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ISEE Middle Level Math : ISEE Middle Level (grades 7-8) Mathematics Achievement

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

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All ISEE Middle Level Math Resources

6 Diagnostic Tests 303 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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Example Question #111 : How To Find The Square Root

Evaluate:

$\frac{25\sqrt{25}}{\sqrt{125}}$

Possible Answers:

$\sqrt{5}$

$25\sqrt{5}$

$5\sqrt{5}$

Correct answer:

$5\sqrt{5}$

Explanation:

Fisrt we need to find the square root of each number and simplify them as follows:

$\sqrt{25}=5$

$\sqrt{125}=\sqrt{25\times 5}=5\sqrt{5}$

Now we can insert the square roots into the equation:

$\frac{25\sqrt{25}}{\sqrt{125}}=\frac{25\times5}{5\sqrt{5}}=\frac{25}{\sqrt{5}}=\frac{25}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\frac{25\sqrt{5}}{5}=5\sqrt{5}$

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Example Question #112 : How To Find The Square Root

If t=5 , evaluate $t(\sqrt{6t^2}\times \sqrt{6t})$ .

Possible Answers:

$25\sqrt{5}$

$150\sqrt{5}$

150

$100\sqrt{5}$

$150\sqrt{6}$

Correct answer:

$150\sqrt{5}$

Explanation:

Since t=5 , we can substitute it into the given equation:

$t(\sqrt{6t^2}\times \sqrt{6t})=5(\sqrt{6\times 5^2}\times \sqrt{6\times 5})$

Now we should find the square root of each number and simplify:

$5(\sqrt{5^2\times 6}\times \sqrt{5\times 6})=5(5\sqrt{5\times 6\times 6})=5\times 5\times 6\times \sqrt{5}=150\sqrt{5}$

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Example Question #113 : How To Find The Square Root

Evaluate:

$6\sqrt{49}+3\sqrt{121}$

Possible Answers:

Correct answer:

Explanation:

First, find the square root of the numbers:

$\sqrt{49}=7$

$\sqrt{121}=11$

Then, solve the equation:

$6\sqrt{49}+3\sqrt{121}=6\times 7+3\times 11=42+33=75$

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Example Question #114 : How To Find The Square Root

Evaluate:

$\frac{\sqrt{36}\times \sqrt{16}\times \sqrt{128}}{4\sqrt{2}}$

Possible Answers:

$24\sqrt{2}$

$48\sqrt{2}$

Correct answer:

Explanation:

First find the square root of each number:

$\sqrt{36}=6$

$\sqrt{16}=4$

$\sqrt{128}=\sqrt{64\times 2}=8\sqrt{2}$

Now we can solve it accordingly:

$\frac{\sqrt{36}\times \sqrt{16}\times \sqrt{128}}{4\sqrt{2}}=\frac{6\times 4\times 8\sqrt{2}}{4\sqrt{2}}=6\times 8=48$

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Example Question #111 : How To Find The Square Root

Which of the following is true about $\sqrt{90}$ ?

Possible Answers:

$9< \sqrt{90}< 10$

$\sqrt{90}> 10$

$11< \sqrt{90}< 12$

$\sqrt{90}= 9$

$10< \sqrt{90}< 11$

Correct answer:

$9< \sqrt{90}< 10$

Explanation:

We know:

9^2=81

10^2=100

Therefore:

$81< 90< 100\Rightarrow \sqrt{81}< \sqrt{90}< \sqrt{100}\Rightarrow 9< \sqrt{90}< 10$

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Example Question #116 : How To Find The Square Root

If is a real number, simplify $\sqrt{-t^2}+\sqrt{t^2}$ .

Possible Answers:

$\sqrt{-t^2}+\sqrt{t^2}$ is undefined for the set of all real numbers.

Correct answer:

$\sqrt{-t^2}+\sqrt{t^2}$ is undefined for the set of all real numbers.

Explanation:

t^2 is always a positive number for the set of real numbers, so (-t^2) is always a negative number for the set of real numbers. Since negative numbers do not have square roots in the real number system, $\sqrt{-t^2}$ is undefined in the set of real numbers and therefore $\sqrt{-t^2}+\sqrt{t^2}$ is undefined as well.

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Example Question #111 : Isee Middle Level (Grades 7 8) Mathematics Achievement

Find :

$\frac{\sqrt{3x^2}+\sqrt{48}}{\sqrt{81}}=\sqrt{75}$

Possible Answers:

$\sqrt{47}$

$\sqrt{40}$

$\sqrt{41}$

Correct answer:

$\sqrt{41}$

Explanation:

First, find the square root of each expression:

$\sqrt{81}=9$

$\sqrt{48}=\sqrt{16\times 3}=4\sqrt{3}$

$\sqrt{75}=\sqrt{25\times 3}=5\sqrt{3}$

Now solve the equation for :

$\frac{\sqrt{3x^2}+\sqrt{48}}{\sqrt{81}}=\sqrt{75}\Rightarrow \frac{\sqrt{3x^2}+4\sqrt{3}}{9}=5\sqrt{3}$

$\Rightarrow \sqrt{3} (\sqrt{x^2})+\sqrt{3}\times 4=\sqrt{3}\times 45\Rightarrow \sqrt{x^2}+4=45\Rightarrow \sqrt{x^2}=41\Rightarrow x=\sqrt{41}$

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Example Question #111 : Isee Middle Level (Grades 7 8) Mathematics Achievement

Find :

$\sqrt{x}+\sqrt{x}=12$

Possible Answers:

$6\sqrt{2}$

$36\sqrt{2}$

Correct answer:

Explanation:

$\sqrt{x}+\sqrt{x}=12\Rightarrow 2\sqrt{x}=12\Rightarrow \sqrt{x}=6\Rightarrow x=36$

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Example Question #112 : Isee Middle Level (Grades 7 8) Mathematics Achievement

Evaluate:

$14+\sqrt{36}+\sqrt{144}$

Possible Answers:

Correct answer:

Explanation:

First find the square roots:

$\sqrt{36}=6$

$\sqrt{144}=12$

Plug back in and add:

$14+\sqrt{36}+\sqrt{144}=14+6+12=32$

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Example Question #120 : How To Find The Square Root

Evaluate:

$\frac{\sqrt{169}\times \sqrt{49}}{\sqrt{169}+\sqrt{49}}$

Possible Answers:

4.5

4.55

4.6

Correct answer:

4.55

Explanation:

First find the square roots:

$\sqrt{169}=13$

$\sqrt{49}=7$

Plug these back into the equation and solve:

$\frac{\sqrt{169}\times \sqrt{49}}{\sqrt{169}+\sqrt{49}}=\frac{13\times 7}{13+7}=\frac{91}{20}=4.55$

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ISEE Middle Level Math : ISEE Middle Level (grades 7-8) Mathematics Achievement

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

All ISEE Middle Level Math Resources

Example Questions

Example Question #111 : How To Find The Square Root

Example Question #112 : How To Find The Square Root

Example Question #113 : How To Find The Square Root

Example Question #114 : How To Find The Square Root

Example Question #111 : How To Find The Square Root

Example Question #116 : How To Find The Square Root

Example Question #111 : Isee Middle Level (Grades 7 8) Mathematics Achievement

Example Question #111 : Isee Middle Level (Grades 7 8) Mathematics Achievement

Example Question #112 : Isee Middle Level (Grades 7 8) Mathematics Achievement

Example Question #120 : How To Find The Square Root

All ISEE Middle Level Math Resources

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