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Answer :
We are given two types of sugar with costs per kilogram of \[tex]$90 and \$[/tex]120, respectively. The businessman wants to sell a kilogram of the mixture at \[tex]$147 with a profit of 50%.
First, we determine the cost price per kilogram of the mixture. Since a 50% profit is made, the relationship between selling price and cost price is given by:
$[/tex][tex]$
1.5 \times (\text{cost price}) = \text{selling price}
$[/tex][tex]$
Substituting the given selling price:
$[/tex][tex]$
1.5 \times (\text{cost price}) = 147 \quad \Longrightarrow \quad \text{cost price} = \frac{147}{1.5} = 98.
$[/tex][tex]$
Thus, the cost of the mixture per kilogram must be \$[/tex]98.
Let [tex]$x$[/tex] be the weight (in kg) of the sugar costing \[tex]$90 and $[/tex]y[tex]$ be the weight (in kg) of the sugar costing \$[/tex]120. The cost per kilogram in the mixture is the weighted average:
[tex]$$
\frac{90x + 120y}{x+y} = 98.
$$[/tex]
Multiply both sides by [tex]$(x+y)$[/tex] to eliminate the denominator:
[tex]$$
90x + 120y = 98(x+y).
$$[/tex]
Expanding the right-hand side, we have:
[tex]$$
90x + 120y = 98x + 98y.
$$[/tex]
Now, rearrange the equation by bringing like terms to one side:
[tex]$$
90x - 98x + 120y - 98y = 0,
$$[/tex]
which simplifies to:
[tex]$$
-8x + 22y = 0.
$$[/tex]
Add [tex]$8x$[/tex] to both sides:
[tex]$$
22y = 8x.
$$[/tex]
Solve for [tex]$y$[/tex] in terms of [tex]$x$[/tex]:
[tex]$$
y = \frac{8}{22}x = \frac{4}{11}x.
$$[/tex]
This shows that the two types of sugar are mixed in the ratio:
[tex]$$
x:y = x : \frac{4}{11}x = 11:4.
$$[/tex]
Therefore, the businessman should mix the sugar costing \[tex]$90 and \$[/tex]120 in the ratio of [tex]$11:4$[/tex]. Additionally, note that the cost price per kilogram for the mixture is \$98.
First, we determine the cost price per kilogram of the mixture. Since a 50% profit is made, the relationship between selling price and cost price is given by:
$[/tex][tex]$
1.5 \times (\text{cost price}) = \text{selling price}
$[/tex][tex]$
Substituting the given selling price:
$[/tex][tex]$
1.5 \times (\text{cost price}) = 147 \quad \Longrightarrow \quad \text{cost price} = \frac{147}{1.5} = 98.
$[/tex][tex]$
Thus, the cost of the mixture per kilogram must be \$[/tex]98.
Let [tex]$x$[/tex] be the weight (in kg) of the sugar costing \[tex]$90 and $[/tex]y[tex]$ be the weight (in kg) of the sugar costing \$[/tex]120. The cost per kilogram in the mixture is the weighted average:
[tex]$$
\frac{90x + 120y}{x+y} = 98.
$$[/tex]
Multiply both sides by [tex]$(x+y)$[/tex] to eliminate the denominator:
[tex]$$
90x + 120y = 98(x+y).
$$[/tex]
Expanding the right-hand side, we have:
[tex]$$
90x + 120y = 98x + 98y.
$$[/tex]
Now, rearrange the equation by bringing like terms to one side:
[tex]$$
90x - 98x + 120y - 98y = 0,
$$[/tex]
which simplifies to:
[tex]$$
-8x + 22y = 0.
$$[/tex]
Add [tex]$8x$[/tex] to both sides:
[tex]$$
22y = 8x.
$$[/tex]
Solve for [tex]$y$[/tex] in terms of [tex]$x$[/tex]:
[tex]$$
y = \frac{8}{22}x = \frac{4}{11}x.
$$[/tex]
This shows that the two types of sugar are mixed in the ratio:
[tex]$$
x:y = x : \frac{4}{11}x = 11:4.
$$[/tex]
Therefore, the businessman should mix the sugar costing \[tex]$90 and \$[/tex]120 in the ratio of [tex]$11:4$[/tex]. Additionally, note that the cost price per kilogram for the mixture is \$98.
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