**Comparing Quantities**- Section 8.1 is about

**Recalling Ratios and Percentages.**Solved examples are given in the chapter for a better explanation. Exercise 8.1 is based on the same concept.

The next section is about

**Finding the Increase or Decrease Percent**. This section is explained with the help of a solved example.

This is followed by the topic

**Finding discounts**

**Discount = Marked price $-$ Sale price****Discount**can also be found when the**discount%**is given.**Estimation in percentages**is also explained briefly in this section.

**Prices Related to Buying and Selling**(

**Profit and Loss**). Once the student has learnt about profit and loss concept, the next section can be taken up which explains about

**Finding cost price/selling price, profit%/loss%**.

The concept of

**Sales Tax/Value Added Tax**is also described.

After solving the questions of exercise 8.2 students will learn about

**Compound Interest**.

- The
**interest**is calculated on the**amount**of the previous year**.**This is known as**interest compounded or Compound Interest (C.I.).** **Compound interest**is the**interest**calculated on the previous year's amount**(A = P + I).**- We can also have
**interest rates compounded half-yearly or quarterly.** - If
**interest is compounded half-yearly,**we compute the**interest**two times. So the**time period**becomes twice and**rate**is taken half.

**Deducing a Formula for Compound Interest**is also explained.

Section 8.8 lays emphasis on

**Rate Compound Annually or Half Yearly(Semi-Annually)**.

After that,

**Applications of Compound Interest**Formula is given in section 8.9.

- There are some situations where we could use the formula for
**calculation of the amount in CI**. Here are a few:

**Increase (or decrease) in population**.

(ii) The

**growth of a bacteria**if the

**rate of growth**is known.

(iii) The value of an item, if its

**price increases or decreases**in the intermediate years

The last exercise is 8.3 which contains 12 questions.

Summarization of the chapter is done in the end.

#### Page No 119:

#### Question 1:

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 Rs 5

#### Answer:

(a) Ratio of the speed of cycle to the speed of scooter

(b) Since 1 km = 1000 m,

Required ratio

(c) Since Re 1 = 100 paise,

Required ratio

#### Page No 119:

#### Question 2:

Convert the following ratios to percentages.

(a) 3:4 (b) 2:3

#### Answer:

(a)

(b)

#### Page No 119:

#### Question 3:

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

#### Answer:

It is given that 72% of 25 students are good in mathematics.

Therefore,

Percentage of students who are not good in mathematics = (100 − 72)%

= 28%

∴Number of students who are not good in mathematics =

= 7

Thus, 7 students are not good in mathematics.

#### Page No 119:

#### Question 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?

#### Answer:

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

It is given that the team won 10 matches and the winning percentage of the team was 40%.

Therefore,

Thus, the team played 25 matches.

##### Video Solution for comparing quantities (Page: 119 , Q.No.: 4)

NCERT Solution for Class 8 math - comparing quantities 119 , Question 4

#### Page No 119:

#### Question 5:

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

#### Answer:

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

It is given that after
spending 75% of Rs *x*, she was left with Rs 600.

Therefore,

(100 − 75)% of *x*
= Rs 600

Or, 25 % of *x* =
Rs 600

Thus, she had Rs 2400 in the beginning.

#### Page No 120:

#### Question 6:

If 60% people in 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 are 50 lakh, find the exact number who like each type of game.

#### Answer:

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

= (100 − 90)% = 10 %

Total number of people = 50 lakh

Therefore, number of people who like cricket = 30 lakh

Number of people who like football = 15 lakh

Number of people who like other games = 5 lakh

##### Video Solution for comparing quantities (Page: 120 , Q.No.: 6)

NCERT Solution for Class 8 math - comparing quantities 120 , Question 6

#### Page No 125:

#### Question 1:

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

#### Answer:

Let the original salary be *x*. It is given that the new salary is Rs 1,54,000.

Original salary + Increment = New salary

However, it is given that the increment is 10% of the original salary.

Therefore,

Thus, the original salary was Rs 1,40,000.

##### Video Solution for comparing quantities (Page: 125 , Q.No.: 1)

NCERT Solution for Class 8 math - comparing quantities 125 , Question 1

#### Page No 125:

#### Question 2:

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?

#### Answer:

It is given that on Sunday, 845 people went to the zoo and on Monday, 169 people went.

Decrease in the number of people = 845 − 169 = 676

Percentage decrease =

#### Page No 125:

#### Question 3:

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

#### Answer:

It is given that the shopkeeper buys 80 articles for Rs 2,400.

Cost of one article =

Profit percent = 16

Selling price of one article = C.P. + Profit = Rs (30 + 4.80) = Rs 34.80

#### Page No 125:

#### Question 4:

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

#### Answer:

Total cost of an article = Cost + Overhead expenses

= Rs 15500 + Rs 450

= Rs 15950

∴Selling price of the article = C.P. + Profit = Rs (15950 + 2392.50)

= Rs 18342.50

#### Page No 125:

#### Question 5:

A VCR and TV were bought for Rs 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.

#### Answer:

C.P. of a VCR = Rs 8000

The shopkeeper made a loss of 4 % on VCR.

This means if C.P. is Rs 100, then S.P. is Rs 96.

When C.P. is Rs 8000, S.P. = = Rs 7680

C.P. of a TV = Rs 8000

The shopkeeper made a profit of 8 % on TV.

This means that if C.P. is Rs 100, then S.P. is Rs 108.

When C.P. is Rs 8000, S.P. = = Rs 8640

Total S.P. = Rs 7680 + Rs 8640 = Rs 16320

Total C.P. = Rs 8000 + Rs 8000 = Rs 16000

Since total S.P.> total C.P., there was a profit.

Profit = Rs 16320 − Rs 16000 = Rs 320

Therefore, the shopkeeper had a gain of 2% on the whole transaction.

#### Page No 125:

#### Question 6:

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 Rs 1450 and two shirts marked at Rs 850 each?

#### Answer:

Total marked price = Rs (1,450 + 2 × 850) = Rs (1,450 +1,700) = Rs 3,150

Given that, discount % = 10%

Discount =

Also, Discount = Marked price − Sale price

Rs 315 = Rs 3150 − Sale price

∴ Sale price = Rs (3150 − 315) = Rs 2835

Thus, the customer will have to pay Rs 2,835.

##### Video Solution for comparing quantities (Page: 125 , Q.No.: 6)

NCERT Solution for Class 8 math - comparing quantities 125 , Question 6

#### Page No 125:

#### Question 7:

A milkman sold two of his buffaloes for Rs 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.

(**Hint:** Find CP of each)

#### Answer:

S.P. of each buffalo = Rs 20000

The milkman made a gain of 5% while selling one buffalo.

This means if C.P. is Rs 100, then S.P. is Rs 105.

C.P. of one buffalo = = Rs 19,047.62

Also, the second buffalo was sold at a loss of 10%.

This means if C.P. is Rs 100, then S.P. is Rs 90.

∴C.P. of other buffalo = = Rs 22222.22

Total C.P. = Rs 19047.62 + Rs 22222.22 = Rs 41269.84

Total S.P. = Rs 20000 + Rs 20000 = Rs 40000

Loss = Rs 41269.84 − Rs 40000 = Rs 1269.84

Thus, the overall loss of milkman was Rs 1,269.84.

##### Video Solution for comparing quantities (Page: 125 , Q.No.: 7)

NCERT Solution for Class 8 math - comparing quantities 125 , Question 7

#### Page No 125:

#### Question 8:

The price of a TV is Rs 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,

#### Answer:

On Rs 100, the tax to be paid = Rs 12

On Rs 13000, the tax to be paid will be

= Rs 1560

Required amount = Cost + Sales Tax = Rs 13000 + Rs 1560

= Rs 14560

Thus, Vinod will have to pay Rs 14,560 for the T.V.

#### Page No 125:

#### Question 9:

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

#### Answer:

Let the marked price be *x*.

Also,

Discount = Marked price − Sale price

Thus, the marked price was Rs 2000.

##### Video Solution for comparing quantities (Page: 125 , Q.No.: 9)

NCERT Solution for Class 8 math - comparing quantities 125 , Question 9

#### Page No 125:

#### Question 10:

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

#### Answer:

The price includes VAT.

Thus, 8% VAT means that if the price without VAT is Rs 100, then price including VAT will be Rs 108.

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

Thus, the price of the hair-dryer before the addition of VAT was Rs 5,000.

#### Page No 133:

#### Question 1:

Calculate the amount and compound interest on

(a) Rs 10800 for 3 years at per annum compounded annually.

(b) Rs 18000 for years at 10% per annum compounded annually.

(c) Rs 62500 for years at 8% per annum compounded half yearly.

(d) Rs 8000 for 1 year at 9% per annum compound half yearly.

(You could use the year by year calculation using SI formula to verify)

(e) Rs 10000 for 1 year at 8% per annum compounded half yearly.

#### Answer:

(a) Principal (P) = Rs 10, 800

Rate (R) = = % (annual)

Number
of years (*n*) = 3

Amount, A =

C.I. = A − P = Rs (15377.34 − 10800) = Rs 4,577.34

(b) Principal (P) = Rs 18,000

Rate (R) = 10% annual

Number
of years (*n*) =

The amount for 2 years and 6 months can be calculated by first calculating the amount for 2 years using the compound interest formula, and then calculating the simple interest for 6 months on the amount obtained at the end of 2 years.

Firstly, the amount for 2 years has to be calculated.

By taking Rs 21780 as principal, the S.I. for the next year will be calculated.

∴ Interest for the first 2 years = Rs (21780 − 18000) = Rs 3780

And interest for the nextyear = Rs 1089

∴ Total C.I. = Rs 3780 + Rs 1089 = Rs 4,869

A = P + C.I. = Rs 18000 + Rs 4869 = Rs 22,869

(c) Principal (P) = Rs 62,500

Rate = 8% per annum or 4% per half year

Number of years =

There will be 3 half years inyears.

C.I. = A − P = Rs 70304 − Rs 62500 = Rs 7,804

(d) Principal (P) = Rs 8000

Rate of interest = 9% per annum or % per half year

Number of years = 1 year

There will be 2 half years in 1 year.

C.I. = A − P = Rs 8736.20 − Rs 8000 = Rs 736.20

(e) Principal (P) = Rs 10,000

Rate = 8% per annum or 4% per half year

Number of years = 1 year

There are 2 half years in 1 year.

C.I. = A − P = Rs 10816 − Rs 10000 = Rs 816

#### Page No 133:

#### Question 2:

Kamala borrowed Rs 26400 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?

(Hint: Find A for 2
years with interest is compounded yearly and then find SI on the 2^{nd}
year amount for
years.)

#### Answer:

Principal (P) = Rs 26,400

Rate (R) = 15% per annum

Number of years (*n*)
=

The amount for 2 years and 4 months can be calculated by first calculating the amount for 2 years using the compound interest formula, and then calculating the simple interest for 4 months on the amount obtained at the end of 2 years.

Firstly, the amount for 2 years has to be calculated.

By taking Rs 34,914 as principal, the S.I. for the nextwill be calculated.

Interest for the first two years = Rs (34914 − 26400) = Rs 8,514

And interest for the nextyear = Rs 1,745.70

Total C.I. = Rs (8514 + Rs 1745.70) = Rs 10,259.70

Amount = P + C.I. = Rs 26400 + Rs 10259.70 = Rs 36,659.70

#### Page No 134:

#### Question 3:

Fabina borrows Rs 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?

#### Answer:

Interest paid by Fabina =

Amount paid by Radha at the end of 3 years = A =

C.I. = A − P = Rs 16637.50 − Rs 12500 = Rs 4,137.50

The interest paid by Fabina is Rs 4,500 and by Radha is Rs 4,137.50.

Thus, Fabina pays more interest.

Rs 4500 − Rs 4137.50 = Rs 362.50

Hence, Fabina will have to pay Rs 362.50 more.

##### Video Solution for comparing quantities (Page: 134 , Q.No.: 3)

NCERT Solution for Class 8 math - comparing quantities 134 , Question 3

#### Page No 134:

#### Question 4:

I borrowed Rs 12000 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?

#### Answer:

P = Rs 12000

R = 6% per annum

T = 2 years

To find the compound interest, the amount (A) has to be calculated.

∴ C.I. = A − P = Rs 13483.20 − Rs 12000 = Rs 1,483.20

C.I. − S.I. = Rs 1,483.20 − Rs 1,440 = Rs 43.20

Thus, the extra amount to be paid is Rs 43.20.

#### Page No 134:

#### Question 5:

Vasudevan invested Rs 60000 at an interest rate of 12% per annum compounded half yearly. What amount would he get

(i) after 6 months?

(ii) after 1 year?

#### Answer:

(i) P = Rs 60,000

Rate = 12% per annum = 6% per half year

*n* = 6 months = 1 half year

(ii) There are 2 half years in 1 year.

*n* = 2

##### Video Solution for comparing quantities (Page: 134 , Q.No.: 5)

NCERT Solution for Class 8 math - comparing quantities 134 , Question 5

#### Page No 134:

#### Question 6:

Arif took a loan of Rs 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after years if the interest is

(i) Compounded annually

(ii) Compounded half yearly

#### Answer:

(i) P = Rs 80,000

R = 10% per annum

*n*
=years

The amount for 1 year and 6 months can be calculated by first calculating the amount for 1 year using the compound interest formula, and then calculating the simple interest for 6 months on the amount obtained at the end of 1 year.

Firstly, the amount for 1 year has to be calculated.

By taking Rs 88,000 as principal, the SI for the next year will be calculated.

Interest for the first year = Rs 88000 − Rs 80000 = Rs 8,000

And interest for the nextyear = Rs 4,400

Total C.I. = Rs 8000 + Rs 4,400 = Rs 1,2400

A = P + C.I. = Rs (80000 + 12400) = Rs 92,400

(ii) The interest is compounded half yearly.

Rate = 10% per annum = 5% per half year

There will be three half years inyears.

Difference between the amounts = Rs 92,610 − Rs 92,400 = Rs 210

#### Page No 134:

#### Question 7:

Maria invested Rs 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 3^{rd} year.

#### Answer:

(i) P = Rs 8,000

R = 5% per annum

*n*
= 2 years

(ii) The interest for the next one year, i.e. the third year, has to be calculated.

By taking Rs 8,820 as principal, the S.I. for the next year will be calculated.

#### Page No 134:

#### Question 8:

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

#### Answer:

P = Rs 10,000

Rate = 10% per annum = 5% per half year

*n* = years

There will be 3 half years inyears.

CI = A − P

= Rs 11576.25 − Rs 10000 = Rs 1,576.25

The amount for 1 year and 6 months can be calculated by first calculating the amount for 1 year using the compound interest formula, then calculating the simple interest for 6 months on the amount obtained at the end of 1 year.

The amount for the first year has to be calculated first.

$\mathrm{A}=\mathrm{Rs}\left[10000{\left(1+\frac{10}{100}\right)}^{1}\right]\phantom{\rule{0ex}{0ex}}=\mathrm{Rs}\left[10000\left(\frac{11}{10}\right)\right]\phantom{\rule{0ex}{0ex}}=\mathrm{Rs}11000$

By taking Rs 11,000 as the principle, the SI for the next $\frac{1}{2}$ year will be calculated.

$\mathrm{S}.\mathrm{I}.=\mathrm{Rs}\left(\frac{11000\times 10\times {\displaystyle \frac{1}{2}}}{100}\right)=\mathrm{Rs}550$

$\therefore $Interest for the first year =$\mathrm{Rs}11000-\mathrm{Rs}10000=\mathrm{Rs}1000$

$\therefore $Total compound interest $=\mathrm{Rs}1000+\mathrm{Rs}550=\mathrm{Rs}1550$

Therefore, the interest would be more when compounded half-yearly than the interest when compounded annually.

#### Page No 134:

#### Question 9:

Find the amount which Ram will get on Rs 4,096, he gave it for 18 months at per annum, interest being compounded half yearly.

#### Answer:

P = Rs 4,096

R = per annum = per half year

*n* = 18 months

There will be 3 half years in 18 months.

Therefore,

Thus, the required amount is Rs 4,913.

#### Page No 134:

#### Question 10:

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?

#### Answer:

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

Therefore,

54000 = (Population in 2001)

Population in 2001 = 48979.59

Thus, the population in the year 2001 was approximately 48,980.

(ii) Population in 2005 =

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

#### Page No 134:

#### Question 11:

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.

#### Answer:

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

Bacteria at the end of 2 hours =

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

##### Video Solution for comparing quantities (Page: 134 , Q.No.: 11)

NCERT Solution for Class 8 math - comparing quantities 134 , Question 11

#### Page No 134:

#### Question 12:

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

#### Answer:

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

Depreciation = 8% of Rs 42,000 per year

Value after 1 year = Rs 42000 − Rs 3360 = Rs 38,640

##### Video Solution for comparing quantities (Page: 134 , Q.No.: 12)

NCERT Solution for Class 8 math - comparing quantities 134 , Question 12

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