NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.3:–
Question 1.
Calculate the amount and compound interest on
(a) ₹ 10,800 for 3 years at 12–1/2 % per annum compounded annually.
Solution:
Given:
P = ₹ 10,800, n = 3 years,
R = 12-1/2 % = 25/2 % pa
A = P ( 1+ R/100)^n
A = 10800(1+25/2×100)³
A = 10800( 9/8)³
A = 10800×9/8×9/8×9/8
A = 492075/32
A = ₹ 15377.34
CI = A – P = ₹ 15,377.35 – ₹ 10,800 = ₹ 4,577.35
Hence amount = ₹ 15,377.34 and CI = ₹ 4,577.34
(b) ₹ 18,000 for 212 years at 10% per annum compounded annually.
Solution :
Given: P = ₹ 18,000, n = 212 years = 52years
R = 10% p.a.
The amount for 212 years, i.e., 2 years and 6 months 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 = 18000( 1+ 10/100)²
A = 18000(1+ 1/10)²
A = 18000( 11/10)²
A = 18000×11/10 ×11/10
A = ₹21780
.: interest after two years = A–P
= 21780– 18000 = ₹ 3780
By taking ₹21780 as a principal amount. The SI for next 1/2 year will be calculated
SI for 1/2 year = P×R×n /100
= 21780 × 10 ×1/ 100×2
= ₹ 1089
Total CI = ₹ 3780 + ₹ 1089 = ₹ 4,869
Amount = P + I = ₹ 21,780 + ₹ 1,089 = ₹ 22,869
Hence, the amount = ₹ 22,869
and CI = ₹ 4,869
(c) ₹ 62,500 for 112 years at 8% per annum compounded half yearly.
Solution :
Given: P = ₹ 62,500, n = 112 years = 32 years per annum compounded half yearly
= 32 × 2 years = 3 half years
R = 8% = 82 % = 4% half yearly
A = P (1 + R /100)^n
A = 62500 ( 1+ 4/100)³
A = 62500 ( 1+ 1/25 )³
A = 62500 ( 26/25 )³
A = 62500 × 26/25 × 26/25 × 26/25
A = ₹ 70304
CI = A – P = ₹ 70,304 – ₹ 62,500 = ₹ 7,804
Hence, amount = ₹ 70304 and CI = ₹ 7804
(d) ₹ 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).
Solution :
Given: P = ₹ 8,000, n = 1 years R = 9% per annum compounded half yearly
Since, the interest is compounded half yearly n = 1 × 2 = 2 half years
R = 9/2 % per half year
A = P ( 1+R/100)^n
A = 8000( 1+ 9/ 2×100)²
A = 8000( 1+9/200)²
A = 8000( 209/200)²
A = 8000× 209/200 ×209/200
A = ₹ 8736.20
CI = A – P = ₹ 8,736.20 – ₹ 8,000 = ₹ 736.20
Hence, the amount = ₹ 8736.20 and CI = ₹ 736.20
(e) ₹ 10,000 for 1 year at 8% per annum compounded half yearly.
Solution :
Given: P = ₹ 10,000, n = 1 year and R = 8% pa compounded half yearly
Since the interest is compounded half yearly n = 1 × 2 = 2 half years
R = 8× 1/2 = 4% per half year
A = P ( 1+R/100)^n
A = 10000(1+ 4/100)²
A = 10000(1+1/25)²
A = 10000( 26/25)²
A = 10000 × 26/25 × 26/25
A = ₹ 10816
CI = A – P = ₹ 10,816 – ₹ 10,000 = ₹ 816
Hence the amount = ₹ 10,816 and Cl = ₹ 816
Question 2.
Kamala borrowed ₹ 26,400 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? (Hint: Find amount for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for 412 years).
Solution:
Given:
P = ₹ 26,400
R = 15% p.a. compounded yearly
n = 2 years and 4 months
Amount for 2 years = P ( 1+R/100)^n
= 26400( 1+ 15/100)²
= 26400( 23/20)²
= 26400 × 23/20 × 23/20
= ₹ 34914
Principal for months i.e.,4 /12 years = ₹34914
.: SI for 4 months = P×R×n/100
= 34914 × 15 × 4/ 100× 12
= ₹ 1745.70
Amount after 2 years and 4 months = ₹ 34,914 + ₹ 1745.70 = ₹ 36,659.70
Hence, the amount to be paid by Kamla = ₹ 36,659.70
Question 3.
Fabina borrows ₹ 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:
For Fabina: P = ₹ 12,500, R = 12% p.a. and n = 3 years
.: SI = P×R×n/100
= 12500 ×12×3/400
= ₹ 4500
For radha : P = ₹12500, R = 10%p.a. and n = 3 years
.: CI = A–P
= P ( 1+R/100)^n – P
= 12500( 1+ 10/100)³ – 12500
= 12500( 1+ 1/10)³ – 12500
= 12500( 11/10)³ – 12500
= 12500× 11/10 ×11/10 ×11/10 –12500
= 12500 × 1331/1000 – 12500
= 12500( 1331/1000 – 1)
= 12500( 1331 – 1000 / 1000)
= 12500 × 331/1000
= ₹ 4137.50
Difference between the two interests = ₹ 4500 – ₹ 4137.50 = ₹ 362.50
Hence, Fabina pays more interest by ₹ 362.50.
Question 4.
I borrowed ₹ 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:
Given: P = ₹ 12,000, R = 6% p.a., n = 2 years
.: SI = P×R×n/100
= 12000 ×6 ×2/100
= ₹ 1440
CI = P ( 1+ R/100)^n – P
= 12000( 1+ 6/100)² – 12000
= 12000( 53/50)² –12000
= 12000( 53 × 53 / 50× 50 – 1)
= 12000 ( 2809 /2500 –1)
= 12000 ( 2809 – 2500 / 2500)
= 12000 ( 309 / 2500)
= ₹ 1483.20
Difference between two interests = ₹ 1483.20 – ₹ 1440 = ₹ 43.20
Hence, the extra amount to be paid = ₹ 43.20
Question 5.
Vasudevan invested ₹ 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) Given: P = ₹ 60,000, R = 12% p.a. compounded half yearly
n = 6 months = 6/12 year = 1/2 year
Simple interest = P×R×n/100
= 60000×12×1/100×2
= ₹ 3600
.: Amount = SI + P
= ₹ 3600 + ₹ 60,000
= ₹ 63600
Hence, the required amount = ₹ 63600
(ii) given P = 60000, R = 12/2% = 6%
n = 1× 2 = 2 half year
.: Amount = P ( 1+ R/100)^n
= 60000(1+6/100)²
= 60000(53/50)²
= 60000×53/50×53/50
= ₹ 67416
Hence, the required amount = ₹ 67416
Question 6.
Arif took a loan of ₹ 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) Given: P = ₹ 80,000
R = 10% p.a.
n = 1–1/2 years
Since the interest is compounded annually
.: SI for 1 year = P×R×n/100
= 80000×10×1/100
= ₹ 8000
Principal for the second year
= P + SI
= 80000 +8000
= 88000
Now interest for 1/2 year
= 88000× 10×1/ 100 ×2
= ₹ 4400
.: Amount = ₹ 88000 + 4400
= ₹ 92400
(ii) if the interest is compounded half yearly
R = 10/2% = 5% half yearly
n = 1-1/2
= 3/2 ×2 = 3 half years
A = P ( 1+ R/100)^n
= 80000( 1+5/100)³
= 80000( 21/20)³
= 80000× 21/20 × 21/20 × 21/20
= ₹ 92610
Difference between the amounts = ₹ 92,610 – ₹ 92,400 = ₹ 210
Question 7.
Maria invested ₹ 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) Given: P = ₹ 8,000, R = 5% p.a.
and n = 2 years
A = P ( 1+ R/100)^n
= 8000(1+5/100)²
= 8000( 21/20)²
= 8000 × 21/20 × 21/20
= ₹ 8820
Thus, the amount credited againsts Maria's name at the end of 2nd year = ₹ 8820
(ii) interest for the 3rd year
= amount after third year – amount after 2 year
= P ( 1+ R/100)^n – ₹ 8820
= 8000( 1+ 5/100)³ – ₹ 8820
= 8000(1+1/20)³ – ₹ 8820
= 8000(21/20)³ – ₹ 8820
= ₹ 9261 – ₹ 8820
= ₹ 441
Hence, interest for the third year = ₹ 441
Question 8.
Find the amount and the compound interest on ₹ 10,000 for 112 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:
Given: P = ₹ 10,000, n = 112 years
R = 10% per annum
Since the interest is compounded half yearly
.: n = 1–1/2
= 3/2 × 2 = 3 half year
R = 10/2 % = 5%
A = P ( 1+ R/100)^n
= 10000( 1+5/100)³
= 10000( 1+1/20)³
= 10000( 21/20)³
= 11576.25
Compound interest = A – P
= 11576.25 – 10000
= 1576.25
If the interest is compound annually, then
n = 1–1/2 , R = 10%
.: SI = P × R × n /100
= 10000 × 10 ×1 /100
= ₹ 1000
Principal of the second year
= ₹ 10000 + ₹ 1000
= ₹ 11000
.: interest for the 1/2 year
= 11000 × 10 ×1 /100 ×2
= ₹ 550
Total interest = ₹ 1,000 + ₹ 550 = ₹ 1,550
Difference between the two interests = ₹ 1,576.25 – ₹ 1,550 = ₹ 26.25
Hence, the interest will be ₹ 26.25 more when compounded half yearly than the interest when compounded annually.
Question 9.
Find the amount which Ram will get on ₹ 4,096, if he gave it for 18 months at 1212 per annum, interest being compounded half yearly.
Solution:
Given: P = ₹ 4,096, R = 1212 % pa, n = 18 months
R = 12–1/2% pa
= 25/2 ×1/2
= 25/4% half yearly
n = 18 months = 18/12 year
= 18/12 ×2 = 3 half year
A = P ( 1+ R/100)^n
A = 4096 (1 + 25/4×100)³
A = 4096 ( 1+1/16)³
A = 4096 ( 17/16)³
A = 4096 × 17/16×17/16×17/16
A = ₹ 4913
Hence, the required amount = ₹ 4913
Question 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) Given: Population in 2003 = 54,000
Rate = 5% pa
Time = 2003 – 2001 = 2 years
Population in 2003 = Population in 2001 ×(1+R/100)^n
54000 = Population in 2001 × (1+5/100)²
54000 = Population in 2001 × (1+1/20)²
54000 = Population in 2001 × (21/20)²
54000 = Population in 2001 × 441 /400
.: Population in 2001 = 54000 × 400/441
= 21600000/ 441
= 48979.59 = 48980
(ii) Population in 2005 = Population in 2003 × (1+R/100)^n
= 54000( 1+5/100)²
= 54000( 1+1/20)²
= 54000(21/20)²
= 54000×21/20×21/20
= 59535
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:
Given: Initial count of bacteria = 5,06,000
Rate = 2.5% per hour
n = 2 hours
Number of bacteria at the end of 2 hours = Number of count of bacteria initially ×(1+R/100)^n
= 506000 (1+ 2.5/100)²
= 506000( 1+1/40)²
= 506000(41/40)²
= 506000×41/40×41/40
= 531616.25
Thus, the number of bacteria after two hours = 5,31,616 (approx).
Question 12.
A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Solution:
Given: Cost price of the scooter = ₹ 42,000
Rate of depreciation = 8% p.a.
Time = 1 year
Final value of the scooter
= present value ×(1– R/100)^n
= 42000(1– 8 /100)
= 42000(23/25)
= 38640
Hence, the value of scooter after 1 year = ₹ 38,640.
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