Selina Concise Solutions for ICSE Class 8 Mathematics Chapter 10 Direct and Inverse Variations

Exercise 10(A)

Question 1. In which of the following tables, x and y vary directly:
(i)

(ii)

(iii)

Answer:
For two quantities \( x \) and \( y \) to vary directly, the ratio \( \frac{x}{y} \) must remain constant for all pairs of values in the table.
(i) Let us check the ratio \( \frac{x}{y} \) for each pair:
\( \frac{x_1}{y_1} = \frac{3}{4.5} = \frac{1}{1.5} \)
\( \frac{x_2}{y_2} = \frac{5}{7.5} = \frac{1}{1.5} \)
\( \frac{x_3}{y_3} = \frac{8}{12} = \frac{1}{1.5} \)
\( \frac{x_4}{y_4} = \frac{11}{16.5} = \frac{1}{1.5} \)
Since all ratios are equal to \( \frac{1}{1.5} \), \( x \) and \( y \) vary directly in this table.
(ii) Let us check the ratio \( \frac{x}{y} \) for each pair:
\( \frac{x_1}{y_1} = \frac{16}{32} = \frac{1}{2} \)
\( \frac{x_2}{y_2} = \frac{30}{60} = \frac{1}{2} \)
\( \frac{x_3}{y_3} = \frac{40}{80} = \frac{1}{2} \)
\( \frac{x_4}{y_4} = \frac{56}{84} = \frac{2}{3} \)
Since \( \frac{1}{2} \neq \frac{2}{3} \), the ratios are not equal. Thus, \( x \) and \( y \) do not vary directly.
(iii) Let us check the ratio \( \frac{x}{y} \) for each pair:
\( \frac{x_1}{y_1} = \frac{27}{81} = \frac{1}{3} \)
\( \frac{x_2}{y_2} = \frac{45}{180} = \frac{1}{4} \)
\( \frac{x_3}{y_3} = \frac{54}{216} = \frac{1}{4} \)
\( \frac{x_4}{y_4} = \frac{75}{225} = \frac{1}{3} \)
Since the ratios are not equal, \( x \) and \( y \) do not vary directly.
In simple words: If two things grow or shrink together at the exact same rate, they vary directly. Here, only the first table has the exact same ratio throughout, so only the first set of numbers shows a direct variation.
Exam Tip: Always check every single pair of values in the table. Even if the first three pairs match, a single mismatch at the end means they do not vary directly.

Question 2. If x and y vary directly, find the values of x, y and z:

Answer:
Since \( x \) and \( y \) vary directly, the ratio of \( \frac{x}{y} \) remains constant throughout the table.
This gives:
\( \frac{3}{36} = \frac{x}{60} = \frac{y}{96} = \frac{10}{z} \)
First, simplify the known ratio:
\( \frac{3}{36} = \frac{1}{12} \)
Now, we can solve for each unknown variable:
1) Finding the value of \( x \):
\( \frac{x}{60} = \frac{1}{12} \)
\( \implies x = \frac{60}{12} = 5 \)
2) Finding the value of \( y \):
\( \frac{y}{96} = \frac{1}{12} \)
\( \implies y = \frac{96}{12} = 8 \)
3) Finding the value of \( z \):
\( \frac{10}{z} = \frac{1}{12} \)
\( \implies z = 10 \times 12 = 120 \)
Therefore, the missing values are \( x = 5 \), \( y = 8 \), and \( z = 120 \).
In simple words: When two things vary directly, their ratios are always equal. We can find the missing numbers by comparing each part to the first known ratio of 3 to 36, which is the same as 1 to 12.
Exam Tip: To avoid calculation errors, simplify the known ratio first (here, 3/36 simplifies to 1/12) before cross-multiplying to find the other values.

Question 3. A truck consumes 28 litres of diesel for moving through a distance of 448 km. How much distance will it cover in 64 litres of diesel?
Answer:
Let the distance covered in 64 litres of diesel be \( x \) km.
Since a larger amount of diesel will allow the truck to cover a greater distance, this is a case of direct variation.
We can set up the proportion:
\( \frac{\text{Diesel (litres)}}{\text{Distance (km)}} = \text{constant} \)
\( \frac{28}{448} = \frac{64}{x} \)
\( \implies 28 \times x = 64 \times 448 \)
\( \implies x = \frac{64 \times 448}{28} \)
Simplify by dividing 448 by 28:
\( 448 \div 28 = 16 \)
\( \implies x = 64 \times 16 = 1024 \text{ km} \)
So, the truck will cover a distance of 1024 km.
In simple words: More diesel means the truck can travel a longer distance. By comparing the ratios, we find that 64 litres of diesel will let the truck go 1024 km.

Exam Tip: Clearly state whether the situation is a direct or inverse variation at the beginning of your answer to earn full method marks.

Question 4. For 100 km, a taxi charges Rs. 1,800. How much will it charge for a journey of 120 km?
Answer:
Let the taxi fare for a journey of 120 km be Rs. \( x \).
As the distance increases, the charges will also increase, which represents a direct variation.
We can set up the ratio:
\( \frac{\text{Distance (km)}}{\text{Fare (Rs.)}} = \text{constant} \)
\( \frac{100}{1800} = \frac{120}{x} \)
\( \implies 100 \times x = 120 \times 1800 \)
\( \implies x = \frac{120 \times 1800}{100} \)
\( \implies x = 120 \times 18 = 2160 \)
Thus, the taxi fare for a 120 km journey is Rs. 2160.
In simple words: Since traveling a longer distance costs more money, we set up a direct ratio. Traveling 120 km will cost Rs. 2160.

Exam Tip: Ensure you include the proper unit (Rs. or km) in your final statement as neglecting units can cost minor marks.

Question 5. If 27 identical articles cost Rs. 1,890, how many articles can be bought for Rs. 1,750?
Answer:
Let the number of articles that can be bought for Rs. 1,750 be \( x \).
Fewer rupees will buy fewer articles, indicating direct variation.
We can write the proportion:
\( \frac{\text{Cost (Rs.)}}{\text{Number of articles}} = \text{constant} \)
\( \frac{1890}{27} = \frac{1750}{x} \)
First, simplify the ratio:
\( \frac{1890}{27} = 70 \)
Now solve for \( x \):
\( 70 = \frac{1750}{x} \)
\( \implies x = \frac{1750}{70} = 25 \)
Therefore, 25 articles can be purchased for Rs. 1,750.
In simple words: Since less money means we can buy fewer items, we use a direct ratio. For Rs. 1,750, we can buy exactly 25 articles.

Exam Tip: Finding the cost of one unit (unitary method, e.g., 1890 / 27 = Rs. 70 per article) is a quick way to double-check your proportion calculations.

Question 6. 7 kg of rice costs Rs. 1,120. How much rice can be bought for Rs. 3,680?
Answer:
Let \( x \) kg of rice be the quantity that can be purchased for Rs. 3,680.
Since a larger amount of money will buy more rice, this is a case of direct variation.
We can express this as:
\( \frac{\text{Quantity of rice (kg)}}{\text{Cost (Rs.)}} = \text{constant} \)
\( \frac{7}{1120} = \frac{x}{3680} \)
\( \implies 1120 \times x = 7 \times 3680 \)
\( \implies x = \frac{7 \times 3680}{1120} \)
Simplify the fraction:
\( x = \frac{3680}{160} \) (since \( 1120 \div 7 = 160 \))
\( \implies x = \frac{368}{16} = 23 \text{ kg} \)
So, 23 kg of rice can be bought for Rs. 3,680.
In simple words: With more money, we can buy more rice. Using the same rate, Rs. 3,680 will buy 23 kg of rice.

Exam Tip: Keep track of large number divisions by canceling out trailing zeros first, which makes the mental math much simpler.

Question 7. 6 note-books cost Rs. 156, find the cost of 54 such note-books.
Answer:
Let the cost of 54 notebooks be Rs. \( x \).
More notebooks will cost more money, which is a direct variation relationship.
We can set up the proportion:
\( \frac{\text{Number of notebooks}}{\text{Cost (Rs.)}} = \text{constant} \)
\( \frac{6}{156} = \frac{54}{x} \)
\( \implies 6 \times x = 54 \times 156 \)
\( \implies x = \frac{54 \times 156}{6} \)
Simplify the calculation:
\( x = 9 \times 156 \) (since \( 54 \div 6 = 9 \))
\( \implies x = 1404 \)
Thus, the cost of 54 notebooks is Rs. 1404.
In simple words: Buying more notebooks means spending more money. Since 54 notebooks is exactly 9 times more than 6 notebooks, the price will also be 9 times higher, which is Rs. 1404.

Exam Tip: Look for quick scale factors (like 54 is 9 times 6) to do the math faster and more reliably.

Question 8. 22 men can dig a 27 m long trench in one day. How many men should be employed for digging 135 m long trench of the same type in one day?
Answer:
Let the number of men required to dig a 135 m long trench in a single day be \( x \).
To dig a longer trench in the same amount of time, more men must be employed. This is a direct variation.
We can write:
\( \frac{\text{Number of men}}{\text{Length of trench (m)}} = \text{constant} \)
\( \frac{22}{27} = \frac{x}{135} \)
\( \implies 27 \times x = 22 \times 135 \)
\( \implies x = \frac{22 \times 135}{27} \)
Simplify:
\( x = 22 \times 5 \) (since \( 135 \div 27 = 5 \))
\( \implies x = 110 \)
So, 110 men should be hired to dig the trench.
In simple words: A longer trench needs more workers. Since a 135-meter trench is 5 times longer than a 27-meter one, we need 5 times as many workers, which is 110 men.

Exam Tip: Pay close attention to the time frame mentioned. Since both scenarios occur "in one day", time is constant and we only compare men and trench length.

Question 9. If the total weight of 11 identical articles is 77 kg, how many articles of the same type would weigh 224 kg?
Answer:
Let the number of articles weighing 224 kg be \( x \).
A heavier total weight means there must be more articles, which represents a direct variation.
We can write the following proportion:
\( \frac{\text{Number of articles}}{\text{Weight (kg)}} = \text{constant} \)
\( \frac{11}{77} = \frac{x}{224} \)
Simplify the left fraction:
\( \frac{11}{77} = \frac{1}{7} \)
Solve for \( x \):
\( \frac{1}{7} = \frac{x}{224} \)
\( \implies x = \frac{224}{7} \)
\( \implies x = 32 \)
Thus, 32 articles of this type would weigh 224 kg.
In simple words: More articles weigh more. Since 1 article weighs 7 kg, we can divide the total weight of 224 kg by 7 to find that there are 32 articles.

Exam Tip: Simplifying the ratio at the beginning makes the division straightforward and avoids large multiplications.

Question 10. A train is moving with uniform speed of 120 km per hour.
(i) How far will it travel in 36 minutes?
(ii) In how much time will it cover 210 km?
Answer:
First, let us express the train's speed in terms of minutes:
Speed = 120 km per hour = 120 km in 60 minutes.
(i) To find the distance covered in 36 minutes:
Since speed is constant, the distance traveled varies directly with the time taken.
\( \frac{\text{Distance (km)}}{\text{Time (minutes)}} = \text{constant} \)
\( \frac{120}{60} = \frac{\text{Distance}}{36} \)
\( \implies 2 = \frac{\text{Distance}}{36} \)
\( \implies \text{Distance} = 2 \times 36 = 72 \text{ km} \)
Therefore, the train will travel 72 km in 36 minutes.
(ii) To find the time taken to travel 210 km:
Let the time taken be \( t \) minutes.
\( \frac{\text{Distance (km)}}{\text{Time (minutes)}} = \text{constant} \)
\( \frac{120}{60} = \frac{210}{t} \)
\( \implies 2 = \frac{210}{t} \)
\( \implies t = \frac{210}{2} = 105 \text{ minutes} \)
Convert 105 minutes into hours and minutes:
105 minutes = 60 minutes + 45 minutes = 1 hour 45 minutes.
Thus, the train will take 1 hour and 45 minutes to cover a distance of 210 km.
In simple words: The train goes 2 km every single minute. So, in 36 minutes it travels 72 km, and to go 210 km it will take half as many minutes, which is 105 minutes (or 1 hour and 45 minutes).

Exam Tip: Always convert mixed units (like hours and minutes) into a single standard unit (usually minutes) before setting up your direct proportion equations.

Exercise 10(B)

Question 1. Check whether x and y vary inversely or not.
(i)

(ii)

(iii)

Answer:
For \( x \) and \( y \) to vary inversely, their product \( x \times y \) must remain constant for all pairs of values in the table.
(i) Let us find the product \( x \times y \) for each column:
\( x_1 y_1 = 4 \times 6 = 24 \)
\( x_2 y_2 = 3 \times 8 = 24 \)
\( x_3 y_3 = 12 \times 2 = 24 \)
\( x_4 y_4 = 1 \times 24 = 24 \)
Since all products are equal to 24, \( x \) and \( y \) vary inversely.
(ii) Let us calculate the product \( x \times y \) for each column:
\( x_1 y_1 = 30 \times 60 = 1800 \)
\( x_2 y_2 = 120 \times 30 = 3600 \)
\( x_3 y_3 = 60 \times 30 = 1800 \)
\( x_4 y_4 = 24 \times 75 = 1800 \)
Since the products are not all equal (\( 1800 \neq 3600 \)), \( x \) and \( y \) do not vary inversely.
(iii) Let us check the product \( x \times y \) for each column:
\( x_1 y_1 = 10 \times 90 = 900 \)
\( x_2 y_2 = 30 \times 30 = 900 \)
\( x_3 y_3 = 60 \times 20 = 1200 \)
\( x_4 y_4 = 10 \times 90 = 900 \)
Since the products are not all equal (\( 900 \neq 1200 \)), \( x \) and \( y \) do not vary inversely.
In simple words: If multiplying the two numbers in each column always gives you the exact same answer, they vary inversely. This only happens in the first table, so only that one shows an inverse variation.
Exam Tip: To prove inverse variation, check that \( x \times y = \text{constant} \) for every column. To prove direct variation, check that \( \frac{x}{y} = \text{constant} \).

Question 2. If x and y vary inversely, find the values of l, m and n :
(i)

(ii)

Answer:
Since \( x \) and \( y \) vary inversely, their product \( x \times y \) must be constant.
(i) For the first table:
The known pair is \( x = 4, y = 4 \).
Constant product \( = 4 \times 4 = 16 \).
We can set up equations for each unknown:
1) Solving for \( l \):
\( 8 \times l = 16 \)
\( \implies l = \frac{16}{8} = 2 \)
2) Solving for \( m \):
\( 2 \times m = 16 \)
\( \implies m = \frac{16}{2} = 8 \)
3) Solving for \( n \):
\( 32 \times n = 16 \)
\( \implies n = \frac{16}{32} = 0.5 \)
So, the values are \( l = 2, m = 8, n = 0.5 \).
(ii) For the second table:
The known pair is \( x = 32, y = 12 \).
Constant product \( = 32 \times 12 = 384 \).
We can set up equations for each unknown:
1) Solving for \( l \):
\( 24 \times l = 384 \)
\( \implies l = \frac{384}{24} = 16 \)
2) Solving for \( m \):
\( m \times 8 = 384 \)
\( \implies m = \frac{384}{8} = 48 \)
3) Solving for \( n \):
\( 16 \times n = 384 \)
\( \implies n = \frac{384}{16} = 24 \)
So, the values are \( l = 16, m = 48, n = 24 \).
In simple words: In an inverse variation, multiplying the top and bottom numbers in each column must always give the same total. For the first table, they must multiply to 16, and for the second table, they must multiply to 384.
Exam Tip: Clearly identify the complete column with no variables first, multiply its values to find the constant product, and use it to solve for the missing variables.

Question 3. 36 men can do a piece of work in 7 days. How many men will do the same work in 42 days?
Answer:
Let the number of men required to complete the work in 42 days be \( x \).
As the number of days allowed for the work increases, fewer men will be needed to complete it. Therefore, this is a case of inverse variation.
We can write:
\( \text{Number of men} \times \text{Number of days} = \text{constant} \)
\( 36 \times 7 = x \times 42 \)
\( \implies x = \frac{36 \times 7}{42} \)
\( \implies x = \frac{36}{6} \) (since \( \frac{7}{42} = \frac{1}{6} \))
\( \implies x = 6 \)
Thus, 6 men will be able to do the same work in 42 days.
In simple words: If we have more days to finish a job, we need fewer workers. Since we have 6 times more days (42 days instead of 7), we only need one-sixth of the workers, which is 6 men.

Exam Tip: Time and work problems are classic examples of inverse variation. Remember that more workers mean less time, and vice-versa.

Question 4. 12 pipes, all of the same size, fill a tank in 42 minutes. How long will it take to fill the same tank, if 21 pipes of the same size are used?
Answer:
Let the time taken by 21 pipes to fill the tank be \( x \) minutes.
If we use more pipes to fill the tank, it will take less time. This represents an inverse variation.
Therefore:
\( \text{Number of pipes} \times \text{Time taken} = \text{constant} \)
\( 12 \times 42 = 21 \times x \)
\( \implies x = \frac{12 \times 42}{21} \)
\( \implies x = 12 \times 2 \) (since \( 42 \div 21 = 2 \))
\( \implies x = 24 \)
Thus, 21 pipes will take 24 minutes to fill the tank.
In simple words: Using more pipes fills the tank faster, so the time goes down. By setting up an inverse relationship, we find that 21 pipes will fill the tank in 24 minutes.

Exam Tip: Ensure you perform the step-by-step division clearly so that you do not lose marks on basic arithmetic in multi-step equations.

Question 5. In a fort 150 men had provisions for 45 days. After 10 days, 25 men left the fort. How long would the food last at the same rate?
Answer:
Originally, the food was enough for 150 men for 45 days.
After 10 days, the food left would have lasted the remaining 150 men for:
\( 45 - 10 = 35 \text{ days} \)
Now, 25 men left the fort, so the number of men remaining is:
\( 150 - 25 = 125 \text{ men} \)
Let the remaining food last for \( x \) days for these 125 men.
Since there are fewer men, the food will last for a longer time. This is an inverse variation.
So:
\( \text{Remaining men} \times \text{Number of days} = \text{constant} \)
\( 150 \times 35 = 125 \times x \)
\( \implies x = \frac{150 \times 35}{125} \)
Simplify the fraction:
\( x = \frac{6 \times 35}{5} \) (dividing 150 and 125 by 25)
\( \implies x = 6 \times 7 \) (since \( 35 \div 5 = 7 \))
\( \implies x = 42 \)
Hence, the food will last for 42 days.
In simple words: After 10 days, the food would last 150 men for another 35 days. Since some men left, leaving only 125 men, the food will stretch further and last for 42 days.

Exam Tip: In "provision" style questions, always calculate the remaining days and the remaining men first before setting up the inverse proportion.

Question 6. 72 men do a piece of work in 25 days. In how many days will 30 men do the same work?
Answer:
Let the time taken by 30 men to complete the work be \( x \) days.
With fewer men working on the job, it will take more days to complete. This is an inverse variation.
We can write:
\( \text{Number of men} \times \text{Number of days} = \text{constant} \)
\( 72 \times 25 = 30 \times x \)
\( \implies x = \frac{72 \times 25}{30} \)
Simplify the fraction:
\( x = \frac{72 \times 5}{6} \) (dividing 25 and 30 by 5)
\( \implies x = 12 \times 5 \) (since \( 72 \div 6 = 12 \))
\( \implies x = 60 \)
Thus, 30 men will complete the same work in 60 days.
In simple words: Fewer workers mean a job will take longer to finish. Since the number of workers drops from 72 to 30, the time needed goes up to 60 days.

Exam Tip: Always double-check your answer's direction: for inverse variation, if one value decreases (men), the other must increase (days).

Question 7. If 56 workers can build a wall in 180 hours, how many workers will be required to do the same work in 70 hours?
Answer:
Let the number of workers required to complete the wall in 70 hours be \( x \).
To finish the building work in less time, we must employ more workers. This is an inverse variation.
So:
\( \text{Number of workers} \times \text{Time (hours)} = \text{constant} \)
\( 56 \times 180 = x \times 70 \)
\( \implies x = \frac{56 \times 180}{70} \)
Simplify the fraction:
\( x = \frac{56 \times 18}{7} \) (canceling the trailing zeros)
\( \implies x = 8 \times 18 \) (since \( 56 \div 7 = 8 \))
\( \implies x = 144 \)
Thus, 144 workers will be needed to complete the work in 70 hours.
In simple words: To build the wall faster (in 70 hours instead of 180), we need more hands. By setting up the ratio, we find we need 144 workers.

Exam Tip: Keep your calculations simple by dividing first (e.g., dividing 56 by 7) rather than performing the large multiplication first.

Question 8. A car takes 6 hours to reach a destination by travelling at the speed of 50 km per hour. How long will it take when the car travels at the speed of 75 km per hour?
Answer:
Let the time taken by the car at a speed of 75 km per hour be \( x \) hours.
If the car travels faster, it will arrive at its destination in less time. This represents an inverse variation.
Therefore:
\( \text{Speed} \times \text{Time} = \text{constant} \)
\( 50 \times 6 = 75 \times x \)
\( \implies 300 = 75 \times x \)
\( \implies x = \frac{300}{75} \)
\( \implies x = 4 \)
Thus, the car will take 4 hours to reach the destination.
In simple words: Driving faster means you get there sooner. Increasing the speed from 50 km/h to 75 km/h cuts the travel time down to 4 hours.

Exam Tip: Remember that the product of speed and time represents distance, which is constant in this problem. Verifying that the distance is 300 km (\( 50 \times 6 \)) and then dividing by the new speed (\( 300 / 75 \)) is a reliable way to solve.

Exercise 10(C)

Question 1. Cost of 24 identical articles is Rs. 108, Find the cost of 40 similar articles.
Answer:
Given:
Cost of 24 articles = Rs. 108.
First, we find the cost of 1 article:
Cost of 1 article = Rs. \( \frac{108}{24} \)
Now, find the cost of 40 such articles:
Cost of 40 articles = Rs. \( \left( \frac{108}{24} \times 40 \right) \)
Simplify the fraction:
\( = \frac{108}{3} \times 5 \) (dividing 40 and 24 by 8)
\( = 36 \times 5 = 180 \)
Thus, the cost of 40 articles is Rs. 180.
In simple words: First find how much a single item costs by dividing Rs. 108 by 24, which is Rs. 4.50. Then, multiply that price by 40 to get a total of Rs. 180.

Exam Tip: This exercise uses the unitary method. Finding the value of a single unit first is highly valued by examiners for partial credit.

Question 2. If 15 men can complete a piece of work in 30 days, in how many days will 18 men complete it?
Answer:
Given:
15 men can finish the work in 30 days.
First, let us calculate the total work in terms of man-days:
1 man can complete the work in \( 30 \times 15 \) days.
Now, find the time taken by 18 men to do the same work:
Time taken by 18 men = \( \frac{30 \times 15}{18} \text{ days} \)
Simplify the expression:
\( = \frac{5 \times 15}{3} \) (dividing 30 and 18 by 6)
\( = 5 \times 5 = 25 \text{ days} \)
Hence, 18 men can complete the work in 25 days.
In simple words: If 15 men take 30 days, one man working alone would need 450 days. If 18 men share the work, they will finish it in 25 days.

Exam Tip: For unitary method work problems, remember that 1 worker takes MORE time, so you multiply first instead of dividing.

Question 3. In order to complete a work in 28 days, 60 men are required. How many men will be required if the same work is to be completed in 40 days ?
Answer:
Given:
60 men are needed to complete the work in 28 days.
First, we find the time 1 man would take:
Time taken by 1 man = \( 28 \times 60 \) days.
To complete the work in 40 days, let the required number of men be \( x \).
\( x = \frac{28 \times 60}{40} \)
Simplify the fraction:
\( x = \frac{28 \times 6}{4} \) (canceling the trailing zeros)
\( x = 7 \times 6 \) (since \( 28 \div 4 = 7 \))
\( x = 42 \)
So, 42 men will be required to finish the work in 40 days.
In simple words: To do the work over a longer period of 40 days, we do not need as many workers. By setting up the ratio, we find we only need 42 men.

Exam Tip: Always clearly label your variables and steps so the examiner can follow your logic easily.

Question 4. A fort had provisions for 450 soldiers for 40 days. After 10 days, 90 more soldiers come to the fort. Find in how many days will the remaining provisions last at the same rate ?
Answer:
Given:
Initially, provisions are sufficient for 450 soldiers for 40 days.
After 10 days, the food left would last the original 450 soldiers for:
\( 40 - 10 = 30 \text{ days} \)
For 1 soldier, these remaining provisions would last:
\( 30 \times 450 \text{ days} \)
The number of soldiers increases by 90, so the total is:
\( 450 + 90 = 540 \text{ soldiers} \)
Now, let us calculate how long the remaining food will last for 540 soldiers:
Number of days \( = \frac{30 \times 450}{540} \)
Simplify this:
\( = \frac{30 \times 50}{60} \) (dividing both 450 and 540 by 9)
\( = \frac{1500}{60} = 25 \text{ days} \)
Therefore, the remaining provisions will last for 25 days.
In simple words: After 10 days, there is enough food for 450 men to eat for 30 days. Since 90 more men join, making a total of 540 men, the food gets eaten faster and only lasts for 25 days.

Exam Tip: Remember that food provisions vary inversely with the number of soldiers. More mouths to feed means the food lasts fewer days.

Question 5. A garrison has sufficient provisions for 480 men for 12 days. If the number of men is reduced by 160; find how long will the provisions last.
Answer:
Given:
A garrison has provisions for 480 men for 12 days.
First, find the duration the provisions would last for 1 man:
Duration for 1 man = \( 480 \times 12 \) days.
The number of men is reduced by 160:
New number of men = \( 480 - 160 = 320 \text{ men} \)
Now, calculate how long the food will last for these 320 men:
Duration for 320 men = \( \frac{480 \times 12}{320} \text{ days} \)
Simplify this calculation:
\( = \frac{48 \times 12}{32} \) (canceling the trailing zeros)
\( = \frac{3 \times 12}{2} \) (dividing 48 and 32 by 16)
\( = \frac{36}{2} = 18 \text{ days} \)
Therefore, the provisions will last for 18 days.
In simple words: If there are fewer people to feed, the food will last longer. Reducing the group from 480 men to 320 men means the food will stretch from 12 days to 18 days.

Exam Tip: Always perform a quick sanity check: if the number of men decreases slightly, the food duration should increase moderately, not grow to an impossible number like 180 days.

Question 6. \( \frac{3}{5} \) quintal of wheat costs Rs. 210. Find the cost of :
(i) 1 quintal of wheat
(ii) 0.4 quintal of wheat
Answer:
Given:
Cost of \( \frac{3}{5} \) quintal of wheat = Rs. 210.
(i) To find the cost of 1 quintal of wheat:
Cost of 1 quintal = Rs. \( \left( 210 \times \frac{5}{3} \right) \)
\( = 70 \times 5 = \text{Rs. } 350 \)
So, 1 quintal of wheat costs Rs. 350.
(ii) To find the cost of 0.4 quintal of wheat:
Using the cost of 1 quintal:
Cost of 0.4 quintal = Rs. \( (350 \times 0.4) \)
\( = 35 \times 4 = \text{Rs. } 140 \)
So, 0.4 quintal of wheat costs Rs. 140.
In simple words: Since three-fifths of a quintal costs Rs. 210, a full quintal costs Rs. 350. Using that price, a smaller amount like 0.4 quintal costs Rs. 140.

Exam Tip: Calculating the cost of a single unit first (unitary method) makes finding the cost of any other fractional amount simple and reliable.

Question 7. If \( \frac{2}{9} \) of a property costs Rs. 2,52,000; find the cost of \( \frac{4}{7} \) of it.
Answer:
Given:
Cost of \( \frac{2}{9} \) of the property = Rs. 2,52,000.
First, let's find the cost of the entire property (1 whole):
Cost of full property = Rs. \( \left( 2,52,000 \times \frac{9}{2} \right) \)
\( = 1,26,000 \times 9 = \text{Rs. } 11,34,000 \)
Now, find the cost of \( \frac{4}{7} \) of this property:
Cost of \( \frac{4}{7} \) of property = Rs. \( \left( 11,34,000 \times \frac{4}{7} \right) \)
\( = 1,62,000 \times 4 = \text{Rs. } 6,48,000 \)
Therefore, the cost of \( \frac{4}{7} \) of the property is Rs. 6,48,000.
In simple words: If two-ninths of the land costs Rs. 2,52,000, then the whole property is worth Rs. 11,34,000. To find the price of four-sevenths of it, we multiply that total value by four-seventves to get Rs. 6,48,000.

Exam Tip: Keep your calculations neat and organized. Dividing the large number first before multiplying helps prevent arithmetic slip-ups.

Question 8. 4 men or 6 women earn Rs. 360 in one day. Find, how much will:
(i) a man earn in one day ?
(ii) a woman earn in one day ?
(iii) 6 men and 4 women earn in one day ?
Answer:
Given:
In one day, either 4 men or 6 women can earn Rs. 360.
(i) To find the daily earnings of 1 man:
Since 4 men earn Rs. 360:
Earnings of 1 man = Rs. \( \frac{360}{4} = \text{Rs. } 90 \)
Thus, a man earns Rs. 90 in one day.
(ii) To find the daily earnings of 1 woman:
Since 6 women earn Rs. 360:
Earnings of 1 woman = Rs. \( \frac{360}{6} = \text{Rs. } 60 \)
Thus, a woman earns Rs. 60 in one day.
(iii) To find the daily earnings of 6 men and 4 women together:
Using the individual rates calculated above:
Earnings of 6 men = \( 6 \times 90 = \text{Rs. } 540 \)
Earnings of 4 women = \( 4 \times 60 = \text{Rs. } 240 \)
Total combined earnings = \( 540 + 240 = \text{Rs. } 780 \)
So, they will earn Rs. 780 in one day.
In simple words: 1 man earns Rs. 90 a day, and 1 woman earns Rs. 60 a day. When 6 men and 4 women work together, their combined daily earnings total Rs. 780.

Exam Tip: The word "or" indicates that the earnings for men and women are independent, which lets us find their separate daily rates first before adding them together for the "and" part.

Question 9. 16 boys went to canteen to have tea and snacks together. The bill amounted to Rs. 114.40. What will be the contribution of a boy who pays for himself and 5 others ?
Answer:
Total cost for 16 boys = Rs. 114.40
Cost for 1 boy = Rs. \( \frac{114.40}{16} \) = Rs. 7.15
The boy is paying for himself and 5 other friends, which makes 6 boys in total.
Therefore, the amount he needs to pay = \( 6 \times \text{Rs. } 7.15 \) = Rs. 42.90.
In simple words: First, find out the bill for just one boy by dividing the total amount. Since one boy pays for 6 people, multiply that single share by 6 to get the final answer.

Exam Tip: Remember to read the phrase "himself and 5 others" carefully. It means you must multiply by 6, not 5.

Question 10. 50 labourers can dig a pond in 16 days. How many labourers will be required to dig an another pond, double in size in 20 days ?
Answer:
Number of workers needed to dig the pond in 16 days = 50
Number of workers needed to dig the pond in 1 day = \( 50 \times 16 \)
Number of workers needed to dig the pond in 20 days = \( \frac{50 \times 16}{20} \)
Since the new pond is twice as large, the work is doubled.
So, workers needed to dig the double-sized pond in 20 days = \( \frac{50 \times 16 \times 2}{20} = 5 \times 8 \times 2 = 80 \)
Hence, 80 labourers are needed.
In simple words: If you have 20 days to dig a pond twice as big, you will need 80 workers. We find this by multiplying the initial workers and days, dividing by the new days, and then doubling the result.

Exam Tip: When the size of the work doubles, remember to multiply the final expression by 2 to account for the extra work.

Question 11. If 12 men or 18 women can complete a piece of work in 7 days, in how many days can 4 men and 8 women complete the same work?
Answer:
We are given that the work of 12 men is equal to the work of 18 women.
Therefore, the work of 1 man = \( \frac{18}{12} \) women
Work of 4 men = \( \frac{18}{12} \times 4 \) = 6 women
Now, the total group consists of 4 men and 8 women.
In terms of women, this is equivalent to: \( 6 \text{ women} + 8 \text{ women} = 14 \text{ women} \)
We know that 18 women can finish the work in 7 days.
Let 14 women take \( x \) days to finish the same work.
Using inverse proportion (fewer women means more days):
\( 18 : 14 = x : 7 \)

\( \implies x = \frac{18 \times 7}{14} = 9 \) days.
Therefore, the work will be completed in 9 days.
In simple words: First, convert the men into an equivalent number of women. Then, use inverse proportion to find how many days this combined group of women will take.

Exam Tip: Always convert either all men to women or all women to men first. This makes it much easier to solve using the unitary method.

Question 12. If 3 men or 6 boys can finish a work in 20 days, how long will 4 men and 12 boys take to finish the same work ?
Answer:
We are given that the work of 3 men is equal to the work of 6 boys.
So, the work of 1 man = \( \frac{6}{3} \) boys = 2 boys
Work of 4 men = \( 4 \times 2 \) = 8 boys
In the second case, we have 4 men and 12 boys.
Total equivalent boys = \( 8 \text{ boys} + 12 \text{ boys} = 20 \text{ boys} \)
We know that 6 boys can finish the work in 20 days.
Let 20 boys complete the work in \( x \) days.
Using inverse proportion (more boys means fewer days):
\( 6 : 20 = x : 20 \)

\( \implies x = \frac{20 \times 6}{20} = 6 \) days.
Hence, they will complete the work in 6 days.
In simple words: Change the men into boys so that you only have boys to calculate. Then, use inverse proportion to find the number of days needed.

Exam Tip: Be careful with the word "or" in the first line. It means 3 men alone or 6 boys alone can do the work in 20 days.

Question 13. A particular work can be completed by 6 men and 6 women in 24 days; whereas the same work can be completed by 8 men and 12 women in 15 days. Find :
(i) according to the amount of work done, one man is equivalent to how many women.
(ii) the time taken by 4 men and 6 women to complete the same work.
Answer:
(i) Let the daily work of a man be \( M \) and that of a woman be \( W \).
Given that 6 men and 6 women can complete the work in 24 days.
To complete the work in 1 day, we would need:
\( 24 \times (6M + 6W) = 144M + 144W \)
Also, 8 men and 12 women can finish the same work in 15 days.
To complete this work in 1 day, we would need:
\( 15 \times (8M + 12W) = 120M + 180W \)
Equating the work of both groups since both finish the same work in 1 day:
\( 144M + 144W = 120M + 180W \)

\( \implies 144M - 120M = 180W - 144W \)

\( \implies 24M = 36W \)

\( \implies M = \frac{36}{24}W = \frac{3}{2}W = 1\frac{1}{2}W \)
Therefore, one man is equivalent to \( 1\frac{1}{2} \) women.

(ii) Let us express both cases in terms of women.
First case:
\( 6 \text{ men} + 6 \text{ women} = 6 \times \frac{3}{2} \text{ women} + 6 \text{ women} = 9 \text{ women} + 6 \text{ women} = 15 \text{ women} \)
So, 15 women can do the work in 24 days.
Second case (target group):
\( 4 \text{ men} + 6 \text{ women} = 4 \times \frac{3}{2} \text{ women} + 6 \text{ women} = 6 \text{ women} + 6 \text{ women} = 12 \text{ women} \)
Using the unitary method:
1 woman can complete the work in \( 24 \times 15 \) days.
Therefore, 12 women can complete the same work in:
\( \frac{24 \times 15}{12} = 2 \times 15 = 30 \) days.
Thus, 4 men and 6 women will take 30 days to finish the work.
In simple words: First, find the relation between a man's work and a woman's work by equating 1 day's total effort. Then convert all workers into women to find the final days using the unitary method.

Exam Tip: Be neat when equating the 1-day work equations, as a small calculation error here will throw off the rest of the steps.

Question 14. If 12 men and 16 boys can do a piece of work in 5 days and, 13 men and 24 boys can do it in 4 days, how long will 7 men and 10 boys take to do it ?
Answer:
Let the daily work of a man be \( M \) and that of a boy be \( B \).
We are given that 12 men and 16 boys can complete the work in 5 days.
To complete this work in 1 day, we would need:
\( 5 \times (12M + 16B) = 60M + 80B \) ... (i)
Also, 13 men and 24 boys can finish the same work in 4 days.
To complete this work in 1 day, we would need:
\( 4 \times (13M + 24B) = 52M + 96B \) ... (ii)
Equating both expressions for 1 day's work:
\( 60M + 80B = 52M + 96B \)

\( \implies 60M - 52M = 96B - 80B \)

\( \implies 8M = 16B \)

\( \implies M = 2B \)
Thus, the work of 1 man is equivalent to the work of 2 boys.
Now, let us convert the groups into equivalent numbers of boys.
First group:
\( 12 \text{ men} + 16 \text{ boys} = 12 \times 2 \text{ boys} + 16 \text{ boys} = 24 + 16 = 40 \text{ boys} \)
So, 40 boys can complete the work in 5 days.
Target group:
\( 7 \text{ men} + 10 \text{ boys} = 7 \times 2 \text{ boys} + 10 \text{ boys} = 14 + 10 = 24 \text{ boys} \)
Using the unitary method:
1 boy can complete the work in \( 5 \times 40 \) days.
Therefore, 24 boys will complete the same work in:
\( \frac{5 \times 40}{24} = \frac{200}{24} = \frac{25}{3} = 8\frac{1}{3} \) days.
Hence, 7 men and 10 boys will take \( 8\frac{1}{3} \) days to complete the work.
In simple words: Equate the total work done in 1 day by both groups to find that 1 man does the work of 2 boys. Then convert all workers into boys and calculate the time.

Exam Tip: Expressing the final answer as a mixed fraction like \( 8\frac{1}{3} \) days is standard practice and keeps your solution precise.

Exercise 10(D)

Question 1. Eight oranges can be bought for Rs. 10.40. How many more can be bought for Rs. 16.90?
Answer:
Number of oranges that can be bought for Rs. 10.40 = 8
Number of oranges that can be bought for Re. 1 = \( \frac{8}{10.40} \)
Number of oranges that can be bought for Rs. 16.90 = \( \frac{8}{10.40} \times 16.90 = \frac{8 \times 1690}{1040} = \frac{13520}{1040} = 13 \)
Therefore, the number of additional oranges that can be bought = \( 13 - 8 = 5 \).
In simple words: Calculate how many oranges you get for Rs. 16.90 using the unitary method, then subtract the initial 8 oranges to see how many more you can buy.

Exam Tip: Always make sure to answer the specific question asked. The question asks for "how many more", so do not forget the final subtraction step.

Question 2. Fifteen men can build a wall in 60 days. How many more men are required to build another wall of same size in 45 days ?
Answer:
To build the wall in 60 days, we need = 15 men
To build the same wall in 1 day, we need = \( 15 \times 60 \) men
To build the wall in 45 days, the number of men required = \( \frac{15 \times 60}{45} = \frac{900}{45} = 20 \) men
Thus, the number of additional men required = \( 20 - 15 = 5 \) men.
In simple words: Find the total number of men needed to finish the work in 45 days. Then, subtract the 15 men we already have to find how many more are needed.

Exam Tip: Be careful with inverse proportion here: as the number of days decreases, the number of men required increases.

Question 3. Six taps can fill an empty cistern in 8 hours. How much more time will be taken, if two taps go out of order ? Assume, all the taps supply water at the same rate.
Answer:
Total number of taps = 6
Taps out of order = 2
Working taps remaining = \( 6 - 2 = 4 \)
Time taken by 6 taps to fill the cistern = 8 hours
Time taken by 1 tap to fill the cistern = \( 6 \times 8 = 48 \) hours
Time taken by 4 taps to fill the cistern = \( \frac{48}{4} = 12 \) hours
Therefore, the extra time taken = \( 12 - 8 = 4 \) hours.
In simple words: Find out how long a single tap would take to fill the tank. Use that to find the time for 4 working taps, then subtract the original time.

Exam Tip: Make sure to subtract the original time (8 hours) from the new time (12 hours) to find the "more time taken".

Question 4. A contractor undertakes to dig a canal, 6 kilometres long, in 35 days and employed 90 men. He finds that after 20 days only 2 km of canal have been completed. How many more men must be employed to finish the work in time ?
Answer:
Total length of the canal = 6 km
Length completed in the first 20 days = 2 km
Remaining length of canal to be dug = \( 6 - 2 = 4 \) km
Total time allowed = 35 days
Remaining time left = \( 35 - 20 = 15 \) days
In 20 days, to dig 2 km of canal, the number of men working = 90
In 1 day, to dig 2 km of canal, the number of men needed = \( 90 \times 20 \)
In 15 days, to dig 2 km of canal, the number of men needed = \( \frac{90 \times 20}{15} \)
In 15 days, to dig 1 km of canal, the number of men needed = \( \frac{90 \times 20}{15 \times 2} \)
In 15 days, to dig the remaining 4 km of canal, the number of men needed = \( \frac{90 \times 20 \times 4}{15 \times 2} = 6 \times 10 \times 4 = 240 \) men
Therefore, the number of additional men to be employed = \( 240 - 90 = 150 \) men.
In simple words: Find the length of the canal left to dig and the remaining days. Calculate how many men are needed to complete this portion in the remaining time, and subtract the current 90 men.

Exam Tip: Break down the compound unitary method steps carefully by dealing with days first, and then with the length of the canal.

Question 5. If 10 horses consume 18 bushels in 36 days. How long will 24 bushels last for 30 horses ?
Answer:
Given that 10 horses consume 18 bushels in 36 days.
So, 1 horse consumes 18 bushels in = \( 36 \times 10 = 360 \) days
Then, 30 horses consume 18 bushels in = \( \frac{360}{30} = 12 \) days
Therefore, 30 horses consume 1 bushel in = \( \frac{12}{18} \) days
Thus, 30 horses consume 24 bushels in = \( \frac{12}{18} \times 24 = \frac{2}{3} \times 24 = 16 \) days.
In simple words: Find the number of days the food lasts for one horse, adjust for 30 horses, and then scale the days according to the new quantity of bushels.

Exam Tip: Carefully identify whether each relationship is direct or inverse. More horses means fewer days, while more bushels means more days.

Question 6. A family of 5 persons can be maintained for 20 days with Rs.2,480. Find, how long Rs.6,944 maintain a family of 8 persons
Answer:
For a family of 5 persons, Rs. 2,480 lasts for = 20 days
So, with Re. 1, a family of 5 persons can be maintained for = \( \frac{20}{2480} \) days
With Rs. 6,944, a family of 5 persons can be maintained for = \( \frac{20}{2480} \times 6,944 = \frac{1}{124} \times 6,944 = 56 \) days
For 1 person, Rs. 6,944 can maintain them for = \( 56 \times 5 \) days
Therefore, for a family of 8 persons, Rs. 6,944 will last for = \( \frac{56 \times 5}{8} = 7 \times 5 = 35 \) days.
In simple words: First, calculate how long the new amount of money lasts for the same family of 5. Then, adjust that time for a larger family of 8.

Exam Tip: Be systematic when changing one variable at a time (first the money, then the number of family members) to prevent confusion.

Question 7. 90 men can complete a work in 24 days working 8 hours a day. How many men are required to complete the same work in 18 days working 7 1/2 hours a day ?
Answer:
To complete the work in 24 days, working 8 hours daily, we need = 90 men
To complete the work in 1 day, working 8 hours daily, we need = \( 90 \times 24 \) men
To complete the work in 18 days, working 8 hours daily, we need = \( \frac{90 \times 24}{18} \) men
To complete the work in 18 days, working 1 hour daily, we need = \( \frac{90 \times 24}{18} \times 8 \) men
Given working hours per day in the second case = \( 7\frac{1}{2} \) hours = \( \frac{15}{2} \) hours
Therefore, to complete the work in 18 days, working \( \frac{15}{2} \) hours daily, we need:
\( \frac{90 \times 24 \times 8}{18} \times \frac{2}{15} = \frac{90 \times 24 \times 8 \times 2}{18 \times 15} = 128 \) men.
Hence, 128 men are required.
In simple words: Find how many men are needed by breaking the problem down step-by-step for days and daily working hours. Less working hours per day means you will need more men.

Exam Tip: Convert the mixed fraction \( 7\frac{1}{2} \) to an improper fraction \( \frac{15}{2} \) before doing the calculations to simplify the division.

Question 8. Twelve typists, all working with same speed, type a certain number of pages in 18 days working 8 hours a day. Find, how many hours per day must sixteen typists work in order to type the same number of pages in 9 days ?
Answer:
For 12 typists to type the pages in 18 days, hours needed per day = 8 hours
For 1 typist to type the pages in 18 days, hours needed per day = \( 8 \times 12 \) hours
For 1 typist to type the pages in 9 days, hours needed per day = \( \frac{18}{9} \times 8 \times 12 = 2 \times 8 \times 12 \) hours
Therefore, for 16 typists to type the pages in 9 days, hours needed per day = \( \frac{2 \times 8 \times 12}{16} = 12 \) hours.
Hence, they must work 12 hours a day.
In simple words: Calculate the total typist-hours required for the task. Then, divide this total by the new number of typists and the new number of days.

Exam Tip: Be sure to write down each step clearly. In inverse variation, as the number of typists increases, the required hours per day will decrease.

Question 9. If 25 horses consume 18 quintal in 36 days, how long will 28 quintal last for 30 horses ?
Answer:
Time 18 quintals last for 25 horses = 36 days
Time 1 quintal lasts for 25 horses = \( \frac{36}{18} = 2 \) days
Time 1 quintal lasts for 1 horse = \( 2 \times 25 = 50 \) days
Time 1 quintal lasts for 30 horses = \( \frac{50}{30} = \frac{5}{3} \) days
Therefore, time 28 quintals last for 30 horses = \( \frac{5}{3} \times 28 = \frac{140}{3} = 46\frac{2}{3} \) days.
Hence, the food will last for \( 46\frac{2}{3} \) days.
In simple words: Find how long one quintal of food lasts for one horse. Use this value to calculate how long 28 quintals will last for 30 horses.

Exam Tip: Carefully identify whether each relationship is direct or inverse. More food (quintals) means more days, but more horses means fewer days.

Question 10. If 70 men dig 15,000 sq. m of a field in 5 days, how many men will dig 22,500 sq. m field in 25 days ?
Answer:
To dig 15,000 sq. m in 5 days, we need = 70 men
To dig 15,000 sq. m in 1 day, we need = \( 70 \times 5 \) men
To dig 1 sq. m in 1 day, we need = \( \frac{70 \times 5}{15000} \) men
Therefore, to dig 22,500 sq. m in 25 days, the number of men required:
\( \frac{70 \times 5 \times 22500}{15000 \times 25} = \frac{70 \times 900}{3000} = \frac{63}{3} = 21 \) men.
Hence, 21 men are required.
In simple words: Find the work rate of one man. Then use that rate to find how many men are needed for the larger area over 25 days.

Exam Tip: Simplify the fraction by cancelling out common factors like 100 or 1000 before multiplying the larger numbers.

Question 11. A contractor undertakes to build a wall 1000 m long in 50 days. He employs 56 men, but at the end of 27 days, he finds that only 448 m of wall is built. How many extra men must the contractor employ so that the wall is completed in time ?
Answer:
Total length of the wall to build = 1000 m
Initial number of men employed = 56 men
Length completed in the first 27 days = 448 m
Remaining length of wall to build = \( 1000 - 448 = 552 \) m
Remaining days to finish the wall = \( 50 - 27 = 23 \) days
To build 448 m wall in 27 days, we need = 56 men
To build 448 m wall in 1 day, we need = \( 56 \times 27 \) men
To build 1 m wall in 1 day, we need = \( \frac{56 \times 27}{448} \) men
Therefore, to build 552 m wall in 23 days, the total men required:
\( \frac{56 \times 27 \times 552}{448 \times 23} = \frac{1 \times 27 \times 24}{8} = 81 \) men.
So, the number of extra men to employ = \( 81 - 56 = 25 \) men.
In simple words: Work out the remaining wall length and the remaining days. Find how many total workers are needed to finish this portion, and then subtract the 56 workers already hired.

Exam Tip: Be very careful when calculating the remaining work and remaining days, as these form the basis of the entire solution.

Question 12. A group of labourers promises to do a piece of work in 10 days, but five of them become absent. If the remaining labourers complete the work in 12 days, find their original number in the group.
Answer:
Let the original number of labourers be \( x \).
Initially, \( x \) labourers were supposed to finish the work in 10 days.
Since 5 labourers were absent, the actual number of labourers is \( x - 5 \).
These remaining labourers finished the work in 12 days.
Since the number of workers and the days they take are in inverse variation:
\( 10 \times x = 12 \times (x - 5) \)

\( \implies 10x = 12x - 60 \)

\( \implies 12x - 10x = 60 \)

\( \implies 2x = 60 \)

\( \implies x = 30 \)
Therefore, the original number of labourers in the group was 30.
In simple words: Set up an equation using the rule that the number of workers multiplied by the days they work remains constant. Solve for the starting number of workers.

Exam Tip: Always define your variable \( x \) clearly at the start so the examiner knows what you are solving for.

Question 13. Ten men, working for 6 days of 10 hours each, finish 5/21 of a piece of work. How many men working at the same rate and for the same number of hours each day, will be required to complete the remaining work in 8 days ?
Answer:
Work completed = \( \frac{5}{21} \)
Remaining work to complete = \( 1 - \frac{5}{21} = \frac{16}{21} \)
To complete \( \frac{5}{21} \) of the work in 6 days, working 10 hours daily, we need = 10 men
To complete the whole work in 6 days, working 10 hours daily, we need = \( \frac{10 \times 21}{5} \) men
To complete the whole work in 1 day, working 10 hours daily, we need = \( \frac{10 \times 21 \times 6}{5} \) men
Therefore, to complete \( \frac{16}{21} \) of the work in 8 days, working 10 hours daily, we need:
\( \frac{10 \times 21 \times 6 \times 16}{5 \times 21 \times 8} = 24 \) men.
Hence, 24 men are required.
In simple words: Determine the remaining fraction of work. Use the unitary method to find how many men are needed to complete this remaining work in 8 days.

Exam Tip: When using fractions in unitary method equations, you can cancel out the common denominators (like 21) to make the calculations easier.

Exercise 10(E)

Question 1. A can do a piece of work in 10 days and B in 15 days. How long will they take together to finish it ?
Answer:
Work completed by A in 1 day = \( \frac{1}{10} \)
Work completed by B in 1 day = \( \frac{1}{15} \)
Work completed by both A and B in 1 day:
\( \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6} \)
Therefore, working together, they will complete the work in 6 days.
In simple words: Find the fraction of work each person does in one day, add them together to get their combined daily work, and flip that fraction to find the total days.

Exam Tip: Remember that the total days taken is always the reciprocal of the work done in one day.

Question 2. A and B together can do a piece of work in 6 2/3 days ; but B alone can do it in 10 days. How long will A take to do it alone ?
Answer:
Time taken by A and B together = \( 6\frac{2}{3} = \frac{20}{3} \) days
Combined work of A and B in 1 day = \( \frac{3}{20} \)
Time taken by B alone = 10 days
Work of B in 1 day = \( \frac{1}{10} \)
Therefore, work of A alone in 1 day:
\( \frac{3}{20} - \frac{1}{10} = \frac{3 - 2}{20} = \frac{1}{20} \)
Thus, A will take 20 days to complete the work alone.
In simple words: Subtract B's daily work fraction from the combined daily work fraction of A and B. This gives A's daily work fraction, which we flip to get the total days.

Exam Tip: Be careful when writing the 1-day work of a group whose total days is a fraction like \( \frac{20}{3} \). Its reciprocal is \( \frac{3}{20} \).

Question 3. A can do a work in 15 days and B in 20 days. If they together work on it for 4 days ; what fraction of the work will be left ?
Answer:
Work completed by A in 1 day = \( \frac{1}{15} \)
Work completed by B in 1 day = \( \frac{1}{20} \)
Combined work of both A and B in 1 day:
\( \frac{1}{15} + \frac{1}{20} = \frac{4 + 3}{60} = \frac{7}{60} \)
Combined work of both in 4 days:
\( \frac{7}{60} \times 4 = \frac{7}{15} \)
Therefore, the fraction of work left:
\( 1 - \frac{7}{15} = \frac{8}{15} \).
In simple words: Find how much of the work they complete together in 4 days. Subtract this fraction from 1 to find the remaining work.

Exam Tip: The total work is always considered as 1. To find what is left, subtract the completed fraction from 1.

Question 4. A, B and C can do a piece of work in 6 days, 12 days and 24 days respectively. In what time will they all together do it ?
Answer:
Work done by A in 1 day = \( \frac{1}{6} \)
Work done by B in 1 day = \( \frac{1}{12} \)
Work done by C in 1 day = \( \frac{1}{24} \)
Combined work done by A, B, and C in 1 day:
\( \frac{1}{6} + \frac{1}{12} + \frac{1}{24} = \frac{4 + 2 + 1}{24} = \frac{7}{24} \)
Therefore, working together, they will finish the work in:
\( \frac{24}{7} = 3\frac{3}{7} \) days.
In simple words: Add the daily work fractions of all three people to find their combined rate per day. Flip this combined fraction to get the total days.

Exam Tip: Express your final answer as a mixed fraction like \( 3\frac{3}{7} \) days, as it is more precise than a decimal approximation.

Question 5. A and B working together can mow a field in 56 days and with the help of C, they could have mowed it in 42 days. How long would C take by himself ?
Answer:
A and B together can finish mowing in 56 days. With C's help, A, B, and C together can complete it in 42 days. Amount of work A and B do in a single day = \( \frac{1}{56} \) Amount of work A, B, and C do in a single day = \( \frac{1}{42} \) Work completed by C in 1 day = \( \frac{1}{42} - \frac{1}{56} \) To subtract these, we calculate the Least Common Multiple (LCM) of 42 and 56. The prime factorization gives LCM \( = 2 \times 7 \times 3 \times 4 = 168 \). So, C's 1-day work = \( \frac{4 - 3}{168} = \frac{1}{168} \) Therefore, C will take 168 days to complete the task alone.
In simple words: First, find how much work they all do in one day. Subtract what A and B do from what A, B, and C do together. This tells you C's daily work, which you flip to get the total days.

Exam Tip: When finding the LCM, show your calculation steps clearly so that the examiner can easily see your work and give you full marks.

Question 6. A can do a piece of work in 24 days, A and B can do it in 16 days and A, B and C in \( 10\frac{2}{3} \) days. In how many days can A and C do it ?
Answer:
Time taken by A alone to do the job = 24 days Time taken by A and B working together = 16 days Time taken by A, B, and C working together = \( 10\frac{2}{3} \) days = \( \frac{32}{3} \) days Now, let us find the 1-day work for each group: A's 1-day work = \( \frac{1}{24} \) Combined 1-day work of A and B = \( \frac{1}{16} \) Combined 1-day work of A, B, and C = \( \frac{3}{32} \) We can find C's 1-day work by subtracting (A+B)'s work from (A+B+C)'s work: C's 1-day work = \( \frac{3}{32} - \frac{1}{16} = \frac{3 - 2}{32} = \frac{1}{32} \) To find the work done by A and C together in 1 day: Combined 1-day work of A and C = A's 1-day work + C's 1-day work Combined 1-day work of A and C = \( \frac{1}{24} + \frac{1}{32} = \frac{4 + 3}{96} = \frac{7}{96} \) So, the time taken by A and C together to finish the work = \( \frac{96}{7} \) days = \( 13\frac{5}{7} \) days.
In simple words: Find out how much work C does in one day by subtracting the work of A and B from the total work of all three. Then, add A's daily work and C's daily work to find their combined speed. Finally, flip this fraction to get the total days.

Exam Tip: Be careful when dealing with mixed fractions like \( 10\frac{2}{3} \). Convert them into improper fractions first before finding their reciprocal for 1-day work.

Question 7. A can do a piece of work in 20 days and B in 15 days. They worked together on it for 6 days and then A left. How long will B take to finish the remaining work ?
Answer:
A can finish the task in = 20 days B can finish the task in = 15 days Work completed by A in 1 day = \( \frac{1}{20} \) Work completed by B in 1 day = \( \frac{1}{15} \) Combined work of A and B in 1 day = \( \frac{1}{20} + \frac{1}{15} \) LCM of 20 and 15 is 60. Combined 1-day work = \( \frac{3 + 4}{60} = \frac{7}{60} \) Since they worked together for 6 days: Work completed in 6 days = \( \frac{7}{60} \times 6 = \frac{7}{10} \) Let us find the remaining portion of work: Remaining work = \( 1 - \frac{7}{10} = \frac{10 - 7}{10} = \frac{3}{10} \) Now, B alone will complete this remaining portion. B completes the entire task in = 15 days Time taken by B to do \( \frac{3}{10} \) of the work = \( 15 \times \frac{3}{10} \) days This equals \( \frac{45}{10} \) days = \( \frac{9}{2} \) days = \( 4\frac{1}{2} \) days.
In simple words: Find how much they both can do in one day and multiply it by 6 days. Subtract this from 1 to see what work is left. Then, multiply this remaining fraction by the total days B needs to do the whole job.

Exam Tip: Always remember to subtract the completed work from 1 to find the remaining work before calculating the individual time taken by the remaining person.

Question 8. A can finish a piece of work in 15 days and B can do it in 10 days. They worked together for 2 days and then B goes away. In how many days will A finish the remaining work?
Answer:
Time taken by A to complete the work = 15 days Time taken by B to complete the work = 10 days Work done by A in 1 day = \( \frac{1}{15} \) Work done by B in 1 day = \( \frac{1}{10} \) Work done by both in 1 day = \( \frac{1}{15} + \frac{1}{10} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6} \) They worked together for a period of 2 days. Work done by both in 2 days = \( \frac{1}{6} \times 2 = \frac{1}{3} \ ") Remaining fraction of work left = \( 1 - \frac{1}{3} = \frac{2}{3} \) This leftover work will be completed by A alone. A completes the entire work in = 15 days Time required by A to finish \( \frac{2}{3} \) of the work = \( 15 \times \frac{2}{3} \) days = 10 days.
In simple words: Find their combined work for 2 days. Subtract this from 1 to find the fraction of work left. Multiply this leftover fraction by the number of days A takes to do the full job alone.

Exam Tip: Make sure to simplify fractions like \( \frac{5}{30} \) to \( \frac{1}{6} \) early on, as this makes your later calculations much simpler and prevents mistakes.

Question 9. A can do a piece of work in 10 days ; B in 18 days; and A, B and C together in 4 days. In what time would C alone do it ?
Answer:
Days taken by A alone = 10 days Days taken by B alone = 18 days Days taken by A, B, and C working together = 4 days A's 1-day work = \( \frac{1}{10} \) B's 1-day work = \( \frac{1}{18} \) Combined 1-day work of A and B = \( \frac{1}{10} + \frac{1}{18} \) LCM of 10 and 18 is 90. Combined 1-day work of A and B = \( \frac{9 + 5}{90} = \frac{14}{90} = \frac{7}{45} \) Total 1-day work of A, B, and C together = \( \frac{1}{4} \) Now, C's 1-day work can be found by subtracting A and B's work from the total: C's 1-day work = \( \frac{1}{4} - \frac{7}{45} \) LCM of 4 and 45 is 180. C's 1-day work = \( \frac{45 - 28}{180} = \frac{17}{180} \) Therefore, C alone can finish the entire job in = \( \frac{180}{17} \) days = \( 10\frac{10}{17} \) days.
In simple words: Find how much A and B can do together in one day. Subtract this from what all three do together in one day. The result is C's daily work, which you then flip to find the total days.

Exam Tip: When subtracting fractions with different denominators, always find the correct Least Common Multiple (LCM) first to ensure your calculation is correct.

Question 10. A can do \( \frac{1}{4} \) of a work in 5 days and B can do \( \frac{1}{3} \) of the same work in 10 days. Find the number of days in which both working together will complete the work.
Answer:
A can complete \( \frac{1}{4} \) of the task in = 5 days So, time taken by A to complete the full task = \( 5 \times 4 = 20 \) days B can complete \( \frac{1}{3} \) of the task in = 10 days So, time taken by B to complete the full task = \( 10 \times 3 = 30 \) days A's 1-day work = \( \frac{1}{20} \) B's 1-day work = \( \frac{1}{30} \) Combined 1-day work of A and B together = \( \frac{1}{20} + \frac{1}{30} \) LCM of 20 and 30 is 60. Combined 1-day work = \( \frac{3 + 2}{60} = \frac{5}{60} = \frac{1}{12} \) Therefore, A and B working together will complete the task in = 12 days.
In simple words: First find how long it takes for each person to do the whole job alone by multiplying their days by the bottom of their fraction. Then, find their combined daily work and flip it to find the total days.

Exam Tip: If a person completes a fraction of work in some days, always multiply the days by the reciprocal of that fraction to find the total time they need for the full job.

Question 11. One tap can fill a cistern in 3 hours and the waste pipe can empty the full cistern in 5 hours. In what time will the empty cistern be full, if the tap and the waste pipe are kept open together ?
Answer:
Time taken by the first tap to fill the cistern = 3 hours Time taken by the waste pipe to empty the cistern = 5 hours Work done by the filling tap in 1 hour = \( \frac{1}{3} \) Work done by the waste pipe in 1 hour = \( \frac{1}{5} \) (as it empties, this is negative) Net work done when both are kept open together for 1 hour = \( \frac{1}{3} - \frac{1}{5} \) Subtracting these fractions: \( \frac{5 - 3}{15} = \frac{2}{15} \) Therefore, the empty cistern will be completely filled in = \( \frac{15}{2} \) hours = \( 7\frac{1}{2} \) hours.
In simple words: The filling tap adds water while the waste pipe lets it out. Subtract the emptying rate from the filling rate to find the net gain per hour. Flip this fraction to get the total hours.

Exam Tip: Remember that a waste pipe or leak empties the cistern, so its rate of work must be subtracted from the rate of the filling tap.

Question 12. A and B can do a work in 8 days; B and C in 12 days, and A and C in 16 days. In what time could they do it, all working together ?
Answer:
A and B can complete the work in = 8 days B and C can complete the work in = 12 days A and C can complete the work in = 16 days Now, let us find their respective 1-day work: Combined 1-day work of A and B = \( \frac{1}{8} \) Combined 1-day work of B and C = \( \frac{1}{12} \) Combined 1-day work of A and C = \( \frac{1}{16} \) Adding these three equations together: [(A+B) + (B+C) + (A+C)]'s 1-day work = \( \frac{1}{8} + \frac{1}{12} + \frac{1}{16} \) This gives: 2(A+B+C)'s 1-day work = \( \frac{6 + 4 + 3}{48} = \frac{13}{48} \) So, (A+B+C)'s 1-day work = \( \frac{13}{48} \times \frac{1}{2} = \frac{13}{96} \) Therefore, A, B, and C working together will complete the work in = \( \frac{96}{13} \) days = \( 7\frac{5}{13} \) days.
In simple words: When you add all the pairs' rates together, you get double the rate of A, B, and C combined. Divide that sum by 2 to get their actual combined daily rate, and then flip it to find the days.

Exam Tip: Don't forget to divide by 2 at the end of adding the pairs' rates, since adding them together counts each person's work twice.

Question 13. A and B complete a piece of work in 24 days. B and C do the same work in 36 days ; and A, B and C together finish it in 18 days. In how many days will
(i) A alone,
(ii) C alone,
(iii) A and C together, complete the work ?
Answer:
Given: A and B can finish the work in = 24 days B and C can finish the work in = 36 days A, B, and C together can finish the work in = 18 days This gives us: Combined 1-day work of A and B = \( \frac{1}{24} \) Combined 1-day work of B and C = \( \frac{1}{36} \) Combined 1-day work of A, B, and C = \( \frac{1}{18} \)
(i) **Time taken by A alone:** We can find A's 1-day work by subtracting (B+C)'s work from (A+B+C)'s work: A's 1-day work = \( \frac{1}{18} - \frac{1}{36} = \frac{2 - 1}{36} = \frac{1}{36} \) So, A alone will complete the work in = 36 days.
(ii) **Time taken by C alone:** We can find C's 1-day work by subtracting (A+B)'s work from (A+B+C)'s work: C's 1-day work = \( \frac{1}{18} - \frac{1}{24} = \frac{4 - 3}{72} = \frac{1}{72} \) So, C alone will complete the work in = 72 days.
(iii) **Time taken by A and C together:** Combined 1-day work of A and C = A's 1-day work + C's 1-day work Combined 1-day work of A and C = \( \frac{1}{36} + \frac{1}{72} = \frac{2 + 1}{72} = \frac{3}{72} = \frac{1}{24} \) So, A and C together will complete the work in = 24 days.
In simple words: To find how long one person takes, subtract the work of the other two from the total work done by all three. For A and C together, add their individual daily work and flip the final fraction.

Exam Tip: Clearly label each sub-part of your answer and write down the subtraction step clearly to avoid simple calculation errors in the exam.

Question 14. A and B can do a piece of work in 40 days; B and C in 30 days; and C and A in 24 days.
(i) How long will it take them to do the work together ?
(ii) In what time can each finish it working alone ?
Answer:
Given details: A and B together can do the work in = 40 days B and C together can do the work in = 30 days C and A together can do the work in = 24 days Therefore: Combined 1-day work of A and B = \( \frac{1}{40} \) Combined 1-day work of B and C = \( \frac{1}{30} \) Combined 1-day work of C and A = \( \frac{1}{24} \)
(i) **Time taken to do the work together:** Adding the three equations: [(A+B) + (B+C) + (C+A)]'s 1-day work = \( \frac{1}{40} + \frac{1}{30} + \frac{1}{24} \) 2(A+B+C)'s 1-day work = \( \frac{3 + 4 + 5}{120} = \frac{12}{120} = \frac{1}{10} \) So, (A+B+C)'s 1-day work = \( \frac{1}{10} \times \frac{1}{2} = \frac{1}{20} \) This means A, B, and C working together can complete the work in = 20 days.
(ii) **Time taken by each working alone:** - **For A:** A's 1-day work = (A+B+C)'s 1-day work - (B+C)'s 1-day work A's 1-day work = \( \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60} \) Hence, A can finish the work in = 60 days. - **For B:** B's 1-day work = (A+B+C)'s 1-day work - (C+A)'s 1-day work B's 1-day work = \( \frac{1}{20} - \frac{1}{24} = \frac{6 - 5}{120} = \frac{1}{120} \) Hence, B can finish the work in = 120 days. - **For C:** C's 1-day work = (A+B+C)'s 1-day work - (A+B)'s 1-day work C's 1-day work = \( \frac{1}{20} - \frac{1}{40} = \frac{2 - 1}{40} = \frac{1}{40} \) Hence, C can finish the work in = 40 days.
In simple words: First add all the pairs' rates to get double the combined rate of all three, then divide by 2 to find their rate together. To find each person's rate alone, subtract the other two's combined rate from this total rate.

Exam Tip: Remember to write down each step clearly for both parts (i) and (ii), as this is a high-mark question that requires showing both the combined and individual calculations.

Question 15. A can do a piece of work in 10 days, B in 12 days and C in 15 days. All begin together but A leaves the work after 2 days and B leaves 3 days before the work is finished. How long did the work last ?
Answer:
Given rates: A's total time = 10 days B's total time = 12 days C's total time = 15 days Thus, their 1-day work rates are: A's 1-day work = \( \frac{1}{10} \) B's 1-day work = \( \frac{1}{12} \) C's 1-day work = \( \frac{1}{15} \) Let the entire task be completed in \( x \) days. According to the question, A works for exactly 2 days, C works for the full \( x \) days, and B leaves 3 days before completion, so B works for \( x - 3 \) days. Sum of their work must equal 1 (the complete task): A's 2 days' work + B's \( (x - 3) \) days' work + C's \( x \) days' work = 1 \[ 2 \times \frac{1}{10} + (x - 3) \times \frac{1}{12} + x \times \frac{1}{15} = 1 \] \[ \frac{1}{5} + \frac{x - 3}{12} + \frac{x}{15} = 1 \] LCM of 5, 12, and 15 is 60. Combining the terms: \[ \frac{12 + 5(x - 3) + 4x}{60} = 1 \] \[ \frac{12 + 5x - 15 + 4x}{60} = 1 \] \[ \frac{9x - 3}{60} = 1 \] Multiplying both sides by 60:
\( \implies 9x - 3 = 60 \)
\( \implies 9x = 63 \)
\( \implies x = \frac{63}{9} = 7 \) Therefore, the total duration of the work is 7 days.
In simple words: Since we do not know the total days, we call it x. We write down how many days each person actually worked, multiply it by their daily rate, and add them up to equal 1. Solve the equation for x to find the total days.

Exam Tip: Be very careful when setting up the time each person worked: A worked 2 days, C worked \( x \) days, and B worked \( x - 3 \) days. Incorrectly writing B's time is a very common mistake.

Question 16. Two pipes P and Q would fill an empty cistern in 24 minutes and 32 minutes respectively. Both the pipes being opened together, find when the first pipe must be turned off so that the empty cistern may be just filled in 16 minutes.
Answer:
Given filling times: P can fill the cistern in = 24 minutes Q can fill the cistern in = 32 minutes So, we have: P's 1-minute work = \( \frac{1}{24} \) Q's 1-minute work = \( \frac{1}{32} \) Let the first pipe (P) be closed after \( x \) minutes. The total time to fill the cistern is exactly 16 minutes, during which Q remains open the whole time. Therefore, P works for \( x \) minutes and Q works for 16 minutes to fill the cistern completely: P's \( x \) minutes' work + Q's 16 minutes' work = 1 \[ \frac{1}{24} \times x + \frac{1}{32} \times 16 = 1 \] \[ \frac{x}{24} + \frac{1}{2} = 1 \] \[ \frac{x}{24} = 1 - \frac{1}{2} \] \[ \frac{x}{24} = \frac{1}{2} \] Multiplying both sides by 24:
\( \implies x = \frac{24}{2} = 12 \) Hence, pipe P must be turned off after 12 minutes.
In simple words: Since the tank is filled in 16 minutes and Q was never turned off, Q worked for all 16 minutes. P worked for x minutes. Add their contributions together to equal 1 and solve for x.

Exam Tip: When one pipe is turned off early, find the total work done by the other pipe which runs for the entire duration first. Subtract this from the whole work to find what the first pipe needs to complete.