Quantitative Aptitude IBPS,SSC,GRE,CAT,CSAT,RRB,SBI POs,MPSC,rpsc

[latexpage]

Example: 1 At what rate percentage compound interest does a sum become nine-fold in two years ?
Example: 2 At what rate percentage (compound interest) will a sum of money become eight times in three years?
Example: 3 At what rate per cent compounded yearly will Rs. 80000 amount to Rs. 88200 in 2 years?
Example: 4 The difference between compound interest and simple interest on an amount of Rs. 15000 for 2 years in Rs. 96. What is the rate of interest per annum ?
Example: 5 At what rate of compound interest per annum will a sum of Rs. 1500 become Rs. 1749.60 in 2 years ?
Example: 6 Determine the effective annual rate of interest corresponding to a nominal rate of 6% per annum payable half-yearly.
Example: 7 In what time will Rs. 30000 amount to Rs. 34347 at 7% compound interest.
Example: 8 In how many years will a sum of Rs. 1600 at 10% per annum compound interest, compounded half yearly become Rs. 1944.81 ?
Example: 9 In what time will Rs. 390625 amount to Rs. 456976 at 4% compound interest ?
Example: 10 A sum placed at compound interest doubles itself in 4 years. In how many years will it amount to eight times itself ?.
Example: 11 Find the least number of complete years in which a sum of money at 20% p.a. Compound interest will be more than double.
Example: 12 A sum of money at compound interest amounts to four times itself in 4 years. In how many years will it be 16 times of itself?
Solution-1

Let the sum be Rs. x and rate be r% per annum,then

$9x=x*(1+\frac{r}{100})^2$

$9=(1+\frac{r}{100})^2$

$3=(1+\frac{r}{100})$

$\frac{r}{100}=2$

r=200%

If a certain sum becomes m times in t years,rate of compound interest r is equal to $100[(m)^1/t-1]$

Solution-2

Rate%=$100[(8)^1/3-1]$

=$[2-1]x100=100%$

Solution-3

We have

$80000*(1+\frac{r}{100})^2=88200$

$(1+\frac{r}{100})^2=\frac{88200}{80000}=\frac{441}{400}=(\frac{21}{20})^2$

$(1+\frac{r}{100})=\frac{21}{20}$

$r=\frac{100}{20}=5%$

Solution-4

$[15000*(1+\frac{R}{100})^2-15000]==\frac{15000*R*2}{100}=96$

$15000*[(1+\frac{R}{100})^2-1-\frac{R}{50}=96$

$15000*[1+(\frac{R}{100})^2+\frac{2R}{100}-1-\frac{R}{50}=96$

$15000*(\frac{R}{100})^2=85$

$R^2=\frac{96*100^2}{15000}$

$R=\sqrt{\frac{96*10^4}{15*10^3}}$

=$\sqrt{\frac{960}{15}}=\sqrt{64}=8$

Solution-5

Let x% be the rate of compound interest per annum

$1500*[1+\frac{x}{100}]^2=1749.60$

$1+(\frac{x}{100})^2=\frac{1749.60}{1500}=1.1664$

x=8%
Solution-6

Let the Principal be Rs.100

∴ Amount at the end of one year=$100*(1+\frac{3}{100})^2$

=$100*(1.03)^2=100*1.0609=106.09$

∴ Effective rate of interest= 6.09%

Solution-7

Let the period be t years

∴ compound interest=$30000*(1+\frac{7}{100})^t=34347$

$(1+\frac{7}{100})^t=\frac{34347}{30000}$

($\frac{107}{100}))^t=\frac{11449}{10000}=(\frac{107}{100})^2$

t=2 years

Solution-8

Let t be the number of years

∴ Number of half years=2t

∴ Amount at the end of t years is :

$1600*(1+\frac{5}{100})^2t=1944.81$

$1600*(\frac{105}{100})^2t=1600*(\frac{21}{20})^2t=1944.81$

$(\frac{21}{20})^2t=\frac{1944.81}{1600}=\frac{194481}{160000}$

$\frac{21}{20}^2t=\sqrt(\frac{194481}{160000})=\frac{441}{400}$

$\frac{21}{20}^2=\frac{441}{400}$

t=2 years

Solution-9

Here$ A*(1+\frac{r}{100})^t=P $

i.e. $390625*(1+\frac{4}{100})^t=456976$

∴ $(1+\frac{1}{25})^t=\frac{456976}{390625}$

i.e. $(\frac{26}{25})^2=\frac{456976}{390625}$

i.e. $(\frac{26}{25})^2=(\frac{26}{25})^4$

∴ t=4

The required time is 4 years

Solution-10

If P is the principal and r ,rate of interest per annum

$P*(1+\frac{r}{100})^4=2P$

$1+\frac{r}{100})^4=2P$

$(1+\frac{r}{100})^12=2^3=8$

$P*(1+\frac{r}{100})^12=8P$

$The required time to become the sum eight times itself =12 years$

Short cut: P becomes 2P in 4 years

2P becomes 4P in next 4 years

4P becomes 8P in next 4 years

P becomes 8P in 4 + 4 + 4= 12 years

Solution-11

we have,

$P*[1+(\frac{20}{100})]^t>2P$

∴ $(\frac{6}{5})^t>2$

By trial,$(\frac{6}{5})^4>2$

∴ The Required time is 4 years

Solution-12

Let the sum be Rs. X

Then, $4x=x*(1+\frac{r}{100})^4$

or $4=(1+\frac{r}{100})^4$

Squaring $4^2=((1+\frac{r}{100})^4)^2$

or $16=(1+\frac{r}{100})^8$

$16x=x*(1+\frac{r}{100})^8$

∴ The sum will be 16 times in 8 Years