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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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All SSAT Middle Level Math Resources

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Example Questions

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Example Question #41 : Ratios & Proportional Relationships

Traffic from the suburbs and farms into a city typically follows an observable pattern. On any given morning there are cars for every trucks. On one particular busy morning there are trucks. How many cars are sitting in traffic?

Possible Answers:

$21\ \text{cars}$

$16\ \text{cars}$

$14\ \text{cars}$

$13\ \text{cars}$

$12\ \text{cars}$

Correct answer:

$14\ \text{cars}$

Explanation:

In order to solve this problem we must make a table of ratios. In the question we are given the base ratio:

$7\ \text{cars}:3\ \text{trucks}$

We can use this ratio to make a table.

Cars

According to the table, there are $14\ \text{cars}$ .

Report an Error

Example Question #61 : Ratio And Proportion

Possible Answers:

$28\ \text{cars}$

$24\ \text{cars}$

$26\ \text{cars}$

$30\ \text{cars}$

$32\ \text{cars}$

Correct answer:

$28\ \text{cars}$

Explanation:

In order to solve this problem we must make a table of ratios. In the question we are given the base ratio:

$7\ \text{cars}:3\ \text{trucks}$

We can use this ratio to make a table.

Cars

According to the table, there are $28\ \text{cars}$ .

Report an Error

Example Question #21 : Make Tables Of Equivalent Ratios, Find Missing Values, And Plot Values On A Coordinate Plane: Ccss.Math.Content.6.Rp.A.3a

Possible Answers:

$40\ \text{cars}$

$25\ \text{cars}$

$35\ \text{cars}$

$30\ \text{cars}$

$45\ \text{cars}$

Correct answer:

$35\ \text{cars}$

Explanation:

In order to solve this problem we must make a table of ratios. In the question we are given the base ratio:

$7\ \text{cars}:3\ \text{trucks}$

We can use this ratio to make a table.

Cars

According to the table, there are $35\ \text{cars}$ .

Report an Error

Example Question #44 : Ratios & Proportional Relationships

Possible Answers:

$49\ \text{cars}$

$35\ \text{cars}$

$42\ \text{cars}$

$28\ \text{cars}$

$41\ \text{cars}$

Correct answer:

$42\ \text{cars}$

Explanation:

In order to solve this problem we must make a table of ratios. In the question we are given the base ratio:

$7\ \text{cars}:3\ \text{trucks}$

We can use this ratio to make a table.

Cars

According to the table, there are $42\ \text{cars}$ .

Report an Error

Example Question #61 : Ratio And Proportion

Possible Answers:

$29\ \text{cars}$

$28\ \text{cars}$

$21\ \text{cars}$

$24\ \text{cars}$

$22\ \text{cars}$

Correct answer:

$21\ \text{cars}$

Explanation:

In order to solve this problem we must make a table of ratios. In the question we are given the base ratio:

$7\ \text{cars}:3\ \text{trucks}$

We can use this ratio to make a table.

Cars

According to the table, there are $21\ \text{cars}$ .

Report an Error

Example Question #1 : 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:

$43\ \text{turnips}$

$39\ \text{turnips}$

$22\ \text{turnips}$

$94\ \text{turnips}$

$93\ \text{turnips}$

Correct answer:

$39\ \text{turnips}$

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.

$\frac{9\ \text{corn}}{T}$

Create a proportion using the two ratios.

$\frac{3\ \text{corn}}{13\ \text{turnips}}=\frac{9\ \text{corn}}{T}$

Cross multiply and solve for .

$9\ \text{corn}\times 13\ \text{turnips}=3\ \text{corn}\times T$

Simplify.

3T=117

Divide both sides of the equation by .

$\frac{3T}{3}=\frac{117}{3}$

Solve.

T=39

The farmer can get $39\ \text{turnips}$ .

Report an Error

Example Question #1 : 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:

$87\ \text{turnips}$

$127\ \text{turnips}$

$111\ \text{turnips}$

$117\ \text{turnips}$

$171\ \text{turnips}$

Correct answer:

$117\ \text{turnips}$

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.

$\frac{27\ \text{corn}}{T}$

Create a proportion using the two ratios.

$\frac{3\ \text{corn}}{13\ \text{turnips}}=\frac{27\ \text{corn}}{T}$

Cross multiply and solve for .

$27\ \text{corn}\times 13\ \text{turnips}=3\ \text{corn}\times T$

Simplify.

3T=351

Divide both sides of the equation by .

$\frac{3T}{3}=\frac{351}{3}$

Solve.

T=117

The farmer can get $117\ \text{turnips}$ .

Report an Error

Example Question #3 : 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:

$12\ \text{turnips}$

$24\ \text{turnips}$

$62\ \text{turnips}$

$26\ \text{turnips}$

$21\ \text{turnips}$

Correct answer:

$26\ \text{turnips}$

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.

$\frac{6\ \text{corn}}{T}$

Create a proportion using the two ratios.

$\frac{3\ \text{corn}}{13\ \text{turnips}}=\frac{6\ \text{corn}}{T}$

Cross multiply and solve for .

$6\ \text{corn}\times 13\ \text{turnips}=3\ \text{corn}\times T$

Simplify.

3T=78

Divide both sides of the equation by .

$\frac{3T}{3}=\frac{78}{3}$

Solve.

T=26

The farmer can get $26\ \text{turnips}$ .

Report an Error

Example Question #4 : 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:

$65\ \text{turnips}$

$60\ \text{turnips}$

$52\ \text{turnips}$

$56\ \text{turnips}$

$57\ \text{turnips}$

Correct answer:

$52\ \text{turnips}$

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.

$\frac{12\ \text{corn}}{T}$

Create a proportion using the two ratios.

$\frac{3\ \text{corn}}{13\ \text{turnips}}=\frac{12\ \text{corn}}{T}$

Cross multiply and solve for .

$12\ \text{corn}\times 13\ \text{turnips}=3\ \text{corn}\times T$

Simplify.

3T=156

Divide both sides of the equation by .

$\frac{3T}{3}=\frac{156}{3}$

Solve.

T=52

The farmer can get $52\ \text{turnips}$ .

Report an Error

Example Question #5 : 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 210 ears of corn, then how many turnips can he get?

Possible Answers:

$980\ \text{turnips}$

$910\ \text{turnips}$

$2100\ \text{turnips}$

$180\ \text{turnips}$

$1009\ \text{turnips}$

Correct answer:

$910\ \text{turnips}$

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 210 ears of corn. Create a ratio with the variable that represents how many turnips he can get.

$\frac{210\ \text{corn}}{T}$

Create a proportion using the two ratios.

$\frac{3\ \text{corn}}{13\ \text{turnips}}=\frac{210\ \text{corn}}{T}$

Cross multiply and solve for .

$210\ \text{corn}\times 13\ \text{turnips}=3\ \text{corn}\times T$

Simplify.

3T=2730

Divide both sides of the equation by .

$\frac{3T}{3}=\frac{2730}{3}$

Solve.

T=910

The farmer can get $910\ \text{turnips}$ .

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 #41 : Ratios & Proportional Relationships

Example Question #61 : Ratio And Proportion

Example Question #21 : Make Tables Of Equivalent Ratios, Find Missing Values, And Plot Values On A Coordinate Plane: Ccss.Math.Content.6.Rp.A.3a

Example Question #44 : Ratios & Proportional Relationships

Example Question #61 : Ratio And Proportion

Example Question #1 : Solve Unit Rate Problems: Ccss.Math.Content.6.Rp.A.3b

Example Question #1 : Solve Unit Rate Problems: Ccss.Math.Content.6.Rp.A.3b

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

Example Question #4 : Solve Unit Rate Problems: Ccss.Math.Content.6.Rp.A.3b

Example Question #5 : Solve Unit Rate Problems: Ccss.Math.Content.6.Rp.A.3b

All SSAT Middle Level Math Resources

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