Let required difference be ? D.
By formula,
D = P x (R/100)2 x [(3 + R)/100]
= 16200 x (25/100)2 x [(3 + 25)/100]
= 16200 x 625/10000 x 13/4
= 162 x 625 x 13/4 x 100 = ? 3290.63
Suppose A gets ? P and B gets ? (25000 - P).
Interest received by A at the rate of 8% per annum CI
= P(1 + 8/100)2 - P
= P(27/25)2 - P
= 104P/625
Interest received by B at the rate of 10% per annum SI
= [(25000 - P) x 10 x 2]/100 = (25000 - P)/5
According to the question,
(25000 - P)/5 = 104P/(625 + 1336)
? P = 10000
SI on ? 65000 at the rate of 10% for 3 yr
= (65000 x 10 x 3)/100 = ? 19500
CI on ? 65000 at the rate of 10% for 3 yr
= 65000(1 + 10/100)2 - 65000
= 65000(11 x 11 x 11 - 10 x 10 x 10)/1000 = ? 21515
? Required gain = 21515 - 19500 = ? 2015
We have (A's present share) (1 + 4/100)7 = (B's present share) (1 + 4/100)9
? A's present share/B's present share = (1 + 4/100)2 = (26/25)2 = 676/625
Dividing Rs. 3903 in the ratio of 676 : 625
? A's present share = 676/(676 + 625) of Rs .3903 = Rs. 2028
B's present share = Rs. 3903 - Rs. 2028 = Rs. 1875
P = Rs. 64000
r = 2.5 paise per rupee per annum (given)
= 0.025 rupee per rupee per annum
= 0.025 x 100 rupee per hundred rupee per annum
= 0.025 x 100 per cent per annum
= 2.5 per cent per annum
t = 3 years
C.I. = 6400[(1 + 2.5/100)3 - 1]
= 64000[(1.025)3 - 1]
= 64000[1.0769 - 1]
= 64000 x 0.0769
= 4921.6
= Rs. 4921
? The compound interest payable is Rs. 4921.
Remaining money = 7044 - 2000 = Rs. 5044
If each installment is of Rs. P
When the amount is Rs. P at the end of first, second and third year at the rate of 5% then principal will be - P/(1 + 5/100), P/(1 +5/100)2 and P/(1 +5/100)3
? P/(1 +5/100) + P/(1 + 5/100)2 + P/(1 +5/100)3 = 5044
? P(20/21) + P(20/21)2 + P(20/21)3 = 5044
? P(20/21) { 1 +(20/21) + (20/21)2} = 5044
? P(20/21) {1 +20/21 +400/441} = 5044
? P(20/21) {441 + 420 + 400/441} =5044
? P(20/21) (1261/441) =5044
? P = (5044 x 21 x 441)/ (20 x 1261) = Rs. 1852.20
Let the first part be ? M .
Then , second part = (2602 - M)
According to the question,
M(1 + 4/100)4 = (2602 - M) (1+4/100)9
? M/(2602 - M) = [(1 + 4/100)9]/[(1 + 4/100)7]
? M/(2602 - M) = (1 + 4/100)2 = (26/25) x (26/25) = 676/625
? 625M = 676 x 2602 - 676M
? 1301M = 676 x 2602
? M = (676 x 2602)/1301 = 676 x 2 = 1352
? The two parts are ? 1352 and ? (2602 - 1352) = ? 1250
According to the question,
1728 = 1331(1 + R/100)3
1728/1331 = (1 + R/100)3
? (12/11)3 = (1 + R/100)3
? 1 + R/100 = 12/11
? R/100 = (12/11) - 1 = 1/11
? R = 100/11= 9.09%
Let the sum be ? P .
Then, CI when compounded half - yearly = [P x (1 + 10/100)4 - P] = 4641P/10000
CI when compounded annually = [ P x (1 + 20/100)2 - P] = 11P/25
According to the question, 4641P/10000 - 11P/25 = 964
? [(4641 - 4400)/10000] x P = 964
? P = (964 x 10000)/241
= ? 40000
Given, P =? 4000,
n = 9 months = 3/4 yr and
CI = ? 630.50
Amount = P + CI = 4000 + 630.50 = ? 4630.50
According to the formula,
? Amount = P[(1+R/(100 x 4)]4n
? 4630.50 = 4000(1+R/400)4 x 3/4
? 4630.50 = 4000[(400 + R)/400]3
? 4630.50/4000 = [(400 + R)/400]3
? 9261/8000 = [(400 + R)/400]3
? (21/20)3 = [(400 + R)/400]3
? [(400 + R)/400]/400 = 21/20
? 400 + R = 21 x 20 = 420
? R = 420 - 400 = 20%
Given, P = ? 4000
R1= 10% (decreased), R2 = 5% (decreases) and R3 = 15% (growth)
? According to the formula,
Income at the end of third year = P(1 - R1/100)(1 - R2/100)(1 + R3/100)
= 4000(1- 10/100) (1 - 5/100) (1 + 15/100)
= 4000 x (9/10) x (19/20) x (23/20)
= 9 x 19 x 23
= ? 3933
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