Simple and Compound Interest Tricks – 2
Read this post carefully and note down all the formulas in a piece of paper for quick revision.
Q. 1) If a sum of money becomes 3 times itself in 20 years at simple interest. What is the rate of interest?
In such questions apply the direct formula
Rate of interest = [100*(Multiple factor – 1)]/T
So R = 100*(3 – 1)/20
Answer : 10%
Note : With this formula you can find Rate if Time is given and Time if rate is given.
Q. 2)
In such questions, just write this line :
1st part : 2nd part : 3rd part = 1/(100+T1 * r) : 1/(100+T2 * r) : 1/(100+T3 * r)
= 1/(100+2 * 5) : 1/(100+3 * 5) : 1/(100+4*5)
= 1/110 : 1/115 : 1/120
= 23*24 : 22*24 : 23*22
Hence 1st part = (23*24)/ (23*24 + 22*24 + 23*22) * 2379
Answer : 828
Note : Surprisingly, such questions when asked mostly have this same data, i.e., R=5% and T1, T2, T3 = 2, 3, 4 years, respectively. Only the Principal is changed. So it would be wise if you can just mug this line :
1st part : 2nd part : 3rd part = 23*24 : 22*24 : 23*22
Based on the above line, you would be able to solve such questions in a jiffy.
But note that it will only work if the question is on Simple Interest. Like the below question appeared in SSC CGL Tier 2
Q. 4
Here the data is same. i.e., R=5% and T1, T2, T3 = 2, 3, 4 years, respectively. So we will write directly –
1st part : 2nd part : 3rd part = 23*24 : 22*24 : 23*22
A received = (23*24)/ (23*24 + 22*24 + 23*22) * 7930
Answer : Rs. 2760
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Q. 5) If a certain sum of money P lent out for a certain time T amounts to P_{1} at R_{1}% per annum and to P_{2} at R_{2}% per annum, then
The above formula is for calculating the Time, if the question asks the rate, then just interchange the rate and time. Hence the formula will become
R = (P_{1 – }P_{2)*100/}P_{2}T_{1 – }P_{1}T_{2}
Apply the formula:
R = (650600)*100/600*6 – 650*4
R = 5%
Alternative Method :
You can solve such questions quickly without mugging the above formula. How?
The sum amounts to Rs. 600 in 4 years and Rs. 650 in 6 years. This means the simple interest is Rs. 50 for 2 years (because the amount increased from Rs. 600 to Rs. 650 in 2 years)
So the SI for 4 years is Rs. 100 (we have seen earlier than SI is proportional. So if SI = 100 for 2 years, then SI = 150 for 3 years, SI = 250 for 5 years and so on)
Now SI = Rs. 100; P = 600100 = Rs. 500; t = 4 years
R = 100*SI/(P*t) = 10000/2000
Answer: 5%
For CI, the formula is different
Difference between CI and SI
This topic is very important from examination point of view. Note the following things
If t=1 year, then SI = CI
If t=2 years then difference between CI and SI can be given by two formulas
If t=3 years then difference between CI and SI can be given by two formulas
In all the above formulas we have assumed that the interest is compounded annually
Let us solve some CGL questions
Q. 6
A = P(1+r/100)^t
Given, A=1.44P t = 2 years
1.44P = P(1 + r/100)^2
r = 20%
Answer : (D)
Q. 7
Here the interest is compounded half yearly, so the formulas we mugged earlier are of no use here. We will have to solve this question manually
SI = P*10*1.5/100 = 0.15P
CI = P(1 + 5/100)^3 – P = P(1.05^3 – 1)
Given CI – SI = 244
P(1.05^3 – 1) – 0.15P = 244
P = Rs. 32000
Answer : (C)
Q. 8
Time = 2 years
Hence apply the formula: Difference(D) = R*SI/200
CI – SI = R*SI/200
CI – SI = (12.5/200)*SI
510 = 1.0625*SI [Since CI = Rs. 510]
SI = Rs. 480
Answer : (D)
Q. 9
CI for 1st year = 10% of 1800 = Rs. 180
CI for 2nd years = 180 + 10% of 180 = Rs. 198
Total = 180+198 = Rs. 378
Hence time = 2 years
Or you can apply the formula
A = P(1+r/100)^t
Answer : (B)
Q. 10
2.5 = P*R*2/100 – P*r*2/100
2.5 = 10R – 10r
R – r = 0.25
Answer : (D)
Q. 11
CI for 1st year = 5% of P = 0.05P
CI for 2nd year = 5% of P + 5% of (5% of P) = 0.05P + 0.0025P = 0.0525P
Total CI = 0.05P + 0.0525P = 0.1025P
Given, 0.1025P = 328
P = Rs. 3200
Answer : (C)
Note : You can solve this question by the formula A = P(1+r/100)^t as well
Q. 12
Note that in this question the CI for 2 years in not given, but the CI for the 2nd year is given.
CI for 2nd year = 10% of P + 10% of (10% of P) = 0.1P + 0.01P = 0.11P
Given, 0.11P = 132
P = Rs. 1200
Answer : (D)
Q. 13
Interest = Re. 1 per day = Rs. 365 for 1 year
SI = P*r*t/100
t=1, r=5%, SI = Rs. 365
So, P = 365*100/5 = Rs. 7300
Answer : (A)
Q. 14
We know
Difference = P(r/100)^2(r/100 + 3)
P = Rs. 10000, r = 5%, t = 3 years
Hence D = Rs. 76.25
Answer : (C)
Q. 15
We know, R = [(y/x)^(1/T2 – T1) – 1]*100
= [(1587/1200)^1/(3 – 1) – 1]*100
= [(1587/1200)^1/2 – 1]*100
= 3/20 * 100
= 15%
Answer : (B)
So this is the end of CI and SI series. If you have any doubt in this topic, please drop a comment…
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