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 did not have any variety of chocolate on that day is: |

✔ A. |
67 |

✖ B. |
68 |

✖ C. |
69 |

✖ D. |
70 |

✖ E. |
Cannot be determined |

**Solution:**

Option(**A**) 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 $f = 23$

The number of students who did not have any variety of chocolate on that day is:

$=44 + 23$

$=67$