Squares and Square Roots, Cube and Cube Roots

Q1: If \(\sqrt{21}=4.578\), then, the value of \(\sqrt{\frac{3}{7}}\) is:

Given that \(\sqrt{21} = 4.578\), we need to find the value of \(\sqrt{\frac{3}{7}}\).
Step 1: To find \(\sqrt{\frac{3}{7}}\), we can break it into two separate square roots: \[ \sqrt{\frac{3}{7}} = \frac{\sqrt{3}}{\sqrt{7}} \]Step 2: We know that the value of \(\sqrt{21} = 4.578\), and we can express 21 as \(3 \times 7\), so: \[ \sqrt{21} = \sqrt{3 \times 7} = \sqrt{3} \times \sqrt{7} \] This means: \[ \sqrt{3} \times \sqrt{7} = 4.578 \] Now, we can solve for \(\sqrt{3}\) and \(\sqrt{7}\): \[ \sqrt{3} = \frac{4.578}{\sqrt{7}} \]Step 3: To approximate \(\sqrt{7}\), we can use a calculator or estimate it based on known values: \[ \sqrt{7} \approx 2.646 \] Now substitute this value into the equation: \[ \sqrt{3} \approx \frac{4.578}{2.646} \approx 1.73 \]Step 4: Now, substitute \(\sqrt{3} \approx 1.73\) and \(\sqrt{7} \approx 2.646\) back into \(\sqrt{\frac{3}{7}} = \frac{\sqrt{3}}{\sqrt{7}}\): \[ \sqrt{\frac{3}{7}} \approx \frac{1.73}{2.646} \approx 0.654 \]Answer: a. 0.54 (approximate value based on the provided options)

Q2: If \(\sqrt{256}\div\sqrt x=2\), then the value of x is:

Step 1: We are given the equation: \[ \frac{\sqrt{256}}{\sqrt{x}} = 2 \] We know that \(\sqrt{256} = 16\), so the equation becomes: \[ \frac{16}{\sqrt{x}} = 2 \]Step 2: Multiply both sides of the equation by \(\sqrt{x}\) to isolate \(\sqrt{x}\) on one side: \[ 16 = 2 \times \sqrt{x} \]Step 3: Now, divide both sides of the equation by 2: \[ \frac{16}{2} = \sqrt{x} \] \[ 8 = \sqrt{x} \]Step 4: Now square both sides to solve for \(x\): \[ 8^2 = x \] \[ x = 64 \]Answer: b. 64

Q3: \(\sqrt{248+\sqrt{52+\sqrt{144}}}\) is equal to:

Step 1: Start with the innermost square root: \[ \sqrt{144} = 12 \] So, the expression becomes: \[ \sqrt{248 + \sqrt{52 + 12}} \]Step 2: Now simplify inside the next square root: \[ 52 + 12 = 64 \] So, the expression becomes: \[ \sqrt{248 + \sqrt{64}} \]Step 3: Now simplify \(\sqrt{64}\): \[ \sqrt{64} = 8 \] So, the expression becomes: \[ \sqrt{248 + 8} \]Step 4: Add the numbers inside the square root: \[ 248 + 8 = 256 \] So, the expression becomes: \[ \sqrt{256} \]Step 5: Finally, simplify \(\sqrt{256}\): \[ \sqrt{256} = 16 \]Answer: c. 16

Q4: For integers x and y, if \(y=x^2\), then \(x=\sqrt y\).

Step 1: We are given the equation \(y = x^2\). We need to determine if \(x = \sqrt{y}\).
Step 2: If \(y = x^2\), then by taking the square root of both sides, we get: \[ \sqrt{y} = \sqrt{x^2} \] \[ \sqrt{x^2} = x \]Step 3: \[ x = \sqrt{y} \].Conclusion: The statement \(x = \sqrt{y}\) is only true if \(x\) is non-negative.
Answer: a. True

Q5: There are 24 non square numbers between two consecutive perfect squares. The sum of these two consecutive perfect squares is:

Step 1: We are given that there are 24 non-square numbers between two consecutive perfect squares. Let the two consecutive perfect squares be \(n^2\) and \((n+1)^2\).
The total number of integers between \(n^2\) and \((n+1)^2\) is given by: \[ (n+1)^2 – n^2 – 1 = 24 \]Step 2: Simplify the equation: \[ (n+1)^2 – n^2 = 24 + 1 \] \[ 2n + 1 = 25 \] \[ 2n = 24 \] \[ n = 12 \]Step 3: Now calculate the two consecutive perfect squares: \[ n^2 = 12^2 = 144 \] \[ (n+1)^2 = 13^2 = 169 \]Step 4: The sum of these two consecutive perfect squares is: \[ 144 + 169 = 313 \]Answer: d. 313

Q6: The cube of a given number can be smaller than the number itself. Is it true?

Step 1: Let’s analyze the cube of a number and its comparison with the number itself.
For any positive number greater than 1, the cube of the number will always be greater than the number itself. For example:
\( 2^3 = 8 \) which is greater than 2.
\( 3^3 = 27 \) which is greater than 3.
Step 2: For any negative number (less than 0), the cube of the number will also be smaller than the number itself. For example:
\( (-2)^3 = -8 \), which is less than -2.
Step 3: For numbers between 0 and 1, the cube of the number is smaller than the number itself. For example:
\( 0.5^3 = 0.125 \), which is smaller than 0.5.
Conclusion: Yes, the cube of a number can be smaller than the number itself for numbers between 0 and 1, or for negative numbers.
Answer: a. Yes

Q7: Product of three distinct non-cube numbers can’t be a perfect cube. This statement is:

Step 1: A cube number is a number that can be expressed as the cube of an integer, i.e., \(n^3\).
For a product of three distinct non-cube numbers to be a perfect cube, their prime factorizations must each contribute powers that are multiples of 3. For example, for three numbers \(a\), \(b\), and \(c\), the product \(a \times b \times c\) is a perfect cube if the combined powers of all prime factors are multiples of 3.
Step 2: Let’s consider three non-cube numbers:
\(2\), \(3\), and \(5\) are non-cube numbers.
If we multiply them: \[ 2 \times 3 \times 5 = 30 \] \(30\) is not a perfect cube, as it can’t be written as \(n^3\).
However, if we consider the numbers:
\(2^2\), \(3^2\), and \(5\), the product is: \[ 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180 \] which is still not a perfect cube, since the powers of the prime factors are not multiples of 3.
Step 3: However, it is possible to find cases where a product of three non-cube numbers results in a perfect cube if the powers of the prime factors in their factorizations combine to be multiples of 3.
Conclusion: The product of three distinct non-cube numbers can be a perfect cube under certain circumstances, so the statement is false.
Answer: b. False

Q8: If x and y are distinct positive rational numbers such that \(x < \frac{1}{x}\) and \(y > \frac{1}{y}\), then \(x^3 – y^3\) will be:

Step 1: We are given that:
\(x < \frac{1}{x}\), which implies that \(x\) is a positive rational number between 0 and 1. This is because if \(x\) is between 0 and 1, then \(\frac{1}{x}\) will be greater than \(x\).
\(y > \frac{1}{y}\), which implies that \(y\) is a positive rational number greater than 1. This is because if \(y\) is greater than 1, then \(\frac{1}{y}\) will be less than \(y\).
Step 2: Now, we are asked to evaluate \(x^3 – y^3\).
Since \(x\) is less than 1 and \(y\) is greater than 1, let’s examine the cubes of \(x\) and \(y\):
\(x^3\) will be a small positive number, less than \(x\) because \(x < 1\).
\(y^3\) will be a larger positive number, greater than \(y\) because \(y > 1\).
Thus, since \(x^3\) is smaller than \(y^3\), we can conclude that: \[ x^3 – y^3 < 0 \]Conclusion: \(x^3 – y^3\) will be negative.
Answer: b. Negative

Q9: If \(x^2=4\times27\times108\times256\), then \(\sqrt[3]{x}\) is equal to:

Step 1: We are given the equation \(x^2 = 4 \times 27 \times 108 \times 256\).
First, simplify the product on the right-hand side: \[ 4 \times 27 = 108 \] \[ 108 \times 108 = 11664 \] \[ 11664 \times 256 = 2985984 \] So, we have: \[ x^2 = 2985984 \]Step 2: Now, we need to find \(x\). To do this, take the square root of both sides: \[ x = \sqrt{2985984} \] \[ x = 1728 \]Step 3: Now, we need to evaluate \(\sqrt[3]{x}\), which is the cube root of 1728: \[ \sqrt[3]{1728} = 12 \]Conclusion: The value of \(\sqrt[3]{x}\) is 12.
Answer: c. 12

Q10: x and y are two numbers \((x,\ y\neq1)\) such that x is a perfect cube and y is a perfect square. Which of the following numbers is a perfect cube?

Step 1: Given that \(x\) is a perfect cube and \(y\) is a perfect square, we need to check each option.

Option i: \(x \times y\)

– Since \(x\) is a perfect cube, we can write \(x = a^3\) for some integer \(a\).
– Since \(y\) is a perfect square, we can write \(y = b^2\) for some integer \(b\).
– So, \(x \times y = a^3 \times b^2\).
– For this to be a perfect cube, the powers of all prime factors must be divisible by 3.
– Since \(a^3\) is already a perfect cube, \(b^2\) is not a perfect cube because its powers are not divisible by 3. Therefore, \(x \times y\) is not a perfect cube.

Option ii: \(x \times y^2\)

– \(x = a^3\) and \(y = b^2\), so \(x \times y^2 = a^3 \times (b^2)^2 = a^3 \times b^4\).
– For this to be a perfect cube, the powers of all prime factors must be divisible by 3.
– \(a^3\) is already a perfect cube, and \(b^4\) has a power divisible by 3 (since \(4 \mod 3 = 1\)).
– Therefore, \(x \times y^2\) is a perfect cube.

Option iii: \(x^2 \times y^2\)

– \(x^2 = (a^3)^2 = a^6\) and \(y^2 = (b^2)^2 = b^4\), so \(x^2 \times y^2 = a^6 \times b^4\).
– For this to be a perfect cube, the powers of all prime factors must be divisible by 3.
– \(a^6\) is a perfect cube, and \(b^4\) is not a perfect cube because its powers are not divisible by 3. Therefore, \(x^2 \times y^2\) is not a perfect cube.

Option iv: \(x^2 \times y^3\)

– \(x^2 = (a^3)^2 = a^6\) and \(y^3 = (b^2)^3 = b^6\), so \(x^2 \times y^3 = a^6 \times b^6 = (a \times b)^6\).
– Since both terms have powers divisible by 3, \(x^2 \times y^3\) is a perfect cube.
Conclusion: The correct options are (ii) and (iv).
Answer: c. only (iv)