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Arithmetic Aptitude :: Compound Interest

Home > Arithmetic Aptitude > Compound Interest > General Questions

1 / 16

 If the simple interest on a sum of money for 2 years at 5% per annum is Rs. 50, what is the compound interest on the same at the same rate and for the same time?

Answer: Option A

Explanation:

Sum = Rs. (50 * 100)/(5*2) = Rs. 500.

Amount = Rs. [500 * (1 + 5/100)2]

= Rs. (500 * 21/20 * 21/20 )
= Rs. 551.25

C.I. = Rs. (551.25 - 500) = Rs. 51.25

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2 / 16

 What is the difference between the compound interests on Rs. 5000 for 1 1/2 years at 4% per annum compounded yearly and half-yearly?

Answer: Option A

Explanation:

C.I. when interest

compounded yearly=rs.[5000*(1 4/100)(1 1/2*4/100)]



= Rs. 5304.



C.I. when interest is

compounded half-yearly=rs.5000(1 2/100)^3



= Rs. 5306.04

Difference = Rs. (5306.04 - 5304) = Rs. 2.04

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3 / 16

 A sum of money amounts to Rs.6690 after 3 years and to Rs.10035 after 6 years on compound interest. Find the sum.

Answer: Option D

Explanation:

Let's assume the Sum = Rs. P.
P [1 + (R/100)]^3 = 6690 -------- (i)
P [1 + (R/100)]^6 = 10035 ----------(ii)
On dividing, we get [1 + (R/100)]^3 = 10035/6690 = 3/2.
P * (3/2) = 6690 or P = 4460.
Hence, the sum is Rs. 4460.

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4 / 16

 The difference between the compound interest and the simple interest on a certain sum at 12% p.a. for two years is Rs.90. What will be the value of the amount at the end of 3 years?

Answer: Option B

Explanation:

difference-between-the-compound-interest-and-the-simple-interest

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5 / 16

 A sum of money doubles itself at C.I. in 15 years. In how many years will it become eight times?

Answer: Option D

Explanation:

P [1 + (R/100)]^15 = 2P => [1 + (R/100)]^15 = 2 ----------- (i)
Let P [1 + (R/100)]^n = 8P => P [1 + (R/100)]^n = 8 = 2^3
= [{1 + (R/100)}^15]^3.
=> [1 + (R/100)]^n = [1 + (R/100)]^45.
=> n = 45.
So, the required time = 45 years.

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6 / 16

 The principal that amounts to Rs. 4913 in 3 years at 6 1/4 % per annum C.I. compounded annually, is?

Answer: Option B

Explanation:

Formula: P = Amount / (1 + R/100)^n
Principal = [4913 / (1 + 25/(4 * 100))3]
=> 4913 * 16/17 * 16/17 * 16/17 = Rs. 4096

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7 / 16

 Anil invests an amount for 2 years at the rate of 15% per annum at simple interest.Had he invested in a scheme in which interest was compounded yearly he would have got Rs.450 more. Find the principal:

Answer: Option C

Explanation:

Lets assume D amount difference between compound and simple interest for 2 years, D=P*(R/100)^2
where P is the principal
R is rate of interest

so here,
450 = P*(15/100)^2 => P= Rs. 20000

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8 / 16

 If the compound interest (compounded Yearly) on a certain sum for 2 years at 3% is Rs.101.50 then what will be the corresponding simple interest?

Answer: Option B

Explanation:

Given, years = 2
3% of 50= (50*(3/100))=Rs.1.5
Simple Interest =50+50=100
Compound Interest=50+50+1.5=101.5

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9 / 16

 If the compound interest on a sum of Rs.5000 at the rate of 10% per annum is Rs.1050, then time period is (Interest compounded yearly):

Answer: Option D

Explanation:

C.I of 5000 at 10% interest for x years is 6050 = Rs. 1050
6050 = 5000 ( 1 + 10/100)^n => 6050/5000 = (11/10)^n
=> 605/500 = (11/10)^n => n = 2 years.

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10 / 16

 At what rate of interest (compounded yearly)will Rs. 10,000 amount to Rs. 12,100 in 2 years?

Answer: Option D

Explanation:

Amount = P(1 + R/100)^n
Where Principal = P, Rate = R% per annum, Time = n years.
12100 = 10000 (1+R/100)^2
=> 121/100 = (1+R/100)^2
=> 11/10 = (1+R/100)=> 11/10 = (100 + R)/100
=> 110 = 100 + R => R = 10%

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