Exercise 2.2
Question 1
Sol :
using the formula , we get
A=5000(1+6100)2 A=P(1+r100)n
=5000×(106100)2
=5000×53×5350×50
= 2×53×53
= 106×53
= 5618
C.I=5618-5000
=618
Question 2
Sol :
Using formula , we get
A=8000(1+10100)4
=8000(1+110)4
=8000(10+110)4
=8000(1110)4
=8000(11×11×11×1110×10×10×)
=585645
=11712.8
C.I=11712.8-8000
=3712.80
Question 3
Sol :
Since, rate of interest per annum is 5%, therefore,
rate of interest per conversion (half yearly)=2.5%
As the money compounded half yearly.
Number of conversion periods=2
A=P(1+r100)n
A=7400(1+2.5100)2
A=7400(1+140)2
A=7400(40+140)2
=7400×41×4140×40
=7774.625
∴Amount =7774.625
Since , rate of interest per annum 10% , therefore rate of interest per conversion=5%
As the money compounded semi annually,
n(number of conversion periods)=3
A=P(1+r100)n
=5000(1+5100)3
=5000(1+120)3
=5000(1+2120)3
=5000×21×21×2120×20×20
=2.5×21×21×214
Compound Interest=5788.125-5000
=788.125
Since, the rate of interest per 12 months (annum) therefore rate of interest per conversion =14×4%
=1 %
As the money compounded quarterly
n(number of conversions)=93
=3
A=P(1+r100)n
=10000(1+1100)3
=10000(100+1100)3
=100000×101×101×101100×100×100
=103030.10
Compound Interest=103030.10-100000
=3030.10
Principal=4800, n=2 and R=5%
Simple Interest=P×n×R100
=4800×2×5100
=480
A=(1+r100)nP
=4800(1+5100)2
=4800(1+120)2
=4800(20+120)2
=4800×21×2120×20
=12×21×21
=5292
C.I=5292-4800=492
∴Difference=492-480
=12(C.I-S.I)
Principal=2500, n=2 and R=4% per annum
Simple Interest=P×n×R100
=2500×2×4100
=200
Given that compound Interest revoked semi annually, therefore n=4 and R=42=2%
A=2500(1+2100)4 A=P(1+r100)n
=2500(1+2100)4
=2500(1+150)4
=2500(5150)4
=2500×51×51×51×5150×50×50×50
=67652012500
=2706.80
Compound Interest=2706.80-2500
=2206.80
∴Difference between simple interest and compound interest is
C.I-S.I=206.080-200
=6.08
Question 8
Sol :
Given that
Principal for the first year=2000
Interest for the first year=2000×1×4100
=80
Principal for the second year=2080
Interest for the second year=2080×1×3100
=62.40
Compound Interest=80+62.40
=142.40
Question 9
Sol :
A=P(1+r1100)(1+r2100)(1+r3100)
=3125(1+4100)(1+5100)(1+6100)
=3125(1+125)(1+120)(1+350)
=3125×26×21×5325×20×50
=144694
=3617.25
∴Compound Interest=3617.25-3125
=492.25
Question 12
Sol :
Given that A=132651, n=3, r=2
A=P(1+r100)n
132651=P(1+r100)3
P(1+150)3=132651
P×51×51×5150×50×50
=132651
P=132651×50×50×5051×51×51
P=132651×125000132651
P=125000
Question 13
Sol :
A=P(1+r100)n
=P(1+4100)2
=P(1+125)2
=P×26×2625×25
C.I.=26×2625×25(P)−P
=676P−625P625
=51P625
5712=51P625
5712×62551=P
P=70000
Question 14
Sol :
A=1275, P=1200, n=1 and R=?
1275=1200(1+r100)n
r=127512−100
r=7512
r=6.25%
C.I for second year=1275×1×6.25100
=7968.75100
=79.68
∴Compound Interest for second year
=80
Question 15
Sol :
Given that A=2500, P=2304, n=2 and R=?
2500=2304(1+r100)2
25002304=(1+r100)2
100+r100=√25002304
100+r100=√(5048)2
100+r=100×5048
r=500048−100
r=5000−48004800
r=2004800
r=124%
r=4.16%
r=416%
Given that
A=2516P , n=2 and r=?
A=P(1+r100)n
P=(1+r100)2=2516P
100+r100=√2516
100+r100=54
r=125-100
r=25%
A=P(1+r100)n
2315.25=2000(1+r100)3
2315.252000=(100+r100)3
3√2.315252.000=100+r100
2315.25×100×5=(100+r)3
3√2315.25×500=100+r
3√1157625=100+r
3√(105)3=100+r
r=105-100=5%
A=48620.25, P=40000, r=R2 , n=2×2=4
A=P(1+r100)n
48620.25=40000(1+R200)4
[(200+R)2]2=48620.25×200×200×200×20040000
(200+R)2=2√(220.5)2(200)2
200+R=2√(220.5)2(200)2
200+R=2√200×220.5
R=210-200
R=10%
G/T A=216125P ,n=3
A=P(1+r100)n
216125P=P(1+r100)3
100+r100=3√216125
100+r=65×100
r=120-100=20%
∴Rate of interest=20×2%
=40%
Given that A=88200, P=80000, n=2
A=P(1+r100)2
8820080000=(100+r100)2
2√21×21×220×20×2=100+r100
21×10020=100+r
r=105-100
r=5%
Principal= 88200
C.I=88200×1×5100
=882×5
=4410
∴Amount to be paid after 3 years=P+C.I
=88200+4410
=92610
∴A=92610
Let the rate of compound interest be r% per annum
The amount afters will be the principal for the third year
5556.50=5292(1+r100)
1+r100=5556.605292
r100=5556.60−52925292
r=264605292
r=5%
∴The rate of compound Interest=5% per annum
Let the original sum be P
5292=P(1+5100)2
P×21×2120×20=5292
P=5292×400441
P=12×400
P=4800
∴Original Sum is 4800
Let the rate of compound Interest be r% per annum
The amount after 3 years will be the principal for the 4th year
=878.46
=798.60(1+r100)
878.46798.60=100+r100
r100=879.46−798.60798.60
r=7986798.60
r=10%
∴The rate of compound interest = 10% per annum
Let the original sum be P
798.6=P=(1+10100)3
798.6=P×1110×1110×1110
798.6×1000121×11=P
P=79860012×11
P=600
∴Original amount =600
A=P(1+r100)n
=17576=15625(1+4100)n
1757615625=(2625)n
(2625)3=(2625)n
Bases are equal then power should be equal n=3
P=12500, C.I.=3246.40
Rate = 8%
A=12500+3246.40
=15746.40
Using 15746.40=12500(1+8100)n
15746.4012500=(2725)n
=(2725)3=(2725)n
∴Bases are equal then powers should be equal
∴Number of years=3
n=3
Here P=16000 , A=18522
As the interest is compounded semi annually
r crate of interest per conversion period 12 of 10%
=5%
Let the number of conversion periods be n, then
18522 =16000(1+5100)n
1852216000=(2120)n
(2120)4=(2120)n
∴Bases are equal then powers should be equal
n=4
Given that,
A=278250,
R1=5 , r2=6
2782.50=P(1+5100)(1+6100)
P=2782.50×20×5021×53
P=10000
∴Sum=10000
CI1=P1×1×r100
P1r=225×100
P1r=22500 P1=22500r
CI2=P2×1×r100
240=22500r×1×r100+r [∴P2=22500r+2]
r=240225
G/I
Compound Interest for the first year CI1=225
Compound Interest for the second year CI2=240
CI1=P1×N×R100
225=P1×1×R100
P1R=22500
∴Principal for the second year
=22500R+CI1
CI2=P2×N×R100
24×1000=P2×1×R
24000=22500R×R+225(R)
225R=24000-22500
R=1500225
(i) Rate pf interest =623 %
(ii) Original Sum
CI1=P1×N×R100
P1=225×100×320
=225×15
∴Original Sum=3375
(iii) Compound Interest for the third year CI3
Principal=3375+225+240
=3840
CI3=3840×20×13×100
Compound Interest for the third year=256
Simple Interest =P×N×R100
=P×2×5100=P10
Compound Interest =P[(1+5100)2−1]
=P[(2120)2−1]
=P[441−400400]
=41P400
Given that
41P400−P10=25
41P−40P400=25
1P=2500×4
P=2500×4
Principal=10000
Simple Interest =P×1×10100=P10
Compound Interest =P[(1+5100)2−1]
=P[(1+120)2−1]
=P[(2120)2−1]
=P[441−400400]=41P400
41P400−P10=15
41P-40P=400×15
P=6000
Given that
Compound Interest for the first year =P[(1+r100)1−1]
1250=P[100+r-100]
1250=Pr
P=1250r
Compound Interest for the second year
=P[(1+r100)2−1]
1375=P[(1+1250P×100)2−1]
1375=P[(P+12.50)2P2−1]
1375=P[P2+25P+156.25P2−P2]
Compound Interest of 2nd conversion period-Compound Interest of 1st conversion=S.I on the Interest of 1st conversion period
∴1375-1250=S.I on Interval on the Interest of 1st Conversion Period
125=1250×1×R100
∴Rate of Interest=10%
Simple Interest for three years=225
x×r×3100=225
x×r=225003 [∵x=Principal and r=rate of Interest]
Compound Interest for two years=153
P[(1+r100)2−1]=153
x[(100+r)2−(100)2(100)2]=153
x[1002+r2+200r-1002]=153×10000
(xr)r+200xr=1530000
225003[r+200]=1530000
r=68×3-200
r=204-200=4
∴Rate of Interest=4%
∴Simple Interest on three years=225
P×3×4100=225
P=25×75
P=1875
Compounded annually
First, we found the amount after one year
A=P(1+r100)n
=P(1+10100)n
=8000(110100)1
=8000×1.1=8800
Principal for the second half year=8800
Interest for the next half year=8800×10×12100
=440
Compounded annually Interest=9240
As the money Invested for 112 years , therefore
n=number of conversion periods=3
A=P(1+r100)n
=8000(1+5100)3
=8000(2120)3
=8000(1.05)3
=9261
Interest for 112 years compounded semi annually=9261
difference between compounded semi annually and annually
=9261-9240
=21
Compounded Annually
Compound Interest for two years
=P[(1+r100)n−1]
=11P25
=0.44P
Compounded half yearly,
Compound Interest for two years =P[(1+r100)n−1]
=P[(1+10100)4−1]
=P[(11100)4−1]
=P[(1.1)4−1]
=P[1.4641-1]
=0.464P
∴Given that 0.464P-0.44P=482
0.024P=482
P=4820.0241
P=20000
∴Sum of money lent out=20000
Amount after one year=P(1+r100)n
13230=P(1+r/2100)2
Amount after 112 year =P(1+r100)n
13891.50=P(1+r/2100)3
13891.50=13230(1+r/2100)2(1+r/2100)2
[∴P=13230(1+r/2100)2]
[P=13230(1+5100)2]
=13230×20×2021×21
=132301.1025
∴Sum=12000
13891.5013230=1+r200
13891.5013230=200+r200
10.5×20=200+r
210=200+r
r=10
∴Rate of Interest=10%
Sum=1323011.10251
=12000
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