#### Comparing Quantities Using Proportion Solution of TS & AP Board Class 8 Mathematics

###### Exercise 5.1

**Question 1.**

Find the ratio of the following

Smita works in office for 6 hours and Kajal works for 8 hours in her office. Find the ratio of their working hours.

**Answer:**

Number of working hours of smita = 6 hours

Number of working hours of kajal = 8 hours

The formula for finding the ratio of their working hours is as follows

∴ The ratio of their working hours is 3:4

**Question 2.**

Find the ratio of the following

One pot contains 8 litre of milk while other contains 750 milliliter.

**Answer:**

One pot of milk = 8 litre

Other pot of milk = 750 milliliter

The formula for finding the ratio is as follows

1 litre = 1000 milliliter

∴ 8 litre = 8000 milliliter

= 32:3

∴ The ratio is 32:3

**Question 3.**

Find the ratio of the following

speed of a cycle is 15km/h and speed of the scooter is 30km/h.

**Answer:**

Speed of a cycle = 15km/h

Speed of the scooter = 30km/h

The formula for finding the ratio between the speed of cycle and the scooter is as follows

∴ The ratio of their working hours is 1:2

**Question 4.**

If the compound ratio of 5:8 and 3:7 is 45:x. Find the value of x.

**Answer:**

The compound ratio of 5:8 and 3:7 = 45:x

Here from the given ratios, a = 5,b = 8,c = 3 and d = 7.Then

If a:b and c:d are any ratios, then their compound ratio =

So, ac:bd

⇒

⇒

⇒

⇒

⇒ X = 168

∴ The Value of X is 168

**Question 5.**

If the compound ratio of 7:5 and 8:x is 84:60. Find x.

**Answer:**

The compound ratio of 7:5 and 8:x = 84:60

Here from the given ratios, a = 7,b = 5,c = 8 and d = X.Then

If a:b and c:d are any ratios, then their compound ratio =

So, ac:bd

⇒

⇒

⇒

⇒

⇒

⇒

⇒ X = 8

∴ The Value of X is 8

**Question 6.**

The compound ratio of 3:4 and the inverse ratio of 4:5 is 45:x. Find x.

**Answer:**

The compound ratio of 3:4 and the inverse ratio of 4:5 = 45:x

The inverse ratio of 4:5 = 5:4

Here from the given ratios, a = 3,b = 4,c = 5 and d = 4.Then

If a:b and c:d are any ratios, then their compound ratio =

So, ac:bd

⇒

⇒

⇒

⇒

⇒ X = 48

∴ The Value of X is 48

**Question 7.**

In a primary school there shall be 3 teachers to 60 students. If there are 400 students enrolled in the school, how many teachers should be there in the school in the same ratio?

**Answer:**

Number of teachers for 60 students = 3

Number of teachers for 400 students = X

The ratio of students = 60:400

The ratio of teachers = 3:X

The ratio of teachers = The ratio of students

60:400 = 3:X

⇒

⇒

⇒X = 20

∴ There are 20 teachers for 400 students in the school

**Question 8.**

In the given figure, ABC is a triangle. Write all possible ratios by taking measures of sides pair wise.

(Hint: Ratio of AB:BC = 8:6)

**Answer:**

In the given triangle, the measurement of the side AB = 8 cm

The measurement of the side BC = 6 cm

The measurement of the side AC = 10 cm

∴ The ratio of AB:BC =

∴ The ratio of AB:AC =

∴ The ratio of BC:AC =

∴ The ratio of BC:AB =

∴ The ratio of AC:AB =

∴ The ratio of AC:AB =

∴The Possible ratios are

**Question 9.**

If 9 out of 24 students scored below 75% marks in a test. Find the ratio of student scored below 75% marks to the student scored 75% and above marks.

**Answer:**

Given that the total number of students = 24

Number of students scored below 75% marks in a test = 9

Number of students scored 75% and above marks in a test = The total number of students - Number of students scored below 75% marks

⇒ 24-9 = 15

∴The ratio of student scored below 75% marks to the student scored 75% and above marks is

**Question 10.**

Find the ratio of number of vowels in the word’ MISSISSIPPI’ to the number of consonants in the simplest form.

**Answer:**

The number of vowels in the word’ MISSISSIPPI’ = 4(I) = 4

The number of consonants in the word’ MISSISSIPPI’ = (1M,4S,2P) = 7

Or

The number of vowels in the word’ MISSISSIPPI’ = (I) = 1

The number of consonants in the word’ MISSISSIPPI’ = (M,S,P) = 3

∴ The ratio is 4:7 or 1:3

**Question 11.**

Rajendra and Rehana own a business. Rehana receives 25% of the profit in each month. If Rehana received 2080 in particular month, what is the total profit in that month?

**Answer:**

The % of profit received by rehana in each month = 25%

The amount received by rehana at particular month = 2080

Let the total profit = X

25% of X = 2080

Here % compares every number to 100,

X = 8320

∴ The total profit in that month is Rs. 8320

**Question 12.**

In triangle ABC, AB = 2.2 cm, BC = 1.5 cm and AC = 2.3 cm. In triangle XYZ, XY = 4.4 cm, YZ = 3 cm and XZ = 4.6 cm. Find the ratio AB:XY, BC:YZ, AC:XZ. Are the lengths of corresponding sides of Î”ABC and Î”XYZ are in proportion?

[Hint: Any two triangles are said to be in proportion, if their corresponding sides are in the same ratio]

**Answer:**

Given that,

In triangle ABC,

AB = 2.2 cm

BC = 1.5 cm

AC = 2.3 cm

In triangle XYZ,

XY = 4.4 cm

YZ = 3 cm

XZ = 4.6 cm

To Find the ratio AB:XY,

∴ The ratio of AB:XY =

To Find the ratio BC:YZ,

∴ The ratio of BC:YZ =

To Find the ratio AC:XZ,

∴ The ratio of AC:XZ =

∴ The ratios of AB:XY, BC:YZ, AC:XZ are , ,

Here, the lengths of corresponding sides of Î”ABC and Î”XYZ are in proportion because their corresponding sides are in the same ratio .

**Question 13.**

Madhuri went to a super market. The price changes are as follows. The price of rice reduced by 5% jam and fruits reduced by 8% and oil and dal increased by 10%. Help Madhuri to find the changed prices in the given table.

**Answer:**

To find the changed price of Rice,

Original price = Rs. 30

Reduction = 5% of 30

= 1.5

Changed Price = Original price – Reduction

= 30-1.5

= Rs. 28.5

∴ The changed price of rice is Rs. 28.5

To find the changed price of Jam,

Original price = Rs. 100

Reduction = 8% of 100

= 8

Changed Price = Original price – Reduction

= 100-8

= Rs. 92

∴ The changed price of jam is Rs. 92

To find the changed price of Apples,

Original price = Rs. 280

Reduction = 8% of 280

= 22.4

Changed Price = Original price – Reduction

= 280-22.4

= Rs. 257.6

∴ The changed price of apples is Rs. 257.6

To find the changed price of Oil,

Original price = Rs. 120

Increased price = 10% of 120

= 12

Changed Price = Original price + Increased price

= 120+12

= Rs. 132

∴ The changed price of oil is Rs. 132

To find the changed price of Dal,

Original price = Rs. 80

Increased price = 10% of 80

= 8

Changed Price = Original price + Increased price

= 80+8

= Rs. 88

∴ The changed price of dal is Rs. 88

∴ The changed prices are as follows

**Question 14.**

There were 2075 members enrolled in the club during last year. This year enrolment is decreased by 4% .

Find the decrease in enrolment.

**Answer:**

Number of members enrolled in the club during last year = 2075

Decreased percentage of enrolment = 4%

Decrease in enrolment = 4% of 2075

= 83

∴ The decrease in enrolment is 83 members

**Question 15.**

There were 2075 members enrolled in the club during last year. This year enrolment is decreased by 4% .

How many members are enrolled during this year?

**Answer:**

Members enrolled this year = Members enrolled last year - decrease in enrolment

Number of members enrolled in the club during last year = 2075

Decrease in enrolment = 83

∴ Members enrolled this year = 2075 – 83

= 1992

∴ Members enrolled this year is 1992 members.

**Question 16.**

A farmer obtained a yielding of 1720 bags of cotton last year. This year she expects her crop to be 20% more. How many bags of cotton does she expect this year?

**Answer:**

Number of bags of cotton yielded last year = 1720

Expected crop in this year in % = 20% of 1720

20 % of 1720

= 344

Expected cotton bags in this year = Number of bags of cotton yielded last year + Expected crop

= 1720+344

= 2064 bags

∴ Expected cotton bags in this year is 2064 bags

**Question 17.**

Points P and Q are both in the line segment AB and on the same side of its midpoint. P divides AB in the ratio 2:3, and Q divides AB in the ratio 3:4. If PQ = 2, then find the length of the line segment AB.

**Answer:**

From the given, draw a line segment below

Let the length of the line segment AB = X

P divides AB in the ratio = 2:3

Thus the length of AP =

The length of PB = AB-AP

∴PB

Q divides AB in the ratio = 3:4

Thus the length of AQ =

The length of QB = AB-AQ

∴QB

The length of PQ in the line segment = AQ-AP

by solving this,

Given, PQ = 2

∴

X = 35×2

∴X = 70

The length of the line segment AB is 70 cm

###### Exercise 5.2

**Question 1.**

In the year 2012, it was estimated that there were 36.4 crore Internet users worldwide. In the next ten years, that number will be increased by 125%. Estimate the number of Internet users worldwide in 2022.

**Answer:**

Internet users in the year 2012 = 36.4 crore

% of Increase in the next ten years = 125%

125% of 36.4

= 45.5

Number of Internet users in the year 2022 = Number of Internet users in the year 2012 + Increased users

= 36.4+45.5

= 81.9 crore

∴ The number of Internet users worldwide in the year 2022 is 81.9 crore

**Question 2.**

A owner increases the rent of his house by 5% at the end of each year. If currently its rent is Rs. 2500 per month, how much will be the rent after 2 years?

**Answer:**

Rent of the house = Rs. 2500

% of Increase in each year = 5%

Rent increase in first year :

5% of 2500

= 125

Rent increase in first year = Rent of the house + increase in %

= 2500+125

= Rs. 2625

Rent increase in second year :

5% of 2625

= 131.25

Rent increase in second year = Rent of the house in first year + increase in %

= 2625+131.25

= Rs. 2756.25

∴ The rent after 2 years is Rs. 2756.25

**Question 3.**

On Monday, the value of a company’s shares was Rs. 7.50. The price increased by 6% on Tuesday, decreased by 1.5% on Wednesday, and decreased by 2% on Thursday. Find the value of each share when trade opened on Friday.

**Answer:**

The value of a company’s shares on Monday = Rs. 7.50

To find the value of a company’s shares on Tuesday,

Value on Monday = Rs. 7.50

% Increase = 6% of 7.50

= 0.45

Value on Tuesday = Value on Monday + value on Increase percentage

= 7.50+0.45

= Rs. 7.95

∴ The Value on Tuesday is Rs. 7.95

To find the value of a company’s shares on Wednesday,

Value on Tuesday = Rs. 7.95

% decrease = 1.5% of 7.95

= 0.11925

Value on Wednesday = Value on Tuesday - value on decrease percentage

= 7.95-0.11925

= Rs. 7.83075

∴ The Value on Wednesday is Rs. 7.83075

Value on Thursday = Rs. 7.83075

% decrease = 2% of 7.83075

= 0.156

Value on Thursday = Value on Wednesday - value on decrease percentage

= 7.83075-0.156

= Rs. 7.674

∴ The Value on Thursday is Rs. 7.674

The value of share when opened on Friday is equal to the Thursday’s closing price.

∴ The opening Value of the share on Friday is Rs. 7.674

**Question 4.**

With most of the Xerox machines, you can reduce or enlarge your original by entering a percentage for the copy. Reshma wanted to enlarge a 2 cm by 4 cm drawing. She set the Xerox machine for 150% and copied her drawing. What will be the dimensions of the copy of the drawing be?

**Answer:**

The original size of the drawing = 2cm × 4cm

Size to be enlarged = 150%

Enlarged new size = 150% of 2cm × 4cm

150% of 2cm =

= 3 cm

150% of 4cm =

= 6 cm

∴ The dimensions of the copy of the drawing is 3 × 6 cm

**Question 5.**

The printed price of a book is Rs. 150. And discount is 15%. Find the actual amount to be paid.

**Answer:**

The printed price of a book = Rs. 150

Discount = 15%

= 15% of 150

= 22.5

The actual amount to pay = printed price of a book – discount

= 150-22.5

= Rs. 127.50

∴ The actual amount to be paid is Rs. 127.50

**Question 6.**

The marked price of an gift item is Rs. 176 and sold it for Rs. 165. Find the discount percent.

**Answer:**

Market price = Rs. 176

Sale price = Rs. 165

Discount = Market price - Sale price

= 176-165

Discount = 11

Discount percent =

= 6.25% or 6%

∴ The discount percent of an gift item is 6%

**Question 7.**

A shop keeper purchased 200 bulbs for Rs. 10 each. However 5 bulbs were fused and put them into scrap. The remaining were sold at Rs. 12 each. Find the gain or loss percent.

**Answer:**

Number of bulbs purchased at Rs. 10 = 200

Purchase price = 200 × 10 = Rs. 2000

If 5 bulbs were defective, remaining bulbs = 195

Sale price = 12

Sale price of 195 bulbs = 195×12 = Rs. 2340

Profit = sale price – purchase price

= 2340 – 2000

= Rs. 340

= 17%

∴ The gain percent is 17%

**Question 8.**

Complete the following table with appropriate entries (Wherever possible)

**Answer:**

1) Given that, cost price = ‘ 750

Expense = Rs. 50

Profit = Rs. 80

Selling price = cost price+ Expense + profit

= 750+50+80

∴ selling price = Rs. 880

Purchase price = cost price+ Expense

= 750+50

= Rs. 800

= 10%

∴ Profit percent = 10%

2) Given that, cost price = ‘ 4500

Expense = Rs. 500

Loss = Rs. 1000

Selling price = cost price+ Expense - loss

= 4500+500-1000

∴ selling price = Rs. 4000

Purchase price = cost price+ Expense

= 4500+500

= Rs. 5000

= 20%

∴ loss percent = 20%

3) Given that, cost price = ‘ 46000

Expense = Rs. 4000

Selling price = Rs. 60000

Purchase price = cost price+ Expense

= 46000+4000

= Rs. 50000

Profit = selling price – purchase price

= 60000- 50000

= Rs. 10000

∴ profit = Rs. 10000

= 20%

∴ Profit percent = 20%

4) Given that, cost price = ‘ 300

Expense = Rs. 50

Profit percent = 12%

Purchase price = cost price+ Expense

= 300+50

= Rs. 350

= Rs. 42

∴ Profit = Rs. 42

Selling price = purchase price + profit

= 350 + 42

= Rs. 392

∴ Selling price = Rs. 392

5) Given that, cost price = ‘ 330

Expense = Rs. 20

Loss percent = 10%

Purchase price = cost price+ Expense

= 330+20

= Rs. 350

= Rs. 35

∴ loss = Rs. 35

Selling price = purchase price - loss

= 350 - 35

= Rs. 315

∴ Selling price = Rs. 315

**Question 9.**

A table was sold for Rs. 2,142 at a gain of 5%. At what price should it be sold to gain 10%.

**Answer:**

Selling price = Rs. 2142

Profit percent = 5%

Let the cost price of the table = X

Selling price = cost price + profit

⇒

⇒

⇒

⇒

∴ cost price = Rs. 2040

Then, For the profit of 10%

= Rs. 204

Selling price = cost price + profit

= 2040+204

= 2244

∴ The selling price of the table at 10% profit isRs. 2244

**Question 10.**

Gopi sold a watch to Ibrahim at 12% gain and Ibrahim sold it to John at a loss of 5%. If John paid Rs. 1,330, then find how much did Gopi sold it?

**Answer:**

Given that,

Profit percent of gopi = 12%

S.P of Ibrahim = Rs. 1330

Loss percent of Ibrahim = 5%

Let the cost price of gopi = X

S.P of Gopi = C.P of Ibrahim = 12%

= 0.12X

Selling price = cost price + profit

⇒

⇒

⇒

∴ S.P of gopi = 1.12X = C.P of Ibrahim

To find the S.P of Ibrahim,

= 0.056X

∴ loss = 0.056X

Selling price = Cost price - loss

= 1.12X – 0.056X

= 1.064X

∴ Selling price of ibrahim = 1.064X

Given, S.P of Ibrahim = Rs. 1330

⇒1.064X = 1330

⇒

X = 1250

∴ The cost price of gopi is Rs. 1250

**Question 11.**

Madhu and Kavitha purchased a new house for Rs. 3,20,000. Due to some economic problems they sold the house for Rs. 2, 80,000.

Find (a) The loss incurred (b) the loss percentage.

**Answer:**

Given that, C.P of house = Rs. 3,20,000

S.P of house = Rs. 2,80,000

(a) loss = cost price – selling price

= 320000 – 280000

= 40000

∴ The loss incurred is Rs. 40000

(b)

= 12.5

∴ The loss percentage is 12.5%

**Question 12.**

A pre-owned car show-room owner bought a second-hand car for Rs. 1,50,000. He spent Rs. 20,000 on repairs and painting, then sold it for Rs. 2,00,000. Find whether he gets profit or loss. If so, what percent?

**Answer:**

Purchase price of a car = Rs. 1,50,000

Repairs and painting = Rs. 20,000

Selling price = Rs. 2,00,000

Total cost price = purchasing price + Repair charges

= Rs. 1,50,000 + Rs. 20,000

= Rs. 1,70,000

∴ cost price = Rs. 1,70,000

Here, selling price > cost price, so there is a profit.

Profit = selling price - cost price

= 2,00,000 - 1,70,000

= 30,000

∴Profit = Rs. 30,000

On cost price of Rs. 1,70,000 profit is 30,000

If cost price is Rs. 100,profit will be?

= 17.65%

∴Profit percent = 17.65%

**Question 13.**

Lalitha took a parcel from a hotel to celebrate her birthday with her friends. It was billed with Rs. 1,450 including 5% VAT. Lalitha asked for some discount, the hotel owner gave

8% discount on the bill amount. Now find the actual amount that lalitha has to pay to the hotel owner

**Answer:**

The cost of parcel including 5% VAT = Rs. 1450

Discount given by the hotel owner = 8%

Actual discount = 8% of 1450

= 116

∴ Actual discount = Rs. 116

The actual amount paid by lalitha = bill amount – actual discount

= 1450 – 116

= Rs. 1334

∴ The actual amount paid by lalitha is Rs. 1334.

**Question 14.**

If VAT is included in the price, find the original price of each of the following.

**Answer:**

(i) Bill amount of diamond = Rs. 10100

VAT = 1%

Original price = Bill amount - VAT

= 10100 - 1% of 10100

∴ The original price of diamond is Rs. 9999

(ii) Bill amount of pressure cooker = Rs. 2940

VAT = 5%

Original price = Bill amount - VAT

= 2940 - 5% of 2940

∴ The original price of pressure cooker is Rs. 2793

(iii) Bill amount of face powder = Rs. 229

VAT = 14.5%

Original price = Bill amount - VAT

= 229 – 14.5% of 229

∴ The original price of face powder is Rs. 195.80(approximate)

**Question 15.**

Find the buying price of each of the following items when a sales tax of 5% is added on them.

(i) a towel of Rs. 50

(ii) Two bars of soap at Rs. 35 each.

**Answer:**

(i) cost of a towel = Rs. 50

Sales tax on a towel = 5% of 50

= Rs. 2.5

∴ sales tax on a towel = Rs. 2.5

Buying price = cost + sales tax

= 50+2.5

= Rs. 52.5

∴ The buying price of a towel = Rs. 52.5

(ii) cost of a soap bar = Rs. 35

Cost of two soap bars = 35×2 = Rs. 70

Sales tax on a soap bars = 5% of 70

= Rs. 3.5

∴ sales tax on a towel = Rs. 3.5

Buying price = cost + sales tax

= 70+3.5

= Rs. 73.5

∴ The buying price of a soap bars = Rs. 73.5

**Question 16.**

A Super-Bazar prices an item in rupees and paise so that when 4% sales tax is added, no rounding is necessary because the result is exactly in ‘n’ rupees, where ‘n’ is a positive integer. Find the smallest value of ‘n’.

**Answer:**

Let the final rupees of an item = n

Let the initial rupees before tax = x

Sales tax = 4%

Initial rupees + sales tax = final rupee

⇒

Here, n should be the factor of 26.so, the factor of 26 are 1,2,13,26.

For X to be terminating decimal, n can be either 13 or 26. 13 is smaller

∴ n = 13

###### Exercise 5.3

**Question 1.**

Sudhakar borrows Rs. 15000 from a bank to renovate his house. He borrows the money at 9% p.a. simple interest over 8 years. What are his monthly repayments?

**Answer:**

Principal (P) = Rs. 15000

Time period (T) = 8

Rate of interest (R) = 9%

I = Rs. 10800

∴ Interest for 8 years is Rs. 10800

Amount to be paid at the end of 8 years = Principal + interest

Amount = 15000+10800

= 25800

∴ Amount to be paid at the end of 8 years is Rs. 25800

⇒ monthly repayment =

= 268.75

∴ Sudhakar pays Rs. 268.75 monthly.

**Question 2.**

A TV was bought at a price of Rs. 21000. After 1 year the value of the TV was depreciated by 5% (Depreciation means reduction of the value due to use and age of the item). Find the value of the TV after 1 year.

**Answer:**

Cost price of TV = Rs. 21000

Depreciation = 5%

Depreciation after 1 year = 5% of 21000

∴ Depreciation after 1 year = Rs. 1050

Value of TV after 1 year = cost price – depreciation

= 21000 – 1050

= Rs. 19950

∴The value of TV after 1 year is Rs. 19,950

**Question 3.**

Find the amount and the compound interest on Rs. 8000 at 5% per annum, for 2 years compounded annually.

**Answer:**

Principal (P) = Rs. 8000

Time period (n) = 2

Rate of interest (R) = 5%

= 8000×1.05^{2}

= 8820

∴ Amount = Rs. 8820

= (8000×1.05^{2})-8000

= 8820-8000

= 820

∴ Compound Interest = Rs. 820

**Question 4.**

Find the amount and the compound interest on Rs. 6500 for 2 years, compounded annually, the rate of interest being 5% per annum during the first year and 6% per annum during the second year.

**Answer:**

Principal (P) for 1^{st} year = Rs. 6500

Rate of interest (R) for first year = 5%

Interest on first year = 5% of 6500

=

= 325

∴ Interest for 1^{st} year = Rs. 325

Principal (P) for second year = Principal (P) for 1^{st} year + Interest for 1^{st} year

Principal (P) for second year = Rs. 6500 + Rs. 325 = Rs. 6825

Rate of interest (R) for second year = 6%

Interest on second year = 6% of 6825

=

= 409.5

∴ Interest for 2^{nd} year = Rs. 409.5

Total interest = Rs. 325+Rs. 409.5 = Rs. 734.5

Amount at second year = Principal (P) for 2^{nd} year + Interest for 2^{nd} year

= 6825+409.5

= 7234.5

∴ Amount at second year = Rs. 7234.5

**Question 5.**

Prathibha borrows Rs. 47000 from a finance company to buy her first car. The rate of simple interest is 17% and she borrows the money over a 5 year period.

Find:

(a) How much amount Prathibha should repay the finance company at the end of five years.

(b) her equal monthly repayments.

**Answer:**

(a) Principal (P) = Rs. 47000

Time period (T) = 5

Rate of interest (R) = 17%

I = Rs. 39950

∴ Interest for 5 years is Rs. 39950

Amount at the end of 5 years = Principal + interest

Amount = 47000+39950

= 86950

∴ Amount at the end of 5 years is Rs. 86950

(b)

⇒ monthly repayment =

= 1449.17

∴ prathibha’s monthly repayment isRs. 1449.17.

**Question 6.**

The population of Hyderabad was 68,09,000 in the year 2011. If it increases at the rate of 4.7% per annum. What will be the population at the end of the year 2015.

**Answer:**

The population of Hyderabad (P) = 68,09,000

Time period (n) = 2015-2011 = 4

Rate of interest (R) = 4.7%

= 6809000×1.047^{4}

= 81821994

∴ The population of Hyderabad at the end of 2015 = 8,18,21,994

**Question 7.**

Find Compound interest paid when a sum of Rs. 10000 is invested for 1 year and 3 months at 8 % per annum compounded annually.

**Answer:**

Principal (P) = Rs. 10000

Rate of interest (R) = 8.5%

Time period (T) = 1year 3 months

For T = 1 year,

I = Rs. 850

∴ Interest for 1 year is Rs. 850

Amount at the end of 1 year = Principal + interest

Amount = 10000+850

= 10850

∴ Amount at the end of 1 year is Rs. 10850

For T = 3 months =

I = Rs. 230.56

∴ Interest for year is Rs. 230.56

Amount at the end of year = Principal + interest

Amount = 10850+230.56

= 11080.56

∴ Amount at the end of year is Rs. 11080.56

Total interest for 1.3 year = Interest for 1 year + Interest for year

= 850+230.56

= 1080.56

∴ The compound interest paid is Rs. 1080.56

**Question 8.**

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 1years, if the interest is compounded annually and compounded half yearly.

**Answer:**

For compounded Annually

Principal (P) = Rs. 80000

Time period (n) = 1

Rate of interest (R) = 10%

= 80000×1.05^{3}

= 88000

∴ Amount for 1 year = Rs. 88000

Interest for remaining 6 months =

= 4400

∴ Amount for 1.5 years = Rs. 88000+4400 = Rs. 92400

= 92400-80000

∴ Compound Interest = Rs. 12400

For compounded half yearly

Principal (P) = Rs. 80000

Time period (n) = 3

Rate of interest (R) for half year = 10%× = 5%

= 80000×1.05^{3}

= 92610

∴ Amount = Rs. 92610

= 92610-80000

∴ Compound Interest = Rs. 12610

∴ the difference in amounts = amount for For compounded half yearly - amount for For compounded annually

= 92610-92400

= Rs. 210

∴ The difference in amounts is Rs. 210

**Question 9.**

I borrowed Rs. 12000 from Prasad at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compounded annually, what extra amount would I have to pay?

**Answer:**

Principal (P) = Rs. 12000

Rate of interest (R) = 6%

Time period (T) = 2years

I = Rs. 1440

∴ Interest for 2 years is Rs. 1440

This sum to be borrowed at 6% per annum compounded annually,

Principal (P) = Rs. 12000

Rate of interest (R) = 6%

Time period (n) = 2years

= 12000×1.06^{2}

= 13483.2

∴ Amount = Rs. 13483.2

= 13483.2-12000

∴ Compound Interest = Rs. 1483.2

∴ The difference in interest = 1483.2 - 1440

= Rs. 43.2

∴ The difference in interest is Rs. 43.2

**Question 10.**

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:**

Principal (P) = 506000

Rate of interest (R) = 2.5%

Time period (n) = 2hours

= 506000×1.025^{2}

= 531616.25

∴ The number of the bacteria at the end of 2 hours is 531616(approximately)

**Question 11.**

Kamala borrowed Rs. 26400 from a bank to buy a scooter at a rate of 15% per annum compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?

**Answer:**

Principal (P) = 26400

Rate of interest (R) = 15%

Time period (T) = 2 years and 4 months

Amount for 2 years,

= 26400×1.15^{2}

= Rs. 34914

∴ Amount for 2 year = Rs. 34914

Interest for remaining 4 months =

= 1745.70

∴ Total amount for 2 years and 4 months = Rs. 34914 + 1745.70

= Rs. 36659.70

∴ The total amount to clear the loan is Rs. 36659.70

**Question 12.**

Bharathi borrows an amount of Rs. 12500 at 12% per annum for 3 years at a simple interest and Madhuri borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?

**Answer:**

Principal (P) = Rs. 12500

Rate of interest (R) = 12%

Time period (T) = 3years

Interest paid by bharathi,

I = Rs. 4500

∴ Interest paid by bharathi is Rs. 4500

Amount paid by madhuri,

Principal (P) = Rs. 12500

Rate of interest (R) = 10%

Time period (n) = 3years

= 12500×1.1^{3}

= Rs. 16637.5

∴ Amount paid by madhuri isRs. 16637.5

Interest = A-P

= Rs. 16637.5 – 12500

= Rs. 4137.5

∴ Interest paid by madhuri is Rs. 4137.5

On comparing the interests paid by bharathi and madhuri,

4500-4137.5 = 362.5

∴ Bharathi paid Rs.362.5 more than by madhuri.

**Question 13.**

Machinery worth Rs. 10000 depreciated by 5%. Find its value after 1 year.

**Answer:**

Principal (P) = 10000

Depreciation (R) = 5%

Time period (n) = 1 year

= Rs. 9500

∴The value of machinery after 1 year is Rs. 9500

**Question 14.**

Find the population of a city after 2 years which is at present 12 lakh, if the rate of increase is 4%.

**Answer:**

Present population (P) = 12 lakh

Rate of interest (R) = 4%

Time period (n) = 2years

= 1200000×1.04^{2}

= 1297920

∴ The population of a city after 2 years is 1297920

**Question 15.**

Calculate compound interest on Rs. 1000 over a period of 1 year at 10% per annum, if interest is compounded quarterly?

**Answer:**

Principal (P) = 1000

Rate of interest (R) = 10%

Time period (n) = 1 year

For quarterly, n = 4

Rate of interest (R) for quarterly =

= 1000

= 1103.81

∴ Amount = Rs. 1103.81

Interest = A-P

= 1103.81-1000

= 103.81

∴ compound interest = Rs. 103.81