Given the values of apples, oranges, and mangoes, how to buy 100 fruits for 100 dollars?

Question

One of my friends gave me this question.

• 1 apple costs \$1

• 20 oranges costs \$1

• 1 mango costs \$5

How many items do you need to buy in order to have a total of 100 items and a value of $100?

I was wondering if there is an algebraic method rather than guess and check.

I tried:

$$ \left\{ \begin{array}{lcl}x+20y+z&=&100,\\ x+y+5z&=&100. \end{array}\right. $$

but no luck.:(

Thanks

Answer 1

Let $x,y,z$ be the number of apples, oranges and mangoes needed to buy. Then $$x+y+z = 100$$ and $$x+ \frac{y}{20} + 5z = 100$$ Subtracting one from another gives $$19y = 80z$$ Since, $y,z$ have to be whole numbers, we want to look for the LCM of $80$ and $19$, which is $80\times 19$. So, $$y=80, z=19$$ and this gives $$x=1$$

Note that we cannot take any higher common multiple of $80,19$ because then the equations won’t be satisfied.

Answer 2

x+ y/20+ 5z= 100 x+ y+ z= 100

Subtract to eliminate x: (19/20)y- 4z= 0 so 4z= (19/20)y

80z= 19y. There are infinitely many real values of y and z that satisfy this but only one pair of positive integer values.

Since 80 and 19 are relatively prime, take z= 19 and y= 80. x+ y+ z= x+ 80+ 19= x+ 99= 100 so x= 1.

1 apple, 80 oranges, and 19 mangoes totals 100 fruits and cost $1+ 80($1/20(+ 19($5)= 1+ 4+ 95= $100.