CBSE Class 8 Mathematics Direct And Inverse Variations Worksheet Set 04

Ch -13 Direct and Inverse proportions

Question 1. In which of the following cases is there direct variation between the two given quantities? (Do not carry out any calculations)
(a) 10 workers finish a job in 6 days. In how many days will 20 workers finish the same job?
(b) A car runs at a uniform speed. If it covers 135 km in 3 hours, how much distance will it cover in 2 hours?
(c) A tap can completely fill a tank in \(1\frac{1}{2}\) hours. How much of it can be filled in \(\frac{3}{4}\) hours?
(d) Curtain cloth was purchased at Rs. 230 per meter. How much will 12 meters of cloth cost?
Answer: Cases (b), (c), and (d) represent direct variation. In case (a), as the number of workers increases, the time taken decreases, which is an inverse variation.
In simple words: Direct variation means as one quantity increases, the other also increases.

Exam Tip: Identify direct variation by checking if the ratio between the two quantities remains constant.

Question 2. If the mass of 35 sheets of paper is 280 g, how many sheets will weigh 7.2 kg?
Answer: Mass of 35 sheets = 280 g
Total mass given = \(7.2 \text{ kg} = 7200 \text{ g}\)
Let the number of sheets be \(x\).
Since the number of sheets and mass are in direct proportion:
\(\frac{35}{280} = \frac{x}{7200}\)
\(x = \frac{35 \times 7200}{280}\)
\(x = \frac{35 \times 720}{28} = 5 \times 180 = 900\)
In simple words: More sheets of paper will weigh more, so this is a direct proportion problem.

Exam Tip: Always convert all quantities to the same units (grams to grams) before setting up the proportion.

Question 3. A family's expenses for 4 weeks 5 days amount to RS1980. What is their expense at the same rate for 6 weeks 2 days?
Answer: Convert time into days (1 week = 7 days).
First period = \(4 \text{ weeks} \times 7 + 5 \text{ days} = 28 + 5 = 33 \text{ days}\)
Second period = \(6 \text{ weeks} \times 7 + 2 \text{ days} = 42 + 2 = 44 \text{ days}\)
Expense for 33 days = Rs. 1980
Let the expense for 44 days be \(x\).
\(\frac{33}{1980} = \frac{44}{x}\)
\(x = \frac{44 \times 1980}{33}\)
\(x = 44 \times 60 = 2640\)
Expense = Rs. 2640
In simple words: Convert weeks and days into total days to make the calculation straightforward.

Exam Tip: Convert mixed time units into a single uniform unit (days) to avoid fractional errors.

Question 4. If 25 workers can finish a job in 40 days, how many workers will complete the same job in 25 days?
Answer: This is an inverse proportion problem because fewer days require more workers.
Let the required number of workers be \(x\).
\(25 \times 40 = x \times 25\)
\(x = \frac{25 \times 40}{25} = 40\)
Workers needed = 40
In simple words: When time decreases, the number of workers must increase in the same ratio.

Exam Tip: For inverse variation, use the product formula \(x_1 y_1 = x_2 y_2\).

Question 5. 40 kg of rice lasts 30 days in a family of 8 persons. If 2 guests stay with the family, how many days will 40 kg of rice last?
Answer: Initial number of persons = 8
New number of persons = \(8 + 2 = 10\)
Amount of rice is constant (40 kg). More people mean the rice will last fewer days (inverse proportion).
Let the number of days be \(d\).
\(8 \times 30 = 10 \times d\)
\(d = \frac{8 \times 30}{10} = 24\)
The rice will last for 24 days.
In simple words: Total consumption rate changes inversely with the number of family members.

Exam Tip: Add the guests to the original family count to find the total persons before applying inverse proportion.

Question 6. If Ram works 10 hours per day, he can finish a job in 28 days. If he has to finish the job in 20 days, how many hours should he work per day?
Answer: This is an inverse variation between hours per day and total days.
Let the required daily hours be \(h\).
\(10 \times 28 = h \times 20\)
\(h = \frac{10 \times 28}{20}\)
\(h = \frac{280}{20} = 14\)
He should work 14 hours per day.
In simple words: Finishing the job in fewer days requires working more hours each day.

Exam Tip: Set up the inverse product equation: \((\text{Hours}_1) \times (\text{Days}_1) = (\text{Hours}_2) \times (\text{Days}_2)\).

Question 7. The cost of 18 notebooks is Rs. 423. Find the cost of 20 notebooks.
Answer: Cost of 18 notebooks = Rs. 423
Cost of 1 notebook = \(\frac{423}{18} = \text{Rs. } 23.50\)
Cost of 20 notebooks = \(20 \times 23.50 = \text{Rs. } 470\)
In simple words: Find the cost of one item first using division, then multiply by the required quantity.

Exam Tip: You can also use direct proportion fractions: \(\frac{423}{18} = \frac{x}{20}\) to solve quickly.

Question 8. 24 oranges can be packed 4 cartons. How many oranges can be packed in 12 cartons?
Answer: Let the number of oranges be \(x\).
Since more cartons hold more oranges, this is direct variation.
\(\frac{24}{4} = \frac{x}{12}\)
\(x = \frac{24 \times 12}{4}\)
\(x = 6 \times 12 = 72\)
Number of oranges = 72
In simple words: More cartons mean space for more oranges.

Exam Tip: Verify direct proportion by checking if the ratio \(\frac{\text{Oranges}}{\text{Cartons}}\) is constant.

Question 9. Check whether the given quantities are in proportion or not:
a) 11, 22, 17 and 36
b) 12, 24, 36 and 48
Answer:
(a) Product of extremes = \(11 \times 36 = 396\)
Product of means = \(22 \times 17 = 374\)
Since \(396 \neq 374\), the given numbers are not in proportion.
(b) Product of extremes = \(12 \times 48 = 576\)
Product of means = \(24 \times 36 = 864\)
Since \(576 \neq 864\), the given numbers are not in proportion.
In simple words: Four numbers are in proportion if the product of the first and last equals the product of the middle two.

Exam Tip: Use the cross-product rule \(a \times d = b \times c\) for numbers \(a, b, c, d\) to test proportion.

Question 10. If 18 notebooks cost RS. 1170. How much will 25 books cost?
Answer: Cost of 18 notebooks = Rs. 1170
Cost of 1 notebook = \(\frac{1170}{18} = \text{Rs. } 65\)
Cost of 25 notebooks = \(25 \times 65 = \text{Rs. } 1625\)
In simple words: Find the unit price and multiply by 25.

Exam Tip: Always double check division steps to ensure the unit cost is exact.

Question 11. In which of the following cases do the two quantities x & y vary inversely?
(a)

(b)

Answer:
For table (a):
\(1 \times 800 = 800\)
\(2 \times 400 = 800\)
\(4 \times 200 = 800\)
\(8 \times 100 = 800\)
\(10 \times 80 = 800\)
Since the product \(x \times y\) is constant (800) in all columns, table (a) varies inversely.
For table (b):
\(2 \times 200 = 400\)
\(4 \times 100 = 400\)
\(5 \times 80 = 400\)
\(8 \times 50 = 400\)
\(10 \times 40 = 400\)
Since the product \(x \times y\) is constant (400) in all columns, table (b) also varies inversely.
In simple words: Both tables show inverse variation because the product of \(x\) and \(y\) is constant in every column.

Exam Tip: Test inverse variation by multiplying corresponding \(x\) and \(y\) values across all columns.

Question 12. X & y vary inversely. Complete the following table.

Answer:
Constant product \(k = X \times y = 10 \times 12 = 120\)
(a) When \(X = 5\), \(y = \frac{120}{5} = 24\)
(b) When \(X = 8\), \(y = \frac{120}{8} = 15\)
(c) When \(X = 20\), \(y = \frac{120}{20} = 6\)
In simple words: Find the constant product from the first complete column, then divide by the given value to find the missing one.

Exam Tip: Always identify the complete pair \((X, y)\) first to determine the constant \(k\).

Question 13. Two quantities x & y vary directly. Complete the table.

Answer:
Ratio \(\frac{x}{y} = \frac{6}{12} = \frac{1}{2}\), so \(y = 2x\) and \(x = \frac{y}{2}\).
(a) Given \(y = 18\), \(x = \frac{18}{2} = 9\)
(b) Given \(y = 24\), \(x = \frac{24}{2} = 12\)
(c) Given \(y = 40\), \(x = \frac{40}{2} = 20\)
(d) Given \(x = 18\), \(y = 18 \times 2 = 36\)
(e) Given \(x = 24\), \(y = 24 \times 2 = 48\)
In simple words: In direct variation, \(y\) is always double of \(x\) based on the first column given.

Exam Tip: Simplify the ratio \(\frac{x}{y}\) from the first complete column to establish the direct relation.

Question 14. Check whether x & y vary inversely or not.
Table 1:

Table 2:

Answer:
For Table 1, let us check products \(x \times y\):
\(2 \times 30 = 60\)
\(12 \times 5 = 60\)
\(60 \times 1 = 60\)
\(4 \times 15 = 60\)
\(5 \times 12 = 60\)
\(7.5 \times 8 = 60\)
\(3 \times 20 = 60\)
Since all products are equal to 60, Table 1 varies inversely.
For Table 2, let us check products \(x \times y\):
\(7 \times 6 = 42\)
\(4 \times 8 = 32\)
Since \(42 \neq 32\), Table 2 does not vary inversely.
In simple words: Table 1 varies inversely because all column products are equal to 60, while Table 2 products are unequal.

Exam Tip: Compute products for all columns; even a single mismatched product proves it is not inverse proportion.

Question 15. Complete the table if x & y vary inversely.

Answer:
Constant product \(k = 3 \times 10 = 30\)
(a) When \(y = 8\), \(X = \frac{30}{8} = 3.75\) (or if table value \(y=8\) corresponds to \(X = \frac{30}{8}\))
(b) When \(X = 5\), \(y = \frac{30}{5} = 6\)
(c) When \(X = 1.5\), \(y = \frac{30}{1.5} = 20\)
(d) When \(X = 10\), \(y = \frac{30}{10} = 3\)
In simple words: Multiply any complete \((X, y)\) pair to get 30, then divide by the given value to find the missing one.

Exam Tip: Double-check decimal divisions when finding missing values in inverse tables.

Question 16. A car travels 286 kms on 26 liters of petrol. How far will it travel on 36 liters.
Answer: Distance traveled on 26 liters = 286 km
Distance traveled on 1 liter = \(\frac{286}{26} = 11 \text{ km}\)
Distance traveled on 36 liters = \(36 \times 11 = 396 \text{ km}\)
In simple words: More petrol allows the car to travel a greater distance (direct variation).

Exam Tip: Set up direct proportion: \(\frac{286}{26} = \frac{x}{36}\).

Question 17. A Map is drawn on a scale of 1cm: 1000km, if the distance on the map between the 2 cities is 5cm. What is the actual distance between them?
Answer: Scale: \(1 \text{ cm} = 1000 \text{ km}\)
Map distance = 5 cm
Actual distance = \(5 \times 1000 = 5000 \text{ km}\)
In simple words: Multiply the map distance by the scale factor to get the actual distance.

Exam Tip: Map scale problems are direct proportion applications.

Question 18. The S.I. on a certain sum is Rs. 300.00 for two years. Find the S.I. on the same sum for 6-years at the same rate.
Answer: Simple Interest for 2 years = Rs. 300
Simple Interest for 1 year = \(\frac{300}{2} = \text{Rs. } 150\)
Simple Interest for 6 years = \(6 \times 150 = \text{Rs. } 900\)
Alternatively, since interest is directly proportional to time:
\(\frac{300}{2} = \frac{x}{6} \implies x = \frac{300 \times 6}{2} = \text{Rs. } 900\)
In simple words: Triple the time period results in triple the simple interest.

Question 19. A family of 16 had enough food to lost them for 20-days. If 4 guests arrived suddenly, for how long will the amount of food lost?
Answer: Initial number of persons = 16
New number of persons = \(16 + 4 = 20\)
Let the number of days the food will last be \(d\).
Since food quantity is fixed, number of persons and days vary inversely:
\(16 \times 20 = 20 \times d\)
\(d = \frac{16 \times 20}{20} = 16 \text{ days}\)
The food will last for 16 days.
In simple words: More people eating the same amount of food means it will finish sooner.

Exam Tip: Always calculate the total total person-days \((16 \times 20 = 320)\) and divide by the new total number of people.

Question 20. 26-men can do a piece of work in 18-days. If the work is to be completed in 13-days, how many more men need to be hired?
Answer: Initial men = 26, Days = 18
Total man-days required = \(26 \times 18 = 468\)
Let the total men required for 13 days be \(M\).
\(M \times 13 = 468\)
\(M = \frac{468}{13} = 36 \text{ men}\)
Additional men to be hired = \(36 - 26 = 10 \text{ men}\)
In simple words: Find the total man-days needed, divide by 13 to find the total workforce needed, then subtract the initial 26 men.

Exam Tip: Read carefully: the question asks for "how many more men", so subtract the initial workers from your final calculated total.

Question 21. At the speed of 18 km/hr. a cyclist covers the distance in 190 minutes. At what speed can he cover the same distance in 3 hrs.
Answer: Speed = 18 km/hr
Time 1 = 190 minutes = \(\frac{190}{60} \text{ hours} = \frac{19}{6} \text{ hours}\)
Distance = \(\text{Speed} \times \text{Time} = 18 \times \frac{19}{6} = 3 \times 19 = 57 \text{ km}\)
Time 2 = 3 hours
New speed = \(\frac{\text{Distance}}{\text{Time}} = \frac{57}{3} = 19 \text{ km/hr}\)
In simple words: First find the total distance covered, then divide by the new time (3 hours) to find the new speed.

Exam Tip: Convert all time units to hours before calculating distance or speed.

Question 22. If thirty men can do a piece of work in 15 days, in how many days will 25 men do it?
Answer: Men and days vary inversely.
Let the required number of days be \(d\).
\(30 \times 15 = 25 \times d\)
\(d = \frac{30 \times 15}{25}\)
\(d = \frac{30 \times 3}{5} = 6 \times 3 = 18 \text{ days}\)
25 men will do it in 18 days.
In simple words: Fewer workers mean it will take more days to complete the same work.

Exam Tip: Use inverse proportion formula: \(M_1 D_1 = M_2 D_2\).

Question 23. A car with the speed of 60 km/hour completes a journey in 3 hrs. If the journey can be completed in 4 hrs., what can be the speed of the car.
Answer: Speed and time vary inversely for a fixed distance.
Let the new speed be \(S\).
\(60 \times 3 = S \times 4\)
\(S = \frac{60 \times 3}{4}\)
\(S = 15 \times 3 = 45 \text{ km/hour}\)
The speed of the car is 45 km/hour.
In simple words: Increasing the travel time means the car must travel at a slower speed.

Exam Tip: Apply constant distance relation: \(S_1 \times T_1 = S_2 \times T_2\).