**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%**

**With this formula you can find Rate if Time is given and Time if rate is given.**

__Note__:

**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.

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-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**

**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**

**T**

**R = (P**P_{1 - }_{2)*100/}P_{2}

_{1 - }**P**T

_{1}

_{2}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=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**

A = P(1+r/100)^t

Given, A=1.44P t = 2 years

1.44P = P(1 + r/100)^2
r = 20%

**Answer : (D)**

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 = 244P = Rs. 32000

**Answer : (C)**

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)**

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)**

2.5 = P*R*2/100 - P*r*2/100

2.5 = 10R - 10r

R - r = 0.25

2.5 = 10R - 10r

R - r = 0.25

**Answer : (D)**

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**

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)**

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)**

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)**

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

ReplyDeleteT = 1.5 years and R = 10%

Deletewhen 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

Sir we can use fromula for 3 yrs here

DeleteThnkyu soo mch sir... :)

ReplyDeleteSir 9th qtn m 378 kese aea pls explain ... ? ? ?

ReplyDeleteCI for 1st year = r% of P

DeleteCI 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

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

ReplyDeleteSir, 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?

ReplyDeleteThanks is advance sir, ur blog is really very helpful for us.

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

DeleteThanks sir

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

Delete(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

Thanks sir..

ReplyDeleteThanks sir..

ReplyDeleteSir, 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?

ReplyDeleteThanks is advance sir, ur blog is really very helpful for us.

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 ?

ReplyDeletePls tell how to solve

When interest is compounded half-yearly

DeleteA = 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...

Ques 15 3/20 kaise aya sir

ReplyDelete(1587/1200)^1/2 = (529/400)^1/2 = 23/20

DeleteKhud se bhi pen chalao bhai :P

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.....:)

Deletehttps://www.dropbox.com/s/m0mq8m8zantz91t/Pic.jpg?dl=0

DeleteThe above image answers your second query :)

For the first one - last year's cutoff for MEA was 540...

Thankyou sir

ReplyDeleteSir out of 600 540 was cuttoff for MEA?

ReplyDeleteOut of 700 (100 marks for interview)

DeleteSir,in very last ques number 15,time t2 will be 3 or 2?

ReplyDeleteT2 is associated with y, hence it is 3 years...

DeleteThanx sir

ReplyDeleteThanx sir

ReplyDeletedear 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

ReplyDeletedear 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

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

ReplyDeleteCI for second conversion period - CI for first conversion period = SI on the interest of the first converSion period

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

ReplyDeleteHi, 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@sohan singh you can remeber instead of above equation...

Delete(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

here p=7930

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

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

ReplyDeleteKeep Shining.God bless you.

Batman...I would prefer Batman :D

DeleteHa ha ha...ok Batman.

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

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

If you have problem solving it, then you can memorize 23*24 + 22*24 + 23*22 = 1586

DeleteIn Q.7, the direct formula cant be applied because interest is compounded "half-yearly" and our difference formula is for the interest compounded annually.

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

Delete(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

Thank you so much for the response.:)

DeletePlease 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.

ReplyDeleteFirst 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..

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...

ReplyDeleteThanks.. #prashantra

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

This comment has been removed by the author.

ReplyDeleteHey, i coudnt figure out how to go ahead in this question.

ReplyDeleteThe 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

SI for 4 years @10% = 40P/100 = 0.4P

DeleteCI for 4 years @10% = P(1.1^4 - 1) = 0.4641P

0.4641P - 0.4P = 1282

P = 20000

You are great...

ReplyDeleteMillion of good wishes to you sir.

You are great...

ReplyDeleteMillion of good wishes to you sir.

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?

ReplyDeleteI havnt taken the pdf print of the application form of cgl..nd now the link has been deactivated..

ReplyDeleteKoi dikkt to nhi na hogi is se!!??.please tell me someone..please

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

DeleteCompound interest of 2 year is rs 156 and 3 year is rs 254. What is rate of interest?

ReplyDeleteSir how to solve this type of questions. ?

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

ReplyDeleteCi for 2 yrs 156 and for 3 yrs 254 .. Find rate

DeleteInterest for 2 years is Rs 156 and interest for 3 years is Rs 254 .find rate ℅

ReplyDeleteInterest for 2 years is Rs 156 and interest for 3 years is Rs 254 .find rate ℅

ReplyDeleteInterest for 2 years is Rs 156 and interest for 3 years is Rs 254 .find rate ℅

ReplyDeleteSolution-

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.

C.I of 2 years is rs 156 and 3 years is rs 254. What is rate?

ReplyDeletemaaf kijiyga sir isko detail me bhj dijiye solve karke..

sir is formule se R = [(y/x)^(1/T2 - T1) - 1]*100

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