Key Concept: Properties of Square Numbers

c) 3

[Solution Description]

Square numbers always end with 0, 1, 4, 5, 6, or 9 at the unit's place. They never end with 2, 3, 7, or 8. Therefore, the digit that cannot be in the unit's place of a square number is 3.

Key Concept: Properties of Square Numbers

d) 9

[Solution Description]

To find the unit digit of the square of a number, we only need to consider the unit digit of the original number. Here, the unit digit of 13 is 3. Now, let us calculate the square of 3:

$3^2 = 9$

Therefore, the unit digit of the square of 13 is 9.

Key Concept: Square Numbers

b) 64

[Solution Description]

The next perfect square after 49 is found by calculating the square of the next integer.

- The square root of 49 is 7.

- The next integer is 8, and its square is $8^2 = 64$.

Therefore, the next perfect square after 49 is 64.

Key Concept: Square Numbers, Square Roots

c) 10

[Solution Description]

Given the equation:

$\sqrt{x^2 + 2x + 1} = 11$

Simplify the expression inside the square root:

$x^2 + 2x + 1 = (x + 1)^2$

So, the equation becomes:

$\sqrt{(x + 1)^2} = 11$

Taking the square root of both sides:

$x + 1 = 11$

Solving for $x$:

$x = 11 - 1 = 10$

Thus, the value of $x$ is 10.

Key Concept: Identifying Square Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

A square number is a number that can be expressed as the product of an integer with itself. Here, 49 can be written as $7 \times 7$, which means it is indeed a square number. Therefore, both the Assertion and Reason are true, and the Reason correctly explains why 49 is a square number.

Key Concept: Identifying Square Numbers

c) 64

[Solution Description]

A square number is a number that can be expressed as $n^2$, where $n$ is a natural number. Let's check each option:

1. 49: $7 \times 7 = 49$. So, 49 is a square number.

2. 50: There is no natural number $n$ such that $n^2 = 50$. So, 50 is not a square number.

3. 64: $8 \times 8 = 64$. So, 64 is a square number.

4. 72: There is no natural number $n$ such that $n^2 = 72$. So, 72 is not a square number.

Therefore, the square numbers among the options are 49 and 64.

Key Concept: Perfect Squares, Units Place of Square Numbers

c) Assertion is true, but Reason is false.

[Solution Description]

The assertion states that 144 is a perfect square because it can be expressed as $12^2$. This is true since $12 \times 12 = 144$.

The reason claims that any number ending with the digit 4 at the units place must be a perfect square. However, this is not always true. For example, the number 14 ends with the digit 4 but it is not a perfect square. Similarly, 24, 34, etc., are not perfect squares.

Therefore, while the assertion is true, the reason is false.

Key Concept: Perfect Squares, Properties of Square Numbers

d) 2

[Solution Description]

To determine which second last digit cannot accompany a last digit of 6 in a perfect square, we analyze the properties of square numbers. Perfect squares ending with 6 must have been formed by squaring numbers ending with 4 or 6. For example, $4^2 = 16$, $6^2 = 36$, $14^2 = 196$, and so on. Observing these examples, the possible second last digits when the last digit is 6 include 1, 3, 5, 7, and 9. Therefore, the second last digit cannot be 2.

Key Concept: Square Numbers, Properties of Square Numbers

c) The square ends with 6

[Solution Description]

To determine the unit digit of the square of a number that ends with 6, we can observe the pattern of squares of numbers ending with 6. For example, $6^2 = 36$, $16^2 = 256$, and $26^2 = 676$. In each case, the square ends with 6. Therefore, if a number ends with 6, its square will also end with 6.

Key Concept: Square Numbers up to 100

c) Assertion is true, but Reason is false.

[Solution Description]

The assertion states that 64 is a perfect square, which is true because $8 \times 8 = 64$. However, the reason claims that any number ending with an even digit in the units place must be a perfect square. This is false because numbers like 18, 26, etc., end with even digits but are not perfect squares. Therefore, the assertion is true, but the reason is false.

Key Concept: Square Numbers, Pythagorean Triples

b) 5

[Solution Description]

Let the lengths of the two shorter sides be $n$ and $(n + 1)$, where $n$ is a positive integer. The hypotenuse $h$ is given by the Pythagorean theorem:

$h^2 = n^2 + (n + 1)^2 \\ \\ h^2 = n^2 + n^2 + 2n + 1 \\ \\ h^2 = 2n^2 + 2n + 1$

For $h$ to be a perfect square number, $2n^2 + 2n + 1$ must be a perfect square. Testing values of $n$:

- For $n = 3$: $h^2 = 2(9) + 2(3) + 1 = 18 + 6 + 1 = 25$, which is a perfect square ($5^2$).

Therefore, the hypotenuse is 5 units long.

Key Concept: Square Numbers, Area of Square

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the area of a square with side length 7 cm is 49 cm$^2$. To verify this, we use the formula for the area of a square: $Area = side \times side$. Substituting the given value, $Area = 7 \times 7 = 49$ cm$^2$. Therefore, the assertion is true.

The reason states that a number that can be expressed as the square of an integer is called a square number. This is a correct definition as per the syllabus. Since 49 can be expressed as $7^2$, it is indeed a square number. Hence, the reason is also true and correctly explains the assertion.

Key Concept: Properties of Square Numbers

c) 5

[Solution Description] To find the unit digit of $25^2$, first calculate $25^2$. Multiply 25 by itself: $25 \times 25 = 625$. The unit digit of 625 is 5. Thus, the unit digit of $25^2$ is 5.

Key Concept: Properties of Square Numbers

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

The assertion states that 169 is a perfect square. This is true because $13^2 = 169$, so 169 is indeed a perfect square. The reason states that 169 ends with the digit 9 at its unit’s place. This is also true, as the last digit of 169 is 9. However, the reason does not explain why 169 is a perfect square; it merely states a property of the number. Therefore, both the assertion and reason are true, but the reason is not the correct explanation of the assertion.

Key Concept: Last Digits of Square Numbers

d) 9

[Solution Description]

To find the one's digit in the square of 26387, we only need to consider the last digit of the number, which is 7. We know that squares of numbers ending with 7 end with 9 because $7^2 = 49$. Therefore, the one's digit in the square of 26387 is 9.

Key Concept: Last Digits of Square Numbers, Real-World Application

c) 7

[Solution Description]

According to the properties of square numbers, the units digit of a square number cannot be 2, 3, 7, or 8. Since the original number ends with 7, its square cannot end with 7. Therefore, the correct option is 7.

Key Concept: Square Numbers and Zeros

b) 8

[Solution Description]

Let the number be $N$ which ends with 4 zeros. When we square $N$, the number of zeros at the end doubles because each zero in the original number contributes two zeros in the square. Therefore, if $N$ has 4 zeros, its square will have $4 \times 2 = 8$ zeros.

Key Concept: Square Numbers and Zeros

b) 8

[Solution Description]

The first number has 3 zeros, so its square will have $2 \times 3 = 6$ zeros. The second number has 7 zeros, so its square will have $2 \times 7 = 14$ zeros. The difference in the number of zeros in their squares is $14 - 6 = 8$.

Key Concept: Number of Zeros in Square

c) 4

[Solution Description]

To find the number of zeros at the end of the square of a number, we can use the property that if a number has $n$ zeros at the end, its square will have $2n$ zeros at the end.

The number 400 has 2 zeros at the end. Therefore, its square will have $2 \times 2 = 4$ zeros at the end.

So, the square of 400 will have 4 zeros at the end.

Key Concept: Unit Digit of Square, Perfect Squares

c) 7928

[Solution Description] A number cannot be a perfect square if its unit digit is 2, 3, 7, or 8. Among the given options, the number ending with 8 cannot be a perfect square because no perfect square ends with 8.

Key Concept: Properties of Square Numbers, Ending Digits

b) 4

[Solution Description]

To determine the possible square root of a number ending with 6, we need to identify which numbers when squared result in a number ending with 6. Let's consider the squares of numbers from 1 to 10:

$1^2 = 1$

$2^2 = 4$

$3^2 = 9$

$4^2 = 16$

$5^2 = 25$

$6^2 = 36$

$7^2 = 49$

$8^2 = 64$

$9^2 = 81$

$10^2 = 100$

From this list, only the square of 4 and 6 end with 6. Therefore, the possible square roots are 4 and 6.

Key Concept: Special Patterns in Square Numbers

b) 4

[Solution Description]

To find the "one's digit" of the square of 52698, we only need to consider the unit's digit of the number, which is 8. Now, we calculate $8^2 = 64$. The one's digit of 64 is 4. Therefore, the one's digit in the square of 52698 is 4.

Key Concept: Sum of First n Odd Numbers

c) 8

[Solution Description]

The sum of the first $n$ odd natural numbers is given by $n^2$. Given that the sum is 64, we have $n^2 = 64$. Solving for $n$, we get $n = \sqrt{64} = 8$.

Key Concept: Sum of Odd Numbers, Perfect Squares

c) 10

[Solution Description]

The sum of the first $n$ odd natural numbers is given by $n^2$. Here, $n = 10$, so the sum is $10^2 = 100$. The square root of 100 is $\sqrt{100} = 10$.

Key Concept: Numbers Between Square Numbers

c) 14

[Solution Description]

We know that between $n^2$ and $(n+1)^2$, there are $2n$ non-square numbers. Here, $n = 7$.

So, the number of non-square numbers between $7^2$ and $8^2$ is:

$2n = 2 \times 7 = 14$

Key Concept: Sum of Consecutive Triangular Numbers

b) 91 and 105

[Solution Description]

Let the two consecutive triangular numbers be $T_n$ and $T_{n+1}$. The sum of two consecutive triangular numbers is a square number. Given that $T_n + T_{n+1} = 196$, and since $196$ is a perfect square ($14^2$), we can find $n$.

Using the formula for triangular numbers:

$T_n + T_{n+1} = (n+1)^2$

Given $(n+1)^2 = 196$, we solve for $n$:

$n+1 = 14 \quad \Rightarrow \quad n = 13$

Therefore, the two triangular numbers are:

$T_{13} = \frac{13 \times 14}{2} = 91 \quad \text{and} \quad T_{14} = \frac{14 \times 15}{2} = 105$

So, the two numbers are 91 and 105.

Key Concept: Adding Consecutive Triangular Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To verify the assertion, we first need to find the 4th and 5th triangular numbers. The formula for the $n^{th}$ triangular number is given by $T_n = \frac{n(n+1)}{2}$.

For the 4th triangular number:

$T_4 = \frac{4(4+1)}{2} = \frac{20}{2} = 10$

For the 5th triangular number:

$T_5 = \frac{5(5+1)}{2} = \frac{30}{2} = 15$

Now, adding these two consecutive triangular numbers:

$T_4 + T_5 = 10 + 15 = 25$

25 is a perfect square since $5^2 = 25$. Therefore, the assertion is true.

The reason states that the sum of two consecutive triangular numbers always results in a perfect square. This is also true based on the general property of triangular numbers where $T_{n} + T_{n+1} = (n+1)^2$, which is always a perfect square.

Since both the assertion and the reason are true, and the reason correctly explains the assertion, the correct answer is option (a).

Key Concept: Adding Consecutive Triangular Numbers

b) 36

[Solution Description]

Let the smaller triangular number be $T_n = \frac{n(n+1)}{2}$. The next triangular number is $T_{n+1} = \frac{(n+1)(n+2)}{2}$. Their sum is $\frac{n(n+1)}{2} + \frac{(n+1)(n+2)}{2} = \frac{(n+1)(n+n+2)}{2} = \frac{(n+1)(2n+2)}{2} = (n+1)(n+1) = (n+1)^2 = 81$. Solving for $n$, we get $n+1 = 9$ so $n = 8$. The smaller triangular number is $T_8 = \frac{8(8+1)}{2} = 36$.

Key Concept: Non-Square Numbers Between Squares

c) 14

[Solution Description]

The square of 7 is $7^2 = 49$. The square of 8 is $8^2 = 64$. The total numbers between 49 and 64 are $64 - 49 - 1 = 14$.

Key Concept: Number of Non-Square Numbers Between Two Square Numbers

b) 16

[Solution Description]

The difference between two consecutive square numbers $(n + 1)^2 - n^2$ is given by $2n + 1$. According to the problem, this difference is 17:

$2n + 1 = 17$

Solving for $n$:

$2n = 17 - 1 = 16 \\ \\ n = 8$

The number of non-square numbers between $n^2$ and $(n + 1)^2$ is $2n = 2 \times 8 = 16$.

Key Concept: Sum of First n Odd Numbers

b) 49

[Solution Description]

The sum of the first n odd numbers is given by $n^2$. Here, n = 7. So, the sum is $7^2 = 49$.

Key Concept: Sum of First n Odd Numbers, Perfect Squares

c) 15

[Solution Description]

We know that the sum of the first $n$ odd numbers is given by $n^2$. Given that the sum is 225, we can write:

$n^2 = 225$

To find the value of $n$, we take the square root of both sides:

$n = \sqrt{225} = 15$

Therefore, the value of $n$ is 15.

Key Concept: Square Numbers as Sum of Two Consecutive Integers

b) 220 and 221

[Solution Description]

To find the two consecutive integers whose sum is $21^2$, we use the formula:

$n + (n + 1) = 21^2$

Simplifying, we get:

$2n + 1 = 441$

Subtracting 1 from both sides:

$2n = 440$

Dividing by 2:

$n = 220$

Therefore, the two consecutive integers are 220 and 221.

Key Concept: Square Numbers, Sum of Two Consecutive Integers

b) 264 and 265

[Solution Description]

To determine the pair of consecutive integers whose sum is $23^2$, let the smaller integer be $n$ and the larger integer be $n+1$. Therefore, we have:

$n + (n + 1) = 23^2$

Simplifying the equation:

$2n + 1 = 529$

Subtracting 1 from both sides:

$2n = 528$

Dividing both sides by 2:

$n = 264$

Thus, the two consecutive integers are 264 and 265.

Key Concept: Patterns in Square Numbers Ending with 5

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To find the square of 95 using the given pattern, we first identify $a$ as 9 since the digits before 5 are 9. According to the formula $(a5)^2 = a(a + 1) \text{ hundred} + 25$, we substitute $a = 9$. Thus, $a(a + 1) = 9 \times 10 = 90$. Therefore, the square of 95 is calculated as $90 \text{ hundred} + 25 = 9025$. The Reason provides the mathematical justification for this pattern, explaining why it works for any number ending with 5. Hence, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Patterns in Square Numbers Ending with 5

c) 13225

[Solution Description]

To find the square of 115 using the pattern for numbers ending with 5, follow these steps:

1. Identify the number before the unit digit 5. In this case, it is 11.

2. Multiply this number by the next consecutive integer: $11 \times 12 = 132$.

3. This product represents the number of hundreds in the result. So, we have 132 hundreds.

4. Add 25 to this result: $13200 + 25 = 13225$.

Therefore, the square of 115 is 13225.

Key Concept: Pythagorean Triplets

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

To verify if 7, 24, 25 is a Pythagorean triplet, we check if $7^2 + 24^2 = 25^2$. Calculating, $49 + 576 = 625$, which simplifies to $625 = 625$. Thus, the assertion is true. Now, consider the reason. For $m = 4$, the formula gives $8$, $15$, and $17$, which is another valid triplet. However, the triplet 7, 24, 25 does not fit this general form because there is no integer $m$ such that $2m = 7$, $m^2 - 1 = 24$, or $m^2 + 1 = 25$. Hence, while both statements are true, the reason does not correctly explain the assertion. The correct answer is (b).

Key Concept: Pythagorean Triplets

b) 6, 7, 8

[Solution Description]

A Pythagorean triplet consists of three natural numbers $a$, $b$, and $c$ such that $a^2 + b^2 = c^2$. Let's check each option:

Option 1: 3, 4, 5

Calculate $3^2 + 4^2 = 9 + 16 = 25$ and $5^2 = 25$. Since $3^2 + 4^2 = 5^2$, this is a Pythagorean triplet.

Option 2: 6, 7, 8

Calculate $6^2 + 7^2 = 36 + 49 = 85$ and $8^2 = 64$. Since $6^2 + 7^2 \neq 8^2$, this is not a Pythagorean triplet.

Option 3: 5, 12, 13

Calculate $5^2 + 12^2 = 25 + 144 = 169$ and $13^2 = 169$. Since $5^2 + 12^2 = 13^2$, this is a Pythagorean triplet.

Option 4: 7, 24, 25

Calculate $7^2 + 24^2 = 49 + 576 = 625$ and $25^2 = 625$. Since $7^2 + 24^2 = 25^2$, this is a Pythagorean triplet.

The correct answer is b) 6, 7, 8 since it does not satisfy the Pythagorean theorem.

Key Concept: Finding the Square of a Number

b) 625

[Solution Description]

To find the square of 25 using the pattern for numbers ending with 5, we use the formula: $a(a + 1)$ hundred + 25. Here, $a = 2$. So, $2(2 + 1) = 6$ hundreds + 25 = 600 + 25 = 625.

Key Concept: Finding the Square of a Number, Decomposition Method

c) 2209

[Solution Description]

We decompose 47 into 40 and 7. Then,

$47^2 = (40 + 7)^2 = 40^2 + 2 \times 40 \times 7 + 7^2$

Calculating each term:

$40^2 = 1600, \quad 2 \times 40 \times 7 = 560, \quad 7^2 = 49$

Adding them together:

$1600 + 560 + 49 = 2209$

Therefore, the square of 47 is 2209.

Key Concept: Squaring a Two-Digit Number

a) 2209

[Solution Description]

We can express 47 as $40 + 7$. Using the formula $(a + b)^2 = a^2 + 2ab + b^2$, we calculate:

$47^2 = (40 + 7)^2 = 40^2 + 2 \times 40 \times 7 + 7^2$

$= 1600 + 560 + 49 = 2209$

Key Concept: Squaring a Two-Digit Number

c) 2209

[Solution Description]

To find the square of 47, we can break it down as follows:

$47 = 40 + 7$

Now, using the formula $(a + b)^2 = a^2 + 2ab + b^2$, substitute $a = 40$ and $b = 7$.

$47^2 = (40 + 7)^2 = 40^2 + 2(40)(7) + 7^2$

Calculate each term step by step:

$40^2 = 1600$,

$2(40)(7) = 560$,

$7^2 = 49$.

Add all the terms together:

$1600 + 560 + 49 = 2209$.

Therefore, the square of 47 is 2209.

Key Concept: Shortcut Methods for Squaring, Other Patterns in Squares

b) 7225

[Solution Description]

Using the pattern for numbers ending with 5: $(a5)^2 = a(a + 1)$ hundred + 25. Here, $a = 8$. So, $85^2 = 8(8 + 1)$ hundred + 25 = $8 \times 9$ hundred + 25 = $7200 + 25 = 7225$. The hundreds part of the result is $72$.

Key Concept: Other patterns in squares

b) 7225

[Solution Description]

For numbers ending with 5, the square can be found using the pattern $a(a + 1)$ hundreds + 25. Here, $a = 8$.

Step 1: Calculate $a(a + 1)$, which is $8 \times 9 = 72$.

Step 2: Multiply by 100 to get the hundreds place: $72 \times 100 = 7200$.

Step 3: Add 25 to the result: $7200 + 25 = 7225$.

Therefore, the square of 85 is 7225.

Key Concept: Using Identities to Find Squares

b) 3136

[Solution Description]

Express 56 as $50 + 6$. Using the identity $(a + b)^2 = a^2 + 2ab + b^2$, we have:

$56^2 = (50 + 6)^2 = 50^2 + 2 \times 50 \times 6 + 6^2$

Step 1: Calculate $50^2 = 2500$.

Step 2: Calculate $2 \times 50 \times 6 = 600$.

Step 3: Calculate $6^2 = 36$.

Step 4: Add the results: $2500 + 600 + 36 = 3136$.

Key Concept: Squares using identities, Multi-step reasoning

b) 2209

[Solution Description]

Let $a = 40$ and $b = 7$. Using the identity $(a + b)^2 = a^2 + 2ab + b^2$, we substitute the values:

$(40 + 7)^2 = 40^2 + 2 \times 40 \times 7 + 7^2$

Calculate each term:

$40^2 = 1600$

$2 \times 40 \times 7 = 560$

$7^2 = 49$

Add them together:

$1600 + 560 + 49 = 2209$

Therefore, the square of 47 is 2209.

Key Concept: Other patterns in squares

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To find the square of 95 using the given pattern, we first remove the last digit (5) to get 9. Then, we multiply 9 by (9+1), which gives 90. According to the pattern, the square of 95 is $(9 \times 10) \text{ hundreds} + 25 = 9000 + 25 = 9025$. Therefore, the assertion is true. The reason explains why this pattern works for numbers ending with 5, as it generalizes the process for any such number. Hence, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Pattern in Squares

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] To find the square of 95 using the given pattern, we first multiply the digits before 5 by its consecutive number: $9 \times 10 = 90$. Then, we add 25 to this product: $90 \times 100 + 25 = 9000 + 25 = 9025$. Thus, the square of 95 is 9025. The pattern simplifies the multiplication process for numbers ending with 5, making the reason a correct explanation for the assertion.

Key Concept: Square Roots, Area of Square

c) 16

[Solution Description]

The area of a square is given by $\text{side}^2$. To find the side length, we take the square root of the area. Given the area is 256, the side length is $\sqrt{256} = 16$.

Key Concept: Square Roots

b) 9

[Solution Description]

To find the square root of 81, we need to find a number whose square is 81. We know that $9^2 = 81$. Therefore, the square root of 81 is 9.

Key Concept: Finding square roots

b) 15

[Solution Description] To find the square root of 225, recall that $15^2 = 225$. Therefore, $\sqrt{225} = 15$.

Key Concept: Repeated Subtraction

c) 12

[Solution Description]

To find the square root of 144 using the repeated subtraction method, we subtract successive odd numbers starting from 1 until we reach 0.

Starting with 144, we subtract 1, then 3, then 5, and so on.

The process will be as follows:

144 - 1 = 143,

143 - 3 = 140,

140 - 5 = 135,

135 - 7 = 128,

128 - 9 = 119,

119 - 11 = 108,

108 - 13 = 95,

95 - 15 = 80,

80 - 17 = 63,

63 - 19 = 44,

44 - 21 = 23,

23 - 23 = 0.

It takes 12 steps to reach 0, so the square root of 144 is 12.

Therefore, the correct answer is 12.

Key Concept: Square Root Definition

b) 15

[Solution Description]

To find the positive square root of 225, we look for a number that when squared equals 225. We know that:

$15^2 = 225$

Therefore, the positive square root of 225 is 15.

Key Concept: Definition of Square Roots

c) Assertion is true, but Reason is false.

[Solution Description]

The assertion states that the square root of 25 is 5, which is correct because $5^2 = 25$. However, the reason claims that the square root of a number is always positive. This is not entirely true because while the principal (positive) square root is considered in most contexts, numbers can have both positive and negative square roots. For example, both 5 and -5 are square roots of 25. Thus, the assertion is true, but the reason is false.

Key Concept: Square roots, quadratic expressions

c) 12

[Solution Description]

To solve the equation $(x - 5)^2 = 49$, take the square root of both sides.

This gives two possible solutions: $x - 5 = 7$ and $x - 5 = -7$.

Solving these, we get $x = 12$ and $x = -2$.

Since we are asked for the positive value, $x = 12$ is the answer.

Key Concept: Relation Between Squares and Square Roots

c) 9

[Solution Description]

The square root of a number is the value that, when multiplied by itself, gives the original number. We know that $9^2 = 81$. Therefore, the positive square root of 81 is 9.

Key Concept: Understanding Square Roots

d) 49

[Solution Description]

Given that $\sqrt{x} = 7$, to find the value of $x$, square both sides of the equation:

$(\sqrt{x})^2 = 7^2$

$x = 49$

Therefore, the value of $x$ is 49.

Key Concept: Properties of Square Roots

c) Assertion is true, but Reason is false.

[Solution Description]

The Assertion states that the square root of 25 is 5, which is correct because $5^2 = 25$. However, the Reason claims that the square root of a number is always positive, which is incorrect. While the principal (positive) square root of a non-negative number is positive, there are also negative square roots. For example, both $5$ and $-5$ are square roots of 25. Therefore, the Assertion is true, but the Reason is false.

Key Concept: Square Roots, Area of a Square

c) 16 units

[Solution Description]

To find the length of one side of the square, we need to calculate the square root of the area. Given that the area $A = 256$ square units, and since the area of a square is given by $A = \text{side}^2$, we set up the equation as follows:

$256 = \text{side}^2$

To find the side, take the square root of both sides:

$\text{side} = \sqrt{256} = 16$

Therefore, the length of one side of the square is 16 units.

Key Concept: Finding Square Root through Prime Factorisation

c) 16

[Solution Description]

To find the square root of 256 using prime factorisation, we first determine its prime factors. The number 256 can be expressed as $256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$. Since each prime factor appears twice in pairs, we take one number from each pair: $2 \times 2 \times 2 \times 2 = 16$. Therefore, $\sqrt{256} = 16$.

Key Concept: Positive Square Root

c) 4

[Solution Description]

The positive square root of a number is denoted by $\sqrt{}$. For example, $\sqrt{16}$ is the positive square root of 16. Since $4^2 = 16$, the positive square root of 16 is 4.

Key Concept: Positive and Negative Square Roots, Perfect Squares

c) $0$

[Solution Description]

Given that $z$ is an integer and $z^2 = 100$. To find the sum of the positive square root of $z$ and the negative square root of $z$, we first solve for $z$.

Step 1: Solve $z^2 = 100$

Taking the square root of both sides, we get $z = \pm10$.

Step 2: Calculate the positive and negative square roots of $z$

The positive square root of $z$ when $z = 10$ is $\sqrt{10}$.

The negative square root of $z$ when $z = -10$ is $-\sqrt{10}$.

Step 3: Sum the two results

$\sqrt{10} + (-\sqrt{10}) = 0$.

The correct option is $0$.

Key Concept: Square Roots, Prime Factorisation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To find the square root of 3600 using prime factorisation, we first factorise 3600 into its prime factors: $3600 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5$. Each prime factor occurs an even number of times, confirming that 3600 is a perfect square. Pairing the prime factors: $(2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (5 \times 5)$. Taking one factor from each pair gives $\sqrt{3600} = 2 \times 2 \times 3 \times 5 = 60$. Therefore, the Assertion is true. The Reason is also true because, for a number to be a perfect square, each prime factor in its prime factorisation must occur an even number of times. Additionally, the Reason correctly explains the Assertion as it justifies why pairing the prime factors allows us to find the square root.

Key Concept: Finding Square Root through Prime Factorisation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] To find the square root of 144, we first perform its prime factorisation. The prime factorisation of 144 is $2 \times 2 \times 2 \times 2 \times 3 \times 3$. By pairing the prime factors, we get $(2 \times 2 \times 3)^2 = 12^2$. Therefore, $\sqrt{144} = 12$. Hence, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Finding square root through repeated subtraction

c) 8

[Solution Description]

Subtract successive odd numbers starting from 1 from 64 until you get 0. The number of steps taken to reach 0 is the square root of 64. Steps:

1. $64 – 1 = 63$

2. $63 – 3 = 60$

3. $60 – 5 = 55$

4. $55 – 7 = 48$

5. $48 – 9 = 39$

6. $39 – 11 = 28$

7. $28 – 13 = 15$

8. $15 – 15 = 0$. It takes 8 steps to reach 0, so the square root of 64 is 8.

Key Concept: Finding square root through repeated subtraction

b) 5

[Solution Description]

Subtract successive odd numbers starting from 1 from 25 until you get 0. The number of steps taken to reach 0 is the square root of 25. Steps:

1. $25 – 1 = 24$

2. $24 – 3 = 21$

3. $21 – 5 = 16$

4. $16 – 7 = 9$

5. $9 – 9 = 0$. It takes 5 steps to reach 0, so $\sqrt{25} = 5$.

Key Concept: Finding Square Root by Division Method

c) 59

[Solution Description]

Step 1: Place bars over every pair of digits starting from the one's place: $\overline{34} \overline{81}$. Since there are two bars, the square root will be a 2-digit number.

Step 2: Find the largest number whose square is less than or equal to 34. $5^2 = 25$ and $6^2 = 36$. So, 5 is the divisor and quotient. The remainder is $34 - 25 = 9$.

Step 3: Bring down the next pair (81) to the right of the remainder. New dividend is 981.

Step 4: Double the quotient (5) to get 10, and leave a blank next to it.

Step 5: Guess the largest possible digit to fill the blank such that when multiplied by the new quotient, the product is $\leq$ 981. $109 \times 9 = 981$. So, the new digit is 9.

Step 6: The remainder is 0, and no digits are left. Thus, $\sqrt{3481} = 59$.

Key Concept: Finding the number of digits in the square root

b) 3

[Solution Description]

According to the syllabus, if a perfect square has an odd number of digits $n$, the number of digits in its square root is $\frac{n+1}{2}$. Here, $n = 5$ (odd), so the number of digits in the square root is $\frac{5+1}{2} = 3$.

Key Concept: Long Division Method, Repeated Subtraction

c) 27

[Solution Description]

To solve this problem, we need to relate the long division method and the repeated subtraction method.

Step 1: From the long division method, we know that the square root of the number is 27. Therefore, the number itself is $27^2 = 729$.

Step 2: In the repeated subtraction method, the number of steps required to reach 0 is equal to the square root of the number. Since $\sqrt{729} = 27$, it will take exactly 27 steps to reach 0.

Therefore, the correct answer is 27.

Key Concept: Finding square root through repeated subtraction

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To find the square root of a number using the repeated subtraction method, we subtract successive odd numbers starting from 1. For example, to find $\sqrt{81}$, we start with 81 and subtract 1, then 3, then 5, and so on, until we reach 0. The number of steps taken to reach 0 is the square root of the number. In this case, it takes 9 steps to reach 0, so $\sqrt{81} = 9$. The reason given is that the sum of the first n odd natural numbers is $n^2$, which is true and explains why this method works. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Repeated Subtraction Method

c) 11

[Solution Description]

To find the square root of 121 using the repeated subtraction method, we subtract successive odd numbers starting from 1 until we reach 0. Starting with 121:

1. $121 - 1 = 120$

2. $120 - 3 = 117$

3. $117 - 5 = 112$

4. $112 - 7 = 105$

5. $105 - 9 = 96$

6. $96 - 11 = 85$

7. $85 - 13 = 72$

8. $72 - 15 = 57$

9. $57 - 17 = 40$

10. $40 - 19 = 21$

11. $21 - 21 = 0$

We reached 0 after subtracting 11 odd numbers. Therefore, $\sqrt{121} = 11$.

Key Concept: Repeated Subtraction Method, Perfect Squares

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] To verify if 225 is a perfect square using the repeated subtraction method, we subtract successive odd numbers starting from 1. The steps are as follows:

1. $225 - 1 = 224$

2. $224 - 3 = 221$

3. $221 - 5 = 216$

4. $216 - 7 = 209$

5. $209 - 9 = 200$

6. $200 - 11 = 189$

7. $189 - 13 = 176$

8. $176 - 15 = 161$

9. $161 - 17 = 144$

10. $144 - 19 = 125$

11. $125 - 21 = 104$

12. $104 - 23 = 81$

13. $81 - 25 = 56$

14. $56 - 27 = 29$

15. $29 - 29 = 0$

After 15 steps, the result is zero. This confirms that 225 is a perfect square, and $\sqrt{225} = 15$. Since the sum of the first $n$ odd numbers equals $n^2$, the reason correctly explains why the assertion is true.

Key Concept: Prime Factorisation Method

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To find the square root of 1296 using the prime factorisation method, we first find its prime factors. The prime factorisation of 1296 is $2^4 \times 3^4$. Pairing the prime factors, we get $(2^2 \times 3^2)^2$. Therefore, the square root of 1296 is $2^2 \times 3^2 = 4 \times 9 = 36$. Both the Assertion and Reason are true, and the Reason correctly explains why the Assertion is true.

Key Concept: Prime Factorization Method

c) Assertion is true, but Reason is false.

[Solution Description]

Let's verify the assertion and reason step by step. First, find the prime factorization of 144: $144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3$. Here, both 2 and 3 occur twice in pairs. Now, pairing the prime factors: $(2 \times 2 \times 3 \times 3) = (2 \times 3)^2 = 6^2 = 36$. However, this contradicts the assertion that the square root of 144 is 12. Actually, pairing the prime factors correctly gives $(2 \times 2 \times 3) = 12$, which is the correct square root. The reason is true as each prime factor occurs twice, but it does not directly explain the assertion due to incorrect pairing in the solution. Therefore, the assertion is true, but the reason is false.

Key Concept: Long Division Method

b) 89

[Solution Description]

Place bars over every pair of digits: $79$ $21$. The largest number whose square is less than or equal to 79 is 8 ($8^2 = 64$). Subtract 64 from 79 to get a remainder of 15. Bring down the next pair of digits (21) to make the new dividend 1521. Double the quotient (8) to get 16 and find a digit such that $16x \times x \leq 1521$. The digit is 9 since $169 \times 9 = 1521$. Therefore, $\sqrt{7921} = 89$.

Key Concept: Long Division Method

c) Assertion is true, but Reason is false.

[Solution Description]

First, consider a 6-digit perfect square number. According to the syllabus, when n is even, the number of digits in the square root is given by $\frac{n}{2}$. Here, n = 6, so the number of digits in the square root should be $\frac{6}{2} = 3$. Thus, the Assertion is true. Next, the Reason states that for any perfect square number with n-digits, the number of digits in its square root is given by $\frac{n+1}{2}$ if n is odd. Since n = 6 is even, this formula does not apply, and the Reason is false. Therefore, the correct answer is (C).

Key Concept: Estimating the Number of Digits in Square Root

c) 4

[Solution Description]

To find the number of digits in the square root of 100000000, we place bars on the number starting from the right. The number 100000000 has 4 bars. Therefore, the square root will have 4 digits.

Key Concept: Estimating the Number of Digits in Square Root

c) 3

[Solution Description]

To find the number of digits in the square root of 36864, we place bars on the number starting from the right. The number 36864 has 3 bars. Therefore, the square root will have 3 digits.

Key Concept: Finding the Smallest Number to be Subtracted to Get a Perfect Square

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To determine the smallest number to be subtracted from 5607 to make it a perfect square, we first find $\sqrt{5607}$ using the long division method. The remainder obtained is 131. This indicates that $74^2 = 5476$ is less than 5607 by 131. Therefore, subtracting 131 from 5607 gives us the perfect square 5476. Hence, the assertion is true. The reason is also true because the remainder obtained when finding $\sqrt{5607}$ is indeed 131, which explains why 131 must be subtracted to get a perfect square.

Key Concept: Finding the Square Root of the Resulting Perfect Square

b) 44

[Solution Description]

Given that the smallest number subtracted is 16, the resulting perfect square is $2000 - 16 = 1984$. To find its square root, note that $44^2 = 1936$ and $45^2 = 2025$. Since 1984 lies between these two squares, it is not a perfect square. However, based on the question's assumption, if 1984 were a perfect square, its square root would be 44.

Key Concept: Square Roots of Decimals, Real-World Application

b) 2.4

[Solution Description]

Let the width of the field be $x$ meters. Then, the length is $4x$ meters.

The area of the rectangle is given by:

$\text{Area} = \text{Length} \times \text{Width}$

Substituting the values:

$23.04 = 4x \times x$

$23.04 = 4x^2$

Solving for $x^2$:

$x^2 = \frac{23.04}{4} = 5.76$

Finding the square root of $5.76$:

$x = \sqrt{5.76} = 2.4$

Therefore, the width of the field is $2.4$ meters.

Key Concept: Square Roots of Decimals

c) $0.8$

[Solution Description]

To find the square root of $0.64$, we can use the method of finding the square root of a decimal number. We know that $0.8 \times 0.8 = 0.64$. Therefore, $\sqrt{0.64} = 0.8$.

Key Concept: Placing Bars, Square Root of Decimals

b) Bars on 45; bars on 67 and 80

[Solution Description]

To place the bars for finding the square root of 45.678, follow these steps:

1. For the integral part (45), start from the unit’s place close to the decimal and move towards the left. The first bar is over 45.

2. For the decimal part (.678), start from the decimal and move towards the right. The first bar is over 67, and since the next digit is 8, we add a 0 to make it .6780 and place the second bar over 80.

Thus, the bars are placed as $\overline{45}.\overline{67}\overline{80}$.

Key Concept: Square Roots of Decimals

b) 4.5

[Solution Description]

Step 1: Place bars on the integral part (20) and the decimal part (25). The number is written as $\overline{20}.\overline{25}$.

Step 2: Find the largest number whose square is less than or equal to 20. The largest number is 4 because $4^2 = 16$. Write 4 in the quotient and subtract 16 from 20 to get a remainder of 4.

Step 3: Bring down the next bar, which is 25, making the new dividend 425.

Step 4: Double the divisor (4) to get 8 and enter it with a blank on its right.

Step 5: Find a digit (x) such that $8x \times x \leq 425$. The digit 5 fits because $85 \times 5 = 425$.

Step 6: Since the remainder is 0 and no bars are left, the square root of 20.25 is 4.5.

Key Concept: Square Roots of Decimals, Long Division Method

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

To find the square root of 17.64 using the long division method:

Step 1: Place bars on the integral part (17) and the decimal part (64) starting from the first decimal place.

Step 2: The largest perfect square less than 17 is 16. So, 4 is placed in the quotient.

Step 3: Subtract 16 from 17, leaving a remainder of 1. Bring down 64 to make 164.

Step 4: Double the divisor (4) to get 8 and place it with a blank on its right. Find a digit (2) such that $82 \times 2 = 164$.

Step 5: Since the remainder is 0 and no more digits are left, the square root of 17.64 is 4.2.

The assertion (A) is true because the square root of 17.64 is indeed 4.2. The reason (R) is also true because the decimal part is divided into pairs from the decimal point towards the right in the long division method. However, the reason (R) does not fully explain the assertion (A), as it only describes the process of pairing digits but does not explain why the square root is 4.2.

Key Concept: Square Roots of Decimals

c) 5.5

[Solution Description]

To find the square root of $30.25$, we follow the steps:

1. Place bars on the integral part (30) and decimal part (25).

2. For 30, $5^2 = 25 < 30 < 6^2 = 36$. So, take 5 as the divisor and 30 as the dividend.

3. The remainder is 5. Bring down the next pair (25) to get 525.

4. Double the divisor (5) to get 10 and place a blank on its right. Since 25 is the decimal part, put a decimal point in the quotient.

5. We know $105 \times 5 = 525$. Therefore, the new digit is 5.

6. The remainder is 0 and no bar is left. Hence, $\sqrt{30.25} = 5.5$.

Key Concept: Application in Practical Problems

c) 48 m

[Solution Description] The area of a square is given by the formula $Area = side^2$. To find the side, we take the square root of the area. So, $side = \sqrt{2304}$. Calculating this, we find that $\sqrt{2304} = 48$. Therefore, the length of one side of the plot is 48 m.

Key Concept: Square Roots of Decimals

c) 11.5 m

[Solution Description]

To find the side length of the square field, we need to take the square root of the area. Given that the area is 132.25 $m^2$, we calculate $\sqrt{132.25}$.

Step 1: We know that $11^2 = 121$ and $12^2 = 144$. Since $121 < 132.25 < 144$, the square root lies between 11 and 12.

Step 2: Let us consider 11.5 as an approximation. Calculating $11.5^2$, we get $11.5 \times 11.5 = 132.25$.

Therefore, the side length of the field is 11.5 meters.

Key Concept: Pythagoras Theorem

b) 10 cm

[Solution Description]

Using Pythagoras theorem, $hypotenuse^2 = side1^2 + side2^2$. Given sides are 6 cm and 8 cm, so $hypotenuse^2 = 6^2 + 8^2 = 36 + 64 = 100$. Taking the square root, $hypotenuse = \sqrt{100} = 10$ cm.

Key Concept: Square Root

c) 16

[Solution Description]

To find the square root of 256, we need a number whose square is 256. We know that $16^2 = 256$. Therefore, the square root of 256 is 16.

Key Concept: Square Root in Geometry

c) $10\sqrt{2}$ cm

[Solution Description]

To find the diagonal of a square, we use the Pythagorean theorem. The diagonal divides the square into two right-angled triangles. Let the side of the square be $AB = BC = 10$ cm. The diagonal $AC$ can be calculated as:

$AC^2 = AB^2 + BC^2 = 10^2 + 10^2 = 100 + 100 = 200$

Taking the square root of both sides:

$AC = \sqrt{200} = 10\sqrt{2} \, \text{cm}$

Key Concept: Pythagoras Theorem

b) 10 cm

[Solution Description]

Let the hypotenuse be $c$. According to Pythagoras theorem:

$c^2 = 6^2 + 8^2 \\ \\ c^2 = 36 + 64 \\ \\ c^2 = 100 \\ \\ c = \sqrt{100} \\ \\ c = 10 \text{ cm}$

Key Concept: Pythagorean Theorem, Square Roots

c) 12 cm

[Solution Description]

Using the Pythagorean theorem, we have:

$\text{Hypotenuse}^2 = \text{Leg}_1^2 + \text{Leg}_2^2$

Given that the hypotenuse is 13 cm and one leg is 5 cm, let the other leg be $x$ cm. Then,

$13^2 = 5^2 + x^2 \\ \\ 169 = 25 + x^2 \\ \\ x^2 = 169 - 25 \\ \\ x^2 = 144 \\ \\ x = \sqrt{144} \\ \\ x = 12$

Therefore, the length of the other leg is 12 cm.

Key Concept: Pythagorean Theorem

b) 10 meters

[Solution Description]

To find the length of the ladder, we use the Pythagorean theorem. Let $L$ be the length of the ladder. The ladder forms a right triangle with the wall and the ground. According to the Pythagorean theorem:

$L^2 = 6^2 + 8^2$

Calculate the squares:

$L^2 = 36 + 64 = 100$

Take the square root of both sides to find $L$:

$L = \sqrt{100} = 10 \text{ meters}$

Therefore, the length of the ladder is 10 meters.

Key Concept: Square Roots

c) 16 cm

[Solution Description] To find the length of the side of the square, we use the formula for the area of a square: $Area = side^2$. Given that the area is 256 cm$^2$, we can set up the equation $side^2 = 256$. To find the side, we take the square root of both sides: $side = \sqrt{256}$. The square root of 256 is 16, so the length of the side is 16 cm.

Key Concept: Square Numbers, Square Roots

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] To find the length of the side of a square when its area is given, we use the formula: $Area = side^2$. Given that the area is 169 cm$^{2}$, we need to find the side length. We know that $\sqrt{169} = 13$. Therefore, the length of the side is indeed 13 cm. Both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Finding the Length of a Square’s Side Using Its Area

c) 14 cm

[Solution Description]

The area of a square is given by the formula $Area = side \times side$. To find the length of the side, we need to find the square root of the area. Here, the area is 196 cm². So, $side = \sqrt{196}$. We know that $14 \times 14 = 196$, so $\sqrt{196} = 14$. Therefore, the length of the side is 14 cm.

Key Concept: Square Formation

c) 42

[Solution Description]

Let the number of rows be $x$. Since the formation is square, the number of columns is also $x$. Therefore, the total number of students is $x^2$.

Given that the total number of students is 1764, we have:

$x^2 = 1764$

To find $x$, take the square root of both sides:

$x = \sqrt{1764} = 42$

Thus, the number of rows is 42.

Key Concept: Square Formation, Left Out Students

b) 24

[Solution Description]

Let the number of rows be $x$.

Since the formation is square, the number of columns is also $x$.

Therefore, the total number of children required is $x^2$.

We need to find the largest integer $x$ such that $x^2 \leq 600$.

Calculating $\sqrt{600}$, we get $x \approx 24.49$.

Rounding down, $x = 24$.

So, $24^2 = 576$ children can be accommodated.

The number of children left out is $600 - 576 = 24$.

Key Concept: Finding the Number of Rows in a Square Formation of Students

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] Let the number of rows be $x$. Since the number of rows equals the number of columns, the total number of students is given by $x^2 = 1764$. To find $x$, take the square root of both sides: $x = \sqrt{1764} = 42$. Therefore, the number of rows is indeed 42. The reason correctly explains the assertion as it provides the method to determine the number of rows using the square root function.