16. A businessman mixes sugar costing Ksh. 90 per kg with another type which costs Ksh. 120 per kg. Find the ratio in which the two types should be mixed so that if a kilogram of the mixture is sold at Ksh. 147, a profit of [tex]50\%[/tex] is realized.

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.