Proportions: Core practice

The greatest common factors are $8$, $12$, and $3$, respectively.

Dividing each pair by its greatest common factor gives

The first two ratios have the same simplest form.

The third differs, since $\frac{3}{5}=\frac{15}{25}$, which is not $\frac{16}{25}$.

With $n$ for the grams of raisins in the larger scoop, writing raisins over peanuts on both sides gives

The peanut amount goes from $6$ to $15$.

The ratio $5:6$ is already in lowest terms and $15$ is not a whole-number multiple of $6$, so there is no whole-number factor to scale by, and cross-multiplying is the way through.

The diagonal you know is

This equals $6$ times $n$, so divide by the $6$ facing the blank.

As a check, the other cross-product $6\times12.5$ is also $75$.

Put wet length on top and dried length below on both sides, with $d$ for the unknown dried length in cm, a positive number.

Since $40$ is not a whole-number multiple of $15$, multiply the known diagonal.

This equals $15$ times $d$, so divide by $15$.

The piece loses the difference of the two lengths.

As a check, $\frac{40}{32}$ and $\frac{15}{12}$ both reduce to $\frac{5}{4}$.

The first machine turns each four centimeters into three.

Its output is

cm.

Those twenty-one centimeters enter the second machine, whose output is

cm.

As a check, $\frac{21}{28}=\frac{3}{4}$ and $\frac{14}{21}=\frac{2}{3}$, matching each machine's stated ratio.

Three prizes at six gold tickets each need

gold tickets.

Writing blue over gold on both sides, with $b$ for the blue tickets, gives

Since $18$ is not a whole-number multiple of $10$, first reduce the exchange ratio by dividing both of its numbers by $5$ (cross-multiplying would also reach $45$).

The gold number now goes from $2$ to $18$, a scale factor of

Multiply the blue number by the same factor.

As a check, $\frac{45}{18}$ and $\frac{25}{10}$ both reduce to $\frac{5}{2}$, so $45$ blue tickets give exactly the $18$ gold tickets needed.

Both sides put mass over dollars, but the left mass is in kilograms and the right mass is in grams, so the record is not set up correctly.

Since $1000$ grams make one kilogram, the parcel's mass is

kg.

For the positive charge $c$, a correct equality is

The mass $1.5$ kg is half of $3$ kg, so the charge is half of 7.50 dollars.

As a check, the cross-products $3\times3.75$ and $7.50\times1.5$ are both $11.25$.

The record as written would instead give $7.50\times1500\div3$, which is 3750 dollars, a thousand times too much.

The denominators are positive, so their product is a valid common denominator.

Multiply the numerator and denominator of the left fraction by $94$, and those of the right fraction by $47$.

Jo has compared the two resulting numerators.

Their equality establishes that the original fractions are equal.

The product calculation itself is correct:

But neither displayed fraction is defined, because both have denominator zero.

Cross-products compare fractions over a shared nonzero denominator.

That requirement fails here, so matching products do not establish the claimed equality.

Dana's cross-products are

They are equal and her denominators are nonzero, so her proportion is true.

Both of her ratios also reduce to $\frac{3}{4}$.

Her proportion has the same two cross-products as the original one, $15\times44$ and $33\times20$.

In general, the cross-products of $\frac{a}{b}=\frac{c}{d}$ are $a\times d$ and $b\times c$.

The cross-products of $\frac{a}{c}=\frac{b}{d}$ are $a\times d$ and $c\times b$, the same pair, since $c\times b$ equals $b\times c$.

The first proportion is true and $b$ and $d$ are nonzero, so $a\times d$ equals $b\times c$.

The swapped proportion then has equal cross-products and nonzero denominators $c$ and $d$, so it is true as well.