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A sum of money under compound interest doubles itself in 4 years. In how many years will it become 16 times itself? 

 

Option: 1

12 years


Option: 2

16 years


Option: 3

8 years


Option: 4

None of these


Find the unknown length ×in the following figures 

 

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Posted by

Kavya

A certain loan amounts, under compound interest, compounded annually earns an interest of Rs.1980 in the second year and Rs.2178 in the third year. How much interest did it earn in the first year?

Option: 1

Rs.1600


Option: 2

Rs.1800


Option: 3

 Rs.1900


Option: 4

None of these

 


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Posted by

Ram

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The interest on a certain sum lent at compound interest, the interest being compounded annually, in the 2nd year is Rs.1200. The interest on it in the 3rd year is Rs 1440. Find the rate of interest per annum.

 

Option: 1

10%


Option: 2

15%


Option: 3

20%


Option: 4

 25% 

 


Interest of 2nd year = 1200

Interest of 3rd year = 1440 

Difference = 240 

In compound interest, interest in next year is calculated on previous year 

So, = 240 / 1200 x 100 = 20 % 

Increase in next year interest is what percent of the interest of previous year is = to the rate of interest

 

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Posted by

shivangi.shekhar

If Rs.2000 amounts to Rs.2880 in 2 years at compound interest, what is the rate of interest per annum if the interest is being compounded annually?

Option: 1

10%


Option: 2

20%


Option: 3

15%


Option: 4

25%


P = 2000, A = 2880, T = 2 years, R =?

A = P [1 + R/100] T

2880 = 2000 [1 + R/100]2

2880/2000 = [1+R/100]2

$\sqrt{ }(144 / 100)=1+\mathrm{R} / 100$

1.2 = 100+R/100

1.2 = 100 + R / 100 

R = 20 %

 

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Posted by

Ritika Harsh

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Find the compound interest earned on Rs.20000 for 2 years at 10% p.a. the interest being compounded annually.

Option: 1

Rs.2100


Option: 2

Rs.4200


Option: 3

Rs.6300


Option: 4

Rs.5600 

 


P = 20000

T = 2 years 

R = 10 % 

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

= 2000 [1+ 10/100]2 – P

24200 – 20000

= 4200 

 

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Posted by

Suraj Bhandari

24. Raju took a loan at 8% per annum simple interest for a period of 5 years. At the end of five years he paid Rs.10640 to clear his loan. How much loan did he take?

Option: 1

Rs.8500


Option: 2

Rs.8000


Option: 3

Rs.7700


Option: 4

Rs.7600


Because S.I is always equal in every year , so in 5 years with 8 % rate S.I 40 % of P

Amount is 40 % of P

Amount is $140 \%$ of $\mathrm{P}$ so, $\mathrm{p}$ is $(10640 / 140) \times 100=7600$

 

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Posted by

shivangi.shekhar

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A sum of money invested at simple interest amounts to Rs 2480 at the end of four years and Rs.4080 at the end of eight years. Find the principal.

Option: 1

Rs.2040


Option: 2

Rs.1480


Option: 3

Rs.1240


Option: 4

Rs.880


Amount after 4 years = 2480 Rs.

Amount after 8 years = 4080 Rs.

Difference = 1600 

Because interest is S.I , so it is equal for every year 

P = A – I 

2480 – 1600 = 880

 

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Posted by

sudhir.kumar

14. Find the present value (in Rs.) of Rs.3000 due after 5 years at 10% p.a. simple interest.

Option: 1

1500


Option: 2

1800


Option: 3

2000


Option: 4

2500


A = 3000 

T = 5 years 

R = 10 % 

Interest for 5 years at 10 % rate is equal to the 50 % of P and amount will be 150 % of P

3000 is 150 %

$3000 / 150 \times 100=2000$

i.e $100 \%$ is equal to the $\mathrm{P}$ which is $2000 \mathrm{Rs}$.

 

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Posted by

Sumit Saini

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If Rs.2000 amounts to Rs.2500 in 2 years at simple interest, what is the rate of interest per annum?

Option: 1

8%


Option: 2

37.5%


Option: 3

25%


Option: 4

12.5%


$\begin{aligned} & \mathrm{I}=\mathrm{A}-\mathrm{P} \\ & =2500-2000=500 \\ & 500 / 2000 \times 100=25 \% \\ & \text { Interest for } 2 \text { years }=25 \% \\ & \text { Interest for } 1 \text { years }=25 \% / 2=12.5 \%\end{aligned}$

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Posted by

Shailly goel

If the difference between compound interest at 8% p.a. and simple interest at 13/2 % p.a. on a certain sum of money for 2 years is Rs. 1820, then find the sum.

Option: 1

Rs. 50000


Option: 2

Rs. 40000


Option: 3

Rs.32000


Option: 4

Rs.25000


Difference between C.I and S.I = 1820
= 1820 

 

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Posted by

Rakesh

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