Arithmetic

Mean, Median & Mode

Hard

In an examination, the average marks of $4$ girls and $6$ boys is $24$ . Each of the girls has the same marks while each of the boys has the same marks. If the marks of any girl is at most double the marks of any boy, but not less than the marks of any boy, then the number of possible distinct integer values of the total marks of $2$ girls and $6$ boys is

$19$

$21$

$22$

$20$

✅ Correct Option: 2

Let marks of each girl be $g$ and each boy be $b$

$\dfrac{4g+6b}{10}=24 \;\;\Rightarrow\;\; 4g+6b=240$

A girl has marks between those of a boy and double of a boy, so $g=kb$ with $1\leq k \leq 2$

$4(kb)+6b=240$

$b(4k+6)=240$

$b(2k+3)=120$

$b=\dfrac{120}{2k+3}$ ... (i)

We need

$2g+6b = 2(kb)+6b = b(2k+6)$

Substitute the value of b from eq. (i):

$= \dfrac{120}{2k+3}(2k+6)$

$=\dfrac{240k+720}{2k+3}$

$= \dfrac{120k + 360+ 360}{2k+3}$

$= \dfrac{120k + 360}{2k+3} + \dfrac{360}{2k+3}$

$= \dfrac{120 (2k+3)}{2k+3} + \dfrac{360}{2k+3}$

$=120+\dfrac{360}{2k+3}$

For integrality,

$\dfrac{360}{2k+3}\in \mathbb{Z}$

Let

$n=\dfrac{360}{2k+3}$

Then

$2g+6b=120+n$

Since $1\leq k \leq 2$

$2\leq 2k \leq 4 \;\;\Rightarrow\;\; 5 \leq 2k+3 \leq 7$

$\dfrac{360}{7}\leq n \leq \dfrac{360}{5}$

$51.42 \leq n \leq 72$

Thus

$52 \leq n \leq 72$

$n$ can take 21 integer values from 52 to 72 inclusive, each giving a distinct $2g+6b$

$\boxed{21}$

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