# Find the value of $m$

This question was asked in CSIR _NET 2014. I cannot figure out what are conditions given:

Lunch-dinner pattern of a person for '$m$' days is given below. He has a choice of a vegeterian food and non-vegetarian food for his lunch or dinner. The pattern below given should be considered.

A. If he takes non-vegetarian lunch then he will have to take only vegetarian for dinner.

B. He takes non-vegetarian dinner for exactly for nine days.

C. He takes vegetarian lunch for exactly $15$ days.

D. He takes a total of $14$ non-vegetarian meals.

Determine the value of $m$.

Denote by $LN$ and $LV$ the numbers of non-vegetarian and vegetarian lunches, and by $DN$ and $DV$ the corresponding numbers of dinners.

• Statement B implies $DN=9$
• Statement C implies $LV=15$
• Statement D impies $LN+DN=14$.

Then $LN=(LN+DN)-DN=14-9=5$. Furthermore,

• the total number of days in this lunch/dinner schedule is $m=LN+LV=5+15=20$,
• and finally we get $DV=m-DN=20-9=11$.

$m=20$

if the person eats lunch and dinner on every day. Otherwise, we can only conclude that

$m\ge20$

Let the number of days in which the person had non-veg lunch and veg dinner be $x$, veg lunch and non-veg dinner $y$ and two veg meals $z$. Then:

$y$ = 9
$y+z$ = 15
$x+y$ = 14

$x+y+z = m$ = 20