On what Principal The amount will becomes Rs 676 in 2 years at the rate 4% pa interest compounded annually?

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:

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

Can anyone help me out , why are we using -x in the step one..

Please help me out
Thanks in advance...

P = 25*25 = 625

p=amount
x=rate

Why are we using 676/625x in the step one?

Please help me.

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

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.

Difference = (D*R^2)/100^2 for 2 year

Difference = {(D*R^2)(300+R)}/100^3 for 3 year.

Rightly said, cleared because of you.

sum = difference*(100/R)^2 for 2 year

sum = {difference*(100^3)}/{R^2*(300+R)} for 3 year

NOTE: (THIS FORMULA IS FOR TWO YEARS ONLY).

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

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;

Principal x (r%/100)^2 = Difference between interests.

THIS FORMULA IS FOR TWO YEARS ONLY.

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.

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.

SI-CI = P*square of (R/100).

Sum = 1(100/4)^2 = 625.

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.

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.

Please tell me how CI rate for 2 years = 8.16%.

Please help me.

From the above step how did you got that P(1/625) = 1.

Can you please explain me?

= (100/4)^2*1 = 625.

p = d*100^2/r^2 (only for 2 yrs).

Where p = principle.
d = difference.
r = rate.

C.I = p[(1+r/100)^n -1].

A = P(1+r/100)n.

CI = P(1+r/100)n-P.

Where in above problem P = x.

[x(1+4/100)2]-x? where and all we can use this?

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.

=> P*(4/100)^2 = 1.

=> P = 625.

Three years = pr2(300+r)/100*100*100.

Can anyone solve this problem.

p.(R/100)^2 = (compound interest - Simple interest) time duration 2 years.

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.

The difference after 2 years = PR^2/100^2.

Here, P - Principal, R - Interest of Rate.

Guys please help me.

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.

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).

4% of x = 0.04x,
4% of 0.04x = (0.04x)4/100 = 1,
x = 625.

(C.I - S.I) = P * R * R/100 * 100.
HERE, 1 = P * 4 * 4/100 * 100.

P = 25 * 25 = 625.

Then, CI=8.16 & SI=8.00,
1÷ 0.16 * 100 = 625.

=>. 1 = p(4/100)square.
1= p(1/25)square,
1= p(1/625),
P = 625.

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.

Clear explanation Thanks.

Principal x (r%/100)^2 = Difference between interests.

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!

Please give me the answer.

Can anyone solve this?

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.

Pr^2 (300+r)/100^3 when the time is given 3yr.

4%..................... (4%)^2 = 8.16.

=>8.16x-8x=0.16%x,
=>0.16%x=1

by solving
You will get x=625.

Am I correct?

Formula :- Sum= Difference(100/r)^2.
Now put the value----- Sum= 1(100/4) ^2
= 1* 625/1
= 625 Ans.

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.

P=(100/R)^t *( C.I - S.I ).

P=[(difference)*(100)^2]/(rate)^2.

and 51x/625-2x/25=1 so x=625 how?

1 * 100 * 100/16 = 625.

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.

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.

Ci;
p=100.
r=4%pa,
t=2 yr.

And; i = 8.16.
0.16 = 100,
1= 100/0.16,
= 625.

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!

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.

C.I = Amount-Principal.
= {X[1+4/100]^2-X}.

S.I = [ X(4)(2)/100].

p = 1 (100/4)^2.
= 1(100 x 100/4 x 4).
= 25 x 25.
= 625.

By using this formula you can directly find the amount i.e. p.

1=p(4/100) ^2.
=> p=625.

Before solving this question you have to be aware on S.I and C.I formula's.

S.I = PTR/100.

Where as AMOUNT = P (1+ R/100) ^T ====> IN C.I.

AMOUNT = P + C.I.

C.I = P (1+ R/100) ^T - P.

In the question, they give the difference between S.I AND C.I is rupee 1.

First Calculate S.I.

Let the PRINCIPAL or SUM be ==> '' P '', TIME = 2, RATE = 4%.

S.I = PTR/100 ==> P*2*4/100 ==> 2P/25.

Now C.I = P (1+R/100) - P ==> P (1+4/100) - P ==> 51P/625.
51P/625 - 2P/25 = 1.
LCM IS 625.
51P - 50P/625 = 1.
P = 625.

I hope this will be helpful.

Thank you.

(100*100*Diff)/Rate * Rate.
here diff =1.
rate=4.

Substitute in above formula u get Rs.625.

Given -> Diff=1 , R =4, Tp find P?

By Formula ,
Diff = PR^2/100^2.
1 = P * 4 * 4/100 * 100.
1= P * 1/20 * 1/20.

Note : (Also for three years difference use this formula -> Diff = PR^2(300 + R)/100^3).