NCERT Solutions for Class 8 Maths

Class 8 NCERT Solutions- Chapter 8 Comparing Quantities – Exercise 8.3

Difficulty Level : Medium
Last Updated : 04 Mar, 2021

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Question 1. Calculate the amount and compound interest on

(i) Rs 10,800 for 3 years at 12 $\mathbf{\frac{1}{2}}$ % per annum compounded annually.

(ii) Rs 18,000 for 2 $\mathbf{\frac{1}{2}}$ years at 10% per annum compounded annually.

(iii) Rs 62,500 for 1 $\mathbf{\frac{1}{2}}$ years at 8% per annum compounded half-yearly.

(iv) Rs 8,000 for 1 year at 9% per annum compounded half-yearly.

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

Solution:

(i) Given values are,
P = Rs 10,800
R = 12 $\frac{1}{2}$ % per annum = $\frac{25}{2}$ %
T = 3 Years
As it is compounded annually then, n = 3 times.
We have,
A = P (1 + $\mathbf{\frac{R}{100}}$ )ⁿ
A = 10,800 (1+ $\frac{25}{2*100}$ )³
A = 10,800 (1+ $\frac{1}{8}$ )³
A = 10,800 ( $\frac{9}{8}$ )³
A = Rs 15,377.34
CI = A – P
CI = 15,377.34 – 10,800
CI = Rs 4,577.34
Hence, the amount = Rs 15,377.34 and
Compound interest = Rs 4,577.34
(ii) Given values are,
P = Rs 18,000
R = 10 % per annum
T = 2 $\frac{1}{2}$ Years
As it is compounded annually then, n = 2 $\frac{1}{2}$ times.
We have,
A = P (1 + $\mathbf{\frac{R}{100}}$ )ⁿ
A = 18,000 (1+ $\frac{10}{100}$ )^2½
What we will do here is Firstly we know 2 $\frac{1}{2}$ Years is 2 years and 6 months which can be calculated by first calculating the amount to 2 years using CI formula and then calculating the simple interest by using SI formula.
The amount for 2 years has to be calculated :
A = 18,000 (1+ $\frac{1}{10}$ )²
A = 18,000 ( $\frac{11}{10}$ )²
A = Rs 21,780
CI = A – P
CI = 21,780 – 18,000
CI = Rs 3,780
Now, The amount for $\frac{1}{2}$ year has to be calculated:
New P is equal to the amount after 2 Years. Hence,
P = Rs 21,780
R = 10 % per annum
T = $\frac{1}{2}$ year
SI = $\frac{PRT}{100}$
SI = $\frac{21,780 × 10 × \frac{1}{2}}{100}$
SI = $\frac{21,780 × 10 × 1}{200}$
SI = Rs 1,089
Hence, the Total amount = A + SI
= 21,780 + 1,809
= Rs 22,869
Total compound interest = CI + SI
= 3,780 + 1,809
= Rs 4,869
(iii) Given values are,
P = Rs 62,500
R = 8 % per annum hence 4% Half Yearly
T = 1 $\frac{1}{2}$ Years
As it is compounded Half yearly then, n = 3 times. (1 $\frac{1}{2}$ Years contains 3 half years)
We have,
A = P (1 + $\mathbf{\frac{R}{100}}$ )ⁿ
A = 62,500 (1+ $\frac{4}{100}$ )³
A = 62,500 (1+ $\frac{1}{25}$ )³
A = 62,500 ( $\frac{26}{25}$ )³
A = Rs 70,304
CI = A – P
CI = 70,304 – 62,500
CI = Rs 7,804
Hence, the amount = Rs 70,304 and
Compound interest = Rs 7,804
(iv) Given values are,
P = Rs 8,000
R = 9 % per annum hence, $\frac{9}{2}$ % Half Yearly
T = 1 Year
As it is compounded Half yearly then, n = 2 times. (1 Year contains 2 half years)
We have,
A = P (1 + $\mathbf{\frac{R}{100}}$ )ⁿ
A = 8,000 (1+ $\frac{9}{2*100}$ )²
A = 8,000 (1+ $\frac{9}{200}$ )²
A = 8,000 ( $\frac{209}{200}$ )²
A = Rs 8,736.20
CI = A – P
CI = 8,736.20 – 8,000
CI = Rs 736.20
Hence, the amount = Rs 8,736.20 and
Compound interest = Rs 736.20
(v) Given values are,
P = Rs 10,000
R = 8 % per annum hence, 4% Half Yearly
T = 1 Year
As it is compounded Half yearly then, n = 2 times. (1 Year contains 2 half years)
We have,
A = P (1+ $\mathbf{\frac{R}{100}}$ )ⁿ
A = 10,000 (1+ ( $\frac{4}{100}$ ))²
A =10,000 (1+ ( $\frac{1}{25}$ ))²
A = 10,000 ( $\frac{26}{25}$ )²
A = Rs 10,816
CI = A – P
CI = 10,816- 10,000
CI = Rs 816
Hence, the amount = Rs 10,816 and
Compound interest = Rs 816

Question 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:

Here, Given values are,
P = Rs 26,400
R = 15 % per annum
T = 2 Years and 4 months, which is 2 $\frac{1}{3}$ years
As it is compounded annually then, n = 2 $\frac{1}{3}$ times
We have,
A = P (1 + $\mathbf{\frac{R}{100}}$ )ⁿ
A = 26,400 (1 + ( $\frac{15}{100}$ )^2(1/3)
What we will do here is Firstly 2 years and 4 months which can be calculated by first calculating the amount to 2 years using CI formula and then calculating the simple interest by using SI formula.
The amount for 2 years has to be calculated:
A = 26,400 (1+ ( $\frac{15}{100}$ )²
A = 26,400 (1+ ( $\frac{3}{20}$ )²
A = 26,400 ( $\frac{23}{20}$ )²
A = Rs 34,914
Now, The amount for (1/3) year (4 months) has to be calculated :
New P is equal to the amount after 2 Years. Hence,
P = Rs 34,914
R = 15 % per annum
T = $\frac{1}{3}$ year
SI = $\frac{P × R × T}{100}$
SI = $\frac{(34,914 × 15 × (1/3)}{100}$
SI = $\frac{(34,914 × 15 × 1)}{300}$
SI = 1,745.70
Hence, the Total amount = A + SI
= 34,914 + 1,745.70
= Rs 36,659.70
Hence, the amount to be paid by Kamla = ₹ 36,659.70

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?

Solution:

Let’s see each case
Fabina Case: at simple interest
P = 12,500
R = 12% per annum
T = 3 Years
SI = $\frac{(P × R × T)}{100}$
SI = $\frac{(12,500 × 12 × 3)}{100}$
SI = Rs 4,500
Radha Case: at compound interest
P = 12,500
R = 10% per annum
T = 3 Years
As it is compounded annually then, n = 3 times
We have,
A = P (1 + $\mathbf{\frac{R}{100}}$ )ⁿ
A = 12,500 (1 + ( $\frac{10}{100}$ ))³
A =12,500 (1 + $\frac{1}{10}$ )³
A = 12,500 ( $\frac{11}{10}$ )³
A = Rs 16,637.5
CI = A – P
CI = 16,637.5 – 12,500
CI = 4,137.5
Clearly we can see that Fabina paid more interest, and she paid
4,500 – 4,137.5 = Rs 362.5 more than Radha

Question 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:

Lets see each case First
At simple interest
P = 12,000
R = 6% per annum
T = 2 Years
SI = $\frac{(P × R × T)}{100}$
SI = $\frac{(12,000 × 6 × 2)}{100}$
SI = Rs 1,440
At compound interest
P = 12,000
R = 6% per annum
T = 2 Years
As it is compounded annually then, n = 2 times
We have,
A = P (1 + $\mathbf{\frac{R}{100}}$ )ⁿ
A = 12,000 (1+ ( $\frac{6}{100}$ ))²
A =12,000 (1+ ( $\frac{3}{50}$ ))²
A = 12,000 ( $\frac{53}{50}$ )²
A = Rs 13,483.2
CI = A – P
CI = 13,483.2 – 12,000
CI = 1,483.2
Clearly we can see that,
1,483.2 – 1,440 = Rs 43.2
Hence, the extra amount to be paid = ₹ 43.20

Question 5. Vasudevan invested Rs 60,000 at an interest rate of 12% per annum compounded half-yearly. What amount would he get

(a) after 6 months?

(b) after 1 year?

Solution:

Let’s see each case
(a)
P = 60,000
R = 12% per annum (6% Half yearly)
T = 6 Months
As it is compounded Half Yearly then, n = 1 times (as 6 months is 1 half year)
We have,
A = P (1 + $\mathbf{\frac{R}{100}}$ )ⁿ
A =60,000 (1+ ( $\frac{6}{100}$ ))¹
A =60,000 (1+ ( $\frac{3}{50}$ ))¹
A = 60,000 ( $\frac{53}{25}$ )¹
A = Rs 63,600
He would get Rs 63,600 after 6 Months.
(b)
P = 60,000
R = 12% per annum (6% Half yearly)
T = 1 Year
As it is compounded Half Yearly then, n = 2 times (as 1 Year is 2 half year)
We have,
A = P (1 + $\mathbf{\frac{R}{100}}$ )ⁿ
A = 60,000 (1+ ( $\frac{6}{100}$ ))²
A = 60,000 (1+ ( $\frac{3}{50}$ ))²
A = 60000 ( $\frac{53}{25}$ )²
A = Rs 67,416
He would get Rs 67,416 after 1 Year.

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 1 $\frac{1}{2}$ years if the interest is

(a) compounded annually.

(b) compounded half-yearly.

Solution:

Let’s see each case
(a) Compounded Annually
P = 80,000
R = 10% per annum
T = 1 $\frac{1}{2}$ Year
As it is compounded annually then, n = 1 $\frac{1}{2}$ times
We have,
A = P (1 + $\mathbf{\frac{R}{100}}$ )ⁿ
A = 80,000 (1 + ( $\frac{10}{100}$ )^1½
What we will do here is Firstly we know 1 $\frac{1}{2}$ Years is 1 year and 6 months which can be calculated by first calculating the amount to 1 year using CI formula and then calculating the simple interest by using SI formula.
The amount for 1 years has to be calculated :
A = 80,000 (1+ ( $\frac{10}{100}$ ))¹
A = 80,000 (1+ ( $\frac{1}{10}$ )¹
A = 80,000 ( $\frac{11}{10}$ )¹
A = Rs 88,000
Now, The amount for $\frac{1}{2}$ Year (6 months) has to be calculated :
New P is equal to the amount after 1 Year. Hence,
P = Rs 88,000
R = 10 % per annum
T = $\frac{1}{2}$ Year
SI = $\frac{(P × R × T)}{ 100}$
SI = $\frac{(88,000 × 10 × \frac{1}{2})}{100}$
SI = $\frac{(88,000 × 10 × 1)}{200}$
SI = 4,400
Hence, the Total amount = A + SI
= 88,000 + 4,400
= Rs 92,400
(b) Compounded Half-yearly
P = 80,000
R = 10% per annum (5 % Half Yearly)
T = 1 $\frac{1}{2}$ Year
As it is compounded annually then, n = 3 times (as 1 $\frac{1}{2}$ Year is 3 half year)
We have,
A = P (1 + $\mathbf{\frac{R}{100}}$ )ⁿ
A = 80,000 (1+ ( $\frac{5}{100}$ )³
A = 80,000 (1+ ( $\frac{1}{20}$ )³
A = 80,000 ( $\frac{21}{20}$ )³
A = Rs 92,610
Hence, the Total amount = Rs 92,610

Question 7. Maria invested Rs 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find

(a) The amount credited against her name at the end of the second year.

(b) The interest for the 3rd year.

Solution:

Let’s see each case
Here,
P = 8,000
R = 5% Per annum
(a) The amount credited against Maria’s name at the end of the second year.
T = 2 Year
As it is compounded annually then, n = 2 times
We have,
A = P (1 + $\mathbf{\frac{R}{100}}$ )ⁿ
A = 8,000 (1+ ( $\frac{5}{100}$ ))²
A = 8,000 (1+ ( $\frac{1}{20}$ ))²
A = 8,000 ( $\frac{21}{20}$ )²
A = Rs 8,820
Hence, the amount credited against Maria’s name at the end of the second year = Rs 8,820
(b) The interest for the 3rd year.
T = 3 Year
As it is compounded annually then, n = 3 times
We have,
A = P (1+ $\mathbf{\frac{R}{100}}$ )ⁿ
A = 8,000 (1+ ( $\frac{5}{100}$ ))³
A = 8,000 (1+ ( $\frac{1}{20}$ ))³
A = 8,000 ( $\frac{21}{20}$ )³
A = Rs 9,261
The interest for the 3rd year = Amount after 3 years – Amount after 2 Years
= 9,261 – 8,820
= Rs 441
Another Solution for (b)
As we can calculate interest of 3^rd year by having 2^nd Year Amount as P.
P = 8,820
R = 5% per annum
T = 1 Year (2^nd to 3^rd year)
SI = $\frac{(P × R × T)}{100}$
SI = $\frac{(8,820 × 5 × 1)}{100}$
SI = Rs 441
The interest for the 3rd year = Rs 441

Question 8. Find the amount and the compound interest on Rs 10,000 for 1 $\frac{1}{2}$ 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:

Let’s see each cases
Compounded Annually
P = 10,000
R = 10% per annum
T = 1 $\frac{1}{2}$ Year
As it is compounded annually then, n = 1 $\frac{1}{2}$ times
We have,
A = P (1 + $\mathbf{\frac{R}{100}}$ )ⁿ
A = 10,000 (1 + ( $\frac{10}{100}$ )^1½
What we will do here is Firstly we know 1½ Years is 1 year and 6 months which can be calculated by first calculating the amount to 1 year using CI formula and then calculating the simple interest by using SI formula.
The amount for 1 year has to be calculated:
A = 10,000 (1 + $\frac{10}{100}$ )¹
A = 10,000 (1+ $\frac{1}{10}$ )¹
A = 10,000 ( $\frac{11}{10}$ )¹
A = Rs 11,000
CI = A – P
CI = 11,000-10,000
CI = 1,000
Now, The amount for $\frac{1}{2}$ Year (6 months) has to be calculated :
New P is equal to the amount after 1 Year. Hence,
P = Rs 11,000
R = 10 % per annum
T = $\frac{1}{2}$ Year
SI = $\frac{(P × R × T)}{100}$
SI = $\frac{(11,000 × 10 × \frac{1}{2})}{100}$
SI = $\frac{11,000 × 10}{200}$
SI = 550
Hence, the Total Interest (compounded annually)= CI + SI
= 1,000 + 550
= Rs 1,550
Compounded Half-yearly
P = 10,000
R = 10% per annum (5 % Half Yearly)
T = 1 $\frac{1}{2}$ Year
As it is compounded annually then, n = 3 times (as 1 $\frac{1}{2}$ Year is 3 half year)
We have,
A = P (1 + $\mathbf{\frac{R}{100}}$ )ⁿ
A = 10,000 (1 + ( $\frac{5}{100}$ )³
A = 10,000 (1+ $\frac{1}{20}$ )³
A = 10,000 ( $\frac{21}{20}$ )³
A = Rs 11,576.25
CI = A – P
CI = 11,576.25 – 10,000
CI = 1,576.25
Hence, the Total Interest (compounded Half Yearly) = Rs 11576.25
Difference between the two interests = 1,576.25 – 1,550 = Rs 26.25
Hence, the interest will be Rs 26.25 more when compounded half-yearly than the interest when compounded annually.

Question 9. Find the amount which Ram will get on Rs 4096, if he gave it for 18 months at 12 $\frac{1}{2}$ % per annum, interest being compounded half-yearly.

Solution:

Let’s see this case
P = Rs 4,096
R = 12 $\frac{1}{2}$ % per annum ( $\frac{25}{4}$ % Half yearly)
T = 18 Months = 1 $\frac{1}{2}$ Year
As it is compounded Half yearly then, n = 3 Times
We have,
A = P (1 + $\mathbf{\frac{R}{100}}$ )ⁿ
A = 4,096 (1+ ( $\frac{\frac{25}{4}}{100}$ )³
A = 4,096 (1+ $\frac{25}{400}$ )³
A = 4,096 (1+ ( $\frac{1}{16}$ )³
A = 4,096 ( $\frac{17}{16}$ )³
A = Rs 4,913
Ram will get the amount = Rs 4,913

Question 10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum

(a) find the population in 2001.

(b) what would be its population in 2005?

Solution:

Here,
P = 54,000 (in 2003)
R = 5% per annum
(a) Population in 2001
T = 2 Years (back)
n = 2
Population in 2003 = Population in 2001 (1 + $\mathbf{\frac{R}{100}}$ )ⁿ
54,000 = P₁(1+( $\frac{5}{100}$ ))²
54,000 = P1 ( $\frac{21}{20}$ )²
54,000 = P1 ( $\frac{441}{400}$ )
P1 = 54,000 ( $\frac{441}{400}$ )
P1 = 48,979.59
P1 = 48,980 (approx.).
Population in 2001 was 48,980 (approx.).
(b) Population in 2005
T = 2 Years
n = 2
We have,
A = P (1 + $\mathbf{\frac{R}{100}}$ )ⁿ
A = 54,000 (1+ $\frac{5}{100}$ )²
A = 54,000 (1+ ( $\frac{1}{20}$ )²
A = 54,000 ( $\frac{21}{20}$ )²
A = 59,535
Population in 2005 will be 59,535

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.

Solution:

Here,
P = 5,06,000
R = 2.5% per hour
T = 2 hours
We have,
A = P (1 + $\mathbf{\frac{R}{100}}$ )ⁿ
A = 5,06,000 (1+ $\frac{2.5}{100}$ )²
A = 5,06,000 (1+ $\frac{25}{1000}$ )²
A = 5,06,000 (1+ $\frac{1}{40}$ )²
A = 5,06,000 ( $\frac{41}{40}$ )²
A = 5,31,616.25
A = 5,31,616 (approx.)
Bacteria at the end of 2 hours = 5,31,616 (approx.)

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.

Solution:

Here,
P = 42,000
R = 8% per annum (depreciated)
T = 1 Year
We have,
A = P (1 + $\mathbf{\frac{R}{100}}$ )ⁿ
A = 42,000 (1- $\frac{8}{100}$ )¹ (negative sign because the price is reduced)
A = 42,000 (1- ( $\frac{2}{25}$ )¹
A = 42,000 ( $\frac{23}{25}$ )¹
A = Rs 38,640
The value of scooter after one year will be = Rs 38,640

My Personal Notes

1. Class 8 NCERT Solutions - Chapter 8 Comparing Quantities - Exercise 8.2

2. Class 8 NCERT Solutions- Chapter 8 Comparing Quantities - Exercise 8.1

3. Percent Change & Discounts - Comparing Quantities | Class 8 Maths

5. Sales Tax, Value Added Tax, and Goods and Services Tax - Comparing Quantities | Class 8 Maths

6. Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Miscellaneous Exercise on Chapter 5

7. Class 11 NCERT Solutions - Chapter 5 Complex Numbers And Quadratic Equations - Miscellaneous Exercise on Chapter 5 | Set 2

8. Class 11 NCERT Solutions- Chapter 14 Mathematical Reasoning - Miscellaneous Exercise on Chapter 14

9. Class 11 NCERT Solutions- Chapter 1 Sets - Miscellaneous Exercise on Chapter 1 | Set 1

10. Class 11 NCERT Solutions- Chapter 1 Sets - Miscellaneous Exercise on Chapter 1 | Set 2

Class 8 NCERT Solutions - Chapter 8 Comparing Quantities - Exercise 8.2

Class 8 NCERT Solutions - Chapter 9 Algebraic Expressions and Identities - Exercise 9.1

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Class 8 NCERT Solutions- Chapter 8 Comparing Quantities – Exercise 8.3

Question 1. Calculate the amount and compound interest on

Question 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?

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?

Question 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?

Question 5. Vasudevan invested Rs 60,000 at an interest rate of 12% per annum compounded half-yearly. What amount would he get

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 1 $\frac{1}{2}$ years if the interest is

Question 7. Maria invested Rs 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find

Question 8. Find the amount and the compound interest on Rs 10,000 for 1 $\frac{1}{2}$ years at 10% per annum, compounded half-yearly. Would this interest be more than the interest he would get if it was compounded annually?

Question 9. Find the amount which Ram will get on Rs 4096, if he gave it for 18 months at 12 $\frac{1}{2}$ % per annum, interest being compounded half-yearly.

Question 10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum

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.

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.

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Class 8 NCERT Solutions- Chapter 8 Comparing Quantities – Exercise 8.3

Question 1. Calculate the amount and compound interest on

Question 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?

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?

Question 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?

Question 5. Vasudevan invested Rs 60,000 at an interest rate of 12% per annum compounded half-yearly. What amount would he get

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 1 years if the interest is

Question 7. Maria invested Rs 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find

Question 8. Find the amount and the compound interest on Rs 10,000 for 1 years at 10% per annum, compounded half-yearly. Would this interest be more than the interest he would get if it was compounded annually?

Question 9. Find the amount which Ram will get on Rs 4096, if he gave it for 18 months at 12 % per annum, interest being compounded half-yearly.

Question 10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum

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.

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.

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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 1 $\frac{1}{2}$ years if the interest is

Question 8. Find the amount and the compound interest on Rs 10,000 for 1 $\frac{1}{2}$ years at 10% per annum, compounded half-yearly. Would this interest be more than the interest he would get if it was compounded annually?

Question 9. Find the amount which Ram will get on Rs 4096, if he gave it for 18 months at 12 $\frac{1}{2}$ % per annum, interest being compounded half-yearly.