Question

The greatest decimal among $3.109,3.019,3.099$ and $3.19$

• A. $3.109$
• B. $3.019$
• C. $3.099$
• D. $3.19$

Answer 1

The correct option is D

$3.19$

Step I: Compare the place value of the very first digit of the numbers in the left most side.

In the given numbers $3.109,3.019,3.099$ and $3.19$.the first digit in the left most side is same for all the numbers which is $3$ and its place value is also same on all the numbers which is at the units place.

Step II: If the place value of the first digit at the left most side is same then check for its face value.

The face value of the left most digits of $3.109,3.019,3.099$ and $3.19$ is same which is $3$.

Step III: Repeat the steps for the next digit until the left most side numbers are sorted or the desired number is filtered out.

In the given numbers $3.109,3.019,3.099$ and $3.19$

$3$ is same in all the numbers, after the decimal the digits are $1,0$ and $9$ the number having $1$as the next digit is greater than the one having $0$.

Going on to the next step the numbers that have the potential to be the greatest are $3.109,3.19$.

Clearly $9>0$, so $3.19>3.109$

Thus the correct option of the above question is D.