Arithmetic: Simple Interest Test – 3

Simple Interest for 6 years is Rs. 1875
Simple Interest for 1 year is : Rs. 1875/6.
The rate of interest is = $\frac{1875\times 100}{6\times 2000}=\frac{125}{8}%~$
The  SI  accrued on Rs. 6,875 @ 125/8 % for 6 years is = Rs. 6445 (approx)

Let the principal be Rs. x and rate = r %.
Amount in 3 years @ r% is $\displaystyle x(1+\frac{3r}{100})$ = Rs. 14160
Amount in 3 years @ (25+r)% is x$\displaystyle \frac{3\left( r+25 \right)}{100}$
=$\displaystyle \frac{\text{x}\left( 1+\frac{3\left( r+25 \right)}{100} \right)}{\text{x}\left( 1+\frac{3\left( r \right)}{100} \right)}=\frac{14700}{14160}$

Let the total sum = y
S.I. of the sum = 16/100 of y
we know that
Rate =Time =x
If x is the rate of interest and x years be the time .
then x²=16. x= 4.
Correct option is (a)

S.I. in 2 years is Rs. 1050.
S.I. in 1 year is Rs. 525.
Rate of interest is$\displaystyle \frac{525} {5000}X100=\frac{105}{10}=10.5~%~$

Let the total number is Rs. 100
He lends Rs. 40 @ 15 %
Rs. 30 @ 10% and
Rs. 30 @ 18%.
Or we can say that Interest earned by each at end of 1 year By First Year
$\displaystyle \Rightarrow \frac{15}{100}\times \frac{40x}{100}=\frac{60}{1000}x$
By second Year
$\displaystyle \,\Rightarrow \frac{10}{100}\times \frac{30x}{100}=\frac{30}{1000}x$
By Third Year
$\displaystyle \,\Rightarrow \frac{18}{100}\times \frac{30x}{100}=\frac{54}{1000}x$
Total Interest = $\displaystyle =\frac{144}{1000}x$
=$\displaystyle =\frac{144}{1000}x$
Therefore Rate % = $\displaystyle \frac{\frac{144x}{1000}}{x}\times 100$
$\displaystyle =14.4%$