Aptitude Discussion

Q. |
A set of cards bearing the number 200-299 is used in a game. if a card is drawn at random, what is the probability that it is divisible by 3: |

✖ A. |
0.66 |

✔ B. |
0.33 |

✖ C. |
0.44 |

✖ D. |
0.55 |

**Solution:**

Option(**B**) is correct

Total numbers of cards = 100 {and not 99}

Multiples of three = 33

**Edit:** To check the detailed calculation on how multiple of three=33 is calculated, check comment by **Shelly. **

The required probability,

$=\dfrac{33}{100}$

$=\textbf{0.33}$

**Edit 2:** For a shortcut alternative solution, check comment by **Sanajit Ghosh.**

**Sanajit Ghosh**

*()
*

**Shelly**

*()
*

Explanation on arriving at the count of multiple of three = 33

All the number divisible by 3 from an A.P. with a common difference, $d$ of 3 as follows.

$201,204, 207,.....297.$

1st term, $a=201$.

If the number of terms in the above A.P. is $n$, we know from the properties of an A.P.

$\text{last term}=a+(n-1)d$

$\Rightarrow 297=201+(n-1)\times3$

$\Rightarrow \dfrac{297-201}{3}=(n-1)$

$\Rightarrow \dfrac{96}{3}=(n-1)$

$\Rightarrow n=32+1$

$\Rightarrow n=\textbf{33}$

**Tejaswini**

*()
*

multiples of three=33

**Shivanand**

*()
*

There are a 100 cards totally. And out of those 100 cards, 33 are divisible by 3. So the answer must be 0.33

**Steve**

*()
*

This question was just stupid didn't state what the author meant.

**Satish Bhat**

*()
*

what is the probability that it is divisible by ?

and

Total numbers of cards = 100 not 200.

Easiest way

$T_n=a+(n-1)d$ where 1st number divisible by 3 is $a=201$,$d=3$,$T_n=297$ we have to find $n$, i.e number div by 3

Solving, we get $n=33$,

Therefore, total no = 200 to 299 i.e 100 nos

So $P=\dfrac{33}{100}$