# Aptitude - Compound Interest - Discussion

### Discussion :: Compound Interest - General Questions (Q.No.2)

2.

The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. The sum (in Rs.) is:

 [A]. 625 [B]. 630 [C]. 640 [D]. 650

Answer: Option A

Explanation:

Let the sum be Rs. x. Then,

 C.I. = x 1 + 4 2 - x = 676 x - x = 51 x. 100 625 625

 S.I. = x x 4 x 2 = 2x . 100 25 51x - 2x = 1 625 25 x = 625.

 Goutham said: (Jul 14, 2010) Hai I am Goutham. Can anyone help me out , why are we using -x in the step one.. Please help me out Thanks in advance...

 Xyz said: (Jul 18, 2010) CI = interest- sum............ observe problem 1 . CI= 3321-3200

 Gutti said: (Aug 1, 2010) You have any shortcut method to this.

 Name said: (Sep 17, 2010) P x 4/100 x 4/100 = 1 P = 25*25 = 625

 Priyanka P. said: (Dec 2, 2010) Its p*(x/100)^2 = ans. p=amount x=rate

 Prudhvi said: (Dec 22, 2010) Hai I'm prudhvi. Can any one help me out? Why are we using 676/625x in the step one? Please help me.

 Krishna said: (Dec 24, 2010) Hello friends! The question is very easy . First know the diff b/w S.I. and C.I. S.I. = P*T*R/100 Amount = S.I. + P C.I. = Amount - P Amount = P*((1 +(R/100))^ n) where (n= no of years) Given that S.I. and C.I diff is 1rupee and also given T=time =2 yrs , R=rate = 4%, and he asked P ie Principal Amount. So S.I. = P*2*4/100; = P*2/25; C.I. = Amount - P = P*(1+ (4/100))^2 - P = P*(1 + (1/25))*2 - P = P*((26/25)^2) - p = P*(676/625) - p = (P*676-P*625)/625 = (P*51)/625 WE KNOW S.I. = (P*2)/25 AND S.I.-C.I. = 1 HENCE (P*2)/25 - (P*51)/625 = 1 BY SOLVING U GET P = 625 RS

 Vivek said: (Feb 3, 2011) How to solve very simple?

 Kusum said: (Feb 12, 2011) Please any one tel me how to solve it in simple method.

 Viji said: (Jul 4, 2011) shotcut method: If you give the difference between si and ci for 2 years means, 4*4/100 =.16 1/.16*100 100/16*100 100/4*25 625

 Sudharsan said: (Sep 18, 2011) For beginners those are asking why they used -x in step one. Let me explain you clearly. First take Amount=Rs.5000 and Rate=2% Years=1 Then calculate SI, SI=PNR/100 So, SI=5000*1*2/100=RS100 [Here you are getting only interest] Now calculate CI, CI=P(1+r)^n So, CI=5000*(1+2/100)^1=RS5100 [Here we are getting interest+Principal amount] That's why there are using -x in the formula. If you use -x(for our example its P) we will get the interest only.

 Vishal Kumar said: (Sep 27, 2011) Shortcut method Difference = (D*R^2)/100^2 for 2 year Difference = {(D*R^2)(300+R)}/100^3 for 3 year.

 Siva said: (Oct 17, 2011) @sudharshan. Rightly said, cleared because of you.

 Atanu said: (Dec 10, 2011) Shortcut Method sum = difference*(100/R)^2 for 2 year sum = {difference*(100^3)}/{R^2*(300+R)} for 3 year

 Rajeev said: (Dec 10, 2011) A=P( 1+R/100) C.I=A-P DIFFERENCE OF CI AND S.I FOR X YEAR= P*(R/100)^X

 Sailaja said: (May 15, 2012) CI-SI=1, T=2 R=4 P=? SHORTCUT IS CI-SI=P(R/1OO)^2 1=P(4/100)^2 P=625

 Priya said: (Jun 21, 2012) C.I-S.I=P(R/100)^2 IS THIS FORMULA CORRECT?

 Siri said: (Aug 3, 2012) Shortcut method: difference= sum (rate/100) ^2.

 Akshay said: (Aug 29, 2012) The correct formula is (ci-si) (100/r) (100/r) = p. NOTE: (THIS FORMULA IS FOR TWO YEARS ONLY).

 Pankaj Parashar said: (Dec 21, 2012) Short cut method: si rate for 2 years=8% ci rate for 2 years=8.16% diff=.16% ie is equal to re 1 so 1/.16*100=625

 Suresh said: (Jan 7, 2013) Please tell me anyone why are subtract X from first step?

 Ankit Gupta said: (Feb 19, 2013) Apply formula : CI-SI= p(r/100)^2 this formula works only for t=2 years; CI-SI = p[(r/100)^3+3(r/100)^2] for t=3 years;

 Srikant Duppada said: (May 3, 2013) FORMULA: Principal x (r%/100)^2 = Difference between interests.

 Manu said: (Sep 29, 2013) The question is to find the sum of CI & SI. The answer we got is the principle, not the sum. Please correct if I am Wrong.

 V Panindraa said: (Nov 16, 2013) Sum = (10,000 * difference) / r*r. THIS FORMULA IS FOR TWO YEARS ONLY.

 Tamanna said: (Dec 14, 2013) How we got 51/625? can any one explain ?

 Dew said: (Mar 15, 2014) @Tamana. Hey its simple sum just Few steps, Step 1: Compound interest = p(1+R/100)^n - p. [Here p is subtracted from total amount at end of annum to find out compound interest]. Step 2: Simple interest= pRn/100. So, Compound interest - simple interest = 1 Rs. p(1+R/100)^n - p - pRn/100 = 1. p(1+4/100)^2 - p - p*4*2/100 = 1. p(1.04)^2 - p - p*.08 = 1. p(1/625) = 1. So p= 625 Answer.

 Ravi said: (Mar 25, 2014) Let's 100. Simple interest = 100*2*4/100 = 8. Total amount = 108. Now find CI = 100*4/100 = 4. 104*4/100 = 4.16. total = 108.16. SI - CI = 108-108.16 = .16 1*100/.16 = 625.

 Manas Biswal said: (Jun 15, 2014) (C.I)2 - (S.I)2 = 1. SUCCESSIVE DISCOUNT(C.I) = 4+4(4*4/100) = 8.16. (S.I) 4+4 = 8. 8.16-8 = 0.16 . 0.16/1*100 = 625.

 Chetan Siyota said: (Jun 18, 2014) Short trick if you have given difference between simple and compound interest, SI-CI = P*square of (R/100).

 Pratik Patel said: (Jul 29, 2014) But. It is for only two years. I think. Not for 4;5;6 years. Try it. @Chetan.

 Subhani said: (Sep 9, 2014) Difference D = P(R*R)/10000.

 Siva Nandi Reddy said: (Sep 18, 2014) For 3 years p=(d*100*100*100)/(r*r(r+300)).

 Amit Kumar said: (Sep 24, 2014) Sum = difference(100/rate)^t (here t=2yrs). Sum = 1(100/4)^2 = 625.

 Tarakeswararao said: (Oct 16, 2014) Hello friend this is Tarak. Now we have formula for finding this type of problems. i.e D = P*(R/100)^2 if the time is 2 years. If time is 3 years now the formula is D = P*(R/100)^2*(3+(R/100)). Where D = Difference. P = Principle amount. R = Interest.

 An Aspirant said: (Nov 12, 2014) Sum = Difference*(100) square/(r)square.

 Prabhjot Singh said: (Apr 3, 2015) Just equate: CI-SI = P(R/100)(R/100) for 2 years. CI-SI = P(R/100)(R/100)(300+R/100) for 3 years. By these formula's it is easier to calculate the result.

 Raja said: (Apr 6, 2015) What is the shortcut formula to find differences of ci and si for 4 years and 5 years?

 Siva said: (Jun 4, 2015) Difference between 2 years = pr^2/100^2.

 Ganga said: (Jun 17, 2015) @Pankaj parashar. Please tell me how CI rate for 2 years = 8.16%. Please help me.

 Aparna said: (Jul 3, 2015) P(1+0.04)^2-P)-P*0.08. From the above step how did you got that P(1/625) = 1. Can you please explain me?

 Hema said: (Jul 17, 2015) Shortcut formula P = (100/R)^T*D, D = Difference here. = (100/4)^2*1 = 625.

 Bhupathi Raju said: (Aug 4, 2015) Any easiest method solve this problem?

 Monalisa said: (Aug 20, 2015) Hey guys, I have a shortcut formula for such problem: p = d*100^2/r^2 (only for 2 yrs). Where p = principle. d = difference. r = rate.

 Rick said: (Sep 5, 2015) You people should add the formula of C.I in the formula section which is missing. C.I = p[(1+r/100)^n -1].

 Eshwar said: (Sep 15, 2015) CI = A-P. A = P(1+r/100)n. CI = P(1+r/100)n-P. Where in above problem P = x.

 \$Weth@ said: (Sep 28, 2015) Is that a formula you used over there? [x(1+4/100)2]-x? where and all we can use this?

 Manaswini said: (Oct 17, 2015) I can't understand this please some one help me to understand this sum.

 Salman said: (Nov 10, 2015) Actually the question is wrong: It should be difference between compound and simple interest. Since compound interest will always be larger than simple interest given the same rate and period.

 Sagy said: (Nov 28, 2015) I did not understand this sum please help give me a simple way to solve this sum.

 Santhosh P said: (Dec 26, 2015) Difference for 2 years = P*(R/100)^2. => P*(4/100)^2 = 1. => P = 625.

 Anu said: (Jan 28, 2016) Krishna please solve last equation also.

 Priya said: (Feb 19, 2016) Should we take x as principle? Or we can take any other variable like P?

 Princess said: (Mar 12, 2016) Can anyone tell me how can we find the sum and rate when compound interest for two successive years are are given?

 Amit said: (Mar 16, 2016) Principal for two years = pr2/100*100. Three years = pr2(300+r)/100*100*100.

 Jyoti said: (Mar 18, 2016) P = D * 100^2/R^2 = 1*100^2/16 = 625.

 Ramesh said: (May 26, 2016) P = D[100/r] * 2. P = 1[100 * 100/4 * 4]. P = 625.

 Rishi said: (Jun 23, 2016) I guess X must be the principal but not the sum.

 Gagan said: (Jul 15, 2016) If the compound interest is 104 and simple interest is 100 for the time of 2 years find the rate of interest. Can anyone solve this problem.

 Raj said: (Aug 8, 2016) Why not R/2?

 Chinmoy Mondal said: (Aug 11, 2016) Very simple formula is; p.(R/100)^2 = (compound interest - Simple interest) time duration 2 years.

 Yeshwanth said: (Aug 19, 2016) @Gagan SI = P * (n * r)/100. CI = P((1 + r/100)^n - 1). Now simplify both equations such that P comes to the left and rest to the right equate both to get r.

 Prasanna Karthik said: (Sep 5, 2016) We can use the direct formula for this one. The difference after 2 years = PR^2/100^2. Here, P - Principal, R - Interest of Rate.

 Mamta Srivastava said: (Sep 14, 2016) How we derive the shortcut trick for 2 or 3 years for the sum of money if the difference given b/w CI and SI? Anyone help me to clear these shortcut to improve myself. Please.

 Ankit said: (Sep 20, 2016) CI for 2 years - SI for 2 years = PR^2/100^2. 1 = P16/10000, 10000/16 = P, P = 625.

 Ankit said: (Sep 20, 2016) Solution for CI on certain amount for 2 years is 2200 and 3 years is 3640. Find the rate? Guys please help me.

 Rajat said: (Sep 26, 2016) Can anybody tell me the formula for finding rate and principle when compound interest and simple interest are given with time 2 years and SI = 800 and CI = 850?

 Dhairya Adhikari said: (Oct 1, 2016) (SI -CI) for 2 year = (r/100)^2 * principal.

 Muthuu said: (Oct 23, 2016) Take as, C.I - S.I = Rs.1. Then, for C.I ->use the Formula[x + y + {(xy)/100}]. For x&y ->put R% value( i.e.,) X=4 &y=4. By this ,we get [4+4+{4*4/100}] = 8.15. For S.I -> R * T = 4 * 2 = 8. So , 8.15 - 8 = RS.1 => 0.16=1. By this, 0.16% = 1 means,then 100%=? So, cross multiply it i.e., 100 * 1/0.16 = 625. I am just trying to explain briefly, hence the sum looks like long . But you all just write the numerical step and look it . It just takes few secs only try it.

 Shankar said: (Oct 25, 2016) SIMPLE STEP. FORMULA : Diff b/w SI and CI when T = 2yrs. P = (D * 100^2)/R^2. D =RS 1. R = 4%. SO, Answer is 625. FOR DIFF B/W CI AND SI FOR 3 YRS FORMULA P = (D * 100^3)/R^2(300 + R).

 Fiza Khan said: (Nov 4, 2016) Can anyone please explain to me how we get 51/625 x in the first answer?

 Krunal Bhut said: (Dec 7, 2016) Interest of interest of amount is difference between SI and CI at rate given, 4% of x = 0.04x, 4% of 0.04x = (0.04x)4/100 = 1, x = 625.

 Anupam Das said: (Dec 14, 2016) We know that, (C.I - S.I) = P * R * R/100 * 100. HERE, 1 = P * 4 * 4/100 * 100. P = 25 * 25 = 625.

 Lithin said: (Dec 28, 2016) Well said, Thank you @Muthuu.

 Revant Borawat said: (Dec 29, 2016) Assume principal 100. Then, CI=8.16 & SI=8.00, 1÷ 0.16 * 100 = 625.

 Pankaj said: (Dec 29, 2016) Hi, one thing I could not understand. In question, it is given that the difference between simple and compound. Not difference between compound and simple. These two questions are distinct in the first, the answer will be in minus. So how can it possible? So please help me please.

 Arun Raj said: (Jan 4, 2017) CI - SI = P(r/100)square P = principal, R = rate = 4. CI - SI = 1. =>. 1 = p(4/100)square. 1= p(1/25)square, 1= p(1/625), P = 625.

 Prasanna Kumar said: (Jan 23, 2017) SI for 2 yrs = 4% +4%. i.e * 8%. And CI for 2 yrs = 4+4+(4*4/100) i.e 8.16% . Then take the options, Given difference is ci - si = 1. 8.16% (625) - 8% (625). ie, 51- 50 = 1.

 Mahan Singh said: (Jan 25, 2017) Its very simple we put -x because the question is asking the difference between Interest but in compound interest formula we also put the principal value together to subtract that we will - x because x is principal value. We don't need principal value only we need interest difference.

 Malinimadhavan said: (Feb 5, 2017) S.I = PNR/100. =(PX2X4)/100, =(PX2X1)/25. THEREFORE S.I = PX2/25 = 0.008P, C.I = P(1+R/100)^N -P, = P(1+4/100)^2 -P, =P(1+1/25)^2 - P, = P(25+1/25)^2 -P, = P(26/25)^2-P, = P(1.04)^2-P, = P(1.0816)-P, =1.0816P-P. C.I =.0816P. DIFFERENCE: CI-SI = 1 0.0816P - 0.08P =1, 0.0016P = 1, P = 1/0.0016 = 625. THEREFORE, SUM IN RUPEES IS 625Rs.

 Karthick said: (Feb 14, 2017) @Sudharsan. Clear explanation Thanks.

 Shivesh said: (Feb 14, 2017) Derivation of this formula. Principal x (r%/100)^2 = Difference between interests.

 Nilesh said: (Mar 9, 2017) How 51/625 comes? Please explain that.

 Prasad Mandati said: (Mar 10, 2017) Note: If, P= Principal sum x= R/100. Now, The difference between Compound and simple interests when compounded annually for two years is Px^2. Similarly if the same is done for 3 years the difference would be P(x^3 + 3x^2). Hope it helps you!

 Yashika said: (Mar 14, 2017) The SI = CI when the principal and rate are the same and time is ________. Please give me the answer.

 Remya said: (May 11, 2017) Anu deposit some money in a bank. After 6 years he got 6500rs, then after 2 years, she got 7800 rs. The how much amount will she get after 10 years? Can anyone solve this?

 Chandan said: (May 12, 2017) Here, sum= (difference * 10000)/rate squre.

 Aditya Verma said: (Jul 6, 2017) Short trick of this sum. If the sum says diff. Of 2 years then apply:: P*R*R/100*100. If the sum says diff. For 3 years then apply:: P*R*R (300+R)/100*100*100.

 Hiten said: (Jul 9, 2017) Diff = pri*(r/100)sqr. 1 = x*(4/100)sqr, 1 = x*(1/25)sqr, 1 = x/625, x=625.

 Hitesh said: (Jul 23, 2017) Why are we subtracting -x in the first part?

 Ritu said: (Jul 26, 2017) Please, someone explains the question. At what % per annum will 625 will amount to 676 in 2 yr at c.i.

 Prince said: (Jul 30, 2017) Easy question try pr^2/100^2 = difference when the time is given 2. Pr^2 (300+r)/100^3 when the time is given 3yr.

 Ravi said: (Sep 9, 2017) x (100/r)^2 = 1(100/4)^2.

 Elisha said: (Sep 15, 2017) Guys you have to take the square of the given percentage. 4%..................... (4%)^2 = 8.16. =>8.16x-8x=0.16%x, =>0.16%x=1 by solving You will get x=625.

 Sneha said: (Sep 20, 2017) Difference formula for 2 years____. Sum = Difference * (100) ^/Rate^ Then, Sum = 1*100*100 / 4*4 =625.

 Revathi said: (Sep 25, 2017) CI-si=p(r/100)^2 this formula applicable for more than 2 yrs also. Am I correct?

 Bunny_Xavi said: (Oct 3, 2017) (R/100)^2= D/P D=difference bw si and ci. 100^2/4*4 = 1/P. P= 10000/16 =>625.

 Md.Iftekhar said: (Oct 8, 2017) Hi, friends we can easily solve this problem with the help of this formula. Formula :- Sum= Difference(100/r)^2. Now put the value----- Sum= 1(100/4) ^2 = 1* 625/1 = 625 Ans.

 Baadshah said: (Nov 1, 2017) Shortcut method: When the difference between SI and CI is x and rate of interest is R then, x(100/R)^n where n is the number of years, So 1(100/4)^2 = (25)^2 = 625.

 Ahana said: (Nov 16, 2017) Is there any formula for diff bw ci and si for 4 years ?

 Vansh Agre said: (Nov 18, 2017) Difference = P * R *R /100*100.

 Shanti Mishra said: (Dec 21, 2017) 1/.16*100=625.

 Arun said: (Dec 29, 2017) 8.16-8=1; 0.16=1, 100=x, x = 625.

 Akshay said: (Jan 2, 2018) You can simple try this formula. P=(100/R)^t *( C.I - S.I ).

 Jojo said: (Feb 3, 2018) For this use equation. P=[(difference)*(100)^2]/(rate)^2.

 Puneet said: (Mar 9, 2018) Anyone can please tell why we have to subtract x from C.I? and 51x/625-2x/25=1 so x=625 how?

 Komal Bishnoi said: (May 4, 2018) D*100^2/r^2 = 1*100*100/4*4 = 10000/16 = 625.

 Rahul said: (Oct 22, 2018) It is 1 * 100^2/4^2 = 625.

 Narendra Singh said: (Dec 28, 2018) DIFF * 100 * 100/R * R. 1 * 100 * 100/16 = 625.

 Ashok said: (Jan 26, 2019) Here is a formula for diff B/w C.I & S.I. Diff between means C.I - S.I = P(r/100)^2. 1 = P(4/100)^2. 1 = P(1/25)^2, 1 =P(1/625), P = 625.

 Sree said: (Apr 16, 2019) Here the formula: P=D^2*100^2÷r^2: This formula applicable for only the difference between 2 yrs. Where; D=1. R=4%. 1^2*100^2÷4^2, = 1*10000÷16, = 625.

 Mjr said: (Apr 21, 2019) Si; p = 100. r = 4%pa, t = 2yr, i = 8, . Ci; p=100. r=4%pa, t=2 yr. And; i = 8.16. 0.16 = 100, 1= 100/0.16, = 625.

 Tarun.P said: (Apr 24, 2019) We have used -x because compound interest formula is Amount - principle. And the formula for Amount is; Amount=P (1+r/100) ^n. First, we will calculate the amount later the total amount can be subtracted by the principle values!

 Nitin said: (Jul 6, 2019) Difference =p*(r/100)^2, 1=p*(4/100)^2, 10000/16=P, 625=P.

 Mukesh Vijey said: (Jul 19, 2019) Where p=x, n=2, r-2. c.i = (p(1+r/100)^n-1). = (x+r/100)^n-x) (//multiply x inside) = (x+2/100)^2-x)(//substitute n and r value) = (676/625)x+x. c.i = (51/625)x. SI = pnr/100. = xnr/100(//p=x). = (x*2*4)/100. SI = 2x/25. Given that , s.i - c.i =1. Therefore, 2x/25-(51x-625)=1. x = 625. Answer is 625.

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