Moderate Bar Charts Solved QuestionData 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. Common Information Question: 4/4 If the number of students who had chocolates of exactly two varieties $D$ and $F$ is $10$ and those who had chocolates of exactly three varieties $D$, $E$ and $F$ is $12$, then the number of students who had chocolates of only variety $F$ is:
 ✖ A. 4 ✔ B. 0 ✖ C. 2 ✖ D. 1 ✖ E. Cannot be determined

Solution:
Option(B) 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 $f = 23$

Number of students who had chocolates of only variety $F$ is:

$=45 – (10 + 12 + 23)$

$=0$

(1) Comment(s)

Flyingdildo
()

how? 45-(10+12+23)...what about those who ate 4 varieties n 5 varieties? can't they include F also?