ML Aggarwal Class 8 Maths Solutions Chapter 09 Direct and Inverse Variation

 

Exercise 9.1

 

Question 1. Observe the following tables and find if x and y are directly proportional:
(i)

(ii)
Answer:
(i) To determine if the variables are directly proportional, we need to check if the ratio x/y stays the same for all values. Computing the ratio for each pair of values:

\( \frac{x}{y} = \frac{5}{15} = \frac{1}{3} \)
\( \frac{x}{y} = \frac{8}{24} = \frac{1}{3} \)
\( \frac{x}{y} = \frac{12}{36} = \frac{1}{3} \)

However, for the remaining pairs:

\( \frac{x}{y} = \frac{15}{60} = \frac{1}{4} \)
\( \frac{x}{y} = \frac{18}{72} = \frac{1}{4} \)
\( \frac{x}{y} = \frac{20}{100} = \frac{1}{5} \)

Since the ratio is not the same throughout, x and y are not directly proportional.

(ii) Checking the ratio x/y for all values:

\( \frac{x}{y} = \frac{3}{9} = \frac{1}{3} \)
\( \frac{x}{y} = \frac{5}{15} = \frac{1}{3} \)
\( \frac{x}{y} = \frac{7}{21} = \frac{1}{3} \)
\( \frac{x}{y} = \frac{9}{27} = \frac{1}{3} \)
\( \frac{x}{y} = \frac{10}{30} = \frac{1}{3} \)

Since the ratio remains constant at \( \frac{1}{3} \), the variables x and y are directly proportional.
In simple words: Two quantities are directly proportional when their ratio stays the same. In part (ii), dividing x by y always gives 1/3, so they are proportional. In part (i), the ratio changes, so they are not.

Exam Tip: Always check the ratio x/y (or y/x) for every pair of values in the table. If all ratios are equal, the variables are directly proportional. If even one ratio differs, they are not proportional.

 

Question 2. If x and y are in direct variation, complete the following tables:
(i)

(ii)
Answer:
(i) First, we find the constant ratio from the known values. Using x = 3 and y = 45:

\( \frac{x}{y} = \frac{3}{45} = \frac{1}{15} \)

Since the ratio is constant at \( \frac{1}{15} \), we can find the missing values:

When x = 5: \( \frac{5}{y} = \frac{1}{15} \) gives y = 5 × 15 = 75
When y = 90: \( \frac{x}{90} = \frac{1}{15} \) gives x = 90 ÷ 15 = 6
When y = 120: \( \frac{x}{120} = \frac{1}{15} \) gives x = 120 ÷ 15 = 8
When x = 10: \( \frac{10}{y} = \frac{1}{15} \) gives y = 10 × 15 = 150

(ii) Using x = 4 and y = 7:

\( \frac{x}{y} = \frac{4}{7} \)

The constant ratio is \( \frac{4}{7} \). Finding the missing values:

When x = 8: \( \frac{8}{y} = \frac{4}{7} \) gives y = (8 × 7) ÷ 4 = 14
When y = 21: \( \frac{x}{21} = \frac{4}{7} \) gives x = (21 × 4) ÷ 7 = 12
When x = 20: \( \frac{20}{y} = \frac{4}{7} \) gives y = (20 × 7) ÷ 4 = 35
When x = 28: \( \frac{28}{y} = \frac{4}{7} \) gives y = (28 × 7) ÷ 4 = 49
In simple words: When two quantities are directly proportional, their ratio stays the same. Find this constant ratio from any known pair, then use it to calculate all the missing numbers in the table.

Exam Tip: Always identify the constant ratio first by dividing one variable by the other using a complete pair. Then apply this ratio to all missing values - this ensures consistency across the entire table.

 

Question 3. If 8 meters cloth costs Rs.250, find the cost of 5.8 meters of the same cloth.
Answer: Let the cost of 5.8 m of cloth be Rs. x. Since the cost and length are directly proportional, we set up a proportion:

\( 8 : 250 = 5.8 : x \)
\( \frac{8}{250} = \frac{5.8}{x} \)
\( x = \frac{5.8 \times 250}{8} \)
\( x = \frac{1450}{8} \)
\( x = \text{Rs.} 181.25 \)
In simple words: When the price per meter stays the same, you can use a simple proportion to find the new cost. Multiply the new length by the price per meter to get the answer.

Exam Tip: Write the proportion in the form "known length : known cost = new length : unknown cost" - this prevents calculation errors. Always cross-multiply to solve.

 

Question 4. If a labourer earns Rs.672 per week, how much will he earn in 18 days?
Answer: Let the labourer earn Rs. x in 18 days. Since earnings and time worked are directly proportional:

\( 7 : 672 = 18 : x \)
\( \frac{7}{672} = \frac{18}{x} \)
\( x = \frac{18 \times 672}{7} \)
\( x = \frac{12096}{7} \)
\( x = \text{Rs.} 1728 \)
In simple words: If a worker earns a fixed amount per day, multiply the daily rate by the number of days to find total earnings for any time period.

Exam Tip: Note that 1 week = 7 days. Always convert all time values to the same unit before setting up the proportion.

 

Question 5. If 175 dollars cost Rs.7350, how many dollars can be purchased in Rs.24024?
Answer: Let 'x' dollars be purchased in Rs.24024. Since the number of dollars and rupees are directly proportional:

\( 7350 : 175 = 24024 : x \)
\( \frac{7350}{175} = \frac{24024}{x} \)
\( x = \frac{24024 \times 175}{7350} \)
\( x = \frac{4204200}{7350} \)
\( x = 572 \text{ Dollars} \)
In simple words: Find how many dollars you can buy per rupee by dividing the rupees by the dollars. Then multiply this rate by the new amount of rupees.

Exam Tip: When dealing with currency conversions, set up the proportion as "first cost in rupees : first amount in dollars = second cost in rupees : second amount in dollars" to avoid mixing up the units.

 

Question 6. If a car travels 67.5 km in 4.5 liters of petrol, how many kilometers will it travel in 26.4 liters of petrol?
Answer: Let the car travel 'x' km in 26.4 liters of petrol. Since distance traveled and petrol used are directly proportional:

\( \frac{x}{26.4} = \frac{67.5}{4.5} \)
\( x = \frac{67.5 \times 26.4}{4.5} \)
\( x = \frac{1782}{4.5} \)
\( x = 396 \text{ km} \)
In simple words: Calculate how far the car travels on one liter of petrol (this is the constant ratio), then multiply by the total liters available to get the total distance.

Exam Tip: In fuel consumption problems, the constant ratio is kilometers per liter. Find this by dividing distance by fuel amount, then use it for new calculations.

 

Question 7. If the thickness of a pile of 12 cardboard sheets is 45 mm, then how many sheets of the same cardboard would be 90 cm thick?
Answer: Given: 12 sheets have a thickness of 45 mm. Let 'x' be the number of sheets for 90 cm thickness. First, convert 90 cm to mm: 90 cm = 900 mm. Since the number of sheets and thickness are directly proportional:

\( \frac{12}{x} = \frac{45}{900} \)
\( x = \frac{12 \times 900}{45} \)
\( x = \frac{10800}{45} \)
\( x = 240 \text{ sheets} \)
In simple words: The thickness of each sheet stays the same. Divide the total thickness needed by the thickness of one sheet to find how many sheets are required.

Exam Tip: Always ensure that measurements are in the same unit (mm with mm, cm with cm) before solving. Converting units first prevents calculation mistakes.

 

Question 8. In a model of a ship, the mast (flagstaff) is 6 cm high, while the mast of the actual ship is 9 m high. If the length of the ship is 33 m, how long is the model of the ship?
Answer: The model dimensions and actual ship dimensions are directly proportional. Let the length of the model be 'x' cm. Setting up the proportion between model mast and actual mast, and between model length and actual ship length:

\( \frac{6}{9} = \frac{x}{33} \)

Converting 9 m to cm for consistency: 9 m = 900 cm, and 33 m = 3300 cm
\( \frac{6}{900} = \frac{x}{3300} \)
\( x = \frac{6 \times 3300}{900} \)
\( x = \frac{19800}{900} \)
\( x = 22 \text{ cm} \)
In simple words: A model of something is smaller in the same proportion as all its parts. If the mast shrinks by a certain factor, the entire length shrinks by that same factor.

Exam Tip: In scale models, all corresponding dimensions must maintain the same ratio. Find this scaling factor from one pair of known measurements, then apply it to find unknown dimensions.

 

Question 9. The mass of an aluminium rod varies directly with its length. If a 16 cm long rod has a mass of 192 g, find the length of the rod whose mass is 105 g.
Answer: Since mass and length vary directly, the ratio of length to mass remains constant. Using the known values:

\( \frac{16}{192} = \frac{x}{105} \)
\( x = \frac{16 \times 105}{192} \)
\( x = \frac{1680}{192} \)
\( x = \frac{35}{4} \)
\( x = 8.75 \text{ cm} \)
In simple words: A longer rod weighs more. Find the weight per cm (the constant), then use it to find what length corresponds to a given weight.

Exam Tip: Always simplify fractions to their lowest form before converting to decimals. This reduces rounding errors and makes the answer clearer.

 

Question 10. Anita has to drive from village A to village B. She measures a distance of 3.5 cm between these villages on the map. What is the actual distance between the villages if the map scale is 1 cm = 20 km?
Answer: The map distance and actual distance are directly proportional. Using the given scale of 1 cm = 20 km:

\( \frac{1}{3.5} = \frac{20}{x} \)
\( x = \frac{20 \times 3.5}{1} \)
\( x = 70 \text{ km} \)
In simple words: A map scale tells you how much real ground each measurement on the map represents. Multiply the map distance by the scale to get the actual distance.

Exam Tip: When using map scales, set up the proportion as "map distance : actual distance = 1 cm : (scale factor)" for clarity. This prevents confusion about which side represents what.

 

Question 11. A 23 m 75 cm high water tank casts a shadow 20 m long. Find at the same time;
(i) the length of the shadow cast by a tree 9 m 50 cm high.
(ii) the height of the tree if the length of the shadow is 12 m.
Answer: When objects cast shadows at the same time, their heights and shadow lengths are directly proportional. Convert the tank height: 23 m 75 cm = 23.75 m = 95/4 m.

(i) For a tree 9 m 50 cm high = 9.5 m = 19/2 m. Let its shadow be 'x' m:

\( \frac{95/4}{19/2} = \frac{20}{x} \)
\( \frac{95}{4} \times \frac{2}{19} = \frac{20}{x} \)
\( \frac{5}{2} = \frac{20}{x} \)
\( x = \frac{20 \times 2}{5} \)
\( x = 8 \text{ m} \)

(ii) For a shadow of 12 m, let the tree height be 'x' m:

\( \frac{95/4}{x} = \frac{20}{12} \)
\( \frac{95}{4x} = \frac{20}{12} \)
\( 4x = \frac{95 \times 12}{20} \)
\( 4x = 57 \)
\( x = \frac{57}{4} = 14.25 \text{ m} = 14 \text{ m } 25 \text{ cm} \)
In simple words: At any given moment, tall objects make longer shadows and short objects make shorter shadows. Their heights and shadows stay in the same proportion.

Exam Tip: Always convert mixed measurements (like 23 m 75 cm) to a single unit before calculating. Working with improper fractions or decimals consistently prevents conversion errors.

 

Question 12. If 5 men or 7 women can earn Rs.525 per day, how much 10 men and 13 women would earn per day.
Answer: First, find the daily earning for one man and one woman.

If 5 men earn Rs.525 per day, then 1 man earns: Rs.525 ÷ 5 = Rs.105 per day
So, 10 men earn: Rs.105 × 10 = Rs.1050 per day

If 7 women earn Rs.525 per day, then 1 woman earns: Rs.525 ÷ 7 = Rs.75 per day
So, 13 women earn: Rs.75 × 13 = Rs.975 per day

Total earning of 10 men and 13 women per day = Rs.1050 + Rs.975 = Rs.2025
In simple words: Find how much one person (man or woman) earns by dividing the group earning by the group size. Then multiply by the new group size to get the new total.

Exam Tip: When dealing with mixed groups, calculate earnings separately for each type of person, then add them together. This prevents confusion and makes the solution clearer.

 

Exercise 9.2

 

Question 1. Which of the following are in inverse variation?
(i) Number of students in a hostel and consumption of food.
(ii) Time taken by a train to cover a fixed distance and the speed of the train.
(iii) Area of land and its cost.
(iv) The number of people working and the time to complete the work.
(v) The quantity of rice and its cost.
Answer: Inverse variation occurs when two quantities multiply to give a constant value (xy = constant). Examining each option:

(i) As more students join the hostel, food consumption increases - this is direct variation, not inverse.
(ii) When a train travels a fixed distance, increasing speed reduces the time taken - this is inverse variation.
(iii) More land area means higher cost - this is direct variation, not inverse.
(iv) When more people work together, the time needed to finish decreases - this is inverse variation.
(v) Buying more rice costs more - this is direct variation, not inverse.

Therefore, (ii) and (iv) are inverse variations.
In simple words: In inverse variation, when one quantity gets bigger, the other gets smaller. As speed goes up, time goes down. As workers increase, time decreases.

Exam Tip: Remember: direct means "both go up together," inverse means "one goes up, the other goes down." Check each option by thinking about what happens when one quantity increases.

 

Question 2. Observe the following tables and find which pair of variables (here x and y) are in inverse variation:
(i)

(ii)
Answer:
(i) For inverse variation, the product xy must be constant for all pairs. Computing xy:

\( xy = 90 \times 10 = 900 \)
\( xy = 60 \times 15 = 900 \)
\( xy = 45 \times 20 = 900 \)
\( xy = 30 \times 30 = 900 \)
\( xy = 20 \times 45 = 900 \)

Since the product is always 900, the variables x and y are in inverse variation.

(ii) Computing xy for each pair:

\( xy = 75 \times 10 = 750 \)
\( xy = 45 \times 30 = 1350 \)
\( xy = 30 \times 25 = 750 \)
\( xy = 20 \times 35 = 700 \)
\( xy = 10 \times 65 = 650 \)

Since the products differ, the variables x and y are not in inverse variation.
In simple words: To test for inverse variation, multiply x and y together for each pair. If the answer is always the same number, they are inversely proportional. If the products change, they are not.

Exam Tip: The key test for inverse variation is checking whether xy = constant. Calculate the product for each row carefully - a single different product means the entire table fails the inverse variation test.

 

Question 3. Under the condition that the temperature remains constant, the volume of gas is inversely proportional to its pressure. If the volume of gas is 630 cubic centimeters at a pressure of 360 mm of mercury, then what will be the pressure of the gas, if its volume is 720 cubic centimeters at the same temperature?
Answer: Since volume and pressure are inversely proportional at constant temperature, their product is constant: V × P = k.

From the initial conditions: 630 × 360 = k
k = 226,800

When the volume changes to 720 cubic cm, let the pressure be 'a' mm:
\( 720 \times a = 226,800 \)
\( a = \frac{226,800}{720} \)
\( a = 315 \text{ mm of mercury} \)
In simple words: When gas is squeezed (volume goes down), the pressure goes up. When gas expands (volume goes up), the pressure goes down. Their product always stays the same.

Exam Tip: In physics problems involving gases, remember that Volume × Pressure = constant (at fixed temperature). Use this relationship to find unknown values - calculate the constant from known data, then apply it to new situations.

 

Question 4. A packet of sweets was distributed among 20 children and each of them received 4 sweets. How many sweets will each child get, if the number of children is reduced by 4?
Answer: The total number of sweets remains constant. When distributed among fewer children, each child gets more. Using the inverse variation principle:

Total sweets = 20 × 4 = 80 sweets

When the number of children is reduced by 4: new number = 20 - 4 = 16 children

\( 20 \times 4 = 16 \times y \)
\( y = \frac{20 \times 4}{16} \)
\( y = \frac{80}{16} \)
\( y = 5 \text{ sweets} \)
In simple words: The same amount of sweets is shared among fewer people, so each person gets more. Divide the total sweets by the new number of children to find the new amount per child.

Exam Tip: In distribution problems, the total amount stays the same. When you use inverse variation (xy = constant), ensure you identify this constant correctly from the initial conditions.

 

Question 5. Pooja has enough money to buy 36 oranges at the rate of Rs.4.50 per orange. How many oranges she can buy if the price of each orange is increased by 90 paisa?
Answer: Pooja's total money is fixed. The number of oranges she can buy varies inversely with the price per orange. First, find her total money:

Total money = 36 × Rs.4.50 = Rs.162

New price per orange = Rs.4.50 + Rs.0.90 = Rs.5.40

Number of oranges at new price = \( \frac{\text{Total money}}{\text{Price per orange}} = \frac{162}{5.40} = 30 \text{ oranges} \)
In simple words: With a fixed amount of money, you can buy fewer oranges if the price goes up. Divide your total money by the new price to find how many you can afford.

Exam Tip: In pricing problems, always identify the fixed quantity (total money, total amount of work, total food) first. Then use inverse variation: as one factor increases, the other decreases proportionally to keep the product constant.

 

Question 6. It takes 8 days for 12 men to construct a wall. How many men should be put on the job if it is required to be constructed in 6 days?
Answer: The amount of work (constructing the wall) is fixed. The number of men and time taken are inversely proportional. First, find the total man-days required:

Total man-days = 12 men × 8 days = 96 man-days

For completion in 6 days, the number of men needed:
\( \text{Number of men} = \frac{96}{6} = 16 \text{ men} \)
In simple words: More workers finish a job faster. The number of people needed is inversely related to the time allowed - more time means fewer workers needed, less time means more workers needed.

Exam Tip: Calculate the total work in "man-days" or "person-days" by multiplying workers by days. This constant total helps you find the new number of workers needed for any different time period.

 

Question 7. Eight taps through which water flows at the same rate can fill a tank in 27 minutes. If two taps go out of order, how long will the remaining taps take to fill the tank?
Answer: The amount of water to fill the tank remains the same. The number of working taps and time taken are inversely proportional. Calculate the total tap-minutes:

Total tap-minutes = 8 taps × 27 minutes = 216 tap-minutes

After 2 taps break, remaining taps = 8 - 2 = 6 taps

Time for 6 taps to fill the tank:
\( \text{Time} = \frac{216}{6} = 36 \text{ minutes} \)
In simple words: With fewer taps working, it takes longer to fill the tank. The total amount of water stays the same - fewer sources just means slower filling.

Exam Tip: In work-rate problems involving multiple workers or sources, multiply the number of workers/sources by the time to get a "capacity measure." Use this to solve for unknowns in inverse variation scenarios.

 

Question 8. A contractor undertook a contract to complete a part of a stadium in 9 months with a team of 560 persons. Later on, it was required to complete the job in 5 months. How many extra persons should he employ to complete the work?
Answer: The amount of work is fixed. The number of workers and time available are inversely proportional. Calculate total person-months of work:

Total person-months = 560 persons × 9 months = 5040 person-months

For completion in 5 months, total persons needed:
\( \text{Number of persons} = \frac{5040}{5} = 1008 \text{ persons} \)

Extra persons to employ = 1008 - 560 = 448 persons
In simple words: To finish the same job in less time, you need more workers. Calculate how many total workers are needed for the new time, then subtract the workers already available to find how many more to hire.

Exam Tip: Always find the total workforce needed for the new timeline first, then subtract the existing workforce to determine "extra persons needed." This two-step approach prevents calculation errors.

 

Question 9. A batch of bottles was packed in 30 boxes with 10 bottles in each box. If the same batch is packed using 12 bottles in each box, how many boxes would be filled?
Answer: The total number of bottles in the batch is fixed. The number of boxes and bottles per box are inversely proportional. Calculate the total number of bottles:

Total bottles = 30 boxes × 10 bottles per box = 300 bottles

When packed with 12 bottles per box, the number of boxes needed:
\( \text{Number of boxes} = \frac{300}{12} = 25 \text{ boxes} \)
In simple words: When each box holds more bottles, fewer boxes are needed for the same total number. Divide the total bottles by the new capacity per box to find the number of boxes.

Exam Tip: In packaging/container problems, the total quantity stays constant. Find this total first, then divide by the new container capacity to get the new number of containers needed.

 

Question 10. Vandana takes 24 minutes to reach her school if she goes at a speed of 5 km/h. If she wants to reach school in 20 minutes, what should be her speed?
Answer: The distance to school remains fixed. Speed and time are inversely proportional - faster speed takes less time. Calculate the fixed distance:

Distance = Speed × Time = 5 km/h × 24 minutes = 5 × (24/60) hours = 2 km

For a 20-minute journey, required speed:
\( \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{2 \text{ km}}{20 \text{ minutes}} = \frac{2}{20/60} = \frac{2 \times 60}{20} = 6 \text{ km/h} \)
In simple words: The distance is fixed. To get there faster, you need to go faster. Calculate the distance first, then divide it by the new time to find the required speed.

Exam Tip: Always keep time units consistent. In this case, convert minutes to hours or find the distance as distance = speed × (time in hours). Consistent units prevent errors.

 

Question 11. A fort is provided with food for 80 soldiers to last for 60 days. Find how long would the food last if 20 additional soldiers join after 15 days.
Answer: Initially, the fort has food for 80 soldiers for 60 days. After 15 days of consumption, the remaining food is for 80 soldiers for (60 - 15) = 45 days. After 15 days, 20 additional soldiers arrive, making the total 80 + 20 = 100 soldiers. The number of soldiers and duration the food lasts are inversely proportional:

Food remaining (in soldier-days) = 80 × 45 = 3600 soldier-days

Duration food will last for 100 soldiers:
\( \text{Duration} = \frac{3600}{100} = 36 \text{ days} \)
In simple words: When more people eat the same food, it runs out faster. Calculate how much food is left (in person-days), then divide by the new number of people to find how long it will last.

Exam Tip: In food/resource problems, first identify how much supply remains after the initial period. Then calculate "supply in person-units" and divide by the new group size. This handles the timing change correctly.

 

Question 12. 1200 soldiers in a fort had enough food for 28 days. After 4 days, some soldiers were sent to another fort and thus, the food lasted for 32 more days. How many soldiers left the fort?
Answer: Initially, 1200 soldiers had food for 28 days. After 4 days, the remaining food was sufficient for 1200 soldiers for (28 - 4) = 24 days. Let 'x' be the number of soldiers remaining. Since the food amount is fixed, the number of soldiers and days the food lasts are inversely proportional:

Original food supply (in soldier-days) = 1200 × 24 = 28,800 soldier-days

After some soldiers leave, the remaining food lasts for 32 days. If 'y' soldiers remain:
\( y \times 32 = 28,800 \)
\( y = \frac{28,800}{32} = 900 \text{ soldiers} \)

Soldiers who left = 1200 - 900 = 300 soldiers
In simple words: With less food remaining but fewer mouths to feed, it lasts longer. Calculate the remaining food supply in person-days, divide by the new duration to find how many people remain, then subtract from the original number.

Exam Tip: In problems where resources deplete over time before a change occurs, always calculate the remaining resource first. Then apply inverse variation using this remainder to find the new group size.

 

Exercise 9.3

 

Question 1. A farmer can reap a field in 10 days while his wife can do it in 8 days (she does not waste time in smoking). If they work together, in how much time can they reap the field?
Answer: When workers collaborate, their work rates add together. The farmer completes 1/10 of the field per day, and his wife completes 1/8 of the field per day. Their combined work rate:

\( \text{Combined rate} = \frac{1}{10} + \frac{1}{8} \)
\( = \frac{4+5}{40} \)
\( = \frac{9}{40} \text{ of the field per day} \)

Time to complete the entire field:
\( \text{Time} = \frac{1}{\frac{9}{40}} = \frac{40}{9} = 4\frac{4}{9} \text{ days} \)
In simple words: Find what fraction of the job each person finishes in one day. Add these fractions to get how much they do together in one day. Then divide 1 by this sum to find total days needed.

Exam Tip: In combined work problems, always express each person's work as "fraction of job per day" (1/time). Add all fractions to get the combined rate. The reciprocal of this combined rate gives the time to finish together.

 

Question 2. A can do 1/5th of a certain work in 2 days and B can do 2/3rd of it in 8 days. In how much time can they together complete the work?
Answer: First, find the daily work rate for each person. If A completes 1/5 of the work in 2 days, then in 1 day A completes:

\( \text{A's daily rate} = \frac{1/5}{2} = \frac{1}{10} \text{ of the work per day} \)

If B completes 2/3 of the work in 8 days, then in 1 day B completes:
\( \text{B's daily rate} = \frac{2/3}{8} = \frac{2}{24} = \frac{1}{12} \text{ of the work per day} \)

Combined daily rate:
\( \frac{1}{10} + \frac{1}{12} = \frac{6+5}{60} = \frac{11}{60} \text{ of the work per day} \)

Time to complete the full work together:
\( \text{Time} = \frac{1}{\frac{11}{60}} = \frac{60}{11} = 5\frac{5}{11} \text{ days} \)
In simple words: When someone completes a fraction of work in multiple days, divide that fraction by the days to get their daily rate. Add all daily rates, then invert to find the time working together.

Exam Tip: Always reduce each person's work rate to "per day" first. This standardizes the calculation and prevents errors when people complete different fractions in different time periods.

 

Question 3. One tap fills a tank in 20 minutes and another tap fills it in 12 minutes. The tank being empty and if both taps are opened together, in how many minutes the tank will be full?
Answer: When both taps work together, their rates of filling add. The first tap fills 1/20 of the tank per minute, and the second tap fills 1/12 of the tank per minute. Combined filling rate:

\( \text{Combined rate} = \frac{1}{20} + \frac{1}{12} \)
\( = \frac{3+5}{60} \)
\( = \frac{8}{60} = \frac{2}{15} \text{ of the tank per minute} \)

Time to fill the entire tank:
\( \text{Time} = \frac{1}{\frac{2}{15}} = \frac{15}{2} = 7\frac{1}{2} \text{ minutes} \)
In simple words: Each tap fills a fraction of the tank per minute. Add these fractions to find how much they fill together per minute. Then divide 1 by this combined rate to get total time.

Exam Tip: In tap/pipe problems, think of each tap as contributing a fraction per unit time. The sum of all fractions is the combined rate - finding the reciprocal of this gives you the answer.

 

Question 4. A can do a work in 6 days and B can do it in 8 days. They worked together for 2 days and then B left the work. How many days will A require to finish the work?
Answer: A's daily work rate = 1/6, and B's daily work rate = 1/8. When working together for 2 days, they complete:

\( \text{Work done in 2 days} = 2 \times \left(\frac{1}{6} + \frac{1}{8}\right) \)
\( = 2 \times \frac{4+3}{24} \)
\( = 2 \times \frac{7}{24} \)
\( = \frac{7}{12} \text{ of the work} \)

Remaining work = \( 1 - \frac{7}{12} = \frac{5}{12} \)

A alone completes 1/6 of the work per day. To complete 5/12 of the work:
\( \text{Days required} = \frac{5/12}{1/6} = \frac{5}{12} \times 6 = \frac{5}{2} = 2\frac{1}{2} \text{ days} \)
In simple words: Calculate how much work they finish together. Find what remains. Then divide the remaining work by the remaining person's daily rate to find how many more days are needed.

Exam Tip: When a person leaves partway through, separate the problem into two phases: (1) work done by all together, and (2) work done by the remaining person alone. Keep track of remaining work after phase 1.

 

Question 5. A can do a piece of work in 40 days. He works at it for 8 days and then B finishes the remaining work in 16 days. How long will they take to complete the work if they do it together?
Answer: A's daily work rate = 1/40. In 8 days, A completes:

\( \text{Work by A} = 8 \times \frac{1}{40} = \frac{8}{40} = \frac{1}{5} \)

Remaining work = \( 1 - \frac{1}{5} = \frac{4}{5} \)

B completes 4/5 of the work in 16 days, so B's daily rate:
\( \text{B's daily rate} = \frac{4/5}{16} = \frac{4}{80} = \frac{1}{20} \text{ per day} \)

Combined daily rate:
\( \frac{1}{40} + \frac{1}{20} = \frac{1+2}{40} = \frac{3}{40} \text{ per day} \)

Time to complete together:
\( \text{Time} = \frac{1}{\frac{3}{40}} = \frac{40}{3} = 13\frac{1}{3} \text{ days} \)
In simple words: From the first scenario (A working alone), find B's work rate by seeing how much B can complete in the given days. Then combine both rates to find the time working together.

Exam Tip: When you need to find someone's work rate from a scenario, use the work equation: (fraction of work) ÷ (days taken) = daily rate. This works even when the fraction is not 1 (the full job).

 

Question 6. A and B separately do a work in 10 and 15 days respectively. They worked together for some days and then A completed the remaining work in 5 days. For how many days had A and B worked together?
Answer: A's daily work rate = 1/10, and B's daily work rate = 1/15. Combined rate:

\( \frac{1}{10} + \frac{1}{15} = \frac{3+2}{30} = \frac{5}{30} = \frac{1}{6} \text{ per day} \)

Let them work together for 'x' days. Work completed together:
\( \text{Work done together} = x \times \frac{1}{6} = \frac{x}{6} \)

Remaining work = \( 1 - \frac{x}{6} \)

A alone completes the remaining work in 5 days:
\( 5 \times \frac{1}{10} = 1 - \frac{x}{6} \)
\( \frac{1}{2} = 1 - \frac{x}{6} \)
\( \frac{x}{6} = 1 - \frac{1}{2} = \frac{1}{2} \)
\( x = 3 \text{ days} \)
In simple words: Set up an equation where the work done together plus the work done alone equals the whole job. Solve for the unknown number of days they worked together.

Exam Tip: Create an equation: (combined rate × days together) + (A's rate × 5 days) = 1 whole job. This translates the word problem directly into math and makes solving straightforward.

 

Question 7. If 3 women or 5 girls take 17 days to complete a piece of work, how long will 7 women and 11 girls working together take to complete the work?
Answer: Since 3 women's work equals 5 girl's work, we can express 1 woman's work as 5/3 girl's work. Therefore, 7 women's work equals (5/3) × 7 = 35/3 girls' work. When we combine 7 women and 11 girls, the total work capacity becomes (35/3) + 11 = 68/3 girls' work. Since 5 girls finish the work in 17 days, 1 girl needs 17 × 5 = 85 days. So 68/3 girls complete it in (85)/(68/3) = (85 × 3)/68 = 255/68 = 3.75 days, which equals 3 ¾ days.
In simple words: First, find how many girls are equal to the 7 women. Then add the 11 girls to get a total. Finally, use the girl-days information to find how long the combined group takes.

Exam Tip: Convert all workers to a single unit (here, girls) before calculating combined work time - this avoids mixing different types of workers.

 

Question 8. A can do a job in 10 days while B can do it in 15 days. If they work together and earn Rs 3500, how should they share the money?
Answer: A's one day work is 1/10 and B's one day work is 1/15. Together in one day they complete 1/10 + 1/15 = (3 + 2)/30 = 5/30 = 1/6 of the work. The earning of Rs 3500 corresponds to 1/6 of the work, so the full work's value is Rs 3500 × 6 = Rs 21,000. A's portion is (1/10) × Rs 21,000 = Rs 2,100 and B's portion is (1/15) × Rs 21,000 = Rs 1,400.
In simple words: The money is split based on how much work each person does. The person who does more work per day gets a larger share.

Exam Tip: Payment distribution in work problems follows the same ratio as their individual work rates - always find the combined rate first.

 

Question 9. A, B and C can separately do a work in 2, 6 and 3 days respectively. Working together, how much time would they require to do it? If the work earns them Rs 960, how should they divide the money?
Answer: A's one day work is 1/2, B's one day work is 1/6, and C's one day work is 1/3. When working together, their combined daily capacity is 1/2 + 1/6 + 1/3 = (3 + 1 + 2)/6 = 6/6 = 1 day. Therefore, all three working together can finish the work in 1 day. For the money division, they should split Rs 960 in the ratio of their work rates: 1/2 : 1/6 : 1/3. Multiplying by 12 gives the ratio 6 : 2 : 4. The sum of ratio terms is 12. A's share is (6/12) × Rs 960 = Rs 480, B's share is (2/12) × Rs 960 = Rs 160, and C's share is (4/12) × Rs 960 = Rs 320.
In simple words: When people work at different speeds, the faster worker gets more money. The money shares match their work-rate shares.

Exam Tip: Always simplify the ratio of work rates before dividing the total payment to avoid calculation errors.

 

Question 10. A, B and C together can do a piece of work in 15 days, B alone can do it in 30 days and C alone can do it in 40 days. In how many days will A alone do the work?
Answer: The combined work rate of A, B, and C is 1/15 per day. B's individual rate is 1/30 per day and C's is 1/40 per day. To find A's rate, subtract B's and C's rates from the combined rate: A's rate = 1/15 - (1/30 + 1/40) = 1/15 - (4 + 3)/120 = 1/15 - 7/120 = (8 - 7)/120 = 1/120. Therefore, A working alone takes 120 days to complete the work.
In simple words: Subtract the rates of the other two workers from the combined rate to find A's individual rate.

Exam Tip: To isolate one person's work rate, always subtract the combined rates of the others from the total combined rate.

 

Question 11. A, B and C working together can plough a field in 4 4/5 days. A and C together can do it in 8 days. How long would B working alone take to plough the field?
Answer: First, convert 4 4/5 days to an improper fraction: 24/5 days. The combined rate of A, B, and C is 5/24 per day. The combined rate of A and C is 1/8 per day. B's individual rate is found by subtracting: B's rate = 5/24 - 1/8 = (5 - 3)/24 = 2/24 = 1/12 per day. Therefore, B working alone takes 12 days to plough the field.
In simple words: Take the rate of all three workers and subtract the rate of A and C together. What remains is B's work rate alone.

Exam Tip: When one person's rate must be found from two group rates, subtract the smaller group's rate from the larger group's rate.

 

Question 12. A and B together can build a wall in 10 days; B and C working together can do it in 15 days; C and A together can do it in 12 days. How long will they take to finish the work, working altogether? Also find the number of days taken by each to do the same work, working alone.
Answer: The combined rate of A and B is 1/10 per day, B and C is 1/15 per day, and C and A is 1/12 per day. Adding all three pairs: (A+B) + (B+C) + (C+A) = 1/10 + 1/15 + 1/12 = (6 + 4 + 5)/60 = 15/60 = 1/4. This equals twice the rate of all three together, so 2(A+B+C) = 1/4, giving (A+B+C) = 1/8. Therefore, all three working together complete the wall in 8 days. To find individual rates: A's rate = 1/8 - 1/15 = (15 - 8)/120 = 7/120 per day, so A takes 120/7 = 17 1/7 days alone. B's rate = 1/8 - 1/12 = (3 - 2)/24 = 1/24 per day, so B takes 24 days alone. C's rate = 1/8 - 1/10 = (5 - 4)/40 = 1/40 per day, so C takes 40 days alone.
In simple words: Add all three pair-rates together and divide by 2 to get the three-person rate. Then subtract pairs from the total to find each person's individual rate.

Exam Tip: When given pair rates, always add all three pairs first, then divide by 2 - this is the key insight that unlocks the problem.

 

Question 13. A pipe can fill a tank in 12 hours. By mistake, a waste pipe in the bottom is left opened and the tank is filled in 16 hours. If the tank is full, how much time will the waste pipe take to empty it?
Answer: The filling pipe alone fills 1/12 of the tank per hour. When the waste pipe is open, the net filling rate is 1/16 per hour. The waste pipe's emptying rate is the difference between the filling pipe's rate and the net rate: 1/12 - 1/16 = (4 - 3)/48 = 1/48 per hour. Therefore, the waste pipe takes 48 hours to empty a full tank.
In simple words: Find how much the filling pipe loses per hour due to the waste pipe. This loss rate tells you how fast the waste pipe empties the tank.

Exam Tip: For pipe problems with filling and emptying, the difference in rates gives the rate of the unopposed action (emptying in this case).