Customer buys some apples, mangos and oranges

[Guest]

a customer buys some apples, mangos and oranges which total comes exactly 100 fruits

the cost of one apple is 5 Rs.
the cost of one mango is 2 Rs.
and four oranges cost 1 Rupee.

the total amount he pays also equal to 100 Rs.

then how many apples, mango and oranges did he buy?

how these kinds of problems solved mathematically?

[shyamjayakannan]

Let the respective numbers of apples, mangoes and oranges be [tex]a[/tex], [tex]m[/tex] and [tex]o[/tex], respectively.

Numbers equation: [tex]a+m+o=100[/tex] and cost equation: [tex]5a+2m+\frac{o}{4}=100[/tex]. We allso know that all three must be whole numbers because we are considering whole fruits.
Multiply the first equation by 2 and subtract the second one to eliminate [tex]m[/tex] from the equations. We get [tex]\frac{7o}{4}-3a=100...(1)[/tex].
Multiply the first equation by 5 and subtract the second one to eliminate [tex]a[/tex] from the equations. We get [tex]3m+\frac{19o}{4}=400...(2)[/tex].

From (1), [tex]\frac{7o}{4}-100=3a\ge0\Rightarrow o\ge\frac{400}{7}=57.14\Rightarrow o\ge58...(3)[/tex].
From (2), [tex]400-\frac{19o}{4}=3m\ge0\Rightarrow o\le\frac{1600}{19}=84.21\Rightarrow o\le84...(4)[/tex].

Also from (1), [tex]\frac{7o}{4}-100=3a[/tex] must be a multiple of 3 [tex]\Rightarrow\frac{o}{4}-1+3\left(\frac{o}{2}-33\right)[/tex] must be a multiple of 3 and so [tex]\frac{o}{4}-1[/tex] must be a multiple of 3.
Also from (2), [tex]400-\frac{19o}{4}=3m[/tex] must be a multiple of 3 [tex]\Rightarrow1-\frac{o}{4}+3\left(133-\frac{3o}{2}\right)[/tex] must be a multiple of 3 and so [tex]1-\frac{o}{4}[/tex] must be a multiple of 3.

From (3) and (4), we have the following options for [tex]o=[60,64,68,72,76,80,84][/tex] and the above requirement brings the list down to [tex]o=[64,76][/tex].

Checking for these two values of [tex]o[/tex], we get [tex]\boxed{o=64,a=4,m=32}[/tex] or [tex]\boxed{o=76,a=11,m=13}[/tex]