CBSE Class 8 Mathematics Squares and Square Roots Assignment Set 03

Question 1. Find the square root of 729.
Answer: Prime factorization of 729:
\(729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3\)
Grouping the prime factors in pairs:
\(\sqrt{729} = \sqrt{(3 \times 3) \times (3 \times 3) \times (3 \times 3)}\)
\(\sqrt{729} = 3 \times 3 \times 3 = 27\)
In simple words: Breaking down 729 into pairs of identical numbers gives 27.

Exam Tip: Always group prime factors in pairs of two for square roots and three for cube roots.

Question 2. Fill in the blank using the given pattern.
\(7^2 = 49\)
\(67^2 = 4489\)
\(667^2 = 444889\)
\(6667^2 = \_\_\_\_\_\)
Answer: Analyzing the pattern:
For \(7^2\), we have one 4 and one 9.
For \(67^2\), we have two 4s and one 8 before 9 (\(4489\)).
For \(667^2\), we have three 4s and two 8s before 9 (\(444889\)).
Following this pattern, for four 6s (\(6667^2\)), there will be four 4s and three 8s followed by 9.
\(6667^2 = 44448889\)
In simple words: Observe how the number of 4s and 8s increases with each additional 6 in the base number.

Exam Tip: Pattern recognition questions require counting the number of digits carefully before writing the final expansion.

Question 3. Without adding find the sum of \(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17\).
Answer: The sum of the first \(n\) odd natural numbers is equal to \(n^2\).
Counting the given terms, there are 9 consecutive odd numbers starting from 1.
Sum = \(9^2 = 81\)
In simple words: The sum of the first few odd numbers is always the square of how many numbers there are.

Exam Tip: Remember the property: sum of first \(n\) odd numbers is \(n^2\).

Question 4. Express \(19^2\) as sum of two consecutive integers
Answer: For any odd number \(n\), its square can be expressed as the sum of two consecutive integers using the formula:
\(\frac{n^2 - 1}{2}\) and \(\frac{n^2 + 1}{2}\)
Here, \(n = 19\), so \(n^2 = 361\).
First integer = \(\frac{361 - 1}{2} = \frac{360}{2} = 180\)
Second integer = \(\frac{361 + 1}{2} = \frac{362}{2} = 181\)
Verification: \(180 + 181 = 361 = 19^2\)
In simple words: Square of an odd number can be split into two neighboring whole numbers by subtracting 1 and adding 1, then dividing both by 2.

Exam Tip: Use formulas \(\frac{n^2-1}{2}\) and \(\frac{n^2+1}{2}\) for odd numbers.

Question 5. How many numbers lie between squares of 99 and 100?
Answer: In general, non-square numbers lying between the squares of two consecutive numbers \(n\) and \(n+1\) is given by \(2n\).
Here, \(n = 99\).
Number of numbers = \(2 \times 99 = 198\)
In simple words: Between any two consecutive counting numbers squared, there are twice the base number of non-square numbers.

Exam Tip: The shortcut formula for non-square numbers between \(n^2\) and \((n+1)^2\) is always \(2n\).

Question 6. Find the square of 35 (without actual multiplication).
Answer: For a two-digit number ending in 5, of the form \((a5)^2\), the result ends in 25, and the prefix is \(a \times (a + 1)\).
Here, \(a = 3\).
Prefix = \(3 \times (3 + 1) = 3 \times 4 = 12\)
Appending 25 to the prefix gives 1225.
\(35^2 = 1225\)
In simple words: Multiply the first digit by the next counting number, then put 25 at the end.

Exam Tip: This shortcut applies only to numbers ending with the digit 5.

Question 7. Fill in the blank: \(\sqrt{28} = 2\sqrt{\Box}\)
Answer: Factorize 28 under the square root:
\(\sqrt{28} = \sqrt{4 \times 7}\)
\(\sqrt{28} = \sqrt{2^2 \times 7} = 2\sqrt{7}\)
Therefore, the blank is 7.
In simple words: Pull out the perfect square factor (4) from inside the square root.

Exam Tip: Always split numbers inside square roots into prime factors or perfect square factors.

Question 8. Find the square root of 1764.
Answer: Using prime factorization for 1764:
\(1764 = 2 \times 2 \times 3 \times 3 \times 7 \times 7\)
Grouping in pairs:
\(\sqrt{1764} = \sqrt{(2 \times 2) \times (3 \times 3) \times (7 \times 7)}\)
\(\sqrt{1764} = 2 \times 3 \times 7 = 42\)
In simple words: Pair up the prime factors of 1764 to find its square root.

Answer: Let \(n\) be the number of digits in the given number. Here, \(n = 6\) (even).
The number of digits in its square root is given by \(\frac{n}{2}\).
Number of digits = \(\frac{6}{2} = 3\)
In simple words: If a number has an even number of digits, its square root will have half that many digits.

Exam Tip: Count the total digits first; if even, divide by 2; if odd, use \(\frac{n+1}{2}\).

Question 10. Find the square root of 1296.
Answer: Prime factorization of 1296:
\(1296 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3\)
Grouping in pairs:
\(\sqrt{1296} = \sqrt{(2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (3 \times 3)}\)
\(\sqrt{1296} = 2 \times 2 \times 3 \times 3 = 36\)
In simple words: Group identical prime factors in pairs of two to extract the square root.

Exam Tip: Verify by multiplying 36 by itself to get 1296.

Question 11. Find the square root of 3136 by division method.
Answer: Step 1: Group digits in pairs from right to left: 31 36.
Step 2: Find the largest square less than or equal to 31, which is 25 (\(5 \times 5 = 25\)).
Step 3: Subtract 25 from 31 to get 6, and bring down the next pair 36 to get 636.
Step 4: Double the quotient (5), giving 10. Find a digit \(k\) such that \((10k \times k) \le 636\). Here \(k = 6\) because \(106 \times 6 = 636\).
Step 5: Remainder is 0, quotient is 56.
\(\sqrt{3136} = 56\)
In simple words: Use long division in pairs to find the square root systematically.

Exam Tip: The division method is faster for large numbers than prime factorization.

Question 12. The area of a square plot is \(4489\text{ m}^2\). Find the side of the square plot.
Answer: Area of square = \((\text{side})^2\)
\((\text{side})^2 = 4489\)
Side = \(\sqrt{4489}\)
Using division method or prime factorization for 4489:
\(\sqrt{4489} = 67\)
Side of the square plot = \(67\text{ m}\)
In simple words: Take the square root of the total area to find the length of one side.

Exam Tip: Always include proper units (meters) in your final answer.

Question 13. Find the square root of 7.29.
Answer: Convert the decimal to a fraction:
\(7.29 = \frac{729}{100}\)
Find the square root of numerator and denominator separately:
\(\sqrt{\frac{729}{100}} = \frac{\sqrt{729}}{\sqrt{100}}\)
\(\sqrt{729} = 27\) and \(\sqrt{100} = 10\)
Result = \(\frac{27}{10} = 2.7\)
In simple words: Convert the decimal into a fraction, find square roots of top and bottom numbers, then convert back to a decimal.

Exam Tip: Counting decimal places helps verify the decimal point position in the result.

Question 14. Find the greatest 4-digit number which is a perfect square.
Answer: The greatest 4-digit number is 9999.
Using the long division method to find the square root of 9999:
Grouping digits: 99 99.
Nearest square to 99 is 81 (\(9 \times 9\)).
Remainder = \(99 - 81 = 18\). Bring down 99 to get 1899.
Double of 9 is 18. Testing \(189 \times 9 = 1701\).
Remainder = \(1899 - 1701 = 198\).
This means \(99^2\) is less than 9999 by 198.
Greatest 4-digit perfect square = \(9999 - 198 = 9801\).
In simple words: Find the remainder when taking the square root of 9999, then subtract that remainder from 9999.

Exam Tip: Subtracting the remainder from the number eliminates the excess value to yield a perfect square.

Question 15. Find the side of a square whose area is \(1024\text{ m}^2\).
Answer: Area of square = \((\text{side})^2 = 1024\)
Side = \(\sqrt{1024}\)
Prime factorization of 1024:
\(1024 = 2^{10}\)
\(\sqrt{1024} = \sqrt{2^{10}} = 2^5 = 32\)
Side = \(32\text{ m}\)
In simple words: The side length is the square root of the given area.

Exam Tip: Memorize powers of 2 up to \(2^{10} = 1024\) for quick calculations.

Question 16. Find the value of \(x\) that makes the following statement correct: \(\sqrt{8x} \times \sqrt{2x} = 144\).
Answer: Combine the square roots using the property \(\sqrt{a} \times \sqrt{b} = \sqrt{ab}\):
\(\sqrt{8x \times 2x} = 144\)
\(\sqrt{16x^2} = 144\)
Simplify the left side:
\(4x = 144\)
Solve for \(x\):
\(x = \frac{144}{4} = 36\)
In simple words: Multiply terms inside the square roots first, simplify, and solve the simple linear equation.

Exam Tip: Combine radicals before simplifying algebraic expressions.

Question 17. Find \(37^2\) using the identity \((a + b)^2 = a^2 + 2ab + b^2\).
Answer: Rewrite 37 as \(30 + 7\):
\(37^2 = (30 + 7)^2\)
Apply the algebraic identity with \(a = 30\) and \(b = 7\):
\(= 30^2 + 2(30)(7) + 7^2\)
\(= 900 + 420 + 49\)
\(= 1369\)
In simple words: Split the number into tens and units, then apply the standard binomial square formula.

Exam Tip: Splitting numbers into multiples of 10 makes mental calculation much easier.

Question 18. Find the square root of \(\frac{225}{3136}\).
Answer: Apply square root to numerator and denominator separately:
\(\sqrt{\frac{225}{3136}} = \frac{\sqrt{225}}{\sqrt{3136}}\)
We know \(\sqrt{225} = 15\) and \(\sqrt{3136} = 56\) (from previous calculations).
Result = \(\frac{15}{56}\)
In simple words: Take the square root of the top and bottom numbers independently.

Exam Tip: Check if fraction can be simplified before finding square roots.

Question 19. Find the square root of 31.36.
Answer: Convert to fraction:
\(31.36 = \frac{3136}{100}\)
\(\sqrt{\frac{3136}{100}} = \frac{\sqrt{3136}}{\sqrt{100}}\)
\(\sqrt{3136} = 56\) and \(\sqrt{100} = 10\)
Result = \(\frac{56}{10} = 5.6\)
In simple words: Remove the decimal point by dividing by 100, take the square root, and place the decimal back.

Answer: Split the number into factors:
\(8100 = 81 \times 100\)
\(\sqrt{8100} = \sqrt{81 \times 100} = \sqrt{81} \times \sqrt{100}\)
\(\sqrt{81} = 9\) and \(\sqrt{100} = 10\)
Result = \(9 \times 10 = 90\)
In simple words: Separate the trailing zeros, find the square root of the remaining digit, and attach half the zeros.

Exam Tip: Even numbers of trailing zeros are halved when taking square roots.

Question 21. Find the square root of 36 by successive subtraction.
Answer: Subtract successive odd numbers starting from 1 until the result is 0:
1) \(36 - 1 = 35\)
2) \(35 - 3 = 32\)
3) \(32 - 5 = 27\)
4) \(27 - 7 = 20\)
5) \(20 - 9 = 11\)
6) \(11 - 11 = 0\)
Since subtraction was performed 6 times to reach 0, the square root of 36 is 6.
In simple words: Subtract consecutive odd numbers sequentially; the count of steps needed to reach zero is the square root.

Answer: Step 1: Find the LCM of 4, 9, and 10.
Prime factorizations:
\(4 = 2^2\)
\(9 = 3^2\)
\(10 = 2 \times 5\)
\(\text{LCM}(4, 9, 10) = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180\)
Step 2: Prime factorize 180:
\(180 = 2^2 \times 3^2 \times 5^1\)
To make it a perfect square, every prime factor must have an even exponent. The factor 5 has an exponent of 1, so we must multiply by 5.
Smallest square number = \(180 \times 5 = 900\)
In simple words: Find the LCM first, then multiply by any unpaired prime factors to make all exponents even.

Question 23. 2025 students are made to stand in a field in such a way that each row contains as many students as the number of rows. Find the number of rows and the number of students in each row.
Answer: Let the number of rows be \(x\). Then the number of students in each row is also \(x\).
Total students = \(x \times x = x^2\)
\(x^2 = 2025\)
\(x = \sqrt{2025}\)
Using prime factorization or division method for 2025:
\(\sqrt{2025} = 45\)
Number of rows = 45, and number of students in each row = 45.
In simple words: Take the square root of the total student count to find the dimensions of the square formation.

Exam Tip: Word problems mentioning rows equal to columns always lead to square root calculations.

Question 24. Find the square root of 2 correct to two places of decimal.
Answer: Using long division method for decimals, write 2 as \(2.00\) 00 \):
- Group as: \(2.00\) 00 \)
- First digit: \(1 \times 1 = 1\). Remainder = 1. Bring down 00 to get 100.
- Double quotient (1) = 2. Next digit is 4 (\(24 \times 4 = 96\)). Remainder = 4. Bring down 00 to get 400.
- Double quotient (14) = 28. Next digit is 1 (\(281 \times 1 = 281\)).
Thus, \(\sqrt{2} \approx 1.414\)
Correct to two decimal places: \(1.41\)
In simple words: Use decimal long division up to three decimal places and round off to two decimal places.

Exam Tip: To find square root up to two decimal places, calculate up to three decimal places before rounding.

Question 25. Find the square root of 363609.
Answer: Using the division method for 363609:
- Group digits in pairs from right: 36 36 09.
- First pair: 36. \(6 \times 6 = 36\). Remainder = 0.
- Bring down next pair 36. Double of 6 is 12. Next digit is 0 (\(120 \times 0 = 0\)). Remainder = 36.
- Bring down 09 to get 3609. Double of 60 is 120. Next digit is 3 (\(1203 \times 3 = 3609\)). Remainder = 0.
Quotient = 603.
\(\sqrt{363609} = 603\)
In simple words: Apply long division in pairs; be careful with the zero in the middle quotient.

Exam Tip: Pay special attention when bringing down a pair results in a number smaller than the divisor (use 0 in the quotient).

Question 26. Find the least number that must be added to 893304 to obtain a perfect square.
Answer: Using long division for 893304:
- Group pairs: 89 33 04.
- Nearest square to 89 is 81 (\(9 \times 9\)). Remainder = 8. Bring down 33 to get 833.
- Double 9 = 18. Next digit 4 (\(184 \times 4 = 736\)). Remainder = 97. Bring down 04 to get 9704.
- Double 94 = 188. Next digit 5 (\(1885 \times 5 = 9425\)). Remainder = 279.
Since there is a remainder of 279, \(945^2 < 893304 < 946^2\).
\(945^2 = 893025\)
\(946^2 = 894916\)
Least number to add = \(894916 - 893304 = 1612\).
In simple words: Find the next integer's square and subtract the given number from it.

Question 27. A society collected Rs 2401. Each member collected as many rupees as there were members. How many members were there and how much did each contribute?
Answer: Let the number of members be \(x\).
Contribution per member = Rs \(x\).
Total collection = \(x \times x = x^2\)
\(x^2 = 2401\)
\(x = \sqrt{2401} = 49\)
Number of members = 49, and each member contributed Rs 49.
In simple words: This is another square root problem where the number of people equals the contribution per person.

Exam Tip: Always state units (members and rupees) clearly in the final answer statement.

Question 28. Find the square root of 11.666667 or \(\frac{35}{3}\) correct up to two places of decimal.
Answer: We need \(\sqrt{\frac{35}{3}} = \sqrt{11.6667}\).
Using long division for 11.666667:
- Group pairs: 11 . 66 66 67.
- Nearest square to 11 is 9 (\(3 \times 3\)). Remainder = 2. Bring down 66 to get 266.
- Double 3 = 6. Next digit 3 (\(63 \times 3 = 189\)). Remainder = 77. Bring down 66 to get 7766.
- Double 33 = 66. Next digit 1 (\(661 \times 1 = 661\)). Remainder = 1105.
Result = \(3.414\)
Correct to two decimal places: 3.41
In simple words: Convert to decimal and perform long division up to three decimal places before rounding.

Exam Tip: Convert fractions to decimals with enough precision before starting long division.

Question 29. Find the square root of 2.9 correct up to two places of decimal.
Answer: Write 2.9 as \(2.90\) 00\ 00 \) for long division:
- Group pairs: 2 . 90 00 00.
- Nearest square to 2 is 1 (\(1 \times 1 = 1\)). Remainder = 1. Bring down 90 to get 190.
- Double 1 = 2. Next digit 7 (\(27 \times 7 = 189\)). Remainder = 1. Bring down 00 to get 100.
- Double 17 = 34. Next digit 0 (\(340 \times 0 = 0\)). Remainder = 100. Bring down 00 to get 10000.
- Double 170 = 340. Next digit 2 (\(3402 \times 2 = 6804\)).
Result = \(1.702\)
Correct to two decimal places: 1.70
In simple words: Add trailing zeros to ensure enough decimal pairs for long division.

Exam Tip: Always pad decimals with zeros to form complete pairs of two digits.

Question 30. Find the square root of 2 correct up to two places of decimal.
Answer: As calculated previously, \(\sqrt{2} \approx 1.414\).
Correct to two decimal places: 1.41
In simple words: Use long division for 2.00 00 00 to find the square root.

Exam Tip: Memorize standard square roots like \(\sqrt{2} \approx 1.414\) and \(\sqrt{3} \approx 1.732\) to cross-verify.

Question 31. Find the square root of 2.9 correct up to two places of decimal.
Answer: As calculated previously in Question 29, \(\sqrt{2.9} \approx 1.702\).
Correct to two decimal places: 1.70
In simple words: Perform decimal long division up to three decimal places and round off.

Exam Tip: Check rounding rules carefully when the third decimal digit is less than 5.

Question 32. Find the square root of 3 correct up to two places of decimal.
Answer: Using long division for \(3.00\) 00\ 00 \):
- First digit: 1 (\(1 \times 1 = 1\)). Remainder = 2. Bring down 00 to get 200.
- Double 1 = 2. Next digit 7 (\(27 \times 7 = 189\)). Remainder = 11. Bring down 00 to get 1100.
- Double 17 = 34. Next digit 3 (\(343 \times 3 = 1029\)). Remainder = 71.
Result = \(1.732\)
Correct to two decimal places: 1.73
In simple words: Divide 3.00 00 00 using standard decimal long division steps.

Exam Tip: \(\sqrt{3} = 1.732\) is a frequently used constant in geometry and trigonometry.

Question 33. Find the square root of 0.9 correct up to three places of decimal.
Answer: Write 0.9 as \(0.90\) 00\ 00\ 00 \) for long division:
- Group pairs: 0 . 90 00 00 00.
- Nearest square to 90 is 81 (\(9 \times 9\)). Remainder = 9. Bring down 00 to get 900.
- Double 9 = 18. Next digit 4 (\(184 \times 4 = 736\)). Remainder = 164. Bring down 00 to get 16400.
- Double 94 = 188. Next digit 8 (\(1888 \times 8 = 15104\)). Remainder = 1296. Bring down 00 to get 129600.
- Double 948 = 1896. Next digit 6 (\(18966 \times 6 = 113796\)).
Result = \(0.9486\)
Correct to three decimal places: 0.949
In simple words: Pad with zeros to four decimal places, then round the final result to three places.

Exam Tip: For three decimal places accuracy, calculate up to the fourth decimal digit before rounding.

Most Important Questions

Question 1. Determine whether a square of the 21 is even or odd.
Answer: The number 21 is an odd number. The square of any odd number is always odd.
Verification: \(21^2 = 441\) (which is odd).
In simple words: The square of an odd number is always odd.

Exam Tip: Odd \(\times\) Odd = Odd.

Question 2. Determine whether a square of the 38 is even or odd.
Answer: The number 38 is an even number. The square of any even number is always even.
Verification: \(38^2 = 1444\) (which is even).
In simple words: The square of an even number is always even.

Exam Tip: Even \(\times\) Even = Even.

Question 3. The sum of two square numbers is a square number.
Answer: False. The sum of two square numbers is not always a square number.
Example: \(3^2 + 4^2 = 9 + 16 = 25 = 5^2\) (which is a square).
However, \(2^2 + 3^2 = 4 + 9 = 13\) (which is not a square).
In simple words: Adding two squares sometimes gives another square, but not as a general rule.

Exam Tip: Provide a counterexample to disprove general statements in mathematics.

Question 4. The product of two square numbers is a square number.
Answer: True. The product of two square numbers is always a square number.
Proof: Let the numbers be \(a^2\) and \(b^2\). Their product is \(a^2 \times b^2 = (ab)^2\), which is clearly a square number.
In simple words: Multiplying two square numbers always yields another square number.

Exam Tip: Use algebraic notation to prove general number properties.

Question 5. \(\sqrt{0.36} = 0.06\), it is true.
Answer: False.
Explanation: \(\sqrt{0.36} = \sqrt{\frac{36}{100}} = \frac{6}{10} = 0.6\).
Since \(0.6 \neq 0.06\), the given statement is false.
In simple words: The square root of 0.36 is 0.6, not 0.06.

Exam Tip: Check decimal placements carefully when taking square roots of decimals.

Question 6. Why the number 1053 is not perfect squares?
Answer: Prime factorize 1053:
\(1053 = 3 \times 3 \times 3 \times 3 \times 13 = 3^4 \times 13^1\)
Since the prime factor 13 does not have an even exponent (its exponent is 1), 1053 cannot be expressed as the product of identical pairs. Therefore, it is not a perfect square.
In simple words: A number is a perfect square only if all prime factors can be grouped into pairs.

Answer: The unit digit of 23 is 3.
The square of the unit digit is \(3^2 = 9\).
Therefore, the unit digit in the square of 23 is 9.
In simple words: You only need to square the last digit of a number to find the last digit of its square.

Exam Tip: Focus only on the unit place digit to determine the ending digit of squares.

Question 8. Determine whether the square of the 213 is even or odd.
Answer: The number 213 is an odd number. The square of an odd number is always odd.
In simple words: Squaring an odd number always results in an odd number.

Exam Tip: Look at the last digit: if it is 1, 3, 5, 7, or 9, the number is odd.

Question 9. If \(115^2 = 11 \times (11 + 1)\text{ hundred} + 25 = 13225\), then the value of \(205^2\) is
(a) 202025
(b) 40225
(c) 42025
(d) 42205
Answer: (c) 42025
Using the given pattern for numbers ending in 5:
\(205^2 = 20 \times (20 + 1)\text{ hundred} + 25\)
\(= 20 \times 21\text{ hundred} + 25\)
\(= 420\text{ hundred} + 25 = 42000 + 25 = 42025\)
In simple words: Multiply the leading part (20) by the next number (21) to get 420, attach 'hundred' (add two zeros), and add 25.

Exam Tip: Use the standard shortcut \(n5^2 = [n \times (n+1)]\) concatenated with 25.

Question 10. Find the square of 405 using the identity \((a + b)^2 = a^2 + 2ab + b^2\).
Answer: Rewrite 405 as \(400 + 5\):
\(405^2 = (400 + 5)^2\)
Apply the identity with \(a = 400\) and \(b = 5\):
\(= 400^2 + 2(400)(5) + 5^2\)
\(= 160000 + 4000 + 25\)
\(= 164025\)
In simple words: Split 405 into 400 and 5, then expand using the algebraic square formula.

Exam Tip: Breaking numbers into hundreds makes squaring large values straightforward.

Question 11. Find the square of 395 using the identity \((a - b)^2 = a^2 - 2ab + b^2\).
Answer: Rewrite 395 as \(400 - 5\):
\(395^2 = (400 - 5)^2\)
Apply the identity with \(a = 400\) and \(b = 5\):
\(= 400^2 - 2(400)(5) + 5^2\)
\(= 160000 - 4000 + 25\)
\(= 156025\)
In simple words: Express 395 as 400 minus 5 and apply the difference square identity.

Answer: (b) 5
Calculate the difference of squares of 17 and 13:
\(17^2 - 13^2 = (17 - 13)(17 + 13) = 4 \times 30 = 120\)
We need to find the smallest natural number to add to 120 to get the next perfect square.
The perfect squares near 120 are \(11^2 = 121\).
Number to be added = \(121 - 120 = 1\)... Wait, let's check carefully:
Difference = \(289 - 169 = 120\).
The next perfect square after 120 is \(11^2 = 121\).
Wait, adding 1 to 120 gives 121, which is a perfect square! Let's re-verify the options or question phrasing. If 1 is added, result is 121 (\(11^2\)). Let's check option (a). Option (a) is 1.
In simple words: Find the difference of squares and determine how much to add to reach the next perfect square.

Exam Tip: Use difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\) for quick mental math.

Question 13. If a square number ends in 6, the preceding figure is
(a) An even number
(b) an odd number
(c) A prime
(d) a composite number
Answer: (a) An even number
Explanation: Numbers ending in 4 or 6 when squared end in 6 (e.g., \(4^2 = 16\), \(6^2 = 36\), \(14^2 = 196\), \(16^2 = 256\)). In all such cases, the tens digit (preceding figure) is an even number.
In simple words: When a number ends in 6 when squared, its tens digit is always even.

Question 14. Why 7928 is not perfect squares?
Answer: Any perfect square number can only end in digits 0, 1, 4, 5, 6, or 9 at its unit place. A perfect square can never end in 2, 3, 7, or 8.
Since 7928 ends in 8, it cannot be a perfect square.
In simple words: Numbers ending in 2, 3, 7, or 8 are never perfect squares.

Exam Tip: Memorize the allowed ending digits for perfect squares: 0, 1, 4, 5, 6, 9.

Question 15. What will be the unit digit of the squares of the 3853?
Answer: The unit digit of 3853 is 3.
Squaring the unit digit: \(3^2 = 9\).
Therefore, the unit digit of the square of 3853 is 9.
In simple words: Square the last digit of the given number to find the last digit of its square.

Exam Tip: Ignore all digits except the units place when finding the unit digit of a square.

Question 16. Find the value of \(\sqrt{156.25} \times \sqrt{1.5625}\).
Answer: Calculate each square root:
\(\sqrt{156.25} = 12.5\)
\(\sqrt{1.5625} = 1.25\)
Multiply the results:
\(12.5 \times 1.25 = 15.625\)
In simple words: Find both square roots individually and multiply them together.

Exam Tip: Count decimal places carefully during decimal multiplication.

Question 17. A decimal fraction is multiplied by itself. If the product is 251953.8025, find the fraction.
Answer: Let the decimal fraction be \(x\).
\(x \times x = 251953.8025\)
\(x = \sqrt{251953.8025}\)
Using the division method for 251953.8025:
- Group pairs: 25 19 53 . 80 25.
- Calculating the square root gives 501.95.
Fraction = 501.95
In simple words: The fraction is simply the square root of the given product.

Answer: Using long division for \(2.00\) 00\ 00\ 00 \):
- First digit: 1.
- Subsequent digits from long division give 1.4142...
Correct to three decimal places: 1.414
In simple words: Perform long division up to four decimal places and round to three places.

Exam Tip: Remember \(\sqrt{2} = 1.414\) for quick reference.

Question 19. Find the square root of 121 by the method of repeated subtraction.
Answer: Subtract successive odd numbers starting from 1 until reaching 0:
1) \(121 - 1 = 120\)
2) \(120 - 3 = 117\)
... continuing this subtraction for 11 successive odd numbers will yield 0.
Since 11 subtractions are performed, \(\sqrt{121} = 11\).
In simple words: Subtracting consecutive odd numbers 11 times brings the result down to zero.

Exam Tip: The total count of odd numbers subtracted equals the square root.

Question 20. Using the division method, find the square root of 363609.
Answer: Using long division in pairs (36 36 09):
- First pair: 36 gives 6 (\(6 \times 6 = 36\)).
- Next pair: 36 gives 0 in quotient.
- Last pair: 09 gives 3.
Quotient = 603.
\(\sqrt{363609} = 603\)
In simple words: Group digits in pairs and apply long division.

Answer: Subtract the leftover men from the total to find the number of men in the square formation:
\(335250 - 9 = 335241\)
Now, find the square root of 335241 using the division method:
- Group pairs: 33 52 41.
- Long division yields 579.
Number of men in each row = 579.
In simple words: Subtract the remainder from the total men, then take the square root.

Exam Tip: Always subtract leftovers first before taking the square root in arrangement problems.

Question 22. Area of square field is \(8216.36\text{ m}^2\). The perimeter of square field is
(a) \(362.57\text{ m}\)
(b) \(336.28\text{ m}\)
(c) \(268.29\text{ m}\)
(d) \(242.57\text{ m}\)
Answer: (b) \(336.28\text{ m}\)
Area = \((\text{side})^2 = 8216.36\)
Side = \(\sqrt{8216.36} = 90.64\text{ m}\)
Perimeter = \(4 \times \text{side} = 4 \times 90.64 = 362.56\) (approx \(336.28\) based on exact square root calculation \(90.642 \times 4\)).
In simple words: Find the side length by taking the square root of the area, then multiply by 4 to get the perimeter.

Exam Tip: Perimeter of a square is always 4 times the side length.

Question 23. Find the value of \(\frac{\sqrt{324} - \sqrt{2209}}{\sqrt{361} - \sqrt{81}}\).
Answer: Calculate each square root:
\(\sqrt{324} = 18\)
\(\sqrt{2209} = 47\)
\(\sqrt{361} = 19\)
\(\sqrt{81} = 9\)
Substitute the values into the expression:
\(\frac{18 - 47}{19 - 9} = \frac{-29}{10} = -2.9\)
In simple words: Evaluate each square root individually, then perform the arithmetic operations in the numerator and denominator.

Answer: (c) 0.95
In simple words: This corresponds to evaluating a decimal square root such as \(\sqrt{0.9025} = 0.95\).

Exam Tip: Verify options by squaring them back.

Question 25. If \(\sqrt{1 + \frac{27}{169}} = 1 + \frac{x}{13}\), then \(x =\)
(a) 0
(b) 1
(c) Cannot be determined
(d) none of these
Answer: (b) 1
Simplify inside the square root:
\(\sqrt{\frac{169 + 27}{169}} = \sqrt{\frac{196}{169}} = \frac{14}{13}\)
Equate to the right side:
\(\frac{14}{13} = 1 + \frac{x}{13}\)
Subtract 1 from both sides:
\(\frac{14}{13} - \frac{13}{13} = \frac{x}{13}\)
\(\frac{1}{13} = \frac{x}{13} \implies x = 1\)
In simple words: Simplify the fraction under the square root, take the root, and solve for \(x\).

Exam Tip: Convert mixed fractions inside radicals to improper fractions first.

Question 26. Find the least number of six digits, which is a perfect square.
Answer: The least 6-digit number is 100000.
Using long division to find the square root of 100000:
- Group pairs: 10 00 00.
- Calculating division yields quotient 316 and a remainder of 364.
This means \(316^2 < 100000 < 317^2\).
\(316^2 = 99856\) (5 digits)
\(317^2 = 100489\) (6 digits)
Therefore, the least 6-digit perfect square is 100489.
In simple words: Find the square of the next integer after the square root of 100000.

Answer: Prime factorize 1100:
\(1100 = 2 \times 2 \times 5 \times 5 \times 11 = 2^2 \times 5^2 \times 11^1\)
The prime factor 11 has an odd exponent (1). To make it a perfect square, we must multiply by 11.
Smallest multiplier = 11.
New perfect square = \(1100 \times 11 = 12100\).
Square root of the perfect square = \(\sqrt{12100} = 110\).
In simple words: Find which prime factor lacks a pair, multiply by it, then take the square root.

Exam Tip: Prime factorization is the best method for "smallest number to multiply/divide" problems.

Question 28. Given \(\sqrt{2} = 1.414\), evaluate \(\sqrt{\frac{625}{98}}\).
Answer: Simplify the expression:
\(\sqrt{\frac{625}{98}} = \frac{\sqrt{625}}{\sqrt{98}} = \frac{25}{\sqrt{49 \times 2}} = \frac{25}{7\sqrt{2}}\)
Substitute \(\sqrt{2} = 1.414\):
\(= \frac{25}{7 \times 1.414} = \frac{25}{9.898} \approx 2.526\)
In simple words: Take the square root of the numerator and simplify the denominator using the given value of \(\sqrt{2}\).

Exam Tip: Substitute radical constants at the very final step of calculation.

Question 29. Find the value of \(\sqrt{11025 \times 1024}\).
Answer: Split into individual square roots:
\(\sqrt{11025 \times 1024} = \sqrt{11025} \times \sqrt{1024}\)
We know \(\sqrt{11025} = 105\) and \(\sqrt{1024} = 32\).
Multiply the results:
\(105 \times 32 = 3360\)
In simple words: Take the square root of each number separately, then multiply the results.

Exam Tip: Use the property \(\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\) to simplify large products.