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#### 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.1600Option: 2 Rs.1800Option: 3  Rs.1900Option: 4 None of these

xyz pqr abc year tt

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• Faculty Support #### 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

#### 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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#### Find the compound interest earned on Rs.20000 for 2 years at 10% p.a. the interest being compounded annually.Option: 1 Rs.2100Option: 2 Rs.4200Option: 3 Rs.6300Option: 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

#### Find the compound interest on Rs. 5000 in 2 years at 4% per annum, the interest being compounded half yearly.Option: 1 R 412.16Option: 2 R 312.16Option: 3 R 400.16Option: 4 R 420.16

Here, Principal P = Rs. 5000
Rate R = 4% pa
Time n = 2 years
Now according to the formula,
$\text { Amount } =\mathrm{P}\left(1+\frac{R}{2 \times 100}\right)^{2 n}=5000\left(1+\frac{4}{200}\right)^4$
$=\left(5000 \times \frac{51}{50} \times \frac{51}{50} \times \frac{51}{50} \times \frac{51}{50}\right)=\left(\frac{51 \times 51 \times 51 \times 51}{1250}\right)$
$=\mathrm{R} 5412.16$
$\therefore Compound Interest = R (5412.16 – 5000) = R 412.16$

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#### Ruchi invested Rs. 1600 at the rate of compound interest for 2 years. She got Rs. 1764 after the specified period. Find the rate of interest.Option: 1 5%Option: 2 3%Option: 3 7%Option: 4 10%

Here, $\mathrm{P}=\mathrm{R} 1600, n=2$ years, $\mathrm{A}=\mathrm{R} 1764$
Now, according to the formula,
$\text { Amount }=\mathrm{P}\left(1+\frac{R}{100}\right)^n$
$1764=1600\left(1+\frac{R}{100}\right)^2$
$\Rightarrow \quad \frac{1764}{1600}=\left(\frac{100+R}{100}\right)^2 \quad \Rightarrow \quad\left(\frac{21}{20}\right)^2=\left(\frac{100+R}{100}\right)^2$
$\Rightarrow \quad \frac{100+R}{100}=\frac{21}{20} \quad \Rightarrow \quad 100+\mathrm{R}=\frac{21}{20} \times 100$
$\Rightarrow 100+R = 105$
$\therefore R = 105-100 = 5%$

#### 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.8500Option: 2 Rs.8000Option: 3 Rs.7700Option: 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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• Faculty Support #### 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.2040Option: 2 Rs.1480Option: 3 Rs.1240Option: 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

#### Find the compound interest on Rs. 8000 at 4% per annum for 2 years compounded annuallyOption: 1  R 652.80Option: 2 R 452.80Option: 3 R 652Option: 4 R 552.80

Here, P = R 8000, R = 4%, Time = 2 years
Now, according to the formula,
$\text { Amount }=\mathrm{P}\left(1+\frac{R}{100}\right)^n=8000\left(1+\frac{4}{100}\right)^2=8000 \times \frac{26}{25} \times \frac{26}{25}=\mathrm{R} 8652.80$
∴ CI = R (8652.80 – 8000) = R 652.80 