Problems on Chain Rule

11. 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 with explanation

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 : $x$

$\Rightarrow 0.128x=25\Rightarrow x=\frac{25}{0.128}=\frac{25000}{128}=\frac{3125}{16}\approx 195$

12. A contract is to be completed in 56 days if 104 persons work, each working at 8 hours a day. After 30 days, $\frac{2}{5}$ of the work is completed. How many additional persons should be deployed so that the work will be completed in the scheduled time,each persons now working 9 hours a day.
A. 160	B. 150
C. 24	D. 56

answer with explanation

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

Persons worked = 104
Number of hours each person worked per day = 8
Number of days they worked = 30
Work completed $=\frac{2}{5}$

Remaining days = 56 - 30 = 26
Remaining Work to be completed $=1-\frac{2}{5}=\frac{3}{5}$
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

$\begin{array}{cc}\text{(days)}& 30:26\\ \text{(hours)}& 8:9\\ \text{(work)}& \frac{3}{5}:\frac{2}{5}\end{array}\}::x:104$

$\Rightarrow 30\times 8\times \frac{3}{5}\times 104$ $=26\times 9\times \frac{2}{5}\times x$

$\Rightarrow x=\frac{30\times 8\times \frac{3}{5}\times 104}{26\times 9\times \frac{2}{5}}=\frac{30\times 8\times 3\times 104}{26\times 9\times 2}=\frac{30\times 8\times 104}{26\times 3\times 2}=\frac{30\times 8\times 4}{3\times 2}=5\times 8\times 4=160$

Number of additional persons required = 160 - 104 = 56

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Solution 2 (Using Time and Work)

Persons worked = 104
Number of hours each person worked per day = 8
Number of days they worked = 30
Work completed $=\frac{2}{5}$

Remaining days = 56 - 30 = 26
Remaining Work to be completed $=1-\frac{2}{5}=\frac{3}{5}$
Let the total number of persons who do the remaining work $=x$
Number of hours each person needs to be work per day = 9

Amount of work 1 person did in 1 day, working 1 hours a day $=\frac{\left(\frac{2}{5}\right)}{104\times 30\times 8}$
Now, the amount of work $x$ person should do in 1 day, working 1 hours a day $=\frac{\left(\frac{3}{5}\right)}{26\times 9}$

$\Rightarrow x=\frac{\frac{\left(\frac{3}{5}\right)}{26\times 9}}{\frac{\left(\frac{2}{5}\right)}{104\times 30\times 8}}=\frac{30\times 8\times \frac{3}{5}\times 104}{26\times 9\times \frac{2}{5}}=\frac{30\times 8\times 3\times 104}{26\times 9\times 2}=\frac{30\times 8\times 104}{26\times 3\times 2}=\frac{30\times 8\times 4}{3\times 2}=5\times 8\times 4=160$

Number of additional persons required = 160 - 104 = 56

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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

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

Persons worked (M₁) = 104
Number of hours each person worked per day (H₁) = 8
Number of days they worked (D₁) = 30
Work completed (W₁)$=\frac{2}{5}$

Remaining days (D₂)= 56 - 30 = 26
Remaining Work to be completed (W₂) $=1-\frac{2}{5}=\frac{3}{5}$
Let the total number of persons who do the remaining work (M₂) $=x$
Number of hours each person needs to be work per day (H₂) = 9

$\frac{M_1{D}_1{H}_1}{W_1}=\frac{M_2{D}_2{H}_2}{W_2}\Rightarrow \frac{104\times 30\times 8}{\left(\frac{2}{5}\right)}=\frac{x\times 26\times 9}{\left(\frac{3}{5}\right)}\Rightarrow \frac{104\times 30\times 8}{2}=\frac{x\times 26\times 9}{3}\Rightarrow 52\times 30\times 8=x\times 26\times 3\Rightarrow 2\times 30\times 8=3x\Rightarrow x=2\times 10\times 8=160$

Number of additional persons required = 160 - 104 = 56

13. $x$ 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. $\frac{x^2}{y^2}$ units	B. $\frac{y^3}{x^2}$ units
C. $\frac{x^3}{y^2}$ units	D. $\frac{y^2}{x^2}$ units

answer with explanation

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

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

$\begin{array}{cc}\text{(men)}& x:y\\ \text{(hours)}& x:y\\ \text{(days)}& x:y\end{array}\}::x:w$

$\Rightarrow x^{3}w=y^{3}x\Rightarrow w=\frac{y^{3}x}{x^3}=\frac{y^3}{x^2}$

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Solution 2 (Using Time and Work)

Amount of work completed by 1 man in 1 day, working 1 hours a day $=\frac{x}{x^3}=\frac{1}{x^2}$

Amount of work completed by $y$ men in $y$ days, working $y$ hours a day $=y^3\times \frac{1}{x^2}=\frac{y^3}{x^2}$

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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 $=x$
H₁ $=x$
D₁ $=x$
W₁ $=x$

M₂ $=y$
D₂ $=y$
H₂ $=y$
Let W₂= w

$\frac{M_1{D}_1{H}_1}{W_1}=\frac{M_2{D}_2{H}_2}{W_2}\Rightarrow \frac{x^3}{x}=\frac{y^3}{w}\Rightarrow x^2=\frac{y^3}{w}\Rightarrow w=\frac{y^3}{x^2}$

14. 21 goats eat as much as 15 cows. How many goats eat as much as 35 cows?
A. 49	B. 32
C. 36	D. 41

answer with explanation

Answer: Option A
Explanation:
15 cows ≡ 21 goats

1 cow ≡ $\frac{21}{15}$ goats

35 cows ≡ $\frac{21\times 35}{15}$ goats
≡ $\frac{21\times 7}{3}$ goats ≡ 7 × 7 goats ≡ 49 goats

15. 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 with explanation

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

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 : $x$

$\Rightarrow 40.25\times x=28.75\times 17.5\Rightarrow x=\frac{28.75\times 17.5}{40.25}=\frac{2875\times 175}{40250}=\frac{2875\times 7}{1610}=\frac{2875}{230}=\frac{575}{46}=12.5$

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Solution 2

shadow of length 40.25 m ≡ 17.5 m high

shadow of length 1 m ≡ $\frac{17.5}{40.25}$ m high

shadow of length 28.75 m ≡ $\frac{28.75\times 17.5}{40.25}$ m high = 12.5 m high

16. 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?
A. 1800	B. 900
C. 2500	D. 2700

answer with explanation

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

More machines, more bottles(direct proportion)
More minutes, more bottles(direct proportion)

Hence we can write as

$\begin{array}{cc}\text{(machines)}& 6:10\\ \text{(minutes)}& 1:4\end{array}\}::270:x$

$\Rightarrow 6\times 1\times x=10\times 4\times 270\Rightarrow x=\frac{10\times 4\times 270}{6}=\frac{10\times 4\times 90}{2}=10\times 4\times 45=1800$

17. A person works on a project and completes $\frac{5}{8}$ of the job in 10 days. At this rate, how many more days will he it take to finish the job?
A. 7	B. 6
C. 5	D. 4

answer with explanation

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

Number of days he worked = 10
Work completed $=\frac{5}{8}$

Let the required number of days $=x$
Remaining Work to be completed $=1-\frac{5}{8}=\frac{3}{8}$

More work, more days(direct proportion)

Hence we can write as
(Work) $\frac{5}{8}$ : $\frac{3}{8}$ :: 10 : $x$

$\Rightarrow \frac{5}{8}\times x=\frac{3}{8}\times 10\Rightarrow 5\times x=3\times 10\Rightarrow x=3\times 2=6$

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Solution 2 (Using Time and Work)

Number of days he worked = 10
Work completed $=\frac{5}{8}$

Let the required number of days $=x$
Remaining Work to be completed $=1-\frac{5}{8}=\frac{3}{8}$

Amount of work 1 person did in 1 day $=\frac{\left(\frac{5}{8}\right)}{10}=\frac{5}{80}$
Now, the amount of work 1 person should do in $x$ days $=\frac{3}{8}$

$\Rightarrow x=\frac{\left(\frac{3}{8}\right)}{\left(\frac{5}{80}\right)}=\frac{3}{8}\times \frac{80}{5}=\frac{3\times 10}{5}=6$

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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

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

Here,
M₁ = M₂
H₁ = H₂

D₁ = 10
W₁ $=\frac{5}{8}$

Let D₂ $=x$
W₂ $=1-\frac{5}{8}=\frac{3}{8}$

Hence, the equation can be written as
$\frac{D_1}{W_1}=\frac{D_2}{W_2}\Rightarrow \frac{10}{\left(\frac{5}{8}\right)}=\frac{x}{\left(\frac{3}{8}\right)}\Rightarrow \frac{10}{5}=\frac{x}{3}\Rightarrow 2=\frac{x}{3}\Rightarrow x=2\times 3=6$

18. A fort had provision of food for 150 men for 45 days. After 10 days, 25 men left the fort. Find out the number of days for which the remaining food will last.
A. 44	B. 42
C. 40	D. 38

answer with explanation

Answer: Option B
Explanation:
Given that fort had provision of food for 150 men for 45 days
Hence, after 10 days, the remaining food is sufficient for 150 men for 35 days

Remaining men after 10 days = 150 - 25 = 125
Assume that after 10 days,the remaining food is sufficient for 125 men for $x$ days

More men, Less days (Indirect Proportion)
(men) 150 : 125 :: $x$ : 35

$\Rightarrow 150\times 35=125x\Rightarrow 6\times 35=5x\Rightarrow x=6\times 7=42$

i.e., the remaining food is sufficient for 125 men for 42 days

19. 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 with explanation

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

Let the required price be $x$

More apples, More price(direct proportion)

Hence we can write as
(apples) 357 : (49 × 12) :: 1517.25 : $x$

$\Rightarrow 357x=(49\times 12)\times 1517.25\Rightarrow x=\frac{49\times 12\times 1517.25}{357}=\frac{7\times 12\times 1517.25}{51}=\frac{7\times 4\times 1517.25}{17}=7\times 4\times 89.25\approx 2500$

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Solution 2

price of 357 apples = Rs.1517.25

price of 1 apple = Rs. $\frac{1517.25}{357}$

price of 49 dozens apples = Rs.$\left(\frac{49\times 12\times 1517.25}{357}\right)\approx \text{Rs. 2500}$

20. On a scale of a map 0.6 cm represents 6.6km. If the distance between two points on the map is 80.5 cm , what is the the actual distance between these points?
A. 885.5 km	B. 860 km
C. 892.5 km	D. 825 km

answer with explanation

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

Let the required actual distance be $x$ km

More scale distance, More actual distance(direct proportion)

Hence we can write as

(scale distance) 0.6 : 80.5 :: 6.6 : $x$

$\Rightarrow 0.6x=80.5\times 6.6\Rightarrow 0.1x=80.5\times 1.1\Rightarrow x=80.5\times 11=885.5$

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Solution 2

0.6 cm in map ≡ actual distance of 6.6 km

1 cm in map ≡ $\frac{6.6}{.6}$ km

80.5 cm in map ≡ $\frac{80.5\times 6.6}{.6}$ km = 885.5 k
m