Monday, 28 March 2016

[Q] Simple and Compound Interest Tricks - 2


In this post I will give some extremely important formulas that will save your time. 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 samei.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

Q. 5) If a certain sum of money P lent out for a certain time T amounts to P1 at R1% per annum and to P2 at R2% 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 = (P1 - P2)*100/P2T1 - P1T2



Apply the formula:
R = (650-600)*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 = 600-100 = 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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62 comments:

  1. Sir pls explain the formula of Q7 ..and thanku sir for these tricks

    ReplyDelete
    Replies
    1. T = 1.5 years and R = 10%
      when interest is compounded half-yearly, then T is multiplied by 2 and R is divided by 2
      CI = Amount - Principal
      = P(1 + R/100)^T - P
      = P(1 + 5/100)^3 - P [Put T = 3 years and R = 5%]
      = P(1.05^3 - 1) [Take P common]
      SI = 0.15P
      CI - SI = 244
      Solve this equation for P

      Delete
    2. Sir we can use fromula for 3 yrs here

      Delete
  2. Sir 9th qtn m 378 kese aea pls explain ... ? ? ?

    ReplyDelete
    Replies
    1. CI for 1st year = r% of P
      CI for 2nd year = CI for 1st year + r% of (CI for 1st year)
      CI for 3rd year = CI for 2nd year + r% of (CI for 2nd year)
      and so on...
      Now,
      CI for 3 years = CI for 1st year + CI for 2nd year + CI for 3rd year

      Delete
  3. Sir 9th qtn m 378 kese aea pls explain ... ? ? ?

    ReplyDelete
  4. Sir, in question 2nd don't you think that calculation will take time even after mugging that line, would it be wise to attempt this type of questions when we have very less time, or is there any way from which we can solve it quickly?
    Thanks is advance sir, ur blog is really very helpful for us.

    ReplyDelete
    Replies
    1. it is not at all time consuming...just remember the sum - 1586

      Delete
    2. @himanshu you can remeber instead of above equation...

      (23*24)/ (23*24 + 22*24 + 23*22)=0.348 for first part
      (22*24)/ (23*24 + 22*24 + 23*22)=0.3329 for second part
      (23*22)/ (23*24 + 22*24 + 23*22)=0.3191 for third part

      just remeber below any two & u easily get the third(1- sum of any two)

      for first part = 0.348*p
      for second part = 0.3329*p
      for third part = {1-(0.348+0.3329) = 0.3191} i.e. .3191*p

      i hope this is easy compare to above

      Delete
  5. Sir, in question 2nd don't you think that calculation will take time even after mugging that line, would it be wise to attempt this type of questions when we have very less time, or is there any way from which we can solve it quickly?
    Thanks is advance sir, ur blog is really very helpful for us.

    ReplyDelete
  6. A sum of money lent at compound interest for 2 years at 20% per annum would fetch re 482 more of interest was payable half yearly than if it was payable annually. The sum is ?
    Pls tell how to solve

    ReplyDelete
    Replies
    1. When interest is compounded half-yearly
      A = P(1 + 10/100)^4
      When interest is compounded annually
      A = P(1 + 20/100)^2
      Given, P(1 + 10/100)^4 - P(1 + 20/100)^2 = 482
      Solve for P
      Answer : Rs. 20000
      Note : In the question, they have given the difference between the interest and we have taken the difference between the Amounts because the difference in amount is equal to the difference in interest...

      Delete
  7. Replies
    1. (1587/1200)^1/2 = (529/400)^1/2 = 23/20
      Khud se bhi pen chalao bhai :P

      Delete
    2. How much score is required to get mea post in cgl . Plz reply ker dena bhai . Mere ek bhi comment ka reply nhi kerte ho aur baki sab ko reply kerte ho. aisa bhedbhav mat Karo bhai.....:)

      Delete
    3. https://www.dropbox.com/s/m0mq8m8zantz91t/Pic.jpg?dl=0

      The above image answers your second query :)
      For the first one - last year's cutoff for MEA was 540...

      Delete
  8. Sir out of 600 540 was cuttoff for MEA?

    ReplyDelete
  9. Sir,in very last ques number 15,time t2 will be 3 or 2?

    ReplyDelete
    Replies
    1. T2 is associated with y, hence it is 3 years...

      Delete
  10. dear sir your blog is very useful for us and a lot of thanks for that .... sir for cbi post how much mark would be needed pls reply

    ReplyDelete
  11. dear sir your blog is very useful for us and a lot of thanks for that .... sir for cbi post how much mark would be needed pls reply

    ReplyDelete
  12. C.I of 2 years is rs 156 and 3 years is rs 254. What is rate? (Any short trick)

    ReplyDelete
    Replies
    1. CI for second conversion period - CI for first conversion period = SI on the interest of the first converSion period

      Delete
  13. C.I of 2 years is rs 156 and 3 years is rs 254. What is rate? (Any short trick)

    ReplyDelete
  14. Hi, Prashant! Your tips are very useful. I have this doubt- Is there any easy to solve this ratio (23*24)/ (23*24 + 22*24 + 23*22) * 7930? Because calculating each part is quite time consuming plus the options are too close, so approximation may not work also.

    ReplyDelete
    Replies
    1. @sohan singh you can remeber instead of above equation...

      (23*24)/ (23*24 + 22*24 + 23*22)=0.348 for first part
      (22*24)/ (23*24 + 22*24 + 23*22)=0.3329 for second part
      (23*22)/ (23*24 + 22*24 + 23*22)=0.3191 for third part

      just remeber below any two & u easily get the third(1- sum of any two)

      for first part = 0.348*p
      for second part = 0.3329*p
      for third part = {1-(0.348+0.3329) = 0.3191} i.e. .3191*p

      i hope this is easy compare to above

      Delete
  15. Thnkuuu for making such a wonderful platform .. Plzz share ur knowledge on a daily basis ..thank a lot Prashant ..

    ReplyDelete
  16. You are no less than a Superman.Thank you so much for the tricks.:)

    Keep Shining.God bless you.

    ReplyDelete
  17. how do you calculate this --> (23*24)/ (23*24 + 22*24 + 23*22)? if any shortcuts please share.

    Also,in Q.7 how did you confirm that simple interest formula will be use??

    ReplyDelete
    Replies
    1. If you have problem solving it, then you can memorize 23*24 + 22*24 + 23*22 = 1586
      In Q.7, the direct formula cant be applied because interest is compounded "half-yearly" and our difference formula is for the interest compounded annually.

      Delete
    2. @tanu you can remeber instead of above equation...

      (23*24)/ (23*24 + 22*24 + 23*22)=0.348 for first part
      (22*24)/ (23*24 + 22*24 + 23*22)=0.3329 for second part
      (23*22)/ (23*24 + 22*24 + 23*22)=0.3191 for third part

      just remeber below any two & u easily get the third(1- sum of any two)

      for first part = 0.348*p
      for second part = 0.3329*p
      for third part = {1-(0.348+0.3329) = 0.3191} i.e. .3191*p

      i hope this is easy compare to above

      Delete
    3. Thank you so much for the response.:)

      Delete
  18. Please reply to my query.. I am in a big mess.I need a authentic verification over my doubt as I think you can do it very well.

    First of all,I have marked "yes"
    In the EQ box of AAO Post.. And I am bsc degree holder..
    Later I came to know something about the specified degrees..
    So I am fearing that I would be disqualified for this..

    So my request is please make it clear that what's gonna happen with me at the time of document verification if I clear ssc..

    ReplyDelete
  19. Don't worry, everything will be sorted out after Tier-2. The posts are assigned after document verification, and there you can easily rectify your mistake...

    ReplyDelete
    Replies
    1. Thanks.. #prashantra

      One more thing,as there will be no interview this time,,so what are its positive and negative consequences over the selection of a canditate??

      Delete
  20. This comment has been removed by the author.

    ReplyDelete
  21. Hey, i coudnt figure out how to go ahead in this question.
    The difference between CI and SI on a certain sum at 10% per annum for 4 years is Rs 1282. Find the sum.
    The answer came out to be Rs20,000

    ReplyDelete
    Replies
    1. SI for 4 years @10% = 40P/100 = 0.4P
      CI for 4 years @10% = P(1.1^4 - 1) = 0.4641P
      0.4641P - 0.4P = 1282
      P = 20000

      Delete
  22. You are great...
    Million of good wishes to you sir.

    ReplyDelete
  23. You are great...
    Million of good wishes to you sir.

    ReplyDelete
  24. A person borrows some of money for 5yrs and loan amount : total interest amount is 5:2. The ratio of loan amount : interest rate is equal to?

    ReplyDelete
  25. I havnt taken the pdf print of the application form of cgl..nd now the link has been deactivated..
    Koi dikkt to nhi na hogi is se!!??.please tell me someone..please

    ReplyDelete
    Replies
    1. No problem at all...you should have your roll no. with you, thats it...

      Delete
  26. Compound interest of 2 year is rs 156 and 3 year is rs 254. What is rate of interest?
    Sir how to solve this type of questions. ?

    ReplyDelete
  27. Sir ye ques ka koi simple method nh mil rh ... Plz solve it

    ReplyDelete
    Replies
    1. Ci for 2 yrs 156 and for 3 yrs 254 .. Find rate

      Delete
  28. Interest for 2 years is Rs 156 and interest for 3 years is Rs 254 .find rate ℅

    ReplyDelete
  29. Interest for 2 years is Rs 156 and interest for 3 years is Rs 254 .find rate ℅

    ReplyDelete
  30. Interest for 2 years is Rs 156 and interest for 3 years is Rs 254 .find rate ℅

    Solution-
    Take difference 254-156= 98
    2*7*7 = 98
    [1/bigger prime number in given factors - 1]100
    1/7-1×100
    1/6×100
    100/6 = 50/3% Ans.

    ReplyDelete
  31. C.I of 2 years is rs 156 and 3 years is rs 254. What is rate?
    maaf kijiyga sir isko detail me bhj dijiye solve karke..
    sir is formule se R = [(y/x)^(1/T2 - T1) - 1]*100

    ReplyDelete
  32. Ankit wrong mathod sabhi q. Me apply nhi hota.eg:- 2 years is rs 820 and 3 years is rs 1261. What is rate?

    ReplyDelete