ncert solutions for class 8 maths chapter 8 deals with Comparing Quantities

ncert solutions for class 8 maths chapter 8 deals with Comparing quantities.

Exercises and solutions of Comparing quantities available here so students can read online view.

mensuration.in provides all exercises with answers for chapter 8 in class 8 maths.

Find solutions to exercise 8.1, exercise 8.2, and 8.3 in chapter 8 in class 8 maths.

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Find solutions to ncert maths class 8 chapter 8 .

class 8 maths chapter 8 exercise 8.1

class 8 maths chapter 8 exercise 8.1 (1)

1 Find the ratio of the following.

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

(b) 5m to 10 km (c) 50 paise to Rs 5

Solution to class 8 maths chapter 8 exercise 8.1 (1).

(a) Speed of cycle = 15 km/hr

Speed of scooter = 30 km/hr

Hence ratio of speed of cycle to that of scooter = 15:30= 1:2

(b) 1km = 1000 m

10 km = 10×1000 = 10000 m

Ratio=5 m: 10000 m=1: 2000

Rs 5 = 5×100=500 paise

Hence ratio = 50 paise : 500 Paise =1: 10

class 8 maths chapter 8 exercise 8.1 (2)

2 Convert the following ratios to percentages.

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

Solution to class 8 maths chapter 8 exercise 8.1 (2).

(a) Percentage of 3 : 4 = (¾)×100 % = 75%

(b) Percentage of 2 : 3 = (⅔)×100% = 66(⅔)%

class 8 maths chapter 8 exercise 8.1 (3)

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

Solution to class 8 maths chapter 8 exercise 8.1 (3).

Total number of students = 25

Number of good students in mathematics = 72% of 25

= (⁷²/₁₀₀)×25 =18.

Number of students not good in mathematics = 25 – 18 = 7

Hence percentage of students not good in mathematics = (⁷/₂₅)×100=28%

class 8 maths chapter 8 exercise 8.1 (4)

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 to class 8 maths chapter 8 exercise 8.1 (4).

Let total number of matches be x

According to question,

40% of total matches = 10

40% of x = 10

⇒(⁴⁰/₁₀₀)×x =10

⇒x=10×(¹⁰⁰/₄₀) =25

Hence total number of matches are 25.

class 8 maths chapter 8 exercise 8.1 (5)

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

Solution to class 8 maths chapter 8 exercise 8.1 (5).

Let her money in the beginning be 2x According to question,

x-75% of x=600

⇒ x – (⁷⁵/₁₀₀)×x =600

⇒ x-(¾) × x=600

⇒ x(1-¾) = 600

⇒ x [⁽⁴⁻³⁾/₄]= 600

⇒ x= 600 × 4= ₹ 2400

Hence the money in the beginning was ₹ 2,400

class 8 maths chapter 8 exercise 8.1 (6)

6. 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 are 50 lakh, find the exact number who like each type of game.

Solution to class 8 maths chapter 8 exercise 8.1 (6).

Number of people who like cricket = 60%

Number of people who like football = 30%

Number of people who like other games = 100% – (60% + 30%) = 10%

Now Number of people who like cricket = 60% of 50,00,000 = (⁶⁰/₁₀₀) × 50,00,000 =30,00,000

And Number of people who like football = 30% of 50,00,000 = (³⁰/₁₀₀)× 50,00,000

=15,00,000

Number of people who like other games = 10% of 50,00,000 = (¹⁰/₁₀₀) × 50,00,000

=5,00,000

Hence, number of people who like other games are 5 lakh.

Exercise 8.2

class 8 maths chapter 8 exercise 8.2 (1)

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

Solution to class 8 maths chapter 8 exercise 8.2 (1).

Let original salary be ‘x’

New salary i.e. 10% increase = x[⁽¹⁰⁰⁺¹⁰⁾/₁₀₀] =x × [⁽¹¹⁰⁾/₁₀₀]

According to problem,

x × [⁽¹¹⁰⁾/₁₀₀] = 1,54,000

⇒ x =1,54,000 × [⁽¹⁰⁰⁾/₁₁₀] = ₹ 1,40,000.

∴ Original salary = ₹ 1,40,000.

class 8 maths chapter 8 exercise 8.2 (2)

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?

Solution to class 8 maths chapter 8 exercise 8.2 (2).

On Sunday, people went to the Zoo = 845

On Monday, people went to the Zoo = 169

Number of decrease in the people = 845 – 169 = 676

Decrease percent = [⁽⁶⁷⁶⁾/₈₄₅] × 100=80%

Hence decrease in the people visiting the Zoo is 80%.

class 8 maths chapter 8 exercise 8.2 (3)

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.

Solution to class 8 maths chapter 8 exercise 8.2 (3).

Number of articles =80

Cost price of articles = ₹ 2,400

And profit percent = 16 %

Selling price of articles= 2,400 × [⁽¹⁰⁰⁺¹⁶⁾/₁₀₀]

=24 × 116 = ₹ 2784.

{∵ If a value increases x % multiply that value by [⁽¹⁰⁰⁺ˣ⁾/₁₀₀] to get new value}

∴ Selling price of 1 article = [⁽²⁷⁸⁴⁾/₈₀]= ₹ 34.80

class 8 maths chapter 8 exercise 8.2 (4)

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.

Solution to class 8 maths chapter 8 exercise 8.2 (4).

Cost price = ₹ 15,500 and cost for repair =₹450

∴ Total cost price = ₹15,500+₹450 =₹15,950

Profit % =15%

∴ Selling price = C.P ×{⁽¹⁰⁰⁺¹⁵⁾/₁₀₀}

{∵ If a value increases x % multiply that value by [⁽¹⁰⁰⁺ˣ⁾/₁₀₀] to get new value}

S.P=15,950×[⁽¹¹⁵⁾/₁₀₀]

S.P=18342.50

class 8 maths chapter 8 exercise 8.2 (5)

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.

Solution to class 8 maths chapter 8 exercise 8.2 (5).

Cost price of VCR = ₹ 8000 and Cost price of TV = ₹8000

Total Cost Price of both articles = ₹8000 + ₹8000 = ₹16,000

Now VCR is sold at 4% loss.

∴ Selling price of VCR =C.P.×{⁽¹⁰⁰⁻⁴⁾/₁₀₀}

S.P of VCR =8000×[⁽⁹⁶⁾/₁₀₀] =80×96=₹7680

{∵ If a value decreases x % multiply that value by [⁽¹⁰⁰⁻ˣ⁾/₁₀₀] to get new value}

And TV is sold at 8% profit,

Then S.P of T.V =C.P×{⁽¹⁰⁰⁺⁸⁾/₁₀₀}

{∵ If a value increases x % multiply that value by [⁽¹⁰⁰⁺ˣ⁾/₁₀₀] to get new value}

S.P of T.V= 8000×[⁽¹⁰⁸⁾/₁₀₀]=₹8,640

Total Selling price =₹7680+₹8640=₹16,320

Since S.P>C.P

∴ Profit =S.P – C.P=16,320-16000=₹320

And profit%=(profit/C.P)×100= [⁽³²⁰⁾/₁₆₀₀₀]×100

=2%

class 8 maths chapter 8 exercise 8.2 (6)

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?

Solution to class 8 maths chapter 8 exercise 8.2 (6).

Rate of discount on all items = 10%

Marked Price of a pair of jeans = ₹1450 and Marked Price of a shirt = ₹850

Discount on pair of jeans= 10% of 1450

=[⁽¹⁰⁾/₁₀₀]×1450

=₹145

∴ Selling price of jeans=₹1450-₹145=₹1305

Marked price of two shirts=2×850=₹1700

Discount on two shirts=10% of ₹1700

=[⁽¹⁰⁾/₁₀₀]×1700

=₹170

∴ Selling price of two shirts=₹1700-₹170=₹1530

∴ The customer had to pay =₹1305+1530=₹2,835

class 8 maths chapter 8 exercise 8.2 (7)

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.

Solution to class 8 maths chapter 8 exercise 8.2 (7).

S.P of each buffalo =₹20,000

S.P of two buffaloes =2 ×₹20,000 =₹40,000.

One buffalo is sold at 5% gain.

C.P=[¹⁰⁰/₍₁₀₀₊ₚ₎]×S.P

{∵ S.P=C.P×[⁽¹⁰⁰⁺ᵖ⁾/₁₀₀] here, p=profit%}

C.P=[¹⁰⁰/₍₁₀₀₊₅₎]×20,000

C.P=[¹⁰⁰/₁₀₅]×20,000=₹19,047.62

Another buffalo is sold at 10% loss.

C.P=[⁽¹⁰⁰⁾/₍₁₀₀₋ₗ₎]×S.P

{∵ S.P=C.P×[⁽¹⁰⁰⁻ˡ⁾/₁₀₀] here, l=loss%}

C.P=[¹⁰⁰/₍₁₀₀₋₁₀₎]×20,000

C.P=[¹⁰⁰/₉₀]×20,000=22,222.22

Total C.P=₹19,047.62+₹22,222.22=₹41,269.84

Since C.P>S.P

∴ Here it is loss.

Loss=C.P-S.P=₹41,269.84-₹40,000=₹1,269.84

Class 8 maths chapter 8 exercise 8.2 (8)

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.

Solution to class 8 maths chapter 8 exercise 8.2 (8).

Cost price=₹13,000 and sales tax rate=12%

Selling price=₹13,000×[⁽¹⁰⁰⁺¹²⁾/₁₀₀]=₹14,560

{∵ If a value increases x % multiply that value by [⁽¹⁰⁰⁺ˣ⁾/₁₀₀] to get new value}

class 8 maths chapter 8 exercise 8.2 (9)

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.

Solution to class 8 maths chapter 8 exercise 8.2 (9).

Selling price=₹1,600. And rate of discount=20%

M.P×[⁽¹⁰⁰⁻ˣ⁾/₁₀₀]=S.P. Here, X=discount percent.

{∵ If a value decreases x% multiply that value by [⁽¹⁰⁰⁻ˣ⁾/₁₀₀] to get new value}

⇒ M.P=S.P×[⁽¹⁰⁰⁾/₍₁₀₀₋ₓ₎]

M.P=marked price.

M.P=1,600×[⁽¹⁰⁰⁾/₍₁₀₀₋₂₀₎]

M.P=1,600×[⁽¹⁰⁰⁾/₍₈₀₎]

M.P=₹2,000

class 8 maths chapter 8 exercise 8.2 (10)

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

Solution to class 8 maths chapter 8 exercise 8.2 (10).

C.P=₹5,400 and rate of VAT=8%

Price including VAT=₹5,400.

VAT=8%

Cost price=Original price×[⁽¹⁰⁰⁺ᵛ⁾/₍₁₀₀₎] ; here,V=VAT%

{∵ If a value increases x% multiply that value by [⁽¹⁰⁰⁺ˣ⁾/₁₀₀] to get new value}

Original price =Price before VAT was added=C.P×[⁽¹⁰⁰⁾/₍₁₀₀₊ᵥ₎]

Original price=5,400×[⁽¹⁰⁰⁾/₍₁₀₀₊₈₎]

Original price=5,400×[¹⁰⁰/₁₀₈]=₹5,000

Exercise-8.3

class 8 maths chapter 8 exercise 8.3 (1)

1. Calculate the amount and compound interest on

(a) Rs 10,800 for 3 years at 12½ % per annum compounded annually.

(b) Rs 18,000 for 2 ½years at 10% per annum compounded annually.

(d) Rs 8,000 for 1 year at 9% per annum compounded half yearly. (You could use the year by year calculation using SI formula to verify).

(e) Rs 10,000 for 1 year at 8% per annum compounded half yearly.

Solution to class 8 maths chapter 8 exercise 8.3 (1).

(a) Here, Principal (P)=₹10,800, Time=3 years.

Rate of interest=(R)=12½%=²⁵/₂%

Amount=(A)=P[1+(ᴿ/₁₀₀)]ⁿ

A=10800×[1+(²⁵/₂ₓ₁₀₀)]³

A=10800×[1+(¹/₈)]³

A=10800×[⁹/₈]³

A=10800× ⁹/₈ × ⁹/₈× ⁹/₈

A=₹15,377.34

Compound interest =A-P=₹10800-₹15,377.34

C.I=₹4577.34

(b) Here, Principal=18,000, Time=n=2½ years, Rate of interest=10% p.a.

Amount=(A)=P[1+(ᴿ/₁₀₀)]ⁿ

Amount=(A)=18000×[1+(¹⁰/₁₀₀)]²

=18000×[1+(¹/₁₀)]²

=18000×[¹¹/₁₀]²

=18000×(¹¹/₁₀)×(¹¹/₁₀)

Amount=(A)=₹21,780

Interest for ½ year on 21,780 at the rate of 10%

=½×{⁽²¹⁷⁸⁰ˣ¹⁰ˣ¹⁾/₁₀₀}=₹1,089

Total amount for 2½ years

=₹21,780+₹1,089=₹22,869

Compound interest=C.I=A-P

=₹22,869-18000=₹4,869

(c)

Principal (P) =₹62,500. Time=1½years=³/₂years=3 half years(compounded half yearly)

Rate of interest=(⁸/₂)%=4%

Amount=(A)=P[1+(ᴿ/₁₀₀)]ⁿ

Amount=(A)=62500[1+(⁴/₁₀₀)]³

=62500[1+(¹/₂₅)]³

=62500×(²⁶/₂₅)³

=2500×(²⁶/₂₅)×(²⁶/₂₅)×(²⁶/₂₅)

=₹70,304

Compound interest=A-P=70,304-62500=₹7,804

Rate of interest=(⁹/₂)% {∵Compounded half yearly}

Amount=(A)=P[1+(ᴿ/₁₀₀)]ⁿ

Amount=(A)=8000[1+(⁹/₍₂ₓ₁₀₀₎)]²

=8000×[²⁰⁹/₂₀₀]²

=8000×[²⁰⁹/₂₀₀]×[²⁰⁹/₂₀₀]

=₹8,736.20

Compound interest=A-P=₹8736.20-₹8000=₹736.20

(d) Principal=10,000 ; Time=1year=2half years(compounded half yearly)

Rate of interest=(⁸/₂)%=4%

Amount=(A)=P[1+(ᴿ/₁₀₀)]ⁿ

Amount=(A)=10000[1+(⁴/₁₀₀)]²

=10000[1+(¹/₂₅)]²

=10000×(²⁶/₂₅)²

=10000×(²⁶/₂₅)×(²⁶/₂₅)

=₹10,816

Compound interest=A-P=10,816-10000=₹816

class 8 maths chapter 8 exercise 8.3 (2)

2. Kamala borrowed Rs 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?

Solution to class 8 maths chapter 8 exercise 8.3 (2).

Principal=₹26,400 ; Time(n)=2years 4months ; Rate of interest(R) =15%

Amount=(A)=P[1+(ᴿ/₁₀₀)]ⁿ

=26,400×[1+(¹⁵/₁₀₀)]²

=26,400×[1+³/₂₀]²

=26,400×(²³/₂₀)×(²³/₂₀)

=₹34,914

Interest for 4 months =⁴/₁₂ years of rate of 15%=⅓ years of rate of 15%

=⅓×[⁽³⁴⁹¹⁴ˣ¹⁵ˣ¹⁾/₁₀₀]=₹1745.70

Total amount=₹34914+₹1745.70=₹36,659.70

class 8 maths chapter 8 exercise 8.3 (3)

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?

Solution to class 8 maths chapter 8 exercise 8.3 (3).

Principal=₹12,500 ; Time(n)=3years ; Rate of interest(R) =12% p.a.

Simple intrest for Fabina=⁽ᴾˣᴿˣᵀ⁾/₁₀₀

=⁽¹²⁵⁰⁰ˣ¹²ˣ³⁾/₁₀₀=125×36=₹4500

Amount for Radha, P=12,500, R=10% and n=3years.

Amount=(A)=P[1+(ᴿ/₁₀₀)]ⁿ

=12500×[1+(¹⁰/₁₀₀)]³

=12500×[1+¹/₁₀]³

=12500×(¹¹/₁₀)×(¹¹/₁₀)×(¹¹/₁₀)

=₹16637.50

C.I of Radha=A-P=₹16637.50-12500=₹4137.50

Here, Fabina pays more intrest=4500-4137.50=₹362.50

class 8 maths chapter 8 exercise 8.3 (4)

4. I borrowed Rs 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?

Solution to class 8 maths chapter 8 exercise 8.3 (4).

Principal=₹12,000 ; Time(n)=2years ; Rate of interest(R) =6% p.a.

Simple intrest =⁽ᴾˣᴿˣᵀ⁾/₁₀₀

=⁽¹²⁰⁰⁰ˣ⁶ˣ²⁾/₁₀₀=₹1440

Had he borrowed this sum at 6%p.a. then,

Compound interest (C.I)=P[1+(ᴿ/₁₀₀)]ⁿ-P

=12000×[1+(⁶/₁₀₀)]²-12000

={12000×[1+³/₅₀]²}-12000

=[12000×(⁵³/₅₀)×(⁵³/₅₀)]-12000

=₹13483.20-₹12000

=₹1483.20

Difference in both interest s =₹1483.20-₹1440.00=₹43.20

class 8 maths chapter 8 exercise 8.3 (5)

5. Vasudevan invested Rs 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?

Solution to class 8 maths chapter 8 exercise 8.3 (5).

(i)Here, Principal (P)=₹60,000, Time(n)=6 months=1 half year.

Rate of interest=(R)=(¹²/₂)%=6%{∵ Compounded half yearly}

Amount=(A)=P[1+(ᴿ/₁₀₀)]ⁿ

A=60000×[1+(⁶/₁₀₀)]¹

A=60000×[1+(³/₅₀)]¹

A=60000×[⁵³/₅₀]

A=60000× (⁵³/₅₀)

A=₹63,600

After 6 months vasudevan would get amount ₹63,600.

(ii) Here, Principal (P)=₹60000, Time=1year.=2 half years.{Compounded half yearly}

Rate of interest=(R)=(¹²/₂)%=6%{∵Compounded half yearly}

Amount=(A)=P[1+(ᴿ/₁₀₀)]ⁿ

A=60000×[1+(⁶/₁₀₀)]²

A=60000×[1+(³/₅₀)]²

A=60000×[⁵³/₅₀]²

A=60000× (⁵³/₅₀)×(⁵³/₅₀)

A=₹67416

After 6 months vasudevan would get amount ₹67,416.

class 8 maths chapter 8 exercise 8.3 (6)

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 1½ years if the interest is

(i) compounded annually.

(ii) compounded half yearly.

Solution to class 8 maths chapter 8 exercise 8.3 (6).

(i)Here, Principal (P) = ₹80,000, Rate of Interest (R) = 10%, Time (n) = 1½ years

A=P[1+(ᴿ/₁₀₀)]ⁿ

A=80000×[1+(¹⁰/₁₀₀)]¹

A=80000×[1+(¹/₁₀)]¹

A=80000×[¹¹/₁₀]

A=80000× (¹¹/₁₀)

A=₹88,000

Intrest for ½ year=[⁽⁸⁸⁰⁰⁰ˣ¹ˣ¹⁰⁾/₍₂ₓ₁₀₀₎]=₹4,400

Total amount = ₹88,000 + ₹4,400 = ₹92,400

(ii) Here, Principal (P) = ₹80,000,

Time (n) =1½ year=3 half-years (compounded half yearly)

Rate of interest (R) =(¹⁰/₂)% = 5% (compounded half yearly)

A=P[1+(ᴿ/₁₀₀)]³

A=80000×[1+(⁵/₁₀₀)]³

A=80000×[1+(¹/₂₀)]³

A=80000×[²¹/₂₀]³

A=80000×[²¹/₂₀]×[²¹/₂₀]×[²¹/₂₀]

A=₹92,610

Difference in amounts =₹ 92,610 – ₹92,400 =₹210

class 8 maths chapter 8 exercise 8.3 (7)

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 3rd year.

Solution to class 8 maths chapter 8 exercise 8.3 (7).

(i)Here, Principal (P) = ₹8000, Rate of Interest (R) = 5%, Time (n) = 2 years

(A)=P[1+(ᴿ/₁₀₀)]ⁿ

A=8000×[1+(⁵/₁₀₀)]²

A=8000×[1+(¹/₂₀)]²

A=8000×[²¹/₂]²

A=8000× (²¹/₂₀)×(²¹/₂₀)

A=₹8,820

(ii) Here, Principal (P) = ₹8000, Rate of Interest (R) = 5%, Time (n) = 3years

(A)=P[1+(ᴿ/₁₀₀)]ⁿ

A=8000×[1+(⁵/₁₀₀)]³

A=8000×[1+(¹/₂₀)]³

A=8000×[²¹/₂]³

A=8000× (²¹/₂₀)×(²¹/₂₀)×(²¹/₂₀)

A=₹9,261

Interest for 3rd year = A P= ₹9,261 -₹8,820 =₹441

class 8 maths chapter 8 exercise 8.3 (8)

8. Find the amount and the compound interest on Rs 10,000 for 1½ years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

Solution to class 8 maths chapter 8 exercise 8.3 (8).

Here, Principal (P) = ₹ 10000, Rate of Interest (R) = 10% = 5% (compounded half yearly) Time (n)=1½ years=3 half (compounded half yearly)

Amount=(A)=P[1+(ᴿ/₁₀₀)]ⁿ

A=10000×[1+(⁵/₁₀₀)]³

A=10000×[1+(¹/₂₀)]³

A=10000×[²¹/₂₀]³

A=10000× (²¹/₂₀)×(²¹/₂₀)×(²¹/₂₀)

A=₹11576.25

Compound Interest (CI) = A- P =₹11,576.25-10,000 = ₹ 1,576.25

If it is compounded annually, then

Here Principal (P) = 10000, Rate of Interest (R) = 10%. Time (n) =1½ years

Amount=(A)=P[1+(ᴿ/₁₀₀)]ⁿ

A=10000×[1+(¹⁰/₁₀₀)]¹

A=10000×[1+(¹/₁₀)]¹

A=10000×[¹¹/₁₀]

A=₹11000

Interest for half year =[⁽¹¹⁰⁰⁰ˣ¹ˣ¹⁰⁾/₍₂ₓ₁₀₀₎] =₹550.

Total amount = ₹11,000 + ₹550 = ₹11,550

C.I. = A – P = ₹11,550 – ₹10,000 = ₹1,550

Yes, interest ₹ 1,576.25 is more than 1,550.

class 8 maths chapter 8 exercise 8.3 (9)

9. Find the amount which Ram will get on Rs 4096, if he gave it for 18 months at 12½% per annum, interest being compounded half yearly.

Solution to class 8 maths chapter 8 exercise 8.3 (9).

Here, Principal (P)=₹4096, Time=18 months=1½ half years.

=3 half years {∵Compounded half yearly}

Rate of interest=(R)=(12½)%=(²⁵/₄)%{∵Compounded half yearly}

Amount=(A)=P[1+(ᴿ/₁₀₀)]ⁿ

A=4096×[1+(²⁵/₄₀₀)]³

A=4096×[1+(¹/₁₆)]³

A=4096×[¹⁷/₁₆]³

A=4096× (¹⁷/₁₆)×(¹⁷/₁₆)×(¹⁷/₁₆)

A=₹4,913

class 8 maths chapter 8 exercise 8.3 (10)

10. 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?

Solution to class 8 maths chapter 8 exercise 8.3 (10).

(i)Here, A₂₀₀₃ = 54,000, R = 5%, n= 2 years Population would be less in 2001 than 2003 in two years. Here population is increasing.

A₂₀₀₃=P₂₀₀₁[1+(ᴿ/₁₀₀)]ⁿ

54000=P₂₀₀₁[1+(⁵/₁₀₀)]²

54000=P₂₀₀₁[1+(¹/₂₀)]²

54000=P₂₀₀₁(²¹/₂₀)²

54000=P₂₀₀₁×(²¹/₂₀)×(²¹/₂₀)

P₂₀₀₁=54000×[⁽²⁰ˣ²⁰⁾/₍₂₁ₓ₂₁₎]

P₂₀₀₁=48,980(approx.)

(ii) According to question, population is increasing.

∴ population in 2005,

A₂₀₀₅=P[1+(ᴿ/₁₀₀)]ⁿ

A₂₀₀₅=54000[1+(⁵/₁₀₀)]²

A₂₀₀₅=54000(²¹/₂₀)²

A₂₀₀₅=54000×(²¹/₂₀)×(²¹/₂₀)=59,535

Hence population in 2005 would be 59,535.

class 8 maths chapter 8 exercise 8.3 (11)

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.

Solution to class 8 maths chapter 8 exercise 8.3 (11).

Here, Principal (P)=₹506000, Time=2 hours.

Rate of interest=(R)=2.5%

After 2 hours number of bacteria,

Amount=(A)=P[1+(ᴿ/₁₀₀)]ⁿ

A=506000×[1+(²·⁵/₁₀₀)]²

A=60000×[1+(²⁵/₁₀₀₀)]²

A=60000×[1+(¹/₄₀)]²

A=60000× (⁴¹/₄₀)×(⁴¹/₄₀)

A=5,31,616.25

Hence number of bacteria after two hours are 5,31,616 (approx.)

class 8 maths chapter 8 exercise 8.3 (12)

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.

Solution to class 8 maths chapter 8 exercise 8.3 (12).

Here, Principal (P)=₹42000, Time=1year.

Rate of interest=(R)=8%

Given scooter value depreciated at the rate of 8% per annum.

Amount=(A)=P[1-(ᴿ/₁₀₀)]ⁿ

A=42000×[1-(⁸/₁₀₀)]¹

A=42000×[1-(²/₂₅)]

A=42000×[²⁷/₂₅]

A=42000× (²⁷/₂₅)

A=₹38,640

Hence, the value of scooter after one year is =₹38,640.

Sudheer Kumar B