CBSE Class 8 Mathematics Comparing Quantities Practice Paper Worksheet Set 08 with Answers (Old NCERT Syllabus)

This Class 8 Maths Comparing Quantities practice paper has 17 questions for 40 marks in Sections A to E, and every question comes with a step-by-step answer. It follows the earlier (old) NCERT Class 8 Maths syllabus, so it is useful for state-board students and for revising this chapter. The questions test ratios and percentages, discount and marked price, sales tax, VAT and GST, successive discounts, compound interest, depreciation and population growth, and reading a pie chart.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 7. The list price of a music system is ₹ 16,925. If GST is charged at the rate of 12%, calculate the amount to be paid by the customer for purchasing it.
Answer:
GST = 12% of ₹ 16,925 = \( \frac{12}{100} \) × 16925 = ₹ 2,031.
Amount to be paid = List price + GST = 16,925 + 2,031 = ₹ 18,956.
In simple words: Add 12% of ₹ 16,925 (which is ₹ 2,031) to the list price.

Teacher's Note:
1. GST is added on top of the list price.
2. Check: 16,925 × 1.12 = 18,956.

Question 8. Anuj played 55 matches of badminton, out of which he won 35 matches. Find the ratio of matches won to matches lost.
Answer:
Matches lost = 55 − 35 = 20.
Won : Lost = 35 : 20.
Divide both by 5: 7 : 4.
In simple words: He lost 20 matches, so won to lost is 35 : 20 = 7 : 4.

Teacher's Note:
1. The ratio asked is won to lost, not won to total.
2. Always reduce a ratio to its simplest form using the HCF (here 5).

Question 9. Ayesha announced a festival discount of 25% on all the items in her mobile phone shop. Ramandeep bought a mobile phone for himself. He got a discount of ₹ 1,960. What was the marked price of the mobile phone?
Answer:
Let the marked price be ₹ x.
25% of x = 1960, so \( \frac{25}{100} \) × x = 1960.
x = 1960 × \( \frac{100}{25} \) = 1960 × 4 = 7840.
The marked price was ₹ 7,840.
In simple words: ₹ 1,960 is one quarter of the marked price, so the marked price is 4 × 1,960 = ₹ 7,840.

Teacher's Note:
1. The discount percent is a percent of the marked price, so set up 25% of MP = discount.
2. Check: 25% of 7840 = 1960.

Section C: Short Answer Questions (3 Marks Each)

Question 10. A shirt was sold for ₹ 598.40 after allowing a discount of 12% on its marked price. Determine the marked price of the shirt.
Answer:
Let the marked price be ₹ x.
Discount = 12% of x, so selling price = x − 0.12x = 0.88x.
0.88x = 598.40
x = 598.40 ÷ 0.88 = 680.
The marked price of the shirt is ₹ 680.
In simple words: After a 12% discount the buyer pays 88% of the marked price; 88% of ₹ 680 is ₹ 598.40.

Teacher's Note:
1. When the selling price after discount is given, work backwards: SP = (100 − d)% of MP.
2. Check: 12% of 680 = 81.60 and 680 − 81.60 = 598.40.

Question 11. A shopkeeper was selling all his items at 25% discount. During the off season, he offered 30% discount over and above the existing discount. If Pragya bought a skirt which was marked for ₹ 1,200, how much did she pay for it?
Answer:
First discount = 25% of ₹ 1200 = ₹ 300, so the price becomes 1200 − 300 = ₹ 900.
Second discount = 30% of ₹ 900 = ₹ 270.
Amount paid = 900 − 270 = ₹ 630.
In simple words: Take 25% off ₹ 1,200 to get ₹ 900, then take 30% off ₹ 900 to get ₹ 630.

Teacher's Note:
1. Successive discounts are not added: 25% + 30% is not a 55% discount.
2. The second discount is calculated on the already reduced price.

Question 12. A car costing ₹ 3,00,000 depreciates each year by 10% of its value at the beginning of the year. Find its value at the end of 3 years.
Answer:
Here P = ₹ 3,00,000, R = 10% (decrease), n = 3.
Value = P\( \left(1 - \frac{R}{100}\right)^n \) = 300000 × \( \left(\frac{9}{10}\right)^3 \)
= 300000 × \( \frac{729}{1000} \) = 218700.
The value after 3 years is ₹ 2,18,700.
In simple words: Each year the car keeps 90% of its value: 3,00,000 → 2,70,000 → 2,43,000 → 2,18,700.

Teacher's Note:
1. Depreciation uses the compound interest formula with a minus sign.
2. Each year's 10% is on the value at the start of that year, not on the original cost.

Question 13. The population of a village is 25,000. It increases at the rate of 2% every year. Find the population at the end of two years.
Answer:
Population after n years = P\( \left(1 + \frac{R}{100}\right)^n \).
= 25000 × \( \left(1 + \frac{2}{100}\right)^2 \) = 25000 × \( \frac{51}{50} \times \frac{51}{50} \)
= 10 × 51 × 51 = 26010.
The population after two years is 26,010 (an increase of 1,010).
In simple words: Add 2% twice: 25,000 → 25,500 → 26,010.

Teacher's Note:
1. Population growth works like compound interest: each year's increase is on the new population.
2. Simple growth would give only 26,000; the extra 10 is the growth on the first year's increase.

Section D: Long Answer Questions (4 Marks Each)

Question 14. Hamid goes to a shop to buy a grinder costing ₹ 2700. The rate of VAT is 8%. He tells the shopkeeper to allow a discount on the price of the grinder to such an extent that he pays ₹ 2700 inclusive of VAT. Find the discount on the price of the grinder.
Answer:
Let the reduced price (after discount) be ₹ x.
VAT = 8% of x, so the amount paid = x + 0.08x = 1.08x.
1.08x = 2700, so x = 2700 ÷ 1.08 = 2500.
Reduced price = ₹ 2500.
Discount = 2700 − 2500 = ₹ 200.
In simple words: The price before VAT must be ₹ 2,500 so that adding 8% makes ₹ 2,700; the discount is ₹ 200.

Teacher's Note:
1. VAT is charged on the discounted price, so set up the equation on the reduced price.
2. Check: 8% of 2500 = 200 and 2500 + 200 = 2700.

Question 15. The population of a village increased to 44100 in 2003 at a rate of 5% per annum. What was the population in 2001?
Answer:
Let the population in 2001 be P. From 2001 to 2003, n = 2 years, R = 5%.
44100 = P\( \left(1 + \frac{5}{100}\right)^2 \) = P × \( \frac{21}{20} \times \frac{21}{20} = P \times \frac{441}{400} \)
P = 44100 × \( \frac{400}{441} \) = 100 × 400 = 40000.
The population in 2001 was 40,000.
In simple words: Undo two years of 5% growth: 44,100 ÷ 1.05 ÷ 1.05 = 40,000.

Teacher's Note:
1. When the final value is given, put it on the left of the growth formula and solve for P.
2. Check: 40,000 → 42,000 → 44,100.

Question 16. Given the principal = ₹ 40,000, rate of interest = 8% p.a. compounded annually. Find
(a) Interest if the period is one year.
(b) Principal for the 2nd year.
(c) Interest for the 2nd year.
(d) Amount if the period is 2 years.
Answer:
(a) Interest for 1 year = \( \frac{40000 \times 8 \times 1}{100} \) = ₹ 3,200
(b) Principal for the 2nd year = 40,000 + 3,200 = ₹ 43,200
(c) Interest for the 2nd year = \( \frac{43200 \times 8 \times 1}{100} \) = ₹ 3,456
(d) Amount after 2 years = 43,200 + 3,456 = ₹ 46,656
Check: 40000 × (1.08)² = 46,656.
In simple words: Find one year's interest, add it to the principal, and repeat for the second year.

Teacher's Note:
1. Compounding annually means the interest is added to the principal at the end of each year.
2. The formula A = P(1 + R/100)ⁿ gives the same answer in one step.

Section E: Case Study Based Question (4 Marks)

Question 17. In a primary school, the parents were asked about the number of hours they spend per day in helping their children to do homework. There were 90 parents who helped for ½ hr to 1½ hr. The distribution of parents according to the time for which they said they helped is given in the adjoining figure: 20% helped for more than 1½ hr per day, 30% helped for ½ hr to 1½ hr, and 50% did not help at all.
Using this, answer the following:
(a) How many parents were surveyed? (2)
(b) How many said that they did not help? (1)
(c) How many said that they helped for more than 1½ hours? (1)
Answer:
(a) Let the number of parents surveyed be x.
30% of x = 90, so \( \frac{30}{100} \) × x = 90 and x = 90 × \( \frac{100}{30} \) = 300.
300 parents were surveyed.
(b) Did not help = 50% of 300 = 150 parents.
(c) Helped for more than 1½ hours = 20% of 300 = 60 parents.
Check: 90 + 150 + 60 = 300.
In simple words: 90 parents are 30% of everyone, so 300 were surveyed; half (150) did not help and one-fifth (60) helped for over 1½ hours.

Teacher's Note:
1. When a part and its percentage are known, divide the part by the percentage to get the whole.
2. Check that the three groups add up to the total surveyed.