4, 14, 24, 34, 44, 54, 64, 74, 84, 94 => has 11 number of (4's)
40, 41, 42, 43, 45, 46, 47, 48, 49 => has 9 number of (4's)
So, totally (11 + 9 = 20) number of 4's in 1 to 100.
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
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After withdrawn ? 5000 from ? 121P/100, the balance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After withdrawn ? 6000 from amount (ii) the balance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After withdrawn ? 10000 from amount (iii) the balance =0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
In the case I,
SI = (P x R x T) /100 = (24200 x 4 x 6) / 100 = ? 5808
? Amount = Principal + SI
SI = 24200 + 5808 = 30008
In the case II,
SI = (30008 x 4 x 4) / 100 = ? 4801.28
Let a, b and c be the amount invested in schemes P, Q and R, respectively.
Then, according to the question,
[(a x 10 x 1)/100] + [(b x 12 x 1)/100] + [(c x 15 x 1)/100] = 3200
? 10a + 12b + 15x = 320000 .....(i)
Now, c = 240% of b = 12b/5 ....(ii)
and c = 150% of a = 3a/2
? a = 2c/3 = (2/3 x 12/5) b = 8b/5 .....(iii)
From Eqs. (i), (ii) and (iii), we get
16b + 12b + 36b = 320000
? 64b = 320000
? b = 5000
? Sum invested in scheme Q = ? 5000
Let the sum be Rs . 100 then
S.I. = Rs.(100 x 5 x 2/100)= Rs. 10
C.I = Rs. [{100 x (1 + 5/100)2} - 100] = Rs. 41/4
? Difference between C.I and S.I = Rs. (41/4 - 10) = Re. 0.25
? 0.25 : 150 : : 100 : P
? P = (1.50 x 100) / 0.25 = Rs. 600
Let the principle is P
so compound interest = p x ( 1 + (12.5/100))sup>2 - p = 170
? p x (112.5/100) x (112.5/100) - p = 170
? P [ 12656.25 - 10000 ] = 170 x 10000
? p = (170 x 10000) / 2656.25
Simple interest SI = (P x T x R )/100
= [{(170 x 10000) / 2656.25} x 2 x 12.5] / 100
= 160
Let the sum be P.
Then, 1352 = P(1 +4/100)2
? 1352 = P x 26/25 x 26/25
? P = (1352 x 25 x 25) / (26 x 26) = 1250
? Principal = Rs. 1250
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.