Rest part = 1 - (1/3 + 1/6) = 1/2
Average rate per cent annum on the total sum
= [(1/3) x 3] + [(1/6) x 6] + [(1/2) x 8] = 6%
? P = (100 x SI)/(R x T) = (100 x 600)/(6 x 2) = ? 5000
Let the first part be ? A.
second part be ? B
and third part be Rs C
According to the question.
(A x 2 x 3)/100 = (B x 3 x 4)/100 = (C x 4 x 5)/ 100
? 3A = 6B = 10C = k
? A = k/3, B = k/100 and C = k /10
Now, A + B + C = 1440
? k/3 + k/6 + k/10 = 1440 ? k = 2400
? so the difference = k/3 - k/10 = 7k/30 = 7/30 x 2400 = ? 560
Let the sum be P.
SI = SI for first 3 years + SI for next 5 years + SI for next 3 years
? P + 6080 = (P x 12 x 3) / 100 + (P x 16 x 5) / 100 + (P x 20 x 3) / 100
? P + 6080 = (36P + 80P + 60P) / 100
? 100 x (P + 6080) = 176P
? P = 608000 / 76 = 8000
Let the principal be Rs. P
? Amount = Rs. 10080
? Simple interest = Amount - Principal = 10080 - P
&because Simple interest = (Principal x Rate x Time) / 100
? 10080 - P = (P x 16 x 5) / 100
? 10080 - P = 4P/5
? 50400 - 5P = 4P
? 50400 = 5P + 4P
? 50400 = 9P
? P = 50400/9 = Rs. 5600
r1 = 5%
r2 = 8%
r3 = 10%
r4 = 12%
t1 = 3 years
t2 = 2 years
t3 = 4 years
t4 = 12 - (3 + 2 + 4) = 3 years
? Principal = ( Interest x 100) / ( r1 t1 + r2 t2 + r3 t3 + r4 t4)
= (8560 x 100) / (5 x 3 + 8 x 2 + 10 x 4 + 12 x 3)
= 856000/107
= Rs. 8000
r1 = 5%,
r2 = 8%,
r3 = 10%,
t1 = 2 years
t2 = 3 years,
t3 = 8 - (2 + 3) = 3 years
? Principal = (Interest x 100) / [(r1 x t1) + (r2 x t2) + (r3 x t3 )]
= (1280 x 100) / (5 x 2 + 8 x 3 + 10 x 3)
= 128000/(10 + 24 + 30)
= 128000/64
=Rs. 2000
Cost of the music system = ? 8000
Money paid at once = ? 3500
Money left = (8000 - 3500) = ? 4500
Time = (18/12) = 3/2 yr and R = 8% per annum
SI = PTR/100 = (4500 x 3/2 x 8)/100 = ? 540
Money to be paid at the end = (4500 + 540) = ? 5040
? Cost of music system = (3500 + 5040) = ? 8540
Let the sum be P and Q, respectively.
Then, (P x 6 x 2)/100 + (Q x 7 x 2)/100 = 792
? 6P + 7Q = 39600 ...(i)
Also P/2 = Q/3 ? 3P = 2Q ... (ii)
On solving Eqs . (i) and (ii), we get P = 2400 and Q = 3600
? Total sum = (2400 + 3600) = ? 6000
Let the money added be ? P, Then,
[(4800 + P) x 12 x 3]/100 - (4800 x 9 x 3)/100 = 720
? (4800 + P) x 36 - (4800 x 27) = 720 x 100
? (4800 + P) x 4 - (4800 x 3) = 8000
? 4800 + P - (1200 x 3 ) = 2000
? P + 1200 = 2000
? P = 800
So, money added is ? 800.
Suppose the person had deposited ? P at the time of opening the account .
? After one year he had P + (P x 10 x 1)/100 = ? 11P/10
After two years, he had
11P/10 + (11P/10 x 10 x 1)/100 = ? 121P/100 ...(i)
After withdrawn ? 5000 from ? 121P/100, the balance
= ? (121P - 500000)/100
After 3 yr, he had
(121P - 500000)/100 + [(121P - 500000)/100 x 10 x 1]/100
= 11(121P - 500000)/100 ... (ii)
After withdrawn ? 6000 from amount (ii) the balance
= (1331P/1000 - 11500)
? After 4 yr, he had ? (1331P - 5500000)/1000 + 10% of ? (1331P - 5500000)/1000
= ? (11/10) x (1331P/1000 - 11500) ... (iii)
After withdrawn ? 10000 from amount (iii) the balance =0
? 11/10(1331P/1000 - 11500) - 10000 = 0
? P = ? 15470
According to the question.
([P + (P x 6 x T)/100] - [P + (P x 1 x T)/100] = 600 - 200
? 5PT/100 = 400
? PT = 8000
Again, for 6% rate,
SI = PTR/100 = (8000 x 6)/100 = ? 480
? Sum = 600 - 480 = ? 120
As we have, PT = 8000
? T = 8000/120 = 200/3 = 662/3 yr
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