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MEDIUM PROBLEMS
1. A sum at Simple interest at 13 1/2% per annum amounts to Rs 2502.50
after 4 years.Find the sum.
Sol:- Let Sum be x. then,
S.I = (P*T*R)/100
= ((x*4*27)/(100*2))
= 27x/100
Amount = (x+(27x)/100)
= 77x/50
77x/50 = 2502.50
x = (2502.50*50)/77
= 1625
Sum = 1625
2. A some of money becomes double of itself in 4 years in 12 years it will become how many times at the same rate.
Sol:- 4 yrs - - - - - - - - - P
12 yrs - - - - - - - - - ?
(12/4)* P =3P
Amount or Sum = P+3P = 4 times
3. A Sum was put at S.I at a certain rate for 3 years.Had it been put at 2% higher rate ,it would have fetched Rs 360 more .Find the Sum.
Sol:- Let Sum =P
original rate = R
T = 3 years
If 2% is more than the original rate ,it would have fetched 360 more ie., R+2
(P*(R+2)*3/100) - (P*R*3)/100 = 360
3PR+ 6P-3PR = 36000
6P = 36000
P = 6000
Sum = 6000.
4.Rs 800 amounts to Rs 920 in 3yrs at S.I.If the interest rate is increased by 3%, it would amount to how much?
Sol:- S.I = 920 - 800 = 120
Rate = (100*120)/(800*3) = 5%
New Rate = 5 + 3 = 8% , Principal = 800 , Time = 3 yrs
S.I = (800*8*3)/100 = 192
New Amount = 800 + 192
= 992
5. Prabhat took a certain amount as a loan from bank at the rate of 8% p.a S.I and gave the same amount to Ashish as a loan at the rate of 12% p.a . If at the end of 12 yrs, he made a profit of Rs. 320 in the deal,What was the original amount?
Sol:- Let the original amount be Rs x.
T = 12
R1 = 8% R2 = 12% Profit = 320
P = x
(P*T*R2)/100 - (P*T*R1)/100 =320
(x*12*12)/100 - (x*8*12)/100 = 320
x = 2000/3
x = Rs.666.67
6. Simple Interest on a certail sum at a certain rate is 9/16 of the sum . if the
number representing rate percent and time in years be equal ,then the rate is.
Sol:- Let Sum = x .Then,
S.I = 9x/16
Let time = n years & rate = n%
n = 100 * 9x/16 * 1/x * 1/n
n * n = 900/16
n = 30/4 = 7 1/2%
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