Data Interpretation Discussion

**Common Information**

One day, a class of $180$ students consumed $342$ chocolates from among the varieties $A, B, C, D, E$ and $F$. No student ate two chocolates of the same variety. The number of students who ate the varieties $D, E$ and $F$ was $73, 93$ and $45$ respectively.

The number of students who did not have any chocolates was equal to the total number of students who either had exactly one of the varieties or all the six varieties. The following bar graph indicates the number of students who consumed exactly $2,3,4$ or $5$ varieties of chocolates.

Q. |
The number of students who had chocolates of only variety $D$ is at least: |

✖ A. |
0 |

✖ B. |
73 |

✖ C. |
44 |

✖ D. |
3 |

✔ E. |
4 |

**Solution:**

Option(**E**) is correct

Let, the number of students who had one, two, three, four, five and six varieties of chocolates be ‘$a$’, ‘$b$’, ‘$c$’, ‘$d$’, ‘$e$’ and ‘$f$’ respectively.

So, the number of students who did not have any chocolate $= (a + f).$

Therefore,

$a + 10 + 12 + 16 + 8 + f = 180 - (a + f)$

$a + 2 × 10 + 3 × 12 + 4 × 16 + 5 × 8 + 6f = 342$

So,

$2a + 2f = 180 - 10 - 12 - 16 - 8 = 134$ and

$a + 6f = 342 - 160$

$= 182$

Solving, the above two equations we get that the value of ‘$a$’ and ‘$f$’ as

$a = 44$ and $f = 23$

In the worst possible case, assume that all the students who ate exactly two, three, four and five varieties of chocolates, definitely had chocolates of variety $D$.

Then, there will be:

$=(10 + 12 + 16 + 8 + 23) = 69$

students who not only had chocolates of variety $D$ but also had chocolates of at least one more variety.

So, there will be at least:

$=73 – 69$

$=4$ students who had chocolates of only variety $D$.