Class 8 Maths - Comparing Quantities NCERT Solutions

Chapter 8: Comparing Quantities (NCERT Solutions)

Exercise 8.1

Q1. Find the ratio of the following:
(a) Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.
(b) 5 m to 10 km
(c) 50 paise to ₹ 5

(a) Ratio of speed of cycle to the speed of scooter = 15 km/hr : 30 km/hr
= 15 / 30 = 1 / 2 = 1 : 2

(b) Since 1 km = 1000 m, therefore, 10 km = 10000 m.
Ratio = 5 m to 10000 m = 5 / 10000 = 1 / 2000 = 1 : 2000

(c) Since ₹ 1 = 100 paise, therefore, ₹ 5 = 500 paise.
Ratio = 50 paise to 500 paise = 50 / 500 = 1 / 10 = 1 : 10

Q2. Convert the following ratios to percentages.
(a) 3 : 4 (b) 2 : 3

(a) Percentage of 3 : 4 = (3 / 4) × 100%
= 3 × 25% = 75%

(b) Percentage of 2 : 3 = (2 / 3) × 100%
= 200% / 3 = 66 ²⁄₃% (or 66.66%)

Q3. 72% of 25 students are interested in mathematics. How many are not interested in mathematics?

Percentage of students interested in mathematics = 72%
Therefore, percentage of students NOT interested in mathematics = 100% - 72% = 28%
Number of students not interested in mathematics = 28% of 25
= (28 / 100) × 25 = 28 / 4 = 7 students

Q4. A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all?

Let the total number of matches played be x.
According to the question, 40% of x = 10
(40 / 100) × x = 10
(2 / 5)x = 10
x = (10 × 5) / 2 = 50 / 2 = 25 matches
Thus, they played 25 matches in all.

Q5. If Chameli had ₹ 600 left after spending 75% of her money, how much did she have in the beginning?

Given, percentage of money spent = 75%
So, percentage of money left = 100% - 75% = 25%
Let the money she had in the beginning be x.
25% of x = 600
(25 / 100) × x = 600
(1 / 4)x = 600
x = 600 × 4 = ₹ 2400
She had ₹ 2400 in the beginning.

Q6. If 60% people in a city like cricket, 30% like football and the remaining like other games, then what per cent of the people like other games? If the total number of people is 50 lakh, find the exact number who like each type of game.

Percentage of people who like cricket = 60%
Percentage of people who like football = 30%
Percentage of people who like other games = 100% - (60% + 30%) = 100% - 90% = 10%

Total number of people = 50 lakh = 50,00,000.
Number of people who like cricket = 60% of 50 lakh
= (60 / 100) × 50 = 30 lakh
Number of people who like football = 30% of 50 lakh
= (30 / 100) × 50 = 15 lakh
Number of people who like other games = 10% of 50 lakh
= (10 / 100) × 50 = 5 lakh

Exercise 8.2

Q1. A man got a 10% increase in his salary. If his new salary is ₹ 1,54,000, find his original salary.

Let original salary be x.
Increase percentage = 10%. Increase in salary = 10% of x = (10/100)x = 0.1x
New salary = Original salary + Increase
1,54,000 = x + 0.1x
1.1x = 1,54,000
x = 1,54,000 / 1.1
x = 1540000 / 11 = ₹ 1,40,000
His original salary is ₹ 1,40,000.

Q2. On Sunday 845 people went to the Zoo. On Monday only 169 people went. What is the per cent decrease in the people visiting the Zoo on Monday?

People who went on Sunday = 845
People who went on Monday = 169
Decrease in number of people = 845 - 169 = 676
Decrease Percentage = (Decrease / Original Number) × 100
= (676 / 845) × 100
= (4 / 5) × 100 (Dividing both 676 and 845 by 169)
= 4 × 20 = 80%
Hence, percentage decrease is 80%.

Q3. A shopkeeper buys 80 articles for ₹ 2,400 and sells them for a profit of 16%. Find the selling price of one article.

Cost Price (CP) of 80 articles = ₹ 2400.
Profit percentage = 16%.
Profit = 16% of CP
= (16 / 100) × 2400 = 16 × 24 = ₹ 384.
Selling Price (SP) of 80 articles = CP + Profit = 2400 + 384 = ₹ 2784.
Selling Price of one article = 2784 / 80 = 278.4 / 8 = ₹ 34.80.

Q4. The cost of an article was ₹ 15,500. ₹ 450 were spent on its repairs. If it is sold for a profit of 15%, find the selling price of the article.

Purchasing cost = ₹ 15,500. Overhead cost (repairs) = ₹ 450.
Total Cost Price (CP) = 15500 + 450 = ₹ 15,950.
Profit % = 15%.
Profit = 15% of CP = (15 / 100) × 15950 = 15 × 159.5 = ₹ 2392.50.
Selling Price (SP) = CP + Profit = 15950 + 2392.50 = ₹ 18,342.50.

Q5. A VCR and TV were bought for ₹ 8,000 each. The shopkeeper made a loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss percent on the whole transaction.

CP of VCR = ₹ 8000, CP of TV = ₹ 8000.
Total CP = 8000 + 8000 = ₹ 16000.
For VCR: Loss = 4%. Loss amount = 4% of 8000 = (4/100) × 8000 = ₹ 320.
SP of VCR = 8000 - 320 = ₹ 7680.
For TV: Profit = 8%. Profit amount = 8% of 8000 = (8/100) × 8000 = ₹ 640.
SP of TV = 8000 + 640 = ₹ 8640.
Total SP = 7680 + 8640 = ₹ 16320.
Since Total SP > Total CP, it is an overall Gain (Profit).
Overall Profit = Total SP - Total CP = 16320 - 16000 = ₹ 320.
Overall Gain % = (Profit / Total CP) × 100 = (320 / 16000) × 100 = 32000 / 16000 = 2%.
Thus, there is a gain of 2% on the whole transaction.

Q6. During a sale, a shop offered a discount of 10% on the marked prices of all the items. What would a customer have to pay for a pair of jeans marked at ₹ 1450 and two shirts marked at ₹ 850 each?

Marked Price of a pair of jeans = ₹ 1450.
Marked Price of two shirts = 2 × 850 = ₹ 1700.
Total Marked Price (MP) = 1450 + 1700 = ₹ 3150.
Discount percentage = 10%.
Discount Amount = 10% of 3150 = (10 / 100) × 3150 = ₹ 315.
Amount customer has to pay (Sale Price) = MP - Discount
= 3150 - 315 = ₹ 2835.

Q7. A milkman sold two of his buffaloes for ₹ 20,000 each. On one he made a gain of 5% and on the other a loss of 10%. Find his overall gain or loss.

SP of each buffalo = ₹ 20,000.
First Buffalo (5% gain): Let CP be x.
SP = CP × (100 + Gain%) / 100
20000 = x × (105 / 100)
x = (20000 × 100) / 105 = 2000000 / 105 = ₹ 19047.62 (approx. CP₁)
Second Buffalo (10% loss): Let CP be y.
SP = CP × (100 - Loss%) / 100
20000 = y × (90 / 100)
y = (20000 × 100) / 90 = 2000000 / 90 = ₹ 22222.22 (approx. CP₂)
Total CP = CP₁ + CP₂ = 19047.62 + 22222.22 = ₹ 41269.84.
Total SP = 20000 + 20000 = ₹ 40000.
Since Total CP > Total SP, it's a loss.
Overall loss = 41269.84 - 40000 = ₹ 1269.84 (approx).

Q8. The price of a TV is ₹ 13,000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it.

Price of TV = ₹ 13,000.
Sales tax rate = 12%.
Sales tax amount = 12% of 13000
= (12 / 100) × 13000 = 12 × 130 = ₹ 1560.
Total amount Vinod has to pay = Price of TV + Sales tax
= 13000 + 1560 = ₹ 14,560.

Q9. Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is ₹ 1,600, find the marked price.

Sale Price (SP) = ₹ 1600. Discount % = 20%.
Let the marked price (MP) be x.
SP = MP - Discount
1600 = x - (20% of x)
1600 = x - 0.20x
1600 = 0.80x
x = 1600 / 0.80 = 160000 / 80 = ₹ 2000.
Thus, marked price was ₹ 2000.

Q10. I purchased a hair-dryer for ₹ 5,400 including 8% VAT. Find the price before VAT was added.

Price including VAT = ₹ 5400. VAT % = 8%.
Let original price be x.
Price including VAT = x + (8% of x)
5400 = x + 0.08x = 1.08x
x = 5400 / 1.08 = 540000 / 108 = ₹ 5000.
Original price before VAT was ₹ 5000.

Q11. An article was purchased for ₹ 1239 including GST of 18%. Find the price of the article before GST was added?

Price including GST = ₹ 1239. GST % = 18%.
Let price before GST be x.
Price including GST = x + 18% of x = 1.18x
1.18x = 1239
x = 1239 / 1.18 = 123900 / 118 = ₹ 1050.
Original price was ₹ 1050.

Exercise 8.3

Q1. Calculate the amount and compound interest on:
(a) ₹ 10,800 for 3 years at 12 ¹⁄₂% per annum compounded annually.
(b) ₹ 18,000 for 2 ¹⁄₂ years at 10% per annum compounded annually.

(a) P = ₹ 10800, R = 12.5% = 25/2 %, n = 3 years.
A = P(1 + R/100)ⁿ = 10800(1 + 25/(2×100))³
= 10800 × (1 + 1/8)³ = 10800 × (9/8) × (9/8) × (9/8)
= 10800 × 729 / 512 = ₹ 15377.34.
Amount = ₹ 15,377.34.
CI = A - P = 15377.34 - 10800 = ₹ 4577.34.

(b) P = ₹ 18000, R = 10%, n = 2 ¹⁄₂ years.
Amount for 2 years (A₂) = 18000(1 + 10/100)²
= 18000(11/10)(11/10) = 180 × 121 = ₹ 21780.
Interest for the half year = (P × R × T) / 100 on principal ₹ 21780.
I = (21780 × 10 × 1/2) / 100 = 21780 × 5 / 100 = ₹ 1089.
Total Amount = A₂ + I = 21780 + 1089 = ₹ 22,869.
CI = A - P = 22869 - 18000 = ₹ 4,869.

Q2. Kamala borrowed ₹ 26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?

P = ₹ 26400, R = 15%, n = 2 years 4 months (2 ⁴⁄₁₂ = 2 ¹⁄₃ years).
For 2 years: A = 26400(1 + 15/100)²
= 26400(115/100)(115/100) = 26400 × (23/20) × (23/20)
= 66 × 529 = ₹ 34914.
For remaining 4 months (1/3 year), SI on ₹ 34914:
SI = (34914 × 15 × 1/3) / 100 = (34914 × 5) / 100 = 174570 / 100 = ₹ 1745.70
Total Amount = 34914 + 1745.70 = ₹ 36,659.70.

Q3. Fabina borrows ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?

Fabina (Simple Interest):
P = 12500, R = 12%, T = 3 years.
SI = (12500 × 12 × 3) / 100 = 125 × 36 = ₹ 4500.

Radha (Compound Interest):
P = 12500, R = 10%, n = 3 years.
A = 12500(1 + 10/100)³ = 12500(11/10)³
= 12500 × (1331 / 1000) = 12.5 × 1331 = ₹ 16637.50.
CI = A - P = 16637.50 - 12500 = ₹ 4137.50.

Fabina pays more interest. Difference = 4500 - 4137.50 = ₹ 362.50.

Q4. I borrowed ₹ 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?

Simple Interest:
P = 12000, R = 6%, T = 2 years.
SI = (12000 × 6 × 2) / 100 = 120 × 12 = ₹ 1440.

Compound Interest:
P = 12000, R = 6%, n = 2 years.
A = 12000(1 + 6/100)² = 12000(106/100)²
= 12000 × (106/100) × (106/100) = 1.2 × 11236 = ₹ 13483.20.
CI = 13483.20 - 12000 = ₹ 1483.20.

Extra amount to pay = CI - SI = 1483.20 - 1440 = ₹ 43.20.

Q5. Vasudevan invested ₹ 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get:
(i) after 6 months?
(ii) after 1 year?

P = ₹ 60000. Rate compounded half yearly = R/2 = 12% / 2 = 6% per half year.

(i) After 6 months (n = 1 half year):
A = P(1 + R/100)ⁿ = 60000(1 + 6/100)¹
= 60000(106/100) = 600 × 106 = ₹ 63,600.

(ii) After 1 year (n = 2 half years):
A = 60000(1 + 6/100)² = 60000(106/100)(106/100)
= 6 × 106 × 106 = 6 × 11236 = ₹ 67,416.

Q6. Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after 1 ¹⁄₂ years if the interest is
(i) compounded annually.
(ii) compounded half yearly.

P = ₹ 80000. R = 10%. Time = 1.5 years.

(i) Compounded Annually:
For 1 year: A₁ = 80000(1 + 10/100)¹ = 80000(11/10) = ₹ 88000.
For next 1/2 year (SI on ₹ 88000): SI = (88000 × 10 × 1/2) / 100 = 880 × 5 = ₹ 4400.
Total Amount (Annually) = 88000 + 4400 = ₹ 92,400.

(ii) Compounded Half Yearly:
Rate per half year = 10% / 2 = 5%. Time (n) = 1.5 years = 3 half years.
A = 80000(1 + 5/100)³ = 80000(105/100)³
= 80000(21/20)(21/20)(21/20) = 80000(9261 / 8000) = 10 × 9261 = ₹ 92,610.

Difference = 92610 - 92400 = ₹ 210.

Q7. Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find:
(i) The amount credited against her name at the end of the second year.
(ii) The interest for the 3rd year.

P = 8000, R = 5%.

(i) Amount at end of 2nd year (n=2):
A₂ = 8000(1 + 5/100)² = 8000(105/100)²
= 8000(21/20)² = 8000 × 441 / 400 = 20 × 441 = ₹ 8,820.

(ii) Interest for 3rd year:
This is SI on the amount at the end of 2nd year.
Principal for 3rd year = ₹ 8820.
Interest = (8820 × 5 × 1) / 100 = (8820 × 5) / 100 = 44100 / 100 = ₹ 441.

Q8. Find the amount and the compound interest on ₹ 10,000 for 1 ¹⁄₂ years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

Compounded half-yearly:
P = 10000. Rate per half year = 10/2 = 5%. n = 1.5 years = 3 half years.
Amount = 10000(1 + 5/100)³ = 10000(21/20)³
= 10000 × 9261 / 8000 = 1.25 × 9261 = ₹ 11,576.25.
CI (half-yearly) = 11576.25 - 10000 = ₹ 1576.25.

Compounded Annually:
For 1 year: A₁ = 10000(1 + 10/100) = ₹ 11000.
For next 1/2 year: SI = (11000 × 10 × 1/2) / 100 = ₹ 550.
Total Amount (Annually) = 11000 + 550 = ₹ 11550.
CI (Annually) = 11550 - 10000 = ₹ 1550.

Since ₹ 1576.25 > ₹ 1550, Yes, the interest compounded half yearly is more.

Q9. Find the amount which Ram will get on ₹ 4096, if he gave it for 18 months at 12 ¹⁄₂% per annum, interest being compounded half yearly.

P = ₹ 4096.
Time = 18 months = 1.5 years = 3 half years (n = 3).
Rate = 12.5% p.a. = 25/2 % p.a. ⇒ Rate per half year = 25/4 %.
A = P(1 + R/100)ⁿ
A = 4096(1 + 25/(4×100))³ = 4096(1 + 1/16)³
= 4096 × (17/16)³ = 4096 × (4913 / 4096)
= ₹ 4,913.

Q10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum
(i) find the population in 2001.
(ii) what would be its population in 2005?

(i) Population in 2001 (Let it be P):
A (in 2003) = 54000, R = 5%, n = 2 years (2003 - 2001).
A = P(1 + R/100)ⁿ
54000 = P(1 + 5/100)² = P(21/20)² = P(441/400)
P = (54000 × 400) / 441 = 21600000 / 441 ≈ 48,980 (approx).

(ii) Population in 2005 (Let 2003 be P):
P = 54000, R = 5%, n = 2 years (2005 - 2003).
A = 54000(1 + 5/100)² = 54000(21/20)²
= 54000 × (441 / 400) = 135 × 441 = 59,535.

Q11. In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.

Initial count (P) = 506000. Rate (R) = 2.5% per hour. Time (n) = 2 hours.
A = P(1 + R/100)ⁿ
A = 506000(1 + 2.5/100)² = 506000(1 + 25/1000)² = 506000(1 + 1/40)²
= 506000 × (41/40)² = 506000 × (1681 / 1600)
= 316.25 × 1681 = 5,31,616.25
Thus, approximate count is 5,31,616.

Q12. A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

P = ₹ 42000. Depreciation Rate (R) = 8%. Time (n) = 1 year.
Value after depreciation = P(1 - R/100)ⁿ
= 42000(1 - 8/100)¹
= 42000(92 / 100) = 420 × 92 = ₹ 38,640.

Class 8 Maths - Comparing Quantities Practice Questions

Chapter 8: Comparing Quantities (Practice Questions)

RD Sharma / Extra Practice Questions

Q1. Find the ratio of 3 km to 300 m.

Both quantities must be in the same unit.
1 km = 1000 m
3 km = 3000 m
Ratio = 3000 m : 300 m = 3000 / 300 = 10 / 1 = 10 : 1.

Q2. Convert the ratio 3 : 4 to percentage.

Percentage = (3/4) × 100%
= 3 × 25% = 75%.

Q3. 60% of students in a class of 50 are boys. Find the number of girls.

Percentage of boys = 60%.
Percentage of girls = 100% - 60% = 40%.
Number of girls = 40% of 50
= (40 / 100) × 50 = (4 / 10) × 50 = 4 × 5 = 20.

Q4. The price of a scooter was ₹ 34,000 last year. It has increased by 20% this year. What is the price now?

Original Price = ₹ 34,000.
Increase = 20% of 34,000
= (20 / 100) × 34000 = 20 × 340 = ₹ 6,800.
New Price = Original Price + Increase
= 34000 + 6800 = ₹ 40,800.

Q5. An item marked at ₹ 840 is sold for ₹ 714. What is the discount and discount percentage?

Marked Price (MP) = ₹ 840. Selling Price (SP) = ₹ 714.
Discount = MP - SP = 840 - 714 = ₹ 126.
Discount % = (Discount / MP) × 100
= (126 / 840) × 100
= (1260 / 84) = 15%.

Q6. A shopkeeper bought a chair for ₹ 375 and sold it for ₹ 400. Find the gain percentage.

Cost Price (CP) = ₹ 375. Selling Price (SP) = ₹ 400.
Since SP > CP, it is a gain (profit).
Gain = SP - CP = 400 - 375 = ₹ 25.
Gain % = (Gain / CP) × 100
= (25 / 375) × 100 = (1 / 15) × 100 = 100 / 15 = 20 / 3 = 6.66% (or 6 ²⁄₃%).

Q7. Cost of an item is ₹ 50. It was sold with a profit of 12%. Find the selling price.

CP = ₹ 50. Profit % = 12%.
Profit = 12% of 50 = (12 / 100) × 50 = 600 / 100 = ₹ 6.
Selling Price (SP) = CP + Profit = 50 + 6 = ₹ 56.

Q8. Rehana bought a dress for ₹ 5400 including a 8% VAT. Find the price of the dress before VAT was added.

Let original price be x.
VAT = 8% of x = 0.08x.
Price including VAT = x + 0.08x = 1.08x.
1.08x = 5400
x = 5400 / 1.08 = 540000 / 108 = ₹ 5000.

Q9. Find Compound Interest on ₹ 12600 for 2 years at 10% per annum compounded annually.

P = ₹ 12600, R = 10%, n = 2 years.
A = P(1 + R/100)ⁿ
A = 12600 × (1 + 10/100)²
A = 12600 × (11/10) × (11/10)
A = 126 × 121 = ₹ 15246.
Compound Interest (CI) = A - P = 15246 - 12600 = ₹ 2646.

Q10. Calculate the amount and CI on ₹ 8000 for 1 year at 9% per annum compounded half yearly.

P = ₹ 8000, R = 9% per annum = 4.5% per half year. Time = 1 year = 2 half years (n=2).
A = P(1 + R/100)ⁿ
A = 8000 × (1 + 4.5/100)² = 8000 × (104.5/100)²
A = 8000 × 1.045 × 1.045 = 8000 × 1.092025 = ₹ 8736.20.
CI = A - P = 8736.20 - 8000 = ₹ 736.20.

Q11. A TV was bought at a price of ₹ 21,000. After one year the value of the TV was depreciated by 5%. Find the value of the TV after one year.

P = ₹ 21000, R = 5%, n = 1 year. (Depreciation means reduction).
Value = P(1 - R/100)ⁿ
= 21000 × (1 - 5/100)¹
= 21000 × (95/100)
= 210 × 95 = ₹ 19,950.

Q12. If 8% VAT is included in the prices, find the original price of a TV bought for ₹ 13,500.

Let original price be ₹ 100.
With 8% VAT, price = ₹ 108.
If price with VAT is 108, original = 100.
If price with VAT is 13500, original = (100 / 108) × 13500
= 100 × 125 = ₹ 12,500.

Q13. Find the saving if 20% discount is offered on a bag marked at ₹ 250.

Saving = Discount amount.
Discount = 20% of Marked Price
= (20 / 100) × 250 = (1 / 5) × 250 = ₹ 50.

Q14. 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 1/2 hour to 1 1/2 hours. The distribution of parents (in %) is: 1.5 hrs or more (20%), 1/2 to 1.5 hrs (30%), till 1/2 hr (50%). How many parents were surveyed?

Parents helping for 1/2 to 1.5 hours = 30%.
It is given that these are 90 parents.
Let total parents be x.
30% of x = 90
(30/100) × x = 90
x = (90 × 100) / 30 = 3 × 100 = 300 parents.

Q15. Out of 15,000 voters in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?

Percentage of voters who voted = 60%.
Percentage of voters who did NOT vote = 100% - 60% = 40%.
Number of voters who did not vote = 40% of 15000
= (40/100) × 15000 = 40 × 150 = 6000 voters.

Q16. Find the simple interest on ₹ 2500 for 2 years 6 months at 6% per annum.

P = ₹ 2500, R = 6% per annum.
Time (T) = 2 years 6 months = 2.5 years (or 5/2 years).
Simple Interest = (P × R × T) / 100
= (2500 × 6 × 2.5) / 100 = 25 × 15 = ₹ 375.

Q17. The population of a city was 20,000 in the year 1997. It increased at the rate of 5% p.a. Find the population at the end of the year 2000.

P = 20000. Rate of increase (R) = 5%. Time (n) = 2000 - 1997 = 3 years.
Population = P(1 + R/100)ⁿ
= 20000 × (1 + 5/100)³
= 20000 × (21/20) × (21/20) × (21/20)
= 20000 × (9261 / 8000)
= 2.5 × 9261 = 23,152.5 (Since population must be an integer, it's approx. 23,153).

Q18. What amount is to be repaid on a loan of ₹ 12000 for 1.5 years at 10% per annum compounded half yearly.

P = ₹ 12000. Rate (R) = 10% p.a. = 5% per half year. Time (n) = 1.5 years = 3 half years.
A = P(1 + R/100)ⁿ
A = 12000 × (1 + 5/100)³
A = 12000 × (21/20) × (21/20) × (21/20)
A = 12000 × (9261 / 8000)
A = 1.5 × 9261 = ₹ 13891.50.

Q19. Find the buying price of a towel when 5% GST is added on the purchase of ₹ 50.

Price (without GST) = ₹ 50.
GST Amount = 5% of ₹ 50 = (5/100) × 50 = 250/100 = ₹ 2.50.
Buying Price = Price + GST Amount
= 50 + 2.50 = ₹ 52.50.

Q20. A shop gives 20% discount. What would the sale price of a dress marked at ₹ 120?

Marked Price (MP) = ₹ 120. Discount % = 20%.
Discount Amount = 20% of 120 = (20/100) × 120 = ₹ 24.
Sale Price (SP) = MP - Discount
= 120 - 24 = ₹ 96.

Class 8 Maths - Comparing Quantities Summary

Chapter 8: Comparing Quantities (Concepts & Formulas)

1. Ratio and Percentage

Ratio: Comparing two quantities of the same kind by division. The ratio of a to b is written as a : b or a/b.
Percentage: Means 'per hundred' or out of 100. It is denoted by the symbol %.
To convert a fraction into a percentage, multiply it by 100.
To convert a percentage into a fraction, divide it by 100.
Percentage Increase/Decrease:
Increase % = (Increase in value / Original value) × 100
Decrease % = (Decrease in value / Original value) × 100

2. Discount

Discount is a reduction given on the Marked Price (MP) of an article.

Discount = Marked Price (MP) - Sale Price (SP)
Discount % = (Discount / Marked Price) × 100
If discount percentage is given: SP = MP - (Discount % of MP)

3. Profit and Loss

Cost Price (CP): The price at which an article is bought. Overhead expenses (like repairs, transport) are added to the CP.
Selling Price (SP): The price at which an article is sold.
If SP > CP, there is a profit. Profit = SP - CP
If CP > SP, there is a loss. Loss = CP - SP
Profit % = (Profit / CP) × 100
Loss % = (Loss / CP) × 100
Note: Profit and loss are always calculated on the Cost Price (CP).

4. Sales Tax / VAT / GST

Sales Tax (ST): Charged by the government on the sale of an item. It is collected by the shopkeeper from the customer and given to the government. It is, therefore, always on the selling price of an item.
Value Added Tax (VAT): Another type of tax included in the prices of goods.
Goods and Services Tax (GST): A tax levied on the supply of goods or services or both.
Bill Amount (or Final Price) = Original Price + Tax Amount

5. Compound Interest

Compound Interest is the interest calculated on the initial principal, which also includes all of the accumulated interest from previous periods.

Amount Formula (Compounded Annually):
A = P(1 + R/100)ⁿ
Where,
A = Amount
P = Principal
R = Rate of interest per annum
n = Time period in years
Compound Interest (CI) = Amount (A) - Principal (P)
Compounded Half Yearly: Rate becomes half (R/2) and Time becomes double (2n).
A = P(1 + R/(2 × 100))²ⁿ

Applications of Compound Interest Formula:

Increase or decrease in population.
Growth of bacteria.
Depreciation in the value of an item: A = P(1 - R/100)ⁿ

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