Question :- 38 fruits, apples, oranges, plums and guavas, are distributed among 4 people A, B, C and D. Each one gets a minimum of one fruits of each type and a maximum of four fruits of each type. There are 12 apples. There are two females, each of whom has the highest number of fruits. D, who has 3 plums, has 5 fruits less than the person who has the highest number of fruits. A has 4 apples and 4 plums and does not have the lowest number of fruits. B has an equal number of apples, oranges and guavas only. C has an equal number of oranges, guavas and plums only. None of the friends has an equal number of fruits of each type.
Question :- What was the total number of plums?
My attempt :-
A has confirmed 4 apples and 4 plums and is not the lowest with the number of fruits
D has confirmed 3 plums
There are 2 females who have the highest number of fruits that means they have equal number of fruits and that value is maximum
For apples to be 12, Apples for B + Apples for C + Apples for D=8
So possible $B,C,D$ apple combinations will be :- $(1,3,4) ; (1,4,3) ; (2,2,4) ; (2,4,2) ; (2,3,3) ; (3,2,3) ; (3,3,2)$
For each value of apple for B, I can get value of oranges and guavas as well since they all are same for B as per the question
I am stuck at this point , how to proceed further ? and is there any better way to approach without making so many cases , any clue which can help me reduce the number of cases ?
The only way to satisfy their criteria for having 38 fruits, with D having at least 6 fruits and 5 fewer than someone else are total number of fruits for each person of:
some ordering of (12,7,12),D=7 or (10,11,11),D= 6 where ($x_1,x_2,x_3$) represent the number of total fruit for A, B, and C in some order (not necessarily the listed order).
D has 3 plums and at least 1 of each other fruit, therefore D has at least 6.
If D had 8 or more fruits, then the two people with the most would each have at least 13. So the three of them would have a total of 13+13+8= 34 fruits, leaving the other person with only 4. But A has at least 10 fruits (4 plums and 4 apples, and at least one of each) and B and C each have at least 5 (at least one of each and not equal number of all 4 fruits).
That narrows the possibilities a lot and from there you can make more deductions. Here is a table you may use:
Other equations to remember: 38= 2*max-5 +3b+p+a+3y, each variable (other than totals and max) must be between 1,2,3, or 4. $b \neq p$, $a\neq y$. Since A doesn't have the least total fruit, they must have either 10,11, or 12, which means $ 2 \leq o+g\leq 4$.
I tried out the possibilities with these 2 scenarios I've laid out. if you assume total number of fruits of 12,12,7,7; there is just no way to make this scenario work out. So, I got that they had 11,11,10,6 fruits respectively. The only way to make this work is with 11 plums, because of the restrictions around B and C having the same numbers of 3 of the 4 fruits.
You may switch the number of oranges and guavas for A, but that's it.
Alternatively you could use this table, which would only have 1 possible answer (where m&n are between 2 and 8 instead of 1 and 4):
If you combine that last table I made, we can show that the 12,12,7,7 model doesn't work.
you get $12 +7 + 3(b+y) + a + p = 38$ and $8 -a-b-m +3 = 7$ (where $m>1$-- because it is the sum of 2 fruits--and $a,b>0$). The latter equation is only satisfied by $m=2$, $a=b=1$. Plugging it back into the first, we get 12 +7+3+3y+1+p =38, which gives $3y+p = 15$. But $3b+p = 3+p = 7 \implies p=4$ Thus $3y+p =3y+4 =15 \implies 3y = 11.$ contradiction. $y$ must be an integer.