times 1\times x\Rightarrow 3\times 2\times 2=x\Rightarrow x=12$

10. 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 with explanation

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

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

$\begin{array}{cc}\text{(persons)}& 39:30\\ \text{(hours)}& 5:6\end{array}\}::x:12$

$\Rightarrow 39\times 5\times 12=30\times 6\times x\Rightarrow 39\times 5\times 2=30\times x\Rightarrow 39=3\times x\Rightarrow x=13$

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Solution 2 (Using Time and Work)

Amount of work 39 persons can do in 1 day, working 5 hours a day $=\frac{1}{12}$

Amount of work 1 person can do in 1 day, working 5 hours a day $=\frac{1}{12\times 39}$

Amount of work 1 person can do in 1 day, working 1 hours a day $=\frac{1}{12\times 39\times 5}$

Amount of work 30 person can do in 1 day, working 1 hours a day $=\frac{30}{12\times 39\times 5}$

Amount of work 30 person can do in 1 day, working 6 hours a day
$=\frac{30\times 6}{12\times 39\times 5}=\frac{30}{2\times 39\times 5}=\frac{3}{39}=\frac{1}{13}$

=> 30 persons can complete the work ,working 6 hours a day in 13 days

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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

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

In this case,
M₁ = 39, M₂ = 30
D₁ = 12, D₂ = $x$
W₁ = W₂
H₁ = 5, H₂ = 6

Hence,
$M_1{D}_1{H}_1=M_2{D}_2{H}_2\Rightarrow 39\times 12\times 5=30\times x\times 6\Rightarrow 39\times 2\times 5=30\times x\Rightarrow 39=3\times x\Rightarrow x=\frac{39}{3}=13$