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COMPLEX PROBLEMS
1. A certain sum of money amounts t 1680 in 3yrs & it becomes 1920
in 7 yrs .What is the sum.
Sol:- 3 yrs - - - - - - - - - - - - - 1680
7 yrs - - - - - - - - - - - - - 1920
then, 4 yrs - - - - - - - - - - - - - 240
1 yr - - - - - - - - - - - - - ?
(1/4) * 240 = 60
S.I in 3 yrs = 3*60 = 18012
Sum = Amount - S.I
= 1680 - 180
= 1500
we get the same amount if we take S.I in 7 yrs
I.e., 7*60 =420
Sum = Amount - S.I
= 1920 - 420
= 1500
2. A Person takes a loan of Rs 200 at 5% simple Interest. He returns Rs.100 at the end of 1 yr. In order to clear his dues at the end of 2yrs ,he would pay:
Sol:- Amount to be paid
= Rs(100 + (200*5*1)/100 + (100*5*1)/100)
= Rs 115
3. A Man borrowed Rs 24000 from two money lenders.For one loan, he
paid 15% per annum and for other 18% per annum .At the end of one
year,he paid Rs.4050.How much did he borrowed at each rate?
Sol:- Let the Sum at 15% be Rs.x
& then at 18% be Rs (24000-x)
P1 = x R1 = 15
P2 = (24000-x) R2 = 18
At the end of ine year T = 1
(P1*T*R1)/100 + (P2*T*R2)/100 = 4050
(x*1*15)/100 + ((24000-x)*1*18)/100 = 4050
15x + 432000 - 18x = 405000
x = 9000
Money borrowed at 15% = 9000
Money borrowed at 18% = (24000 - 9000)
= 15000
4. What annual instalment will discharge a debt of Rs. 1092 due in 3 years
at 12% Simple Interest ?
Sol:- Let each instalment be Rs x
(x + (x * 12 * 1)/100) + (x + (x * 12 * 2)/100) + x = 1092
28x/25 + 31x/25 + x =1092
(28x +31x + 25x) = (1092 * 25)
84x = 1092 * 25
x = (1092*25)/84 = 325
Each instalement = 325
5. If x,y,z are three sums of money such that y is the simple interest on x,z
is the simple interest on y for the same time and at the same rate of
interest ,then we have:
Sol:- y is simple interest on x, means
y = (x*R*T)/100
RT = 100y/x
z is simple interest on y,
z = (y*R*T)/100
RT = 100z/y
100y/x = 100z/y
y * y = xz
6.A Sum of Rs.1550 was lent partly at 5% and partly at 5% and partly at 8% p.a Simple interest .The total interest received after 3 years was Rs.300.The ratio of the money lent at 5% to that lent at 8% is:
Sol:- Let the Sum at 5% be Rs x
at 8% be Rs(1550-x)
(x*5*3)/100 + ((1500-x)*8*3)/100 = 300
15x + 1500 * 24 - 24x = 30000
x = 800
Money at 5%/ Money at 8% = 800/(1550 - 800)
= 800/750 = 16/15
7. A Man invests a certain sum of money at 6% p.a Simple interest and another sum at 7% p.a Simple interest. His income from interest after 2 years was Rs 354 .one fourth of the first sum is equal to one fifth of the second sum.The total sum invested was:
Sol:- Let the sums be x & y
R1 = 6 R2 = 7
T = 2
(P1*R1*T)/100 + (P2*R2*T)/100 = 354
(x * 6 * 2)/100 + (y * 7 * 2)/100 = 354
6x + 7y = 17700 ———(1)
also one fourth of the first sum is equal to one
fifth of the second sum
x/4 = y/5 => 5x - 4y = 0 —— (2)
By solving 1 & 2 we get,
x = 1200 y = 1500
Total sum = 1200 +1500
= 2700
8. Rs 2189 are divided into three parts such that their
amounts after 1,2& 3 years respectively may be equal ,the rate of
S.I being 4% p.a in all cases. The Smallest part is:
Sol:- Let these parts be x,y and[2189-(x+y)] then,
(x*1*4)/100 = (y*2*4)/100 = (2189-(x+y))*3*4/100
4x/100 = 8y/100
x = 2y
By substituting values
(2y*1*4)/100 = (2189-3y)*3*4/100
44y = 2189 *12
y = 597
Smallest Part = 597
9. A man invested 3/3 of his capital at 7% , 1/4 at 8% and the remainder at 10%
If his annual income is Rs.561. The capital is:
Sol:- Let the capital be Rs.x
Then, (x/3 * 7/100 * 1) + ( x/4 * 8/100 * 1)
+ (5x/12 * 10/100 * 1) = 561
7x/300 + x/50 + x/24 = 561
51x = 561 * 600
x = 6600
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