DAV Class 8 Maths Solution Chapter 6 compound interest
DAV School Books Class 8 Maths Solution Chapter 6 compound interest all Question Answer.
DAV Class 8 6th chapter compound interest full Chapter explanation is provided by expert teacher.
DAV School Books Class 8 Maths Solution Chapter 6 compound interest:
Worksheet 1
1) Find the compound interest on ₹25000 at the rate of 12% per annum for 3 years.
Ans:
For the first year
Here,
Principal (P) = ₹25000
Time (T) = 1 year
Rate (R) = 12%
∴ Simple interest (S.I) = (P×R×T)/100 = (25000×12×1)/100 = 3000.
∴ Interest for the first year = ₹3000
So, Amount = P + S.I = 25000 + 3000 = ₹28000
For the second year
In the second year, the amount in the first year becomes the principal in second year.
So, we have,
Principal (P) = ₹28000
Time (T) = 1 year
Rate (R) = 12%
∴ Simple interest (S.I) = (P×R×T)/100 = (28000×12×1)/100 = 3360.
∴ Interest for the second year = ₹3360
So, Amount = 28000 + 3360 = ₹31360
For the Third year
In the third year, the amount in the second year becomes the principal in the third year.
So, we have,
Principal (P) = ₹31360
Time (T) = 1 year
Rate (R) = 12%
∴ Simple interest (S.I) = (P×R×T)/100 = (31360×12×1)/100 = 3763.20.
∴ Interest for the third year = ₹3763.20
So, the compound interest for 3 years = ₹ (3000+3360+3763.20) = ₹10,123.2
2) Find the compound interest on ₹6500 for 2 years at 9% per annum.
Ans:
For the first year
Here,
Principal (P) = ₹6500
Time (T) = 1 year
Rate (R) = 9%
∴ Simple interest (S.I) = (P×R×T)/100 = (6500×9×1)/100 = 585.
∴ Interest for the first year = ₹585
So, Amount = P + S.I = 6500 + 585 = ₹7085
For the second year
In the second year, the amount in the first year becomes the principal in second year.
So, we have,
Principal (P) = ₹7085
Time (T) = 1 year
Rate (R) = 9%
∴ Simple interest (S.I) = (P×R×T)/100 = (7085×9×1)/100 = 637.65
∴ Interest for the second year = ₹ 637.65
So, the compound interest for 2 years = ₹ (585 + 637.65) = ₹1222.65
3) Find the amount and compound interest on a sum of ₹8000 at 5% per annum for 3 years compounded annually.
Ans:
For the first year
Here,
Principal (P) = ₹8000
Time (T) = 1 year
Rate (R) = 5%
∴ Simple interest (S.I) = (P×R×T)/100 = (8000×5×1)/100 = 400.
∴ Interest for the first year = ₹400
So, Amount = P + S.I = 8000 + 400 = ₹8400
For the second year
In the second year, the amount in the first year becomes the principal in second year.
So, we have,
Principal (P) = ₹8400
Time (T) = 1 year
Rate (R) = 5%
∴ Simple interest (S.I) = (P×R×T)/100 = (8400×5×1)/100 = 420.
∴ Interest for the second year = ₹420
So, Amount after second year= ₹ (8400 + 420) = ₹8820
For the Third year
In the third year, the amount in the second year becomes the principal in the third year.
So, we have,
Principal (P) = ₹8820
Time (T) = 1 year
Rate (R) = 5%
∴ Simple interest (S.I) = (P×R×T)/100 = (8820×5×1)/100 = 441.
∴ Interest for the third year = ₹441
So, the compound interest for 3 years = ₹ (400+420+441) = ₹1261
So the amount after three years will be ₹ (8000+1261) = ₹9261
4) Harvinder deposited ₹40000 in a post office for a period of 3 years. The post office credits the interest yearly in his account at 7% per annum compounded annually. Find the balance in his account after three years.
Ans:
For the first year
Here,
Principal (P) = ₹40000
Time (T) = 1 year
Rate (R) = 7%
∴ Simple interest (S.I) = (P×R×T)/100 = (40000×7×1)/100 = 2800.
∴ Interest for the first year = ₹2800
So, Amount = P + S.I = 40000 + 2800 = ₹42800
For the second year
In the second year, the amount in the first year becomes the principal in second year.
So, we have,
Principal (P) = ₹42800
Time (T) = 1 year
Rate (R) = 7%
∴ Simple interest (S.I) = (P×R×T)/100 = (42800×7×1)/100 = 2996.
∴ Interest for the second year = ₹2996.
So, Amount after second year= ₹ (42800 + 2996) = ₹ 45796
For the Third year
In the third year, the amount in the second year becomes the principal in the third year.
So, we have,
Principal (P) = ₹45796
Time (T) = 1 year
Rate (R) = 7%
∴ Simple interest (S.I) = (P×R×T)/100 = (45796×7×1)/100 = 3205.72.
∴ Interest for the third year = ₹3205.72
So the balance in the account of Harvinder after three years will be ₹ (45796+3205.72) = ₹49001.72
5) Monika borrowed ₹4096 from Shalini for 3 years at 6(1/4)% per annum. Find the amount and the compound interest paid by her to shalini after 3 years if the interest is compounded annually.
Ans:
For the first year
Here,
Principal (P) = ₹4096
Time (T) = 1 year
Rate (R) = 6(1/4)% = 6.25%
∴ Simple interest (S.I) = (P×R×T)/100 = (4096×6.25×1)/100 = 256.
∴ Interest for the first year = ₹256
So, Amount = P + S.I = 4096 + 256 = ₹4352
For the second year
In the second year, the amount in the first year becomes the principal in second year.
So, we have,
Principal (P) = ₹4352
Time (T) = 1 year
Rate (R) = 6.25%
∴ Simple interest (S.I) = (P×R×T)/100 = (4352×6.25×1)/100 = 272.
∴ Interest for the second year = ₹272.
So, Amount after second year= ₹ (4352 + 272) = ₹ 4624
For the Third year
In the third year, the amount in the second year becomes the principal in the third year.
So, we have,
Principal (P) = ₹4624
Time (T) = 1 year
Rate (R) = 6.25%
∴ Simple interest (S.I) = (P×R×T)/100 = (4624×6.25×1)/100 = 289.
∴ Interest for the third year = ₹289
Amount after 3 years ₹ (4624+289) = ₹4913
The interest for 3 years is ₹ (256+272+289) = ₹817
So after 3 years monika will return ₹4913 to Shalini and the interest paid by her will be ₹817.
6) Ravi purchased a house from DDA on credit. If the cost of the house in 750000 and DDA charges interest at 6% per annum compounded annually, find the interest paid by Ravi if he makes payment to DDA after 3 years.
Ans:
For the first year
Here,
Principal (P) = ₹750000
Time (T) = 1 year
Rate (R) = 6%
∴ Simple interest (S.I) = (P×R×T)/100 = (750000×6×1)/100 = 45000
∴ Interest for the first year = ₹45000
So, Amount = P + S.I = 750000 + 4500 = ₹ 795000
For the second year
In the second year, the amount in the first year becomes the principal in second year.
So, we have,
Principal (P) = ₹ 795000
Time (T) = 1 year
Rate (R) = 6%
∴ Simple interest (S.I) = (P×R×T)/100 = (795000×6×1) /100 = 47700.
∴ Interest for the second year = ₹47700.
So, Amount after second year= ₹ (795000 + 47700) = ₹ 842700
For the Third year
In the third year, the amount in the second year becomes the principal in the third year.
So, we have,
Principal (P) = ₹842700
Time (T) = 1 year
Rate (R) = 6%
∴ Simple interest (S.I) = (P×R×T)/100 = (842700×6×1)/100 = 50562
∴ Interest for the third year = ₹50562.
The amount after 3 years ₹ (842700+50562) = ₹893262
So the amount Ravi has to pay DDA after 3 years is ₹893262.
However there is a formula for obtaining compound interest directly.
Worksheet 2
1) Compute the compound interest on ₹5000 for 1(1/2) years at 16% p.a compounded half yearly.
Ans:
Rate of interest (R) = 16% per annum = 8 % per half-yearly
Time period (T) = 1(1/2) years = 3 half-years
Principal for the first half year (P) = ₹5000
∴ Interest for the first half-year = ₹ [(5000×1×8)/100] = ₹400
So, amount for the first half-year = ₹ (5000+400) = ₹5400
So, the principal for the second half-year = ₹5400
∴ Interest for the second half-year = ₹ [(5400×1×8)/100] = ₹432
So, the amount for the second half year = ₹ (5400+432) = ₹ 5832
So, the principal for the third half-year = ₹5832
∴ Interest for the third half-year = ₹ [(5832×1×8)/100] = ₹466.56
So, the amount for third half-year = ₹ (5832+466.56) = ₹6298.56
The total compound interest on ₹5000 for 1(1/2) years at 16% interest compounded half-yearly is = ₹ (6298.56 – 5000) = ₹1298.56
2) Find the compound interest on ₹15625 at 16% p.a for 9 months when compounded quarterly.
Ans:
Rate of interest (R) = 16% per annum = 4% per quarter.
Time period (T) = 9 months = 3 quarters
Principal for the first quarter (P) = ₹15625
∴ Interest for the first quarter = ₹ [(15625×1×4)/100] = ₹625
So, amount for the first quarter = ₹ (15625+625) = ₹16250
So, principal for the second quarter = ₹16250
∴ Interest for the second quarter = ₹ [(16250×1×4)/100] = ₹650
So, amount for the second quarter = ₹ (16250+650) = ₹16900
So, principal for the third quarter = ₹16900
∴ Interest for the third quarter = ₹ [(16900×1×4)/100] = ₹676
So, the amount for the third quarter = ₹ (16900+676) = ₹17576
∴ The compound interest on ₹15625 for 9 months at 16% p.a compounded quarterly is ₹ (17576-15625) = ₹1951
3) Rohit deposited ₹10000 in a bank for 6 months. If the bank pays compound interest at 12% p.a reckoned quarterly, find the amount to be received by him on maturity.
Ans:
Rate of interest (R) = 12% per annum = 3% per quarterly
Time period (T) = 6 months = 2 quarters
Principal for first quarter (P) = ₹10000
∴ Interest for first quarter = ₹ [(10000×1×3)/100] = ₹300
So, the amount for the first quarter = ₹ [10000+300] = ₹10300
So, the principal for the second quarter = ₹10300
∴ Interest for the second quarter = ₹ [(10300×1×3)/100] = ₹ 309
So, amount after second quarter is = ₹ (10300+309) = ₹10609
So Rohit will receive ₹10609 after 6 months.
4) Find the difference between the compound interest on ₹25000 at 16% p.a for 6 months compounded half yearly and quarterly respectively. Which option is better?
Ans:
From the above we can see that the interest earned in the case where the interest is compounded quarterly is more than the other.
5) Bela borrowed ₹25000 from a finance company to start her boutique at 20% p.a compounded half yearly. What amount of money will clear her debt after 1(1/2) years?
Ans:
Rate of interest (R) = 20% per annum = 10% per half yearly
Time period (T) = 1(1/2) years = 3 half years.
Principal for the first half year (P) = ₹25000
∴ Interest earned in the first half year = ₹ [(25000×1×10)/100] = ₹2500
So, the amount for the first half year = ₹ [25000+2500] = ₹27500
So, the principal for the second half year = ₹27500
∴Interest for the second half year = ₹ [(27500×1×10)/100] = ₹2750
So, amount after second half-year ₹ (27500+2750) = ₹30250
So, the principal for third half-year = ₹30250
∴ Interest for third half-year = ₹ [(30250×1×10)/100] = ₹3025
So, amount after third year will be ₹(30250+3025) = 33275
∴ Bela has to pay ₹33275 after 1(1/2) years to clear her debt.
Worksheet 3
1) Find the amount for ₹15000 at 8% per annum compounded annually for 2 years.
Ans:
We know that
Amount A = P[1+(r/n)]nt
Where P = principal = ₹15000
r = rate of interest = 8%
t = time period in years = 2 years
n = number of time interest compounded in a year = 1 (as the interest is compounded yearly)
∴ A = 15000× [1+ (8%/1)] 1×2
Or, A = 15000× [1+ (8/100)] 2
Or, A = 15000× (108/100)2
Or, A = 17496
So, the amount for ₹15000 at 8% per annum compounded annually for 2 years is ₹17496.
2) Find the compound interest on ₹11200 at 17.5% per annum for 2 years.
Ans:
We know that
Amount A = P[1+(r/n)]nt
Where P = principal = ₹11200
r = rate of interest = 17.5%
t = time period in years = 2 years
n = number of time interest compounded in a year = 1 (as the interest is compounded yearly)
∴ A = 11200× [1+ (17.5%/1)] 1×2
Or, A = 11200× [1+ (17.5/100)] 2
Or, A = 11200× (117.5/100)2
Or, A = 15463
So, the amount for ₹11200 at 17.5% per annum compounded annually for 2 years is ₹15463.
3) Ram borrowed a sum of ₹30000 from shyam for 3 years. If the rate of interest is 6% per annum compounded annually, find the interest paid by Ram to Shyam after 3 years.
Ans:
We have to find out the compound interest of ₹30000 at 6% per annum for 3 years.
We know that,
Amount A = P[1+(r/n)]nt
Where P = principal = ₹30000
r = rate of interest = 6%
t = time period in years = 3 years
n = number of time interest compounded in a year = 1 (as the interest is compounded yearly)
∴ A = 30000× [1+ (6%/1)] 1×3
Or, A = 30000× [1+ (6/100)] 3
Or, A = 30000× (106/100)3
Or, A = 35730.48
So, the amount for ₹30000 at 6% per annum compounded annually for 3 years is ₹35730.48.
4) Nidhi deposited ₹7500 in a bank which pays her 4% interest per annum compounded annually. Find the amount and the interest received by her after 3 years.
Ans:
We have to find out the compound interest of ₹7500 at 4% per annum for 3 years.
We know that,
Amount A = P[1+(r/n)]nt
Where P = principal = ₹7500
r = rate of interest = 4%
t = time period in years = 3 years
n = number of time interest compounded in a year = 1 (as the interest is compounded yearly)
∴ A = 7500× [1+ (4%/1)] 1×3
Or, A = 7500× [1+ (4/100)] 3
Or, A = 7500× (104/100)3
Or, A = 8436.48
So, the amount for ₹7500 at 4% per annum compounded annually for 3 years is ₹8436.48.
The interest received by her is ₹ (8436.48 – 7500) = ₹936.48
5) Find the difference between the compound interest and the simple interest on ₹30000 at 7% per annum for 3 years.
Ans:
The simple interest of ₹30000 at 7% per annum for 3 years will be = (30000×7×3)/100 = ₹6300
Now we will calculate the compound interest of ₹30000 at 7% per annum for 3 years.
We know that,
Amount A = P[1+(r/n)]nt
Where P = principal = ₹30000
r = rate of interest = 7%
t = time period in years = 3 years
n = number of time interest compounded in a year = 1 (as the interest is compounded yearly)
∴ A = 30000× [1+ (7%/1)] 1×3
Or, A = 30000× [1+ (7/100)] 3
Or, A = 30000× (107/100)3
Or, A = 36751.29
So, the amount for ₹30000 at 4% per annum compounded annually for 3 years is ₹36751.29.
So the compound interest is ₹ (36751.29 – 30000) = ₹6751.29
The difference between the compound interest and simple interest on ₹30000 at 7% per annum for 3 years is ₹ (6751.29 – 6300) = ₹451.29
6) Aman borrows ₹14500 at 11% per annum for 3 years at simple interest and Tarun borrows the same amount at 10% for same time compounded annually. Who pays more interest and by how much?
Ans:
To find out how much Aman has to pay we have to find the simple interest of ₹14500 at 11% per annum for 3 years.
Simple interest = (14500×3×11)/100 = ₹4785
Now, we have to find out how much Tarun has to pay for that we will calculate the compound interest on ₹14500 at 11% per annum for 3 years.
We know that,
Amount A = P[1+(r/n)]nt
Where P = principal = ₹14500
r = rate of interest = 11%
t = time period in years = 3 years
n = number of time interest compounded in a year = 1 (as the interest is compounded yearly)
∴ A = 14500× [1+ (11%/1)] 1×3
Or, A = 14500× [1+ (11/100)] 3
Or, A = 14500× (111/100)3
Or, A = 19830.64
So, the amount for ₹14500 at 11% per annum compounded annually for 3 years is ₹19830.64.
So, the interest Tarun has to pay ₹ (19830.64-14500) = ₹5330.64
Tarun has to pay more than Aman by ₹ (5330.64 – 4785) = ₹545.64
7) The simple interest on a certain sum of money for 2 years at 5.5% is ₹6600. What will be the compound interest on that sum at the same rate for the same time period?
Ans:
Let the principal be P.
∴ (P×5.5×2)/100 = 6600
Or, P = 60000
∴ The principal is ₹60000
For calculating compound interest we have
Compound interest (C.I) = A – P = P [1+(r/n)]nt – P
Where A = amount
P = principal = ₹60000
r = rate of interest = 5.5%
t = time period in years = 2 years
n = number of time interest compounded in a year = 1 (as the interest is compounded yearly)
∴ C.I = 60000× [1+ (5.5%/1)] 1×2 – 600000
Or, C.I = 60000× [1+ (5.5/100)] 2– 600000
Or, C.I = 60000× (105.5/100)2 – 600000
Or, C.I = 70454.48 – 600000
Or, C.I = 10454.28
So, the compound interest for ₹60000 at 5.5% per annum compounded annually for 2 years is ₹10454.28.
8) A certain sum amounts to ₹2970.25 in 2 years at 9% per annum compounded annually. Find the sum.
Ans:
Let’s assume the sum to be P
We know that,
Amount A = P[1+(r/n)]nt
Where, A = Amount = ₹2970.25
P = principal
r = rate of interest = 9%
t = time period in years = 2 years
n = number of time interest compounded in a year = 1 (as the interest is compounded yearly)
∴ A = P[1+(r/n)]nt
Or, 2970.25 = P× [1+ (9%/1)] 1×2
Or, 2970.25 = P× [1+ (9/100)] 2
Or, 2970.25 = P× (109/100)3
Or, P = 2500
The sum is ₹2500.
9) On what sum will the compound interest at 7.5% per annum for 3 years compounded annually be ₹3101.40?
Ans:
Let’s assume the sum to be P
We know that,
Amount A = P[1+(r/n)]nt
Where, A = Amount = ₹3101.40
P = principal
r = rate of interest = 7.5%
t = time period in years = 3 years
n = number of time interest compounded in a year = 1 (as the interest is compounded yearly)
∴ A = P[1+(r/n)]nt
Or, 3101.40 = P× [1+ (7.5%/1)] 1×3
Or, 3101.40 = P× [1+ (7.5/100)] 3
Or, 3101.40 = P× (107.5/100)3
Or, P = 2496.50
The sum is ₹2496.50
10) At what rate per cent will a sum of ₹640 be compounded to ₹774.40 in 2 years?
Ans:
Let us consider the rate of interest = r%
Here Principal P = ₹640
Amount A = ₹774.40
Time t = 2 years
n = number of time interest compounded in a year = 1 (as the interest is compounded yearly)
∴ A = P[1+(r%/n)]nt
Or, 774.40 = 640×[1+ (r%/1)]1×2
Or, 774.40 = 640× [1+ (r%)]2
Or, (774.40/640) = [1+ (r%)]2
Or, 1.21 = [1+ (r%)]2
Or, 1.1 = [1+ (r%)]
Or, r% = 0.1
Or, r = 10
The required rate of interest in this case is 10%.
11) At what rate per cent will a sum of ₹64000 be compounded to ₹68921 in 3 years?
Ans:
Let us consider the Rate of interest = r%
Here Principal P = ₹64000
Amount A = ₹68921
Time t = 3 years
n = number of time interest compounded in a year = 1 (as the interest is compounded yearly)
∴ A = P[1+(r%/n)]nt
Or, 68921 = 64000×[1+ (r%/1)]1×3
Or, 68921 = 64000× [1+ (r%)]3
Or, (68921/64000) = [1+ (r%)]3
Or, ∛(68921/64000) = [1+ (r%)]
Or, 1.025 = [1+ (r%)]
Or, r% = 0.025
Or, r = 2.5
The required rate of interest in this case is 2.5%.
12) In how many years will ₹8000 amount to ₹9261 at 5% per annum compounded annually?
Ans:
Let’s assume that the time period = t
Here Principal P = ₹8000
Amount A = ₹9261
Rate of interest = 5%
n = number of time interest compounded in a year = 1 (as the interest is compounded yearly)
∴ A = P [1+(r%/n)]nt
Or, 9261 = 8000× [1+ (5%/1)]1×t
Or, 9261 = 8000× [1+ (5%)]t
Or, (9261/8000) = [1+ (5/100)]t
Or, (9261/8000) = [21/20]t
Or, (21/20)3 = (21/20)t
Or, t = 3
∴ The time period required for the given problem t = 3
13) In what time will a sum of ₹3750 at 20% per annum compounded annually amount to ₹6480?
Ans:
Let’s assume that the time period = t
Here Principal P = ₹3750
Amount A = ₹6480
Rate of interest = 20%
n = number of time interest compounded in a year = 1 (as the interest is compounded yearly)
∴ A = P [1+(r%/n)]nt
Or, 6480 = 3750× [1+ (20%/1)]1×t
Or, 6480 = 3750× [1+ (20%)]t
Or, (6480/3750) = [1+ (20/100)]t
Or, (216/125) = [6/5] t
Or, (6/5)3 = (6/5) t
Or, t = 3
∴ The time period required for the given problem t = 3
14) The difference between the compound interest and simple interest on a certain sum of money at 6(2/3) % per annum for 3 years is ₹46. Find the sum.
Ans:
Let’s assume that the sum is ₹100.
Here, r = 6(2/3)% =(20/3)%, t = 3 years
We know that
Compound interest (C.I) = A – P = P [1+(r/n)]nt – P
∴ C.I = P [{1+(r/n)}nt – 1]
Or, C.I = 100[{1+(20/3)%}3 – 1]
Or, C.I = 100[{1+(1/15)}3 – 1]
Or, C.I = 100[(16/15)3 – 1]
Or, C.I = 21.36
∴ Compound interest = ₹21.36
Also, Simple interest = [100×(20/3)×3]/100
S.I = 20
∴ Simple interest = ₹20
Difference between S.I and C.I = ₹ (21.36 – 20) = ₹1.36
Now, when the difference between S.I and C.I is ₹1.36 than the sum is ₹100.
When the difference between S.I and C.I is ₹1 than the sum is ₹ (100/1.36)
When the difference between S.I and C.I is ₹46 than the sum is ₹ (100/1.36) ×46 = ₹3382.35
[Short trick
The difference between the S.I and C.I for three years on principal P on rate of interest r is
Difference = 3P (r/100)2+P(r/100)3
With this we can find out the difference very easily]
15) The difference between the compound interest and the simple interest on a certain sum of money 15% per annum for 3 years is ₹283.50. Find the sum.
Ans:
Let’s assume that the sum is ₹100.
Here, r = 15% , t = 3 years
We know that
Compound interest (C.I) = A – P = P [1+(r/n)]nt – P
∴ C.I = P [{1+(r/n)}nt – 1]
Or, C.I = 100[{1+15%}3 – 1]
Or, C.I = 100[{1+(15/100)}3 – 1]
Or, C.I = 100[{1+(3/20)}3 – 1]
Or, C.I = 100[(23/20)3 – 1]
Or, C.I = 52.09
∴ Compound interest = ₹52.09
Also, Simple interest = [100×15×3]/100
S.I = 45
∴ Simple interest = ₹45
Difference between S.I and C.I = ₹ (52.09 – 45) = ₹7.09
Now, when the difference between S.I and C.I is ₹7.09 than the sum is ₹100.
When the difference between S.I and C.I is ₹1 than the sum is ₹ (100/7.09)
When the difference between S.I and C.I is ₹283.50 than the sum is ₹ (100/7.09) ×283.50 = ₹3998.6
16) Find the amount and the compound interest on ₹12800 for 1 year at 7.5% p.a compounded semi-annually.
Ans:
We have to find out the compound interest of ₹12800 at 7.5% per annum for 1 year compounded semi annually
We know that,
Amount A = P[1+(r/n)]nt
Where P = principal = ₹12800
r = rate of interest = 7.5%
t = time period in years = 1 years
n = number of time interest compounded in a year = 2 (as the interest is compounded semiannually)
∴ A = 12800[1+(7.5%/2)]2×1
Or, A = 12800× [1+ (7.5/200)] 2
Or, A = 12800× (207.5/200)2
Or, A = 13778
So, the amount for ₹12800 at 7.5% per annum compounded semi-annually for 1 year is ₹13778.
The compound interest is ₹ (13778– 12800) = ₹978
17) Mr. Arora borrowed ₹40960 from a bank to start a play school. If the bank charges 12.5% p.a compounded half-yearly, what amount will he have to pay after 1.5 years?
Ans:
To find out the amount Mr. Arora has to pay after 1.5 years we have to find out the compound interest of ₹40960 at 12.5% per annum for 1.5 year compounded semi annually
We know that,
Amount A = P[1+(r/n)]nt
Where P = principal = ₹40960
r = rate of interest = 12.5%
t = time period in years = 1.5 years = (3/2) years
n = number of time interest compounded in a year = 2 (as the interest is compounded half yearly)
∴ A = 40960[1+ (12.5%/2)]2×(3/2)
Or, A = 40960× [1+ (12.5/200)] 3
Or, A = 40960× (212.5/200)3
Or, A = 49130
So, the amount for ₹40960 at 12.5% per annum compounded semi-annually for 1 year is ₹49130.
The interest received by Mr. Arora is ₹ (49130– 40960) = ₹8170
18) Meera lent out ₹20000 for 9 months at 20% p.a compounded quarterly to Mrs Sharma. What amount will she get after the expiry of the period?
Ans:
To find out the amount Meera will receive after the expiry of the period we have to find out the compound interest of ₹20000 at 20% per annum for 9 months compounded quarterly.
We know that,
Amount A = P[1+(r/n)]nt
Where P = principal = ₹20000
r = rate of interest = 20%
t = time period in years = 9 months = (9/12) years = (3/4) years
n = number of time interest compounded in a year = 4 (as the interest is compounded quarterly i.e. in every 3 months)
∴ A = 20000[1+ (20%/4)]4×(3/2)
Or, A = 20000× [1+ (20/400)] 3
Or, A = 20000× [1+ (1/20)] 3
Or, A = 20000× (21/20)3
Or, A = 23152.5
So, the amount for ₹20000 at 20% per annum compounded quarterly for 9 months is ₹23152.5.
The interest received by meera is ₹ (23152.5 – 20000) = ₹3152.5
19) Find the amount and the compound interest on ₹24000 for 6 months if the interest is payable quarterly at the rate of 20 paise a rupee per annum.
Ans:
The rate of interest is 20 paise a rupee. This means the rate of interest is 20% per annum
We have to find out the compound interest of ₹24000 at 20% per annum for 6 months compounded quarterly.
We know that,
Amount A = P[1+(r/n)]nt
Where P = principal = ₹24000
r = rate of interest = 20%
t = time period in years = 6 months = (6/12) years = (1/2) years
n = number of time interest compounded in a year = 4 (as the interest is compounded quarterly i.e. in every 3 months)
∴ A = 24000[1+ (20%/4)]4×(1/2)
Or, A = 24000× [1+ (20/400)] 2
Or, A = 24000× [1+ (1/20)] 2
Or, A = 24000× (21/20)2
Or, A = 26460
So, the amount for ₹24000 at 20% per annum compounded quarterly for 6 months is ₹26460.
The compound interest is ₹ (26460 – 24000) = ₹2460
Worksheet 4
1) The population of a town is increasing at the rate of 8% p.a. What will be the population of the town after 2 years if the present population is 12500?
Ans:
Present population of the town (P) = 12500
The rate of growth per year (r) = 8%
Time period (t) = 2 years
Population after 2 years will be
= 12500× [1+ (8/100)]2
= 12500× [108/100]2
= 14580
So, the population of the town after 2 years will be 14580.
2) Three years ago the population of a town was 50000. If the annual increase during three successive years was at the rate 4%, 5% and 4% p.a respectively, find the present population.
Ans:
Three years ago population of the town was = 50000
The rate of growth in the first year = 4%
The rate of growth in the second year = 5%
The rate of growth in the third year = 4%
So, the population after 3 years
= 50000× [1+ (4/100)] × [1+ (5/100)] × [1+ (4/100)]
= 50000 × (104/100) × (105/100) × (104/100)
= 56784
So, the population after 3 years will be 56784.
3) The profits of a firm were ₹72000 in the year 2005. During the next year, it increased by 7% and it decreased by 5% in the following year. What are the profits of the firm after 2 years?
Ans:
Profit in 2005 was ₹72000
The rate of growth in the first year = 7%
The rate of depreciation in the second year = 5%
So the profit after 2 years will be
= 72000× [1+ (7/100)] × [1 – (5/100)]
= 72000× (107/100) × (95/100)
= 73188
Therefore, the profit of the farm after 2 years will be ₹73188.
4) The population of a town is 64000. If the annual birth rate is 10.7% and the annual death rate is 3.2%, calculate the population after 3 years. [Hint: Net growth rate = (Birth rate – Death rate)%]
Ans:
We know that the,
Net growth rate % = (Birth rate – Death rate)%]
Net growth rate % = (10.7 – 3.2) % = 7.5%
The current population of the town is 64000.
The population of the town after 3 years will be
= 64000 × [1+ (7.5/100)] 3
= 79507
So, the population after 3 years will be 79507.
5) In a factory, the production of motor bikes was 40000 in a particular year, which rose to 48400 in 2 years. Find the rate of growth per annum, if it was uniform during 2 years.
Ans:
Let’s assume that the rate of growth per annum is r%.
the number of motorcycle produced in a certain year is 40000. And the number of motorcycles produces after two years is 48400.
According to the question
48400 = 40000× [1+ (r/100)]2
Or, (48400/40000) = [1+ (r/100)]2
Or, (48400/40000) = [1+ (r/100)]2
Or, (220/200)2 = [(1+r)/100]2
Or, 220/200 = (100+r) /100
Or, 110 = 100 + r
∴ r = 10
The rate of growth during these two years was 10% per annum.
6) Madhu bought a house for ₹13125000. If its value depreciates at the rate of 10% p.a., what will be its sale price after 3 years?
Ans:
The current price of the house is ₹13125000
The rate of depreciation is 10% per annum.
The sale price of the house after 3 years will be
= 13125000× [1 – (10/100)]3
= 13125000× [9/10]3
=9568125
The sale price of the house after 3 years will be ₹9568125
7) A builder employed 4000 workers to work on a residential project. At the end of the first year, 10% workers were removed, at the end of the second year, 5% of those working at that time were retrenched. However, to complete the project in time, the number of workers was increased by 15% at the end of the third year. How many workers were working during the fourth year?
Ans:
The number of employees on the beginning of the work is 4000.
At the end of the first year 10% workers were removed.
At the end of the second year 5% workers were removed.
At the end of the third year 15% workers were increased.
So the number of workers during the 4th year
= 4000× [1 – (10/100)] × [1 – (5/100)] × [1+ (15/100)]
= 4000 × (9/10) × (95/100) × (115/100)
= 3933
The number of workers working during 4th year is 3933.
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