Question

If $\left[x\right]$ denotes the greatest integer less than or equal to $x$ then evaluate
$[\left[\frac{3}{4}\right]+[\frac{3}{4}+\frac{1}{100}]+[\frac{3}{4}+\frac{2}{100}]+\cdots +[\frac{3}{4}+\frac{99}{100}]]$

• A. 100
• B. 75
• C. 76
• D. 74

Answer 1

The correct option is B 75

Using the definition that the greatest integer function is the function that returns the greatest integer value of a real number which is just less than or equal to the given real number.
We have

$\left[\frac{3}{4}\right]=\left[0.75\right]=0,[\frac{3}{4}+\frac{1}{100}]=\left[0.76\right]=0,[\frac{3}{4}+\frac{2}{100}]=\left[0.77\right]=0,\cdots [\frac{3}{4}+\frac{24}{100}]=\left[0.99\right]=0[\frac{3}{4}+\frac{25}{100}]=\left[1\right]=1[\frac{3}{4}+\frac{26}{100}]=\left[1.01\right]=1\cdots [\frac{3}{4}+\frac{99}{100}]=\left[1.74\right]=1$
since there are 75 terms from $[\frac{3}{4}+\frac{25}{100}]$ to $[\frac{3}{4}+\frac{99}{100}]$ which will be equal to $1$ because these $75$ terms lies between $1and2$
First 24 terms will be $0$ because they lie between $0$ and $1.$

So, Adding all above equations, we get
$[\left[\frac{3}{4}\right]+[\frac{3}{4}+\frac{1}{100}]+[\frac{3}{4}+\frac{2}{100}]+\cdots +[\frac{3}{4}+\frac{99}{100}]]==[0+0+\cdots +1+1+\cdots +1]=\left[75\right]=75$