SSAT Middle Level Quantitative - How to find a proportion

How many are in $15\ feet?$

Explanation

To solve this problem we can make proportions.

We know that $1\ foot=12\ inches$ , and we can use as our unknown.

$\frac{1\ foot}{12\ inches}=\frac{15\ feet}{x\ inches}$

Next, we want to cross multiply and divide to isolate the on one side.

$12\ inches\times 15\ feet= 1\ foot \times x\ inches$

$\frac{12\ inches\times 15\ feet}{1\ foot}= 180\ inches$

The $\yards$ $\ feet$ will cancel and we are left with $180\ inches$

If the ratio of boys to girls in a classroom is , and there are a total of of students in the classroom, how many boys are in the classroom?

Explanation

If the ratio of boys to girls in a classroom is , that means that there are boys for every girls. Thus, when there are students in a classroom, the breakdown will be boys and girls. If there are students in a classroom, the breakdown will be boys and girls, which translates to a ratio of , or .

Thus, if there are students, will be boys.

How many $\textup{centimeters}$ are in $13\textup { meters?}$

$1\textup,300\textup { centimeters}$

$130\textup { centimeters}$

$1.3\textup{ centimeters}$

$.0013\textup { centimeters}$

$13\textup,000\textup { centimeters}$

Explanation

To solve this problem we can make proportions.

We know that $1\ centimeter= .01\ meters$ and we can use as our unknown.

$\frac{1\ centimeter}{.01\ meter}=\frac{x\ centimeters}{13\ meters}$

Next, we want to cross multiply and divide to isolate the on one side.

$1\ centimeter\times 13\ meters=.01\ meter\times x\ centimeters$

$\frac{1\ centimeter\times 13\ meters}{.01\ meter}= x\ centimeters$

The will cancel and we are left with $1,300\ centimeters$

Give the value of that makes this proportion statement correct:

$\frac{100}{N} = \frac{40}{30}$

$N = 133 \frac{1}{3}$

$N = 83 \frac{1}{3}$

$N = 66 \frac{2}{3}$

Explanation

Cross-multiply, then solve for :

$\frac{100}{N} = \frac{40}{30}$

$40 N = 100 \cdot 30$

$40 N \div 40= 3,000\div 40$

The distance between Youngston and Wynne is 240 miles in reality and three inches on a map. On the same map, Charlesville and Petersburg are one and three-fourths inches apart. How far apart are they in reality?

$140 \textrm{ mi}$

$120 \textrm{ mi}$

$128 \textrm{ mi}$

$144 \textrm{ mi}$

Explanation

240 real miles is represented by 3 map inches, making this a ratio of $240 \div 3 = 80$ real miles per map inch.

Therefore, one and three-fourths inches represents

$80 \times 1 \frac{3}{4} = \frac{80}{1} \times \frac{7}{4} = \frac{20}{1} \times \frac{7}{1} = 140$ miles, the distance between Charlesville and Petersburg.

Today Jenny worked on homework for minutes when she got home from school and then worked on homework for hours after dinner. In minutes, what was the total time that Jenny spent working on homework?

$240\ \space minutes$

$230\ \space minutes$

$220\ \space minutes$

$210\ \space minutes$

$200\ \space minutes$

Explanation

The first thing we need to do is put our hours into minutes because our final answer is asking for minutes. We know that . We can set up a proportion and cross multiply to solve.

$\frac{1hr}{60min}=\frac{2hr}{x}$

Then we can divide to isolate the .

$\frac{1hr(x)}{1hr}=\frac{60min(2hr)}{1hr}$

Now we can add our minutes from before and after dinner together to find our total.

Give the value of that makes this proportion statement correct:

$\frac{N}{80} = \frac{40}{30}$

$N = 106 \frac{2}{3}$

$N = 133 \frac{1}{3}$

Explanation

Multiply both sides by 80 and solve for :

$\frac{N}{80} = \frac{40}{30}$

$\frac{N}{80} \cdot80 = \frac{40}{30} \cdot80$

$N = \frac{40 \cdot 80}{30} = 106 \frac{2}{3}$

Jessica ran sprints for seconds, and Megan ran sprints for minutes. How many more minutes did Megan run than Jessica?

$5\ \space minutes$

$4\ \space minutes$

$3\ \space minutes$

$2\ \space minutes$

$1\ \space minutes$

Explanation

The first thing we need to do is put our seconds into minutes because our final answer is asking for minutes. We know that . We can set up a proportion and cross multiply to solve.