## NCERT Solutions for Chapter 8 Comparing Quantities Class 8 Mathematics

Page No: 119**Exercise 8.1**

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

(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 = 15/30 = 1/2= 1 : 2

(b) ∵ 1 km = 1000 m

∴ 10 km = 10×1000 = 10000 m

∴ Ratio = 5m/10000m = 1/2000 = 1 : 2000

(c) ∵ Rs 1 = 100 paise

∴ Rs 5 = 5× 100 = 500 paise

Hence Ratio = 50 paise/ 500 paise = 1/10 = 1 : 10

2. Convert the following ratios to percentages:

(a) 3 : 4

(c) 50 paise to Rs. 5

**Solution**(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 = 15/30 = 1/2= 1 : 2

(b) ∵ 1 km = 1000 m

∴ 10 km = 10×1000 = 10000 m

∴ Ratio = 5m/10000m = 1/2000 = 1 : 2000

(c) ∵ Rs 1 = 100 paise

∴ Rs 5 = 5× 100 = 500 paise

Hence Ratio = 50 paise/ 500 paise = 1/10 = 1 : 10

2. Convert the following ratios to percentages:

(a) 3 : 4

(b) 2 : 3

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

(b) Percentage of 2 : 3 = 2/3 × 100% = 66.2/3%

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

(b) Percentage of 2 : 3 = 2/3 × 100% = 66.2/3%

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

**Solution**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%

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 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.

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

**Solution**

Total percentage of money she didn't spent = 100% - 75%=25%

According to question,

25% = 600

1% = 600/25

100% = 600/ 25 × 100

Hence the money in the beginning was Rs 2,400.

6. If 60% people in a city like cricket, 30% like football and the remaining like other games, then what percent 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**

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.

Pg No. 125

**Exercise 8.2**

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

**Solution**

Let original salary be Rs.100.

Therefore New salary i.e., 10% increase

= 100 + 10 = Rs.110

∵ New salary is Rs.110, when original salary = Rs.100

∴ New salary is Rs.1, when original salary = 100/110

∴ New salary is Rs.1,54,000, when original salary = 100/110 × 154000 = Rs.1,40,000

Hence original salary is Rs. 1,40,000.

2. On Sunday 845 people went to the Zoo. On Monday only 169 people went. What is the percent decrease in the people visiting the Zoo on Monday?

**Solution**

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%.

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**

No. of articles = 80

Cost Price of articles = Rs. 2,400

And Profit = 16%

∵ Cost price of articles is Rs.100, then selling price = 100 + 16 = Rs.116

∴ Cost price of articles is Rs.1, then selling price = 116/100

∴ Cost price of articles is Rs.2400, then selling price = 116/100 × 2400 = Rs.2784

Hence, Selling Price of 80 articles = Rs.2784

Therefore Selling Price of 1 article = 2784/80 = Rs.34.80

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

**Solution**

Here, C.P. = Rs.15,500 and Repair cost = Rs.450

Therefore Total Cost Price = 15500 + 450 = Rs.15,950

Let C.P. be Rs.100, then S.P. = 100 + 15 = Rs.115

When C.P. is Rs.100, then S.P. = Rs.115

∴ When C.P. is Rs.1, then S.P. = 115/100

∴ When C.P. is Rs.15950, then S.P. =115/100 × 15950 = Rs.18,342.50

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**

Cost price of VCR = Rs.8000 and Cost price of TV = Rs.8000

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

Now, VCR is sold at 4% loss.

Let C.P. of each article be Rs.100, then S.P. of VCR = 100 – 4 = Rs.96

When C.P. is Rs.100, then S.P. = Rs.96

∴ When C.P. is Rs.1, then S.P. = 96/100

∴ When C.P. is Rs.8000, then S.P. = 96/100 ×8000 = Rs.7,680

And TV is sold at 8% profit, then S.P. of TV = 100 + 8 = Rs.108

When C.P. is Rs.100, then S.P. = Rs.108

∴ When C.P. is Rs.1, then S.P. = 108/100

∴ When C.P. is Rs.8000, then S.P. = 108/100 × 8000 = Rs.8,640

Then, Total S.P. = Rs.7,680 + Rs.8,640 = Rs. 16,320

Since S.P. >C.P.,

Therefore Profit = S.P. – C.P. = 16320 – 16000 = Rs.320

And Profit% = Profit/ cost price × 100 = 320/16000 × 100 = 2%

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

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.1450and two shirts marked at Rs.850 each?

**Solution**

Rate of discount on all items = 10%

Marked Price of a pair of jeans = Rs.1450 and Marked Price of a shirt = Rs.850

Discount on a pair of jeans = (Rate × M.P)/100 = (10×1450)/100 = Rs.145

∴ S.P. of a pair of jeans = Rs.1450 – Rs.145 = Rs.1305

Marked Price of two shirts = 2× 850 = Rs.1700

Discount on two shirts = (Rate × M.P)/100 = (10 × 1700)/100 = Rs.170

∴ S.P. of two shirts = Rs.1700 – Rs.170 = Rs.1530

Therefore the customer had to pay = 1305 + 1530

Therefore the customer had to pay = 1305 + 1530

= Discount on a pair of jeans

= (Rate × M.P)/100

= (10×1450)/100 = Rs.145

∴ S.P. of a pair of jeans = Rs.1450 – Rs.145 = Rs.2,835

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

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)

∴ S.P. of a pair of jeans = Rs.1450 – Rs.145 = Rs.2,835

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

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)

**Solution**
S.P. of each buffalo = Rs.20,000

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

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

One buffalo is sold at 5% gain.

Let C.P. be Rs.100, then S.P. = 100 + 5 = Rs.105

When S.P. is Rs.105, then C.P. = Rs.100

∴ When S.P. is Rs.1, then C.P. = 100/105

∴ When S.P. is Rs.20,000, then C.P. = 100/105 × 20000 = Rs.19,047.62

Another buffalo is sold at 10% loss.

Let C.P. be Rs.100, then S.P. = 100 – 10 = Rs.90

When S.P. is Rs.90, then C.P. = Rs.100

∴ When S.P. is Rs.1, then C.P. = 100/90

∴ When S.P. is Rs.20,000, then C.P. = 100/90 × 20000 = Rs.22,222.22

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

Since C.P. >S.P.

Therefore here it is loss.

Loss = C.P. – S.P. = Rs.41,269.84 – Rs. 40,000.00 = Rs.1,269.84

The overall loss of milkman was Rs.1269.84

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.

Let C.P. be Rs.100, then S.P. = 100 + 5 = Rs.105

When S.P. is Rs.105, then C.P. = Rs.100

∴ When S.P. is Rs.1, then C.P. = 100/105

∴ When S.P. is Rs.20,000, then C.P. = 100/105 × 20000 = Rs.19,047.62

Another buffalo is sold at 10% loss.

Let C.P. be Rs.100, then S.P. = 100 – 10 = Rs.90

When S.P. is Rs.90, then C.P. = Rs.100

∴ When S.P. is Rs.1, then C.P. = 100/90

∴ When S.P. is Rs.20,000, then C.P. = 100/90 × 20000 = Rs.22,222.22

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

Since C.P. >S.P.

Therefore here it is loss.

Loss = C.P. – S.P. = Rs.41,269.84 – Rs. 40,000.00 = Rs.1,269.84

The overall loss of milkman was Rs.1269.84

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**
C.P. = Rs.13,000 and S.T. rate = 12%

Let C.P. be Rs.100, then S.P. for purchaser = 100 + 12 = Rs.112

When C.P. is Rs.100, then S.P. = Rs.112

∴ When C.P. is Rs.1, then S.P. = 112/100

∴ When C.P. is Rs.13,000, then S.P. = 112/100 × 13000 = Rs.14,560

He will have to pay Rs.14,560

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.

S.P. = Rs.1,600 and Rate of discount = 20%

Let M.P. be Rs.100, then S.P. for customer = 100 – 20 = Rs.80

When S.P. is Rs.80, then M.P. = Rs.100

∴ When S.P. is Rs.1, then M.P. = 100/80

∴ When S.P. is Rs.1600, then M.P. = 100/ 80 × 1600 = Rs.2,000

Thus, the marked price was Rs. 2,000

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

C.P. = Rs.5,400 and Rate of VAT = 8%

Let C.P. without VAT is Rs. 100, then price including VAT = 100 + 8 = Rs.108

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

∴ When price including VAT is Rs.1, then original price = 100/108

∴ When price including VAT is Rs.5400, then original price = 100/108 × 5400 = Rs.5000

Thus, the price of Hair Dryer before the addition of VAT was Rs 5000

Let C.P. be Rs.100, then S.P. for purchaser = 100 + 12 = Rs.112

When C.P. is Rs.100, then S.P. = Rs.112

∴ When C.P. is Rs.1, then S.P. = 112/100

∴ When C.P. is Rs.13,000, then S.P. = 112/100 × 13000 = Rs.14,560

He will have to pay Rs.14,560

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**S.P. = Rs.1,600 and Rate of discount = 20%

Let M.P. be Rs.100, then S.P. for customer = 100 – 20 = Rs.80

When S.P. is Rs.80, then M.P. = Rs.100

∴ When S.P. is Rs.1, then M.P. = 100/80

∴ When S.P. is Rs.1600, then M.P. = 100/ 80 × 1600 = Rs.2,000

Thus, the marked price was Rs. 2,000

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

**Solution**C.P. = Rs.5,400 and Rate of VAT = 8%

Let C.P. without VAT is Rs. 100, then price including VAT = 100 + 8 = Rs.108

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

∴ When price including VAT is Rs.1, then original price = 100/108

∴ When price including VAT is Rs.5400, then original price = 100/108 × 5400 = Rs.5000

Thus, the price of Hair Dryer before the addition of VAT was Rs 5000

Pg No. 133

**Exercise 8.3**

1. Calculate the amount and compound interest on:

(a) Rs.10,800 for 3 years at 12.1/2% per annum compounded annually.

(b) Rs.18,000 for 2.1/2 years at 10% per annum compounded annually.

(c) Rs.62,500 for 1.1/2 years at 8% per annum compounded annually.

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

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

**Solution**

(a) Here, Principal (P) = Rs. 10800,

Time(n) = 3 years,

Rate of interest (R) = 12.1/2% = 25/2 %

Amount (A) =

= =

= =

= 10800 × 9/8 × 9/8 × 9/8

= Rs. 15,377.34 (approx.)

Compound Interest (C.I.) = A – P

= Rs. 10800 – Rs. 15377.34 = Rs. 4,577.34

(b) Here, Principal (P) = Rs. 18,000, Time (n) = 2.1/2 years, Rate of interest (R) = 10% p.a.

Amount (A) =

= =

= 18000(11/10)

^{2}= 18000 × 11/10 ×11/10

= Rs. 21,780

Interest for 1/2 years on Rs. 21,780 at rate of 10% = (21780×10×1)/100 = Rs. 1,089

Total amount for 2.1/2 years = Rs. 21,780 + Rs. 1089 = Rs. 22,869

Compound Interest (C.I.) = A – P = Rs. 22869 – Rs. 18000 = Rs. 4,869

(c) Here, Principal (P) = Rs. 62500,

Time (n) = 1.1/2 = 3/2 years = 3 years (compounded half yearly)

Rate of interest (R) = 8% = 4% (compounded half yearly)

Amount (A) =

=

=

= 62500 (26/25)

^{3}

= 62500 × 26/25 × 26/25 × 26/25

= Rs. 70,304

Compound Interest (C.I.) = A – P = Rs. 70304 – Rs. 62500 = Rs. 7,804

(d) Here, Principal (P) = Rs. 8000,

Time (n) = 1 years = 2 years(compounded half yearly)

Rate of interest (R) = 9% = 9/2 % (compounded half yearly)

Amount (A) =

=

=

= 8000 (209/200)

^{2}

= 8000 × 209/200 × 209/200

= Rs. 8,736.20

Compound Interest (C.I.) = A – P = Rs. 8736.20 – Rs. 8000 = Rs. 736.20

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

Time (n) = 1 years = 2 years (compounded half yearly)

Rate of interest (R) = 8% = 4% (compounded half yearly)

Amount (A) =

=

=

= 10000 (26/25)

^{2}

= 10000 × 26/26 × 26/25

= Rs. 10,816

Compound Interest (C.I.) = A – P = Rs. 10,816 – Rs. 10,000 = Rs. 816

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?

(Hint: Find A for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for 4/12 years).

**Solution**

Here, Principal (P) = Rs. 26,400,

Time(n) = 2 years 4 months,

Rate of interest (R) = 15% p.a.

Amount for 2 years (A) =

= =

= 26400(23/20)

^{2}= 26400 × 23/20 × 23/20

= Rs. 34,914

Interest for 4 months = 4/12 = 1/3 years at the rate of 15% = (34914 × 15×1)/100

= Rs. 1745.70

∴Total amount = Rs. 34,914 + Rs. 1,745.70 = Rs. 36,659.70

3. Fabina borrows Rs.12,500 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**

Here, Principal (P) = Rs.12,500,

Time (T) = 3 years,

Rate of interest (R) = 12% p.a.

Simple Interest for Fabina = = = Rs. 4,500

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

Amount (A) =

= =

= =

= Rs. 16,637.50

C.I. for Radha = A – P

= Rs. 16,637.50 – Rs. 12,500 = Rs. 4,137.50

Thus, Fabina pays more interest = Rs. 4,500 – Rs. 4,137.50 = Rs. 362.50

4. I borrows 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**

Here, Principal (P) = Rs.12,000,

Time (T) = 2 years,

Rate of interest (R) = 6% p.a.

Simple Interest = (P×R×T)/100

= (12000 × 6× 2)/100 = Rs. 1,440

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

Compound Interest =- P

=

=

= 12000(53/50)

= 12000 × 53/50 × 53/50 - 12000

= Rs. 13,483.20 – Rs. 12,000

= Rs. 1,483.20

Difference in both interests = Rs. 1,483.20 – Rs. 1,440.00 = Rs. 43.20

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

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?

(i) Here, Principal (P) = Rs. 60,000,

Time (n)= 6 months = 1 year(compounded half yearly)

Rate of interest (R) = 12% = 6% (compounded half yearly)

Amount (A) =

=

=

= 60000 (53/50)

= 60000 × 53/50

= Rs. 63,600

After 6 months Vasudevan would get amount Rs. 63,600.

(ii) Here, Principal (P) = Rs. 60,000,

Time (n) = 1 year = 2 year(compounded half yearly)

Rate of interest (R) = 12% = 6% (compounded half yearly)

Amount (A) =

=

=

= 60000(53/50)

= 60000 × 53/50 × 53/50

= Rs. 67,416

After 1 year Vasudevan would get amount Rs. 67,416.

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 1.1/2 years if the interest is:

(i) compounded annually.

(ii) compounded half yearly.

(i) Here, Principal (P) = Rs. 80,000

Simple Interest = (P×R×T)/100

= (12000 × 6× 2)/100 = Rs. 1,440

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

Compound Interest =- P

=

=

= 12000(53/50)

^{2}- 12000= 12000 × 53/50 × 53/50 - 12000

= Rs. 13,483.20 – Rs. 12,000

= Rs. 1,483.20

Difference in both interests = Rs. 1,483.20 – Rs. 1,440.00 = Rs. 43.20

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

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**(i) Here, Principal (P) = Rs. 60,000,

Time (n)= 6 months = 1 year(compounded half yearly)

Rate of interest (R) = 12% = 6% (compounded half yearly)

Amount (A) =

=

=

= 60000 (53/50)

^{1}= 60000 × 53/50

= Rs. 63,600

After 6 months Vasudevan would get amount Rs. 63,600.

(ii) Here, Principal (P) = Rs. 60,000,

Time (n) = 1 year = 2 year(compounded half yearly)

Rate of interest (R) = 12% = 6% (compounded half yearly)

Amount (A) =

=

=

= 60000(53/50)

^{2}= 60000 × 53/50 × 53/50

= Rs. 67,416

After 1 year Vasudevan would get amount Rs. 67,416.

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 1.1/2 years if the interest is:

(i) compounded annually.

(ii) compounded half yearly.

**Solution**(i) Here, Principal (P) = Rs. 80,000

Time (n)= 1.1/2 years,

Rate of interest (R) = 10%

Amount for 1 year (A) =

=

=

= 80000 (11/10)

= Rs. 88,000

Interest for 1/2 year = (88000×10×1)/(100×2) = Rs. 4,400

Total amount = Rs. 88,000 + Rs. 4,400 = Rs. 92,400

(ii) Here, Principal (P) = Rs.80,000,

Time (n) = 1.1/2 year = 3/2 years (compounded half yearly)

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

Amount (A) =

=

=

= 80000(21/20)

= 80000 × 21/20 × 21/20 × 21/20

= Rs. 92,610

Difference in amounts = Rs. 92,610 – Rs. 92,400 = Rs. 210

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 third year.

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

Amount (A) =

=

=

= 8000(21/20)

= 8000 × 21/20 × 21/20

= Rs. 8,820

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

Amount (A) =

=

=

= 8000(21/20)

= 8000 × 21/20 × 21/20 × 21/20

= Rs. 9,261

Interest for 3rd year = A – P = Rs. 9,261 – Rs. 8,820 = Rs. 441

8. Find the amount and the compound interest on Rs.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?

Here, Principal (P) = Rs. 10000,

Amount for 1 year (A) =

=

=

= 80000 (11/10)

^{1}= Rs. 88,000

Interest for 1/2 year = (88000×10×1)/(100×2) = Rs. 4,400

Total amount = Rs. 88,000 + Rs. 4,400 = Rs. 92,400

(ii) Here, Principal (P) = Rs.80,000,

Time (n) = 1.1/2 year = 3/2 years (compounded half yearly)

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

Amount (A) =

=

=

= 80000(21/20)

^{3}= 80000 × 21/20 × 21/20 × 21/20

= Rs. 92,610

Difference in amounts = Rs. 92,610 – Rs. 92,400 = Rs. 210

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 third year.

**Solution**(i) Here, Principal (P) = Rs. 8000, Rate of Interest (R) = 5%, Time (n) = 2 years

Amount (A) =

=

=

= 8000(21/20)

^{2}= 8000 × 21/20 × 21/20

= Rs. 8,820

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

Amount (A) =

=

=

= 8000(21/20)

^{3}= 8000 × 21/20 × 21/20 × 21/20

= Rs. 9,261

Interest for 3rd year = A – P = Rs. 9,261 – Rs. 8,820 = Rs. 441

8. Find the amount and the compound interest on Rs.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?

**Solution**Here, Principal (P) = Rs. 10000,

Rate of Interest (R) = 10% = 5% (compounded half yearly)

Time (n) = 1.1/2 years = 3 years (compounded half yearly)

Amount (A) =

=

=

= 10000 (21/20)

= 10000 × 21/20 × 21/20 × 21/20

= Rs. 11,576.25

Compound Interest (C.I.) = A – P = Rs. 11,576.25 – Rs. 10,000 = Rs. 1,576.25

Time (n) = 1.1/2 years = 3 years (compounded half yearly)

Amount (A) =

=

=

= 10000 (21/20)

^{3}= 10000 × 21/20 × 21/20 × 21/20

= Rs. 11,576.25

Compound Interest (C.I.) = A – P = Rs. 11,576.25 – Rs. 10,000 = Rs. 1,576.25

If it is compounded annually, then

Here, Principal (P) = Rs. 10000

Here, Principal (P) = Rs. 10000

Rate of Interest (R) = 10%,

Time (n) = 1.1/2years

Amount (A) for 1 year =

=

=

= 10000(11/10)

= 10000 × 11/10

= Rs. 11,000

Interest for 1/2 year = (11000 × 1 × 10)/(2×100) = Rs. 550

∴ Total amount = Rs. 11,000 + Rs. 550 = Rs. 11,550

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

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

9. Find the amount which Ram will get on Rs.4,096, if he gave it for 18 months at 12.1/2% per annum, interest being compounded half yearly.

Here, Principal (P) = Rs. 4096,

Rate of Interest (R) = 12.1/2 = 25/2% = 25/4% (compounded half yearly)

Time (n)= 18 months = 1.1/2 years = 3 years (compounded half yearly)

Amount (A) =

=

=

= 4096(17/16)

=

= 10000(11/10)

^{1}= 10000 × 11/10

= Rs. 11,000

Interest for 1/2 year = (11000 × 1 × 10)/(2×100) = Rs. 550

∴ Total amount = Rs. 11,000 + Rs. 550 = Rs. 11,550

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

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

9. Find the amount which Ram will get on Rs.4,096, if he gave it for 18 months at 12.1/2% per annum, interest being compounded half yearly.

**Solution**Here, Principal (P) = Rs. 4096,

Rate of Interest (R) = 12.1/2 = 25/2% = 25/4% (compounded half yearly)

Time (n)= 18 months = 1.1/2 years = 3 years (compounded half yearly)

Amount (A) =

=

=

= 4096(17/16)

^{3}
= 4096 × 17/16 × 17/16 × 17/16

= Rs. 4,913

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?

(i) Here, A

= Rs. 4,913

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**(i) Here, A

_{2003}= Rs. 54,000,
R = 5%,

n = 2 years

Population would be less in 2001 than 2003 in two years.

Here population is increasing.

∴ A

Population would be less in 2001 than 2003 in two years.

Here population is increasing.

∴ A

_{2003}=
⇒ 54000 =

⇒ 54000 =

⇒ 54000 = P

⇒ 54000 = P

⇒ P

⇒ P

(ii) According to question, population is increasing.

Therefore population in 2005,

A

= 54000 (1 + 5/100)

= 54000(1+1/20)

= 54000(21/20)

= 54000 × 21/20 × 21/20

= 59,535

Hence population in 2005 would be 59,535.

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.

Here, Principal (P) = 5,06,000,

Rate of Interest (R) = 2.5%,

Time (n) = 2 hours

After 2 hours, number of bacteria,

Amount (A) =

= 506000(1+ 2.5/100)

= 506000(1+25/1000)

= 506000(1+1/40)

= 506000(41/40)

= 506000 × 41/40 × 41/40

= 5,31,616.25

Hence, number of bacteria after two hours are 531616 (approx.).

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.

Here, Principal (P) = Rs. 42,000,

Rate of Interest (R) = 8%,

Time (n) = 1 years

Amount (A) = P ( 1 - R/100)

= 42000(1 - 8/100)

= 42000(1+ 2/25)

= 42000 (27/25)

= 42000 × 27/25

= Rs. 38,640

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

⇒ 54000 =

⇒ 54000 = P

_{2001}(21/20)^{2}⇒ 54000 = P

_{2001}×21/20 × 21/20⇒ P

_{2001}= (54000 × 20 × 20)/(21 × 21) =48,979.5⇒ P

_{2001}= 48,980 (approx.)(ii) According to question, population is increasing.

Therefore population in 2005,

A

_{2015}== 54000 (1 + 5/100)

^{2}= 54000(1+1/20)

^{2}= 54000(21/20)

^{2}= 54000 × 21/20 × 21/20

= 59,535

Hence population in 2005 would be 59,535.

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**Here, Principal (P) = 5,06,000,

Rate of Interest (R) = 2.5%,

Time (n) = 2 hours

After 2 hours, number of bacteria,

Amount (A) =

= 506000(1+ 2.5/100)

^{2}= 506000(1+25/1000)

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

^{2}= 506000(41/40)

^{2}= 506000 × 41/40 × 41/40

= 5,31,616.25

Hence, number of bacteria after two hours are 531616 (approx.).

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**Here, Principal (P) = Rs. 42,000,

Rate of Interest (R) = 8%,

Time (n) = 1 years

Amount (A) = P ( 1 - R/100)

^{n}= 42000(1 - 8/100)

^{1}= 42000(1+ 2/25)

^{1}= 42000 (27/25)

^{1}= 42000 × 27/25

= Rs. 38,640

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

**Go to Chapters list NCERT Solutions of Class 8 Maths**