1040.If a certain sum of money invested at a certain rate of compound interest doubles in 6 years. In how many years will it become 8 times?
18 years.
15 years.
14 years.
13 years.
Explanation:
Since, 2323 = 8.
Therefore, The amount will become 8 times in 3 x 6 = 18 years.
Since, 2323 = 8.
Therefore, The amount will become 8 times in 3 x 6 = 18 years.
1041.A sum of money becomes Rs.6690 after three years and Rs.10,035 after 6 years on compound interest. The sum is
Rs.4400
Rs.4445
Rs.4460
Rs.4520
Explanation:
Let the sum be P.
Then, P [1+R100]3=6690[1+R100]3=6690.........(i)
and P [1+R100]6=10,035[1+R100]6=10,035 ..... (ii)
Dividing (ii) by (i), we get
(1+R100)3=100356690=32(1+R100)3=100356690=32
P=(6690×23)(6690×23)=Rs.4460
Let the sum be P.
Then, P [1+R100]3=6690[1+R100]3=6690.........(i)
and P [1+R100]6=10,035[1+R100]6=10,035 ..... (ii)
Dividing (ii) by (i), we get
(1+R100)3=100356690=32(1+R100)3=100356690=32
P=(6690×23)(6690×23)=Rs.4460
1042.Rs.1600 at 10% per annum compound interest compound half-yearly amount to Rs.1944.81 in
2 years
3 years
5 years
6 years
Explanation:
1600(1+5100)T=1944.81(1+5100)T=1944.81
⇒(2120)r=1944.811600.00=194481160000⇒(2120)r=1944.811600.00=194481160000
=(441400)2=(2120)4(441400)2=(2120)4
T = 4 (Half - years) or T = 2 years
1600(1+5100)T=1944.81(1+5100)T=1944.81
⇒(2120)r=1944.811600.00=194481160000⇒(2120)r=1944.811600.00=194481160000
=(441400)2=(2120)4(441400)2=(2120)4
T = 4 (Half - years) or T = 2 years
1043.The difference between simple interest and compound interest on a sum for 2 years at 8%, when the interest is compounded annually Rs.16. If the interest was compounded half-yearly, the difference in two interests would be nearly :
Rs.16
Rs.16.80
Rs.21.85
Rs.24.64
Explanation:
For 1st year, S.I = C.I
Thus, Rs.16 is the S.I on S.I for 1 year, which at 8% is thus Rs.200
i.e.S.I on the principal for 1 year is Rs.200
Principal = Rs. (100×2008×1)=(100×2008×1)=Rs.2500
Amount for 2 years, compounded half-yearly
=Rs.[2500×(1+4100)4][2500×(1+4100)4]=Rs.2924.64
C.I = Rs.424.64
Also, S.I = Rs.(2500×8×2100)(2500×8×2100)=Rs.400
Hence, [(C.I)-(S.I)] = Rs. (424.64-400)=Rs.24.64
For 1st year, S.I = C.I
Thus, Rs.16 is the S.I on S.I for 1 year, which at 8% is thus Rs.200
i.e.S.I on the principal for 1 year is Rs.200
Principal = Rs. (100×2008×1)=(100×2008×1)=Rs.2500
Amount for 2 years, compounded half-yearly
=Rs.[2500×(1+4100)4][2500×(1+4100)4]=Rs.2924.64
C.I = Rs.424.64
Also, S.I = Rs.(2500×8×2100)(2500×8×2100)=Rs.400
Hence, [(C.I)-(S.I)] = Rs. (424.64-400)=Rs.24.64
1044.The difference in C.I and S.I for 2 years on a sum of money is Rs.160. If the S.I for 2 years be Rs.2880, the rate percent is :
5%
115%
1119%
19%
Explanation:
S.I for 1 year = Rs.1440
S.I on Rs.1440 for 1 year = Rs.160
Hence, rate percent = (100×1601440×1)(100×1601440×1) = 1119%
S.I for 1 year = Rs.1440
S.I on Rs.1440 for 1 year = Rs.160
Hence, rate percent = (100×1601440×1)(100×1601440×1) = 1119%
1045.The value k of a machine depreciates every year at the rate of 10% on its value at the beginning of that year. If the present value of the machine is Rs.729, its worth 3 years ago was :
Rs.947.10
Rs.800
Rs.1000
Rs.750.87
Explanation:
P=(1−10100)3=729(1−10100)3=729
P=Rs.(729×10×10×109×9×9)(729×10×10×109×9×9)=Rs.1000
P=(1−10100)3=729(1−10100)3=729
P=Rs.(729×10×10×109×9×9)(729×10×10×109×9×9)=Rs.1000
1046.The least number of complete years in which a sum of money put out at 20% C.I. will be more than doubled is :
3
4
5
6
Explanation:
x(1+20100)n>2xx(1+20100)n>2x or (65)n>2(65)n>2
Now, (65×65×65×65)>2(65×65×65×65)>2
n = 4 years
x(1+20100)n>2xx(1+20100)n>2x or (65)n>2(65)n>2
Now, (65×65×65×65)>2(65×65×65×65)>2
n = 4 years
1047.A sum of Rs.550 was taken a loan. This is to be repaid in two equal annual instalments. If the rate of interest be 20% compounded annually, then the value of each instalment is :
Rs.421
Rs.396
Rs.360
Rs.350
Explanation:
Let the value of each instalment be Rs.x.
Then, x(1+20100)+x(1+20100)2=550x(1+20100)+x(1+20100)2=550
or 5x6+25x36=5505x6+25x36=550 or x = 360
Let the value of each instalment be Rs.x.
Then, x(1+20100)+x(1+20100)2=550x(1+20100)+x(1+20100)2=550
or 5x6+25x36=5505x6+25x36=550 or x = 360
1048.A loan was repaid in two annual instalments of Rs.112 each. If the rate of interest be 10% per annum compounded annually, the sum borrowed was :
Rs.200
Rs.210
Rs.217.80
Rs.280
Explanation:
Principal = (Present value of Rs.121 due 1 year hence ) + (Present value of Rs.121 due 2 years hence )
= Rs. 121(1+10100)+121(1+10100)2121(1+10100)+121(1+10100)2 =Rs.210
Principal = (Present value of Rs.121 due 1 year hence ) + (Present value of Rs.121 due 2 years hence )
= Rs. 121(1+10100)+121(1+10100)2121(1+10100)+121(1+10100)2 =Rs.210
1049.A sum amounts to Rs.2916 in 2 years and to Rs.3149.28 in 3 years at compound interest. The sum is :
Rs.1500
Rs.2000
Rs.2500
Rs.3000
Explanation:
Let P be the principal and R% per annum be rate.
Then P (1+R100)3(1+R100)3=3149.28 ........ (i)
and P (1+R100)2(1+R100)2=2916 ..........(ii)
On dividing (i) and (ii) we get
(1+R100)(1+R100) = 3149.2829163149.282916
or 100=233.282916100=233.282916 or R = 233.282916×100=8%233.282916×100=8%
Now P (1+8100)2=2916(1+8100)2=2916
or P ×2725×2725=2916×2725×2725=2916
or P = 2916×25×2527×272916×25×2527×27=Rs. 2500
Let P be the principal and R% per annum be rate.
Then P (1+R100)3(1+R100)3=3149.28 ........ (i)
and P (1+R100)2(1+R100)2=2916 ..........(ii)
On dividing (i) and (ii) we get
(1+R100)(1+R100) = 3149.2829163149.282916
or 100=233.282916100=233.282916 or R = 233.282916×100=8%233.282916×100=8%
Now P (1+8100)2=2916(1+8100)2=2916
or P ×2725×2725=2916×2725×2725=2916
or P = 2916×25×2527×272916×25×2527×27=Rs. 2500
1050.A sum of money amounts to Rs.10648 in 3 years and Rs.9680 in 2 years. The rate of interest is :
5%
10%
15%
20%
Explanation:
Let P be the principal and R% annum be the rate.
Then P (1+R100)3=10648(1+R100)3=10648...(i)
and P (1+R100)2=9680(1+R100)2=9680 ....(ii)
On dividing (i) by (ii), we have
(1+R100)=106489680(1+R100)=106489680
or R100=9689680=110R100=9689680=110
or R = 110×100=10%110×100=10%
Let P be the principal and R% annum be the rate.
Then P (1+R100)3=10648(1+R100)3=10648...(i)
and P (1+R100)2=9680(1+R100)2=9680 ....(ii)
On dividing (i) by (ii), we have
(1+R100)=106489680(1+R100)=106489680
or R100=9689680=110R100=9689680=110
or R = 110×100=10%110×100=10%
1051.The difference between simple interest and compound interest at the same rate for Rs.5000 for 2 years is Rs.72. The rate of interest is :
10%
12%
6%
8%
Explanation:
[5000×(1+R1002)−5000]−5000×2×R100=72[5000×(1+R1002)−5000]−5000×2×R100
=72⇒5000[(1+R1002)−1−R50]=72⇒5000[(1+R1002)−1−R50]=72
⇒1+R2100+2R100−1−R50=725000⇒1+R2100+2R100−1−R50=725000
⇒R2=(725000×10000)=144⇒R2=(725000×10000)=144 or R = 12%
[5000×(1+R1002)−5000]−5000×2×R100=72[5000×(1+R1002)−5000]−5000×2×R100
=72⇒5000[(1+R1002)−1−R50]=72⇒5000[(1+R1002)−1−R50]=72
⇒1+R2100+2R100−1−R50=725000⇒1+R2100+2R100−1−R50=725000
⇒R2=(725000×10000)=144⇒R2=(725000×10000)=144 or R = 12%
1052.The compound interest on a certain sum of money for 2 years at 10% per annum is Rs.420. The simple interest on the same sum at the same rate and for the same time will be :
Rs.350
Rs.375
Rs.380
Rs.400
Explanation:
Let principal be P. Then, P(1+P1002)−P=420⇒PP(1+P1002)−P=420⇒P=Rs.2000
S.I = Rs.2000×2×101002000×2×10100= Rs.400
Let principal be P. Then, P(1+P1002)−P=420⇒PP(1+P1002)−P=420⇒P=Rs.2000
S.I = Rs.2000×2×101002000×2×10100= Rs.400
1053.The difference between the compound interest and simple interest on a certain sum at 5% per annum for 2 years is Rs.1.50. The sum is :
Rs.600
Rs.500
Rs.400
Rs.300
Explanation:
Let the sum be Rs. 100. Then.
S.I = Rs. (100×5×2100)(100×5×2100) = Rs.10
C.I = Rs.[{100×(1+5100)2}−100][{100×(1+5100)2}−100] = Rs. 414414
Difference between C.I and S.I. = Rs. (414−10)(414−10)=Rs.0.25
0.25 : 1.50 : : 100 : x
x = 1.50×1000.251.50×1000.25= Rs.600
Let the sum be Rs. 100. Then.
S.I = Rs. (100×5×2100)(100×5×2100) = Rs.10
C.I = Rs.[{100×(1+5100)2}−100][{100×(1+5100)2}−100] = Rs. 414414
Difference between C.I and S.I. = Rs. (414−10)(414−10)=Rs.0.25
0.25 : 1.50 : : 100 : x
x = 1.50×1000.251.50×1000.25= Rs.600
1054.A sum of money placed at C.I doubles itself in 5 years. It will amount to eight times itself in :
15 years
20 years
12 years
10 years
Explanation:
Let the principal P and rate be r% . Then, 2P = P (1+r100)5(1+r100)5 or
(1+r100)5(1+r100)5 = 2
Let it be 8 times in t years . Then, 8p = p (1+r100)t(1+r100)t
or (1+r100)t(1+r100)t=8=(2)3(2)3 = ((1+r100)5)3=(1+r100)15((1+r100)5)3=(1+r100)15
t = 15 years
Let the principal P and rate be r% . Then, 2P = P (1+r100)5(1+r100)5 or
(1+r100)5(1+r100)5 = 2
Let it be 8 times in t years . Then, 8p = p (1+r100)t(1+r100)t
or (1+r100)t(1+r100)t=8=(2)3(2)3 = ((1+r100)5)3=(1+r100)15((1+r100)5)3=(1+r100)15
t = 15 years
1055.The simple interest on a certain sum for 2 years at 10% per annum is Rs.90. The corresponding compound interest is:
Rs.99
Rs.95.60
Rs.94.50
Rs.108
Explanation:
Sum = Rs. (100×1902×10)(100×1902×10) = Rs.450
C.I = Rs. [450×(1+10100)2−450][450×(1+10100)2−450] = Rs.94.50
Sum = Rs. (100×1902×10)(100×1902×10) = Rs.450
C.I = Rs. [450×(1+10100)2−450][450×(1+10100)2−450] = Rs.94.50
What is the principal amount which earns Rs.132 as compound interest for the second year at 10% per annum ?
bRs.1200
A sum amounts to Rs.1352 in 2 years at 4% compound interest. The sum is:
bRs.1200
A sum amounts to Rs.1352 in 2 years at 4% compound interest. The sum is:
Solution:
Let x be the principal at the end of first year.
So interest on first year ending amount
= x×10×1100=132⇒x=1320x×10×1100=132⇒x=1320
Let y be the original principal. Then one year interest + principal = 1320
So, y + y×10×1100=1320⇒y=1200y×10×1100=1320⇒y=1200
Let x be the principal at the end of first year.
So interest on first year ending amount
= x×10×1100=132⇒x=1320x×10×1100=132⇒x=1320
Let y be the original principal. Then one year interest + principal = 1320
So, y + y×10×1100=1320⇒y=1200y×10×1100=1320⇒y=1200
1057.The compound interest on Rs.30000 at 7% per annum for a certain time is Rs.4347. The time is :
Rs.1250
Rs.1260
Rs.1200
2 years
Explanation:
Let the sum be P . Then, 1352= P (1+4100)2(1+4100)2
⇒1352=P×2625×2625⇒1352=P×2625×2625
⇒P=1352×25×2526×26=1250⇒P=1352×25×2526×26=1250
Let the sum be P . Then, 1352= P (1+4100)2(1+4100)2
⇒1352=P×2625×2625⇒1352=P×2625×2625
⇒P=1352×25×2526×26=1250⇒P=1352×25×2526×26=1250
Rs.800 at 5% per annum compound interest will amount to Rs.882 in:
Solution:
30000×(1+7100)t=30000+434730000×(1+7100)t=30000+4347
or (107100)t=3434730000=1144910000=(107100)2(107100)t=3434730000=114491000
=(107100)2
Time = 2 years
30000×(1+7100)t=30000+434730000×(1+7100)t=30000+4347
or (107100)t=3434730000=1144910000=(107100)2(107100)t=3434730000=114491000
=(107100)2
Time = 2 years
1059.Simple interest on a sum at 4% per annum for 2 years is Rs.80.The compound interest on the same sum for the same period is:
2 years
3 years
4 years
Rs.81.60
Explanation:
Let time be t years
882=800(1+5100)t=882800=(2120)t882=800(1+5100)t=882800=(2120)t
= (2120)2=(2120)t⇒t=2(2120)2=(2120)t⇒t=2
time = 2 years
Let time be t years
882=800(1+5100)t=882800=(2120)t882=800(1+5100)t=882800=(2120)t
= (2120)2=(2120)t⇒t=2(2120)2=(2120)t⇒t=2
time = 2 years