# NCERT Solutions for Class 8 Maths Chapter 7 Comparing Quantities (Updated Pattern)

Chapter 7 of NCERT Solutions for Class 8 Maths is really important to learn as the basic knowledge of tax, interest, profit, loss and much more are conveyed in this chapter. The concepts discussed in this chapter are:

- Discount is a reduction given on marked price. Discount = Marked Price – Sale Price.
- Discounts can be calculated when the discount percentage is given. Discount = Discount % of the Marked Price
- Additional expenses made after buying an article are included in the cost price and are known as overhead expenses. C.P. = Buying price + Overhead expenses
- Sales tax is charged on the sale of an item by the government and is added to the Bill Amount. Sales tax = Tax% of Bill Amount.
- GST stands for Goods and Services Tax and is levied on the supply of goods or services or both.
- Compound interest is the interest calculated on the previous year’s amount (A = P + I)

**———————–Exercise-7.1——————————–**

**1. Find the ratio of the following:**

**(a) Speed of a cycle 15 km per hour to the speed of a scooter 30 km per hour.**

**(b) 5 m to 10 km**

**(c) 50 paise to ₹ 5**

**Solution:**

a) Ratio of the speed of the cycle to the speed of the scooter = 15/30 = ½ = 1:2

b) Since 1 km = 1000 m

5m/10 km = 5 m/(10 x 1000)m = 5/10000 = 1/2000 = 1:2000

The required ratio is 1:2000

c) Since, ₹1 = 100 paise

50 paise/₹5 = 50/(5 x 100) = 50/500 = 1/10 = 1:10

The required ratio is 1:10

**2. Convert the following ratio to percentages:**

**a) 3:4**

**b) 2:3**

**Solution:**

a) 3:4 = ¾ = ¾ x 100% = 0.75 x 100% = 75%

b) 2:3 = 2/3 = 2/3 x 100% = 0.666 x 100% = 66.66% = 66⅔%

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**3. 72% of 25 students are good in mathematics. How many are not good in mathematics?**

**Solution:**

It’s given that 72% of 25 students are good in mathematics

So, the percentage of students who are not good in mathematics = (100 – 72)%

= 28%

Here, the number of students who are good in mathematics = 72/100 x 25 = 18

Thus, the number of students who are not good in mathematics = 25 – 18 = 7[Also, 28% of 25 = 28/100 x 25 = 7]

Therefore, 7 students are not good in mathematics.

**4. 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?**

**Solution:**

Let the total number of matches played by the team be x.

Given that the team won 10 matches and the winning percentage of the team was 40%.

⇒ 40/100 × x = 10

40x = 10 × 100

40x = 1000

x = 1000/40

= 100/4

= 25

Therefore, the team played 25 matches.

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

**Solution:**

Let the amount of money which Chameli had, in the beginning, be x

Given that, after spending 75% of ₹x, she was left with ₹600

So, (100 – 75)% of x = ₹600

Or, 25% of x = ₹600

25/100 × x = ₹600

x = ₹600 × 4

= ₹2400

Therefore, Chameli had ₹2400 in the beginning.

**6. If 60% of people in the 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 lakhs, find the exact number who like each type of game.**

**Solution:**

Percentage of people who like other games = (100 – 60 – 30)%

= (100 – 90)%

= 10%

Total number of people = 50 lakhs

So,

Number of people who like cricket = 60/100 x 50 = 30 lakhs

Number of people who like football = 30/100 x 50 = 15 lakhs

Number of people who like other games = 10/100 x 50 = 5 lakhs

**———————–Exercise-7.2——————————–**

**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?**

**Solution:**

Total market price = ₹ (1,450 + 2 × 850)

= ₹ (1,450 +1,700)

= ₹ 3,150

Given that, the discount percentage = 10%

Discount = ₹ (10/100 x 3150) = ₹ 315

Also, Discount = Marked price − Sale price

₹ 315 = ₹ 3150 − Sale price

∴ Sale price = ₹ (3150 − 315)

= ₹ 2835

Therefore, the customer will have to pay ₹ 2,835.

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**2.** **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.**

**Solution:**

On ₹ 100, the tax to be paid = ₹ 12

Here, on ₹ 13000, the tax to be paid will be = 12/100 × 13000

= ₹ 1560

Required amount = Cost + Sales Tax

= ₹ 13000 + ₹ 1560

= ₹ 14560

Therefore, Vinod will have to pay ₹ 14,560 for the TV.

**3. 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.**

**Solution:**

Let the marked price be x

Discount percent = Discount/Marked Price x 100

20 = Discount/x × 100

Discount = 20/100 × x

= x/5

Also,

Discount = Marked price – Sale price

x/5 = x – ₹ 1600

x – x/5 = 1600

4x/5 = 1600

x = 1600 x 5/4

= 2000

Therefore, the marked price was ₹ 2000.

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

**Solution:**

The price includes VAT

So, 8% VAT means that if the price without VAT is ₹ 100,

Then, the price including VAT will be ₹ 108

When price including VAT is ₹ 108, original price = ₹ 100

When price including VAT is ₹ 5400, original price = ₹ (100/108 × 5400)

= ₹ 5000

Therefore, the price of the hair dryer before the addition of VAT was ₹ 5,000.

**5. An article was purchased for ₹ 1239 including a GST of 18%. Find the price of thearticle before GST was added?**

**Solution:** Let the original price of the article be 100.

GST = 18%.

Price after GST is included = (100+18) = 118

When the selling price is 118 then the original price = 100.

When the selling price is 1239, then the original price = 100/118 × 1239 = 1050

**———————–Exercise-7.3——————————–**

**The population of a place increased to 54000 in 2003 at a rate of 5% per annum**

**(i) find the population in 2001**

**(ii) what would be its population in 2005?**

**Solution:**

(i) It’s given that the population in the year 2003 = 54,000

54,000 = (Population in 2001) (1 + 5/100)^{2}

54,000 = (Population in 2001) (105/100)^{2}

Population in 2001 = 54000 x (100/105)^{2}

= 48979.59

Therefore, the population in the year 2001 was approximately 48,980

(ii) Population in 2005 = 54000(1 + 5/100)^{2}

= 54000(105/100)^{2}

= 54000(21/20)^{2}

= 59535

Therefore, the population in the year 2005 would be 59,535.

**2. 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.**

**Solution:**

The initial count of bacteria is given as 5,06,000

Bacteria at the end of 2 hours = 506000(1 + 2.5/100)^{2}

= 506000(1 + 1/40)^{2}

= 506000(41/40)^{2}

= 531616.25

Therefore, the count of bacteria at the end of 2 hours will be 5,31,616 (approx.).

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

**Solution:**

Principal = Cost price of the scooter = ₹ 42,000

Depreciation = 8% of ₹ 42,000 per year

= (P x R x T)/100

= (42000 x 8 x 1)/100

= ₹ 3360

Thus, the value after 1 year = ₹ 42000 − ₹ 3360 = ₹ 38,640.