GSEB Class 8 Maths Solutions Chapter 6 Square and Square Roots Exercise 6.1

Question 1. What will be the unit digit of the squares of the following numbers?
1. 81
2. 272
3. 799
4. 3853
5. 1234
6. 26387
7. 52698
8. 99880
9. 12796
10. 55555
Answer:
1. The unit digit of 81 is 1. When we square 1, we get \( 1 \times 1 = 1 \). Therefore, the unit digit of \( (81)^2 \) will be 1.
2. The unit digit of 272 is 2. When we square 2, we get \( 2 \times 2 = 4 \). Thus, the unit digit of \( (272)^2 \) will be 4.
3. The unit digit of 799 is 9. When we square 9, we get \( 9 \times 9 = 81 \). So, the unit digit of \( (799)^2 \) will be 1.
4. The unit digit of 3853 is 3. When we square 3, we get \( 3 \times 3 = 9 \). Hence, the unit digit of \( (3853)^2 \) will be 9.
5. The unit digit of 1234 is 4. When we square 4, we get \( 4 \times 4 = 16 \). Consequently, the unit digit of \( (1234)^2 \) will be 6.
6. The unit digit of 26387 is 7. When we square 7, we get \( 7 \times 7 = 49 \). As a result, the unit digit of \( (26387)^2 \) will be 9.
7. The unit digit of 52698 is 8. When we square 8, we get \( 8 \times 8 = 64 \). Therefore, the unit digit of \( (52698)^2 \) will be 4.
8. The unit digit of 99880 is 0. When we square 0, we get \( 0 \times 0 = 0 \). Hence, the unit digit of \( (99880)^2 \) will be 0.
9. The unit digit of 12796 is 6. When we square 6, we get \( 6 \times 6 = 36 \). So, the unit digit of \( (12796)^2 \) will be 6.
10. The unit digit of 55555 is 5. When we square 5, we get \( 5 \times 5 = 25 \). Thus, the unit digit of \( (55555)^2 \) will be 5.
In simple words: To find the unit digit of a squared number, just look at the unit digit of the original number. Then, square that single digit. The unit digit of that result will be the unit digit of the larger number's square.

Exam Tip: Remember that the unit digit of a square number is determined solely by the unit digit of the original number, simplifying the calculation for large numbers.

Question 2. The following numbers are obviously not perfect squares. Give reason?
1. 1057
2. 23453
3. 7928
4. 222222
5. 64000
6. 89722
7. 222000
8. 505050
Answer:
1. The number 1057 ends with the digit 7. Numbers ending with 7 are never perfect squares. Therefore, 1057 is not a perfect square.
2. The number 23453 ends with the digit 3. Numbers ending with 3 are never perfect squares. So, 23453 is not a perfect square.
3. The number 7928 ends with the digit 8. Numbers ending with 8 are never perfect squares. Thus, 7928 is not a perfect square.
4. The number 222222 ends with the digit 2. Numbers ending with 2 are never perfect squares. Hence, 222222 is not a perfect square.
5. The number 64000 has an odd number of zeros at its end (three zeros). Perfect squares always have an even number of zeros at their end. Consequently, 64000 is not a perfect square.
6. The number 89722 ends with the digit 2. Numbers ending with 2 are never perfect squares. As a result, 89722 is not a perfect square.
7. The number 222000 has an odd number of zeros at its end (three zeros). Perfect squares always have an even number of zeros at their end. Therefore, 222000 is not a perfect square.
8. The number 505050 ends with an odd number of zeros. Perfect squares must have an even number of zeros. So, 505050 cannot be a perfect square.
In simple words: Perfect squares only end in 0, 1, 4, 5, 6, or 9. Also, if a number ends in zeros, it must have an even count of zeros to be a perfect square. If a number does not meet these conditions, it is not a perfect square.

Exam Tip: Memorize the unit digits that perfect squares can end with (0, 1, 4, 5, 6, 9) and the rule about even numbers of zeros to quickly identify non-perfect squares.

Question 3. The squares of which of the following would be odd numbers?
1. 431
2. 2826
3. 7779
4. 82004
Answer:
We know that the square of an odd natural number is always an odd number, and the square of an even natural number is always an even number.
1. The number 431 is an odd number. Therefore, its square will also be an odd number.
2. The number 2826 is an even number. So, its square will be an even number.
3. The number 7779 is an odd number. Thus, its square will also be an odd number.
4. The number 82004 is an even number. Hence, its square will be an even number.
In simple words: If you take an odd number and multiply it by itself, the answer will always be odd. If you take an even number and multiply it by itself, the answer will always be even. We use this simple rule to tell if the square will be odd or even.

Exam Tip: A quick way to determine if a number's square is odd or even is to check its unit digit. If the unit digit is odd, the square will be odd; if the unit digit is even, the square will be even.

Question 4. Observe the following pattern and find the missing digits?
\( 11^2 = 121 \)
\( 101^2 = 10201 \)
\( 10101^2 = 1002001 \)
\( 1010101^2 = \) ______ \( 2 \) ______ \( 1 \)
\( 10000001^2 = \) ___________
Answer:
By carefully observing the given pattern, we can fill in the missing digits:
\( 11^2 = 121 \)
\( 101^2 = 10201 \)
\( 10101^2 = 1002001 \)
\( 1010101^2 = 1002003002001 \)
\( 10000001^2 = 100000020000001 \)
In simple words: Look at how the numbers change. When you add a zero in the middle, the middle digit goes up, and you get more zeros around it. The number of ones in the original number tells you how high the middle digit will go.

Exam Tip: Patterns in squares often involve the number of digits or the arrangement of zeros and ones. Look for systematic changes in the output for each added digit or zero.

Question 5. Observe the following pattern and supply the missing number?
\( 11^2 = 121 \)
\( 101^2 = 10201 \)
\( 10101^2 = 102030201 \)
\( 1010101^2 = \) ___________
__________ \( ^2 = 10203040504030201 \)
Answer:
Observing the above pattern helps us find the missing numbers:
\( 11^2 = 121 \)
\( 101^2 = 10201 \)
\( 10101^2 = 102030201 \)
\( 1010101^2 = 1020304030201 \)
\( (101010101)^2 = 10203040504030201 \)
In simple words: This pattern shows that if you have numbers like 1, 101, 10101, etc., their squares increase by adding more digits in the middle. The highest middle digit shows how many '1's were in the original number. For example, 1010101 has four '1's, so its square goes up to 4 in the middle.

Exam Tip: When dealing with such number patterns, count the occurrences of the repeating digit (like '1's in '10101') to predict the peak digit in the square. Each '1' corresponds to an ascending number in the center of the square result.

Question 6. Using the given pattern, find the missing numbers?
\( 1^2 + 2^2 + 2^2 = 3^2 \)
\( 2^2 + 3^2 + 6^2 = 7^2 \)
\( 3^2 + 4^2 + 12^2 = 13^2 \)
\( 4^2 + 5^2 + \) ______ \( ^2 = 21^2 \)
\( 5^2 + \) ______ \( ^2 + 30^2 = 31^2 \)
\( 6^2 + 7^2 + \) ______ \( ^2 = \) ______ \( ^2 \)
Answer:
To find the missing numbers, we need to observe the relationships between the numbers in each equation:
The third number is the product of the first two numbers. For example, in \( 1^2 + 2^2 + 2^2 = 3^2 \), the third number (2) is \( 1 \times 2 \). In \( 2^2 + 3^2 + 6^2 = 7^2 \), the third number (6) is \( 2 \times 3 \).
The fourth number (on the right side of the equation) is one more than the third number. For example, if the third number is 2, the fourth is \( 2+1=3 \). If the third number is 6, the fourth is \( 6+1=7 \).
Using these rules, the missing numbers are:
1. \( 4^2 + 5^2 + (4 \times 5)^2 = 20^2 \). Thus, \( 4^2 + 5^2 + 20^2 = 21^2 \).
2. For \( 5^2 + \) ______ \( ^2 + 30^2 = 31^2 \), the second number must be \( 30 \div 5 = 6 \). So, \( 5^2 + 6^2 + 30^2 = 31^2 \).
3. For \( 6^2 + 7^2 + \) ______ \( ^2 = \) ______ \( ^2 \), the third number is \( 6 \times 7 = 42 \). The fourth number is \( 42+1 = 43 \). So, \( 6^2 + 7^2 + 42^2 = 43^2 \).
In simple words: This pattern involves three squared numbers on one side equaling one squared number on the other. The key is that the third number on the left is always the result of multiplying the first two numbers. Then, the number on the right side of the equals sign is always one greater than that third number.

Exam Tip: When analyzing number patterns, try to find relationships between consecutive numbers or positions. Look for arithmetic operations (addition, subtraction, multiplication, division) or sequential increments that hold true across all given examples.

Question 7. Without adding, find the sum,
1. \( 1+3+5+7+9 \)
2. \( 1+3+5+7+9+11+13+15+17 \)
3. \( 1+3+5+7+9+11+13+15+17+19+21+23 \)
Answer:
The sum of the first 'n' odd natural numbers is always equal to \( n^2 \). We can use this rule to find the sum without actually adding.
1. The series \( 1+3+5+7+9 \) contains the first 5 odd numbers. So, \( n=5 \). The sum is \( 5^2 = 25 \).
2. The series \( 1+3+5+7+9+11+13+15+17 \) contains the first 9 odd numbers. So, \( n=9 \). The sum is \( 9^2 = 81 \).
3. The series \( 1+3+5+7+9+11+13+15+17+19+21+23 \) contains the first 12 odd numbers. So, \( n=12 \). The sum is \( 12^2 = 144 \).
In simple words: If you add up a string of odd numbers starting from 1, the total will always be the square of how many odd numbers you added. Just count the odd numbers, then multiply that count by itself.

Exam Tip: Remember the formula for the sum of the first 'n' odd natural numbers: \( \text{Sum} = n^2 \). This shortcut saves time and prevents calculation errors in tests.

Question 8.
1. Express 49 as the sums of 7 odd numbers.
2. Express 121 as the sums of 11 odd numbers.
Answer:
1. We know that the sum of the first 'n' odd numbers is \( n^2 \). Since \( 49 = 7^2 \), we need to express 49 as the sum of the first 7 odd numbers.
\( 49 = 1+3+5+7+9+11+13 \)
2. Similarly, since \( 121 = 11^2 \), we need to express 121 as the sum of the first 11 odd numbers.
\( 121 = 1+3+5+7+9+11+13+15+17+19+21 \)
In simple words: To write a perfect square like 49 or 121 as a sum of odd numbers, just count out that many odd numbers starting from 1. For 49 (which is 7 squared), list the first 7 odd numbers and add them up. For 121 (which is 11 squared), list the first 11 odd numbers and add them.

Exam Tip: This question tests your understanding of the relationship between perfect squares and the sum of consecutive odd numbers. Ensure you start with 1 and include all odd numbers up to the required count.

Question 9. How many numbers lie between squares of the following numbers?
1. 12 and 13
2. 25 and 26
3. 99 and 100
Answer:
The number of non-perfect square numbers between \( n^2 \) and \( (n+1)^2 \) is always \( 2n \). We will use this rule for all parts.
1. For numbers 12 and 13, we have \( n=12 \). The number of non-square numbers between \( 12^2 \) and \( 13^2 \) is \( 2 \times 12 = 24 \) numbers.
2. For numbers 25 and 26, we have \( n=25 \). The number of non-square numbers between \( 25^2 \) and \( 26^2 \) is \( 2 \times 25 = 50 \) numbers.
3. For numbers 99 and 100, we have \( n=99 \). The number of non-square numbers between \( 99^2 \) and \( 100^2 \) is \( 2 \times 99 = 198 \) numbers.
In simple words: To find how many numbers are between two perfect squares that are right next to each other (like 12 squared and 13 squared), just double the smaller number. For example, between \( 12^2 \) and \( 13^2 \), you double 12 to get 24 numbers.

Exam Tip: Remember the formula \( 2n \) for finding the count of non-square numbers between consecutive squares \( n^2 \) and \( (n+1)^2 \). Always use the smaller of the two numbers for 'n'.