# Aptitude - Compound Interest - Discussion

Discussion Forum : 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:

Answer: Option

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.

Discussion:

135 comments Page 1 of 14.
SICI said:
4 weeks ago

In C. I = where -x is come from? Please explain me.

Joa said:
4 weeks ago

Thanks everyone for explaining the answer.

Saudagar Rokade said:
1 month ago

Diff = PR^²/10000.

1 = P×4^²/10000,

10000 = P×16,

P = 10000/16,

P = 625.

1 = P×4^²/10000,

10000 = P×16,

P = 10000/16,

P = 625.

(1)

Vanhel said:
2 months ago

SI = 8%.

CI = 8.16%.

CI - SI = 1.

8.16% - 8% = 1.

0.16% = 1.

Sum of the difference principal = 1/0.16 * 100 = 625.

CI = 8.16%.

CI - SI = 1.

8.16% - 8% = 1.

0.16% = 1.

Sum of the difference principal = 1/0.16 * 100 = 625.

(2)

P manoj said:
1 year ago

It is simply derived by using this formula => Diff =PR^2/100^2 -> It is only for 2 years difference.

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).

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).

(5)

Naveen said:
2 years ago

1=10000*x/4 * 4.

= 625.

= 625.

Kurdush said:
2 years ago

Shortcut to solve :

(100*100*Diff)/Rate * Rate.

here diff =1.

rate=4.

Substitute in above formula u get Rs.625.

(100*100*Diff)/Rate * Rate.

here diff =1.

rate=4.

Substitute in above formula u get Rs.625.

(3)

Madhav said:
2 years ago

Hi.

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.

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.

(23)

Mohit said:
2 years ago

How 51/625x came? Please explain.

(3)

Sumit said:
3 years ago

Diff = p(R/100) ^n.

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

1=p(4/100) ^2.

=> p=625.

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

1=p(4/100) ^2.

=> p=625.

(2)

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