A sum of money under compound interest doubles itself in 4 years. In how many years will it become 16 times itself?
12 years
16 years
8 years
None of these
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View Full Answer(11)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?
Rs.1600
Rs.1800
Rs.1900
None of these
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View Full Answer(3)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.
10%
15%
20%
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
View Full Answer(1)
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?
10%
20%
15%
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 %
View Full Answer(1)
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Find the compound interest earned on Rs.20000 for 2 years at 10% p.a. the interest being compounded annually.
Rs.2100
Rs.4200
Rs.6300
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
View Full Answer(1)
Find the compound interest on Rs. 5000 in 2 years at 4% per annum, the interest being compounded half yearly.
R 412.16
R 312.16
R 400.16
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)$
$=\mathrm{R} 5412.16$
$\therefore Compound Interest = R (5412.16 – 5000) = R 412.16$
View Full Answer(1)
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.
5%
3%
7%
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?
Rs.8500
Rs.8000
Rs.7700
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 $(10640 / 140) \times 100=7600$
View Full Answer(1)
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.
Rs.2040
Rs.1480
Rs.1240
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
View Full Answer(1)
Find the compound interest on Rs. 8000 at 4% per annum for 2 years compounded annually
R 652.80
R 452.80
R 652
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
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