GSEB Class 8 Maths Solutions Chapter 6 Square and Square Roots Exercise 6.3

Question 1. What could be the ones' digit of the square root of each of the following numbers?
1. 9801
2. 99856
3. 998001
4. 657666025
Answer:
The possible digit at the ones' place of the square root is as follows:
1. For 9801, the number ends with 1, so its square root's ones' digit can be 1 or 9. This is because \( 1 \times 1 = 1 \) and \( 9 \times 9 = 81 \).
2. For 99856, the number ends with 6, so its square root's ones' digit can be 4 or 6. This is because \( 4 \times 4 = 16 \) and \( 6 \times 6 = 36 \).
3. For 998001, the number ends with 1, so its square root's ones' digit can be 1 or 9. This is because \( 1 \times 1 = 1 \) and \( 9 \times 9 = 81 \).
4. For 657666025, the number ends with 5, so its square root's ones' digit can be 5. This is because \( 5 \times 5 = 25 \).
In simple words: Look at the last digit of the number. If it ends in 1, the square root's last digit is 1 or 9. If it ends in 6, it's 4 or 6. If it ends in 5, it's 5. This helps us predict without doing the full calculation.

Exam Tip: Remember the squares of digits 0-9 to quickly determine the possible unit digit of a square root.

Question 2. Without doing any calculation, find the numbers which are surely not perfect squares?
1. 152
2. 257
3. 408
4. 441
Answer:
We understand that a perfect square's last digit can only be 0, 1, 4, 5, 6, or 9. Consequently, any number ending in 2, 3, 7, or 8 cannot ever be a perfect square.
1. 152 ends in 2, so it cannot be a perfect square.
2. 257 ends in 7, so it cannot be a perfect square.
3. 408 ends in 8, so it cannot be a perfect square.
4. 441 ends in 1, so it can be a perfect square (in this case, \( 21 \times 21 = 441 \)).
Thus, the numbers 152, 257, and 408 are definitely not perfect squares.
In simple words: Numbers that end with 2, 3, 7, or 8 can never be perfect squares. We check the last digit of each given number to find the ones that are not perfect squares.

Exam Tip: Memorize the list of possible unit digits for perfect squares (0, 1, 4, 5, 6, 9) to quickly identify non-perfect squares. This saves time in calculations.

Question 3. Find the square roots of 100 and 169 by the method of repeated subtraction?
1. \( \sqrt{100} \)
2. \( \sqrt{169} \)
Answer:
1. To find \( \sqrt{100} \):
We perform repeated subtraction of consecutive odd numbers starting from 1:
\( 100 - 1 = 99 \)
\( 99 - 3 = 96 \)
\( 96 - 5 = 91 \)
\( 91 - 7 = 84 \)
\( 84 - 9 = 75 \)
\( 75 - 11 = 64 \)
\( 64 - 13 = 51 \)
\( 51 - 15 = 36 \)
\( 36 - 17 = 19 \)
\( 19 - 19 = 0 \)
We reach 0 by successively subtracting 10 odd numbers.
\( \implies \)\( \sqrt{100} = 10 \)
2. To find \( \sqrt{169} \):
We perform repeated subtraction of consecutive odd numbers starting from 1:
\( 169 - 1 = 168 \)
\( 168 - 3 = 165 \)
\( 165 - 5 = 160 \)
\( 160 - 7 = 153 \)
\( 153 - 9 = 144 \)
\( 144 - 11 = 133 \)
\( 133 - 13 = 120 \)
\( 120 - 15 = 105 \)
\( 105 - 17 = 88 \)
\( 88 - 19 = 69 \)
\( 69 - 21 = 48 \)
\( 48 - 23 = 25 \)
\( 25 - 25 = 0 \)
We reach 0 by successively subtracting 13 odd numbers.
\( \implies \)\( \sqrt{169} = 13 \)
In simple words: To find a square root using repeated subtraction, keep subtracting odd numbers (1, 3, 5, etc.) until you get zero. The count of how many odd numbers you subtracted tells you the square root.

Exam Tip: This method is effective for smaller perfect squares but becomes lengthy for larger numbers. Ensure you subtract consecutive odd numbers correctly.

Question 4. Find the square root of the following numbers by the Prime Factorisation Method?
1. 729
2. 400
3. 1764
4. 4096
5. 7744
6. 9604
7. 5929
8. 9216
9. 529
10. 8100
Answer:
1. For 729:
We can express 729 as a product of its prime factors: \( 729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \)
To find the square root, we take one factor from each pair:
\( \sqrt{729} = 3 \times 3 \times 3 = 27 \)
Thus, the square root of 729 is 27.
2. For 400:
We can express 400 as a product of its prime factors: \( 400 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 \)
To find the square root, we take one factor from each pair:
\( \sqrt{400} = 2 \times 2 \times 5 = 20 \)
Thus, the square root of 400 is 20.
3. For 1764:
We can express 1764 as a product of its prime factors: \( 1764 = 2 \times 2 \times 3 \times 3 \times 7 \times 7 \)
To find the square root, we take one factor from each pair:
\( \sqrt{1764} = 2 \times 3 \times 7 = 42 \)
Thus, the square root of 1764 is 42.
4. For 4096:
We can express 4096 as a product of its prime factors: \( 4096 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \)
To find the square root, we take one factor from each pair:
\( \sqrt{4096} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64 \)
Thus, the square root of 4096 is 64.
5. For 7744:
We can express 7744 as a product of its prime factors: \( 7744 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11 \times 11 \)
To find the square root, we take one factor from each pair:
\( \sqrt{7744} = 2 \times 2 \times 2 \times 11 = 88 \)
Thus, the square root of 7744 is 88.
6. For 9604:
We can express 9604 as a product of its prime factors: \( 9604 = 2 \times 2 \times 7 \times 7 \times 7 \times 7 \)
To find the square root, we take one factor from each pair:
\( \sqrt{9604} = 2 \times 7 \times 7 = 98 \)
Thus, the square root of 9604 is 98.
7. For 5929:
We can express 5929 as a product of its prime factors: \( 5929 = 7 \times 7 \times 11 \times 11 \)
To find the square root, we take one factor from each pair:
\( \sqrt{5929} = 7 \times 11 = 77 \)
Thus, the square root of 5929 is 77.
8. For 9216:
We can express 9216 as a product of its prime factors: \( 9216 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \)
To find the square root, we take one factor from each pair:
\( \sqrt{9216} = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 96 \)
Thus, the square root of 9216 is 96.
9. For 529:
We can express 529 as a product of its prime factors: \( 529 = 23 \times 23 \)
To find the square root, we take one factor from each pair:
\( \sqrt{529} = 23 \)
Thus, the square root of 529 is 23.
10. For 8100:
We can express 8100 as a product of its prime factors: \( 8100 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \)
To find the square root, we take one factor from each pair:
\( \sqrt{8100} = 2 \times 3 \times 3 \times 5 = 90 \)
Thus, the square root of 8100 is 90.
In simple words: To find the square root using prime factorization, first break the number down into its prime factors. Then, group identical prime factors into pairs. For each pair, take one factor. Multiply all these single factors together to get the square root.

Exam Tip: Be careful when performing prime factorization, especially for larger numbers. Double-check your divisions to avoid errors. Ensure all prime factors are paired before taking the square root.

Question 5. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained?
1. 252
2. 180
3. 1008
4. 2028
5. 1458
6. 768
Answer:
1. For 252:
The prime factorization of 252 is \( 252 = 2 \times 2 \times 3 \times 3 \times 7 \).
Here, the prime factor 7 does not have a pair.
To make 252 a perfect square, we must multiply it by 7.
New number: \( 252 \times 7 = 1764 \).
The prime factorization of 1764 is \( 2 \times 2 \times 3 \times 3 \times 7 \times 7 \).
The square root of 1764 is \( \sqrt{1764} = 2 \times 3 \times 7 = 42 \).
Thus, the smallest whole number to multiply by is 7, and the square root of the new number is 42.
2. For 180:
The prime factorization of 180 is \( 180 = 2 \times 2 \times 3 \times 3 \times 5 \).
Here, the prime factor 5 does not have a pair.
To make 180 a perfect square, we must multiply it by 5.
New number: \( 180 \times 5 = 900 \).
The prime factorization of 900 is \( 2 \times 2 \times 3 \times 3 \times 5 \times 5 \).
The square root of 900 is \( \sqrt{900} = 2 \times 3 \times 5 = 30 \).
Thus, the smallest whole number to multiply by is 5, and the square root of the new number is 30.
3. For 1008:
The prime factorization of 1008 is \( 1008 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 \).
Here, the prime factor 7 does not have a pair.
To make 1008 a perfect square, we must multiply it by 7.
New number: \( 1008 \times 7 = 7056 \).
The prime factorization of 7056 is \( 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 \times 7 \).
The square root of 7056 is \( \sqrt{7056} = 2 \times 2 \times 3 \times 7 = 84 \).
Thus, the smallest whole number to multiply by is 7, and the square root of the new number is 84.
4. For 2028:
The prime factorization of 2028 is \( 2028 = 2 \times 2 \times 3 \times 13 \times 13 \).
Here, the prime factor 3 does not have a pair.
To make 2028 a perfect square, we must multiply it by 3.
New number: \( 2028 \times 3 = 6084 \).
The prime factorization of 6084 is \( 2 \times 2 \times 3 \times 3 \times 13 \times 13 \).
The square root of 6084 is \( \sqrt{6084} = 2 \times 3 \times 13 = 78 \).
Thus, the smallest whole number to multiply by is 3, and the square root of the new number is 78.
5. For 1458:
The prime factorization of 1458 is \( 1458 = 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \).
Here, the prime factor 2 does not have a pair.
To make 1458 a perfect square, we must multiply it by 2.
New number: \( 1458 \times 2 = 2916 \).
The prime factorization of 2916 is \( 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \).
The square root of 2916 is \( \sqrt{2916} = 2 \times 3 \times 3 \times 3 = 54 \).
Thus, the smallest whole number to multiply by is 2, and the square root of the new number is 54.
6. For 768:
The prime factorization of 768 is \( 768 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \).
Here, the prime factor 3 does not have a pair.
To make 768 a perfect square, we must multiply it by 3.
New number: \( 768 \times 3 = 2304 \).
The prime factorization of 2304 is \( 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \).
The square root of 2304 is \( \sqrt{2304} = 2 \times 2 \times 2 \times 2 \times 3 = 48 \).
Thus, the smallest whole number to multiply by is 3, and the square root of the new number is 48.
In simple words: First, find the prime factors of the number. Look for any factor that doesn't have a pair. That unpaired factor is the smallest number you need to multiply by to make it a perfect square. Once you get the new perfect square, find its square root by taking one factor from each pair.

Exam Tip: When using the prime factorization method, make sure to list all prime factors and clearly identify any that are not part of a pair. This determines the multiplier or divisor.

Question 6. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained?
1. 252
2. 2925
3. 396
4. 2645
5. 2800
6. 1620
Answer:
1. For 252:
The prime factorization of 252 is \( 252 = 2 \times 2 \times 3 \times 3 \times 7 \).
Here, the prime factor 7 does not have a pair.
To make 252 a perfect square, we must divide it by 7.
New number: \( \frac{252}{7} = 36 \).
The prime factorization of 36 is \( 2 \times 2 \times 3 \times 3 \).
The square root of 36 is \( \sqrt{36} = 2 \times 3 = 6 \).
Thus, the smallest whole number to divide by is 7, and the square root of the new number is 6.
2. For 2925:
The prime factorization of 2925 is \( 2925 = 3 \times 3 \times 5 \times 5 \times 13 \).
Here, the prime factor 13 does not have a pair.
To make 2925 a perfect square, we must divide it by 13.
New number: \( \frac{2925}{13} = 225 \).
The prime factorization of 225 is \( 3 \times 3 \times 5 \times 5 \).
The square root of 225 is \( \sqrt{225} = 3 \times 5 = 15 \).
Thus, the smallest whole number to divide by is 13, and the square root of the new number is 15.
3. For 396:
The prime factorization of 396 is \( 396 = 2 \times 2 \times 3 \times 3 \times 11 \).
Here, the prime factor 11 does not have a pair.
To make 396 a perfect square, we must divide it by 11.
New number: \( \frac{396}{11} = 36 \).
The prime factorization of 36 is \( 2 \times 2 \times 3 \times 3 \).
The square root of 36 is \( \sqrt{36} = 2 \times 3 = 6 \).
Thus, the smallest whole number to divide by is 11, and the square root of the new number is 6.
4. For 2645:
The prime factorization of 2645 is \( 2645 = 5 \times 23 \times 23 \).
Here, the prime factor 5 does not have a pair.
To make 2645 a perfect square, we must divide it by 5.
New number: \( \frac{2645}{5} = 529 \).
The prime factorization of 529 is \( 23 \times 23 \).
The square root of 529 is \( \sqrt{529} = 23 \).
Thus, the smallest whole number to divide by is 5, and the square root of the new number is 23.
5. For 2800:
The prime factorization of 2800 is \( 2800 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 7 \).
Here, the prime factor 7 does not have a pair.
To make 2800 a perfect square, we must divide it by 7.
New number: \( \frac{2800}{7} = 400 \).
The prime factorization of 400 is \( 2 \times 2 \times 2 \times 2 \times 5 \times 5 \).
The square root of 400 is \( \sqrt{400} = 2 \times 2 \times 5 = 20 \).
Thus, the smallest whole number to divide by is 7, and the square root of the new number is 20.
6. For 1620:
The prime factorization of 1620 is \( 1620 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 \).
Here, the prime factor 5 does not have a pair.
To make 1620 a perfect square, we must divide it by 5.
New number: \( \frac{1620}{5} = 324 \).
The prime factorization of 324 is \( 2 \times 2 \times 3 \times 3 \times 3 \times 3 \).
The square root of 324 is \( \sqrt{324} = 2 \times 3 \times 3 = 18 \).
Thus, the smallest whole number to divide by is 5, and the square root of the new number is 18.
In simple words: Find the prime factors of the number. If any factor is alone (not in a pair), divide the original number by that factor. The result will be a perfect square, and then you can find its square root by pairing up the remaining factors.

Exam Tip: This method for division is similar to multiplication. Always identify the unpaired prime factor and either multiply or divide by it to achieve a perfect square. Make sure to clearly show the prime factorization steps.

Question 7. The students of Class VIII of a school donated Rs.2401 in all, for Prime Minister's National Relief Fund Each student donated as many rupees as the number of students in the class. Find the number of students in the class?
Answer:
Let 'x' represent the total number of students in the class.
According to the problem, each student donated 'x' rupees.
The total amount donated by all students is \( x \times x = x^2 \).
We are given that the total amount donated is Rs.2401.
So, \( x^2 = 2401 \).
To find the number of students, we need to calculate the square root of 2401:
\( x = \sqrt{2401} \)
Let's find the prime factors of 2401:
\( 2401 = 7 \times 7 \times 7 \times 7 \)
So, \( \sqrt{2401} = \sqrt{7 \times 7 \times 7 \times 7} = 7 \times 7 = 49 \).
Therefore, the total number of students in the class is 49.
In simple words: We know the total money donated and that each student gave an amount equal to the number of students. So, if there are 'x' students, 'x' rupees were given by each, making the total \( x^2 \). We find 'x' by calculating the square root of the total donated money.

Exam Tip: In problems where the number of items and the value per item are equal, the total is the square of that number. To find the number, calculate the square root of the total.

Question 8. 2025 plants we so be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row?
Answer:
Let 'x' represent the number of rows in the garden.
According to the problem, each row contains 'x' plants.
The total number of plants is given by the number of rows multiplied by the number of plants per row, which is \( x \times x = x^2 \).
We are given that the total number of plants to be planted is 2025.
So, \( x^2 = 2025 \).
To find the number of rows (and plants per row), we need to calculate the square root of 2025:
\( x = \sqrt{2025} \)
Let's find the prime factors of 2025:
\( 2025 = 3 \times 3 \times 3 \times 3 \times 5 \times 5 \)
So, \( \sqrt{2025} = \sqrt{3 \times 3 \times 3 \times 3 \times 5 \times 5} = 3 \times 3 \times 5 = 45 \).
Therefore, the required number of rows is 45.
Also, the number of plants in each row is 45.
In simple words: Since the number of rows and plants in each row are the same, the total number of plants is a perfect square. To find this number, simply calculate the square root of the total plants.

Exam Tip: This type of problem is a direct application of square roots. Clearly define the variable 'x' and set up the equation, then solve for 'x'.

Question 9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.
Answer:
To find the smallest number divisible by 4, 9, and 10, we first need to determine their Least Common Multiple (LCM).
The prime factorization of each number is:
\( 4 = 2 \times 2 \)
\( 9 = 3 \times 3 \)
\( 10 = 2 \times 5 \)
To find the LCM, we take the highest power of each prime factor:
\( LCM(4, 9, 10) = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180 \).
So, 180 is the smallest number divisible by 4, 9, and 10.
Now, we need to find the smallest *square* number that is divisible by these numbers. This means 180 itself might not be a perfect square.
The prime factors of 180 are \( 2 \times 2 \times 3 \times 3 \times 5 \).
Here, the prime factor 5 is unpaired.
To make 180 a perfect square, we need to multiply it by 5.
New square number: \( 180 \times 5 = 900 \).
The prime factors of 900 are \( 2 \times 2 \times 3 \times 3 \times 5 \times 5 \). All prime factors are now paired.
Therefore, 900 is the smallest square number that is divisible by 4, 9, and 10.
In simple words: First, find the smallest number that 4, 9, and 10 can all divide into evenly (this is the LCM). Then, look at the prime factors of that LCM. If any factor is alone, multiply the LCM by that factor to make all factors paired. The new number will be the smallest square number divisible by all three.

Exam Tip: Remember that a square number must have all its prime factors in pairs. After finding the LCM, always check its prime factorization to identify any unpaired factors that need to be multiplied to form a perfect square.

Question 10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20?
Answer:
To find the smallest number divisible by 8, 15, and 20, we first need to determine their Least Common Multiple (LCM).
The prime factorization of each number is:
\( 8 = 2 \times 2 \times 2 = 2^3 \)
\( 15 = 3 \times 5 \)
\( 20 = 2 \times 2 \times 5 = 2^2 \times 5 \)
To find the LCM, we take the highest power of each prime factor present in any of the numbers:
\( LCM(8, 15, 20) = 2^3 \times 3 \times 5 = 8 \times 3 \times 5 = 120 \).
So, 120 is the smallest number divisible by 8, 15, and 20.
Now, we need to find the smallest *square* number that is divisible by these numbers. This means 120 itself might not be a perfect square.
The prime factors of 120 are \( 2 \times 2 \times 2 \times 3 \times 5 \).
Here, the prime factors 2 (one of them), 3, and 5 are unpaired.
To make 120 a perfect square, we need to multiply it by the unpaired factors, which are \( 2 \times 3 \times 5 = 30 \).
New square number: \( 120 \times 30 = 3600 \).
The prime factors of 3600 are \( (2 \times 2 \times 2 \times 3 \times 5) \times (2 \times 3 \times 5) = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 \). All prime factors are now paired.
Therefore, 3600 is the smallest square number that is divisible by 8, 15, and 20.
In simple words: First, find the smallest number divisible by 8, 15, and 20 by getting their LCM. Then, break down that LCM into its prime factors. Identify any prime factors that are not in pairs. Multiply the LCM by those missing factors to create pairs, and the resulting number will be the smallest perfect square divisible by all three original numbers.

Exam Tip: After calculating the LCM, always write out its prime factorization. This visual representation helps to easily spot any factors that are not in pairs, indicating what you need to multiply by to form a perfect square.