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SSAT Middle Level Math : Ratio and Proportion

Study concepts, example questions & explanations for SSAT Middle Level Math

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Example Questions

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Example Question #6 : Solve Unit Rate Problems: Ccss.Math.Content.6.Rp.A.3b

At a local market, farmers trade produce to obtain a more diverse crop. A farmer will trade turnips for ears of corn. If a man has 150 ears of corn, then how many turnips can he get?

Possible Answers:

$560\ \text{turnips}$

$550\ \text{turnips}$

$750\ \text{turnips}$

$650\ \text{turnips}$

$860\ \text{turnips}$

Correct answer:

$650\ \text{turnips}$

Explanation:

Ratios can be written in the following format:

$A:B\rightarrow \frac{A}{B}$

Using this format, substitute the given information to create a ratio.

$3\ \text{corn}: 13\ \text{turnips}$

Rewrite the ratio as a fraction.

$3\ \text{corn}: 13\ \text{turnips}\rightarrow \frac{3\ \text{corn}}{13\ \text{turnips}}$

We know that the farmer has 150 ears of corn. Create a ratio with the variable that represents how many turnips he can get.

$\frac{150\ \text{corn}}{T}$

Create a proportion using the two ratios.

$\frac{3\ \text{corn}}{13\ \text{turnips}}=\frac{150\ \text{corn}}{T}$

Cross multiply and solve for .

$150\ \text{corn}\times 13\ \text{turnips}=3\ \text{corn}\times T$

Simplify.

3T=1950

Divide both sides of the equation by .

$\frac{3T}{3}=\frac{1950}{3}$

Solve.

T=650

The farmer can get $650\ \text{turnips}$ .

Report an Error

Example Question #71 : How To Find A Ratio

At a local market, farmers trade produce to obtain a more diverse crop. A farmer will trade turnips for ears of corn. If a man has 180 ears of corn, then how many turnips can he get?

Possible Answers:

$878\ \text{turnips}$

$787\ \text{turnips}$

$870\ \text{turnips}$

$877\ \text{turnips}$

$780\ \text{turnips}$

Correct answer:

$780\ \text{turnips}$

Explanation:

Ratios can be written in the following format:

$A:B\rightarrow \frac{A}{B}$

Using this format, substitute the given information to create a ratio.

$3\ \text{corn}: 13\ \text{turnips}$

Rewrite the ratio as a fraction.

$3\ \text{corn}: 13\ \text{turnips}\rightarrow \frac{3\ \text{corn}}{13\ \text{turnips}}$

We know that the farmer has 180 ears of corn. Create a ratio with the variable that represents how many turnips he can get.

$\frac{180\ \text{corn}}{T}$

Create a proportion using the two ratios.

$\frac{3\ \text{corn}}{13\ \text{turnips}}=\frac{180\ \text{corn}}{T}$

Cross multiply and solve for .

$180\ \text{corn}\times 13\ \text{turnips}=3\ \text{corn}\times T$

Simplify.

3T=2340

Divide both sides of the equation by .

$\frac{3T}{3}=\frac{2340}{3}$

Solve.

T=780

The farmer can get $780\ \text{turnips}$ .

Report an Error

Example Question #8 : Solve Unit Rate Problems: Ccss.Math.Content.6.Rp.A.3b

At a local market, farmers trade produce to obtain a more diverse crop. A farmer will trade turnips for ears of corn. If a man has ears of corn, then how many turnips can he get?

Possible Answers:

$429\ \text{turnips}$

$249\ \text{turnips}$

$483\ \text{turnips}$

$449\ \text{turnips}$

$924\ \text{turnips}$

Correct answer:

$429\ \text{turnips}$

Explanation:

Ratios can be written in the following format:

$A:B\rightarrow \frac{A}{B}$

Using this format, substitute the given information to create a ratio.

$3\ \text{corn}: 13\ \text{turnips}$

Rewrite the ratio as a fraction.

$3\ \text{corn}: 13\ \text{turnips}\rightarrow \frac{3\ \text{corn}}{13\ \text{turnips}}$

We know that the farmer has ears of corn. Create a ratio with the variable that represents how many turnips he can get.

$\frac{99\ \text{corn}}{T}$

Create a proportion using the two ratios.

$\frac{3\ \text{corn}}{13\ \text{turnips}}=\frac{99\ \text{corn}}{T}$

Cross multiply and solve for .

$99\ \text{corn}\times 13\ \text{turnips}=3\ \text{corn}\times T$

Simplify.

3T=1287

Divide both sides of the equation by .

$\frac{3T}{3}=\frac{1287}{3}$

Solve.

T=429

The farmer can get $429\ \text{turnips}$ .

Report an Error

Example Question #9 : Solve Unit Rate Problems: Ccss.Math.Content.6.Rp.A.3b

At a local market, farmers trade produce to obtain a more diverse crop. A farmer will trade turnips for ears of corn. If a man has ears of corn, then how many turnips can he get?

Possible Answers:

$156\ \text{turnips}$

$166\ \text{turnips}$

$516\ \text{turnips}$

$165\ \text{turnips}$

$159\ \text{turnips}$

Correct answer:

$156\ \text{turnips}$

Explanation:

Ratios can be written in the following format:

$A:B\rightarrow \frac{A}{B}$

Using this format, substitute the given information to create a ratio.

$3\ \text{corn}: 13\ \text{turnips}$

Rewrite the ratio as a fraction.

$3\ \text{corn}: 13\ \text{turnips}\rightarrow \frac{3\ \text{corn}}{13\ \text{turnips}}$

We know that the farmer has ears of corn. Create a ratio with the variable that represents how many turnips he can get.

$\frac{36\ \text{corn}}{T}$

Create a proportion using the two ratios.

$\frac{3\ \text{corn}}{13\ \text{turnips}}=\frac{36\ \text{corn}}{T}$

Cross multiply and solve for .

$36\ \text{corn}\times 13\ \text{turnips}=3\ \text{corn}\times T$

Simplify.

3T=468

Divide both sides of the equation by .

$\frac{3T}{3}=\frac{468}{3}$

Solve.

T=156

The farmer can get $156\ \text{turnips}$ .

Report an Error

Example Question #10 : Solve Unit Rate Problems: Ccss.Math.Content.6.Rp.A.3b

At a local market, farmers trade produce to obtain a more diverse crop. A farmer will trade turnips for ears of corn. If a man has 3000 ears of corn, then how many turnips can he get?

Possible Answers:

$31000\ \text{turnips}$

$13013\ \text{turnips}$

$33000\ \text{turnips}$

$13000\ \text{turnips}$

$13113\ \text{turnips}$

Correct answer:

$13000\ \text{turnips}$

Explanation:

Ratios can be written in the following format:

$A:B\rightarrow \frac{A}{B}$

Using this format, substitute the given information to create a ratio.

$3\ \text{corn}: 13\ \text{turnips}$

Rewrite the ratio as a fraction.

$3\ \text{corn}: 13\ \text{turnips}\rightarrow \frac{3\ \text{corn}}{13\ \text{turnips}}$

We know that the farmer has 3000 ears of corn. Create a ratio with the variable that represents how many turnips he can get.

$\frac{3000\ \text{corn}}{T}$

Create a proportion using the two ratios.

$\frac{3\ \text{corn}}{13\ \text{turnips}}=\frac{3000\ \text{corn}}{T}$

Cross multiply and solve for .

$3000\ \text{corn}\times 13\ \text{turnips}=3\ \text{corn}\times T$

Simplify.

3T=39000

Divide both sides of the equation by .

$\frac{3T}{3}=\frac{39000}{3}$

Solve.

T=13000

The farmer can get $13000\ \text{turnips}$ .

Report an Error

Example Question #11 : Solve Unit Rate Problems: Ccss.Math.Content.6.Rp.A.3b

At a local market, farmers trade produce to obtain a more diverse crop. A farmer will trade turnips for ears of corn. If a man has ears of corn, then how many turnips can he get?

Possible Answers:

$353\ \text{turnips}$

$531\ \text{turnips}$

$533\ \text{turnips}$

$351\ \text{turnips}$

$153\ \text{turnips}$

Correct answer:

$351\ \text{turnips}$

Explanation:

Ratios can be written in the following format:

$A:B\rightarrow \frac{A}{B}$

Using this format, substitute the given information to create a ratio.

$3\ \text{corn}: 13\ \text{turnips}$

Rewrite the ratio as a fraction.

$3\ \text{corn}: 13\ \text{turnips}\rightarrow \frac{3\ \text{corn}}{13\ \text{turnips}}$

We know that the farmer has ears of corn. Create a ratio with the variable that represents how many turnips he can get.

$\frac{81\ \text{corn}}{T}$

Create a proportion using the two ratios.

$\frac{3\ \text{corn}}{13\ \text{turnips}}=\frac{81\ \text{corn}}{T}$

Cross multiply and solve for .

$81\ \text{corn}\times 13\ \text{turnips}=3\ \text{corn}\times T$

Simplify.

3T=1053

Divide both sides of the equation by .

$\frac{3T}{3}=\frac{1053}{3}$

Solve.

T=351

The farmer can get $351\ \text{turnips}$ .

Report an Error

Example Question #12 : Solve Unit Rate Problems: Ccss.Math.Content.6.Rp.A.3b

At a local market, farmers trade produce to obtain a more diverse crop. A farmer will trade turnips for ears of corn. If a man has 126 ears of corn, then how many turnips can he get?

Possible Answers:

$645\ \text{turnips}$

$456\ \text{turnips}$

$546\ \text{turnips}$

$564\ \text{turnips}$

$654\ \text{turnips}$

Correct answer:

$546\ \text{turnips}$

Explanation:

Ratios can be written in the following format:

$A:B\rightarrow \frac{A}{B}$

Using this format, substitute the given information to create a ratio.

$3\ \text{corn}: 13\ \text{turnips}$

Rewrite the ratio as a fraction.

$3\ \text{corn}: 13\ \text{turnips}\rightarrow \frac{3\ \text{corn}}{13\ \text{turnips}}$

We know that the farmer has 126 ears of corn. Create a ratio with the variable that represents how many turnips he can get.

$\frac{126\ \text{corn}}{T}$

Create a proportion using the two ratios.

$\frac{3\ \text{corn}}{13\ \text{turnips}}=\frac{126\ \text{corn}}{T}$

Cross multiply and solve for .

$126\ \text{corn}\times 13\ \text{turnips}=3\ \text{corn}\times T$

Simplify.

3T=1638

Divide both sides of the equation by .

$\frac{3T}{3}=\frac{1638}{3}$

Solve.

T=546

The farmer can get $546\ \text{turnips}$ .

Report an Error

Example Question #1 : Find A Percent Of A Quantity As A Rate Per 100: Ccss.Math.Content.6.Rp.A.3c

Red is a very popular car color. A production company manufactures cars and parks them in a lot behind the plant. There are 100 cars in the parking lot and $60 \%$ of them are red. How many red cars are in the parking lot?

Possible Answers:

Correct answer:

Explanation:

We can use ratios and proportions to solve this problem. Percentages can be written as ratios. The word “percent” means for every hundred. In the problem, we are told that $60\%$ of the cars are red. In other words, for every hundred cars of them are red. We can write the following ratio:

$60:100\rightarrow\frac{60}{100}$

Reduce.

$\frac{60}{100}\rightarrow \frac{6}{10}\rightarrow \frac{3}{5}$

We know that there are 100 cars in the parking lot. We can write the following ratio by substituting the variable Red for the number of red cars:

$Red:100\rightarrow \frac{Red}{100}$

Now, we can create a proportion using our two ratios.

$\frac{3}{5}=\frac{Red}{100}$

Cross multiply and solve for Red .

$5 \times Red=3\times100$

Simplify.

5 Red=300

Divide both sides of the equation by .

$\frac{5Red}{5}=\frac{300}{5}$

Solve.

Red=60

There are red cars in the parking lot.

Report an Error

Example Question #2 : Find A Percent Of A Quantity As A Rate Per 100: Ccss.Math.Content.6.Rp.A.3c

Red is a very popular car color. A production company manufactures cars and parks them in a lot behind the plant. There are 250 cars in the parking lot and $10 \%$ of them are red. How many red cars are in the parking lot?

Possible Answers:

Correct answer:

Explanation:

We can use ratios and proportions to solve this problem. Percentages can be written as ratios. The word “percent” means for every hundred. In the problem, we are told that $10\%$ of the cars are red. In other words, for every hundred cars of them are red. We can write the following ratio:

$10:100\rightarrow\frac{10}{100}$

Reduce.

$\frac{10}{100}\rightarrow \frac{1}{10}$

We know that there are 250 cars in the parking lot. We can write the following ratio by substituting the variable Red for the number of red cars:

$Red:250\rightarrow \frac{Red}{250}$

Now, we can create a proportion using our two ratios.

$\frac{1}{10}=\frac{Red}{250}$

Cross multiply and solve for Red .

$10 \times Red=1\times 250$

Simplify.

10 Red=250

Divide both sides of the equation by .

$\frac{10Red}{10}=\frac{250}{10}$

Solve.

Red=25

There are red cars in the parking lot.

Report an Error

Example Question #1 : Find A Percent Of A Quantity As A Rate Per 100: Ccss.Math.Content.6.Rp.A.3c

Red is a very popular car color. A production company manufactures cars and parks them in a lot behind the plant. There are 136 cars in the parking lot and $25 \%$ of them are red. How many red cars are in the parking lot?

Possible Answers:

Correct answer:

Explanation:

We can use ratios and proportions to solve this problem. Percentages can be written as ratios. The word “percent” means for every hundred. In the problem, we are told that $25\%$ of the cars are red. In other words, for every hundred cars of them are red. We can write the following ratio:

$25:100\rightarrow\frac{25}{100}$

Reduce.

$\frac{25}{100}\rightarrow \frac{1}{4}$

We know that there are 136 cars in the parking lot. We can write the following ratio by substituting the variable Red for the number of red cars:

$Red:136\rightarrow \frac{Red}{136}$

Now, we can create a proportion using our two ratios.

$\frac{1}{4}=\frac{Red}{136}$

Cross multiply and solve for Red .

$4 \times Red=1\times136$

Simplify.

4 Red=136

Divide both sides of the equation by .

$\frac{4Red}{4}=\frac{136}{4}$

Solve.

Red=34

There are red cars in the parking lot.

Report an Error

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SSAT Middle Level Math : Ratio and Proportion

Study concepts, example questions & explanations for SSAT Middle Level Math

All SSAT Middle Level Math Resources

Example Questions

Example Question #6 : Solve Unit Rate Problems: Ccss.Math.Content.6.Rp.A.3b

Example Question #71 : How To Find A Ratio

Example Question #8 : Solve Unit Rate Problems: Ccss.Math.Content.6.Rp.A.3b

Example Question #9 : Solve Unit Rate Problems: Ccss.Math.Content.6.Rp.A.3b

Example Question #10 : Solve Unit Rate Problems: Ccss.Math.Content.6.Rp.A.3b

Example Question #11 : Solve Unit Rate Problems: Ccss.Math.Content.6.Rp.A.3b

Example Question #12 : Solve Unit Rate Problems: Ccss.Math.Content.6.Rp.A.3b

Example Question #1 : Find A Percent Of A Quantity As A Rate Per 100: Ccss.Math.Content.6.Rp.A.3c

Example Question #2 : Find A Percent Of A Quantity As A Rate Per 100: Ccss.Math.Content.6.Rp.A.3c

Example Question #1 : Find A Percent Of A Quantity As A Rate Per 100: Ccss.Math.Content.6.Rp.A.3c

All SSAT Middle Level Math Resources

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