CBSE Class 8 Mathematics Square And Square Roots Worksheet Set 07

Question 1. Find the square root of 729.
Answer: Prime factorization of 729:
\(729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6\)
\(\sqrt{729} = \sqrt{3^6} = 3^3 = 27\)
In simple words: The square root of 729 is 27 because 27 multiplied by itself gives 729.

Exam Tip: Memorize squares up to 30 to quickly identify square roots of common numbers.

Question 2. Fill in the blank using the given pattern
\(7^2 = 49\)
\(67^2 = 4489\)
\(667^2 = 444889\)
\(6667^2 =\) _____
Answer: Observing the pattern:
For n digits of 6 followed by 7, the result starts with n digits of 4, followed by 8, then n digits of 9.
For \(6667^2\), there are three 6s and one 7 (total 4 digits).
Thus, the answer is 44448889.
In simple words: Follow the sequence of repeating 4s and 8s based on the number of sixes.

Exam Tip: Pattern-based questions test observation skills; count the repeating digits carefully.

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 given by \(n^2\).
Counting the terms, there are 9 consecutive odd numbers starting from 1.
Sum = \(9^2 = 81\)
In simple words: Count how many odd numbers are added and square that count.

Exam Tip: This property applies only to consecutive odd numbers starting from 1.

Question 4. Find the square root of 1296
Answer: Using prime factorization:
\(1296 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 = 2^4 \times 3^4\)
\(\sqrt{1296} = 2^2 \times 3^2 = 4 \times 9 = 36\)
In simple words: Break the number into prime factors and take half of each factor's exponent.

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 = \(\frac{361 + 1}{2} = \frac{362}{2} = 181\)
\(19^2 = 180 + 181\)
In simple words: Subtract 1 and add 1 to the square, then divide both results by 2.

Exam Tip: This shortcut works for the square of any odd number.

Question 6. How many numbers lie between squares of 99 and 100.
Answer: Non-perfect square numbers between \(n^2\) and \((n+1)^2\) is given by \(2n\).
Here \(n = 99\).
Number of non-perfect square numbers = \(2 \times 99 = 198\)
In simple words: Multiply the smaller base number by 2 to find how many numbers lie between their squares.

Exam Tip: Remember the formula \(2n\) for numbers between consecutive squares.

Question 7. Find the square of 35 (without actual multiplication).
Answer: For a number ending in 5, i.e., \((a5)^2\), the result ends in 25, and the prefix is \(a \times (a + 1)\).
Here \(a = 3\), so \(a + 1 = 4\).
Prefix = \(3 \times 4 = 12\)
Square = 1225
In simple words: Multiply the tens digit by the next integer and attach 25 at the end.

Exam Tip: This trick is exclusively for numbers ending in 5.

Question 8. Numbers like 1, 4, 9, 16, 25,... are known as:
(A) Set of whole numbers
(B) Square numbers
(C) Integers
(D) None
Answer: (B) Square numbers
These numbers are products of an integer with itself (\(1^2, 2^2, 3^2, \dots\)).
In simple words: Numbers obtained by multiplying an integer by itself are square numbers.

Exam Tip: Also known as perfect squares.

Question 9. Find perfect square between 30 and 40.
(A) 32
(B) 35
(C) 36
(D) 39
Answer: (C) 36
36 is equal to \(6^2\), which lies between 30 and 40.
In simple words: Check which option is the square of a whole number (\(6^2 = 36\)).

Exam Tip: Scan options for known square values.

Question 10. Square number does not end with at units place.
(A) 2, 3, 7, 8
(B) 1, 2, 3, 4
(C) 4, 5, 6, 7
(D) 1, 4, 6, 8
Answer: (A) 2, 3, 7, 8
A perfect square never ends in 2, 3, 7, or 8.
In simple words: Perfect squares never have 2, 3, 7, or 8 in their ones place.

Exam Tip: Use this rule to quickly reject numbers that cannot be perfect squares.

Question 11. Which of the following would end with digit 1.
(A) \(123^2\)
(B) \(77^2\)
(C) \(82^2\)
(D) \(109^2\)
Answer: (D) \(109^2\)
The units digit of \(109^2\) depends on \(9^2 = 81\), which ends in 1.
In simple words: Look at the last digit of each number; \(9 \times 9 = 81\), ending in 1.

Exam Tip: Only square the units digit to find the ending digit of the square.

Question 12. If a number has 1 or 9 in the units place, then its square end with digit:
(A) 9
(B) 1
(C) 3
(D) 2
Answer: (B) 1
\(1^2 = 1\) and \(9^2 = 81\) (ends in 1).
In simple words: Squaring a number ending in 1 or 9 always results in a number ending in 1.

Exam Tip: Remember units digit pairings: (1,9) gives 1, (2,8) gives 4, (3,7) gives 9, (4,6) gives 6.

Question 13. If a number has 4 or 6 in its unit's place, then the square of that number ends with the digit:
(A) 4
(B) 2
(C) 6
(D) None
Answer: (C) 6
\(4^2 = 16\) and \(6^2 = 36\), both end in 6.
In simple words: Squaring numbers ending in 4 or 6 always gives 6 in the units place.

Exam Tip: Keep units digit rules handy for quick objective answers.

Question 14. What will be the "one's digit" in the square of 26387.
(A) 7
(B) 9
(C) 3
(D) 8
Answer: (B) 9
The units digit of 26387 is 7. \(7^2 = 49\), which ends in 9.
In simple words: Square the last digit 7 to get 49, so the ones digit is 9.

Exam Tip: Ignore all digits except the ones digit.

Question 15. What will be the number of zeros in the square of 400.
(A) 2
(B) 4
(C) 8
(D) 6
Answer: (B) 4
When squaring a number ending in zeros, the number of trailing zeros doubles. 400 has 2 zeros, so its square has \(2 \times 2 = 4\) zeros.
In simple words: Double the number of zeros at the end of the number when squaring it.

Exam Tip: \((a \times 10^n)^2 = a^2 \times 10^{2n}\).

Question 16. Non perfect square numbers between the squares of the numbers (n) and (n + 1) is:
(A) None
(B) 2
(C) Infinits
(D) 2n
Answer: (D) 2n
Standard algebraic property for consecutive integer squares.
In simple words: There are always \(2n\) non-square numbers between \(n^2\) and \((n+1)^2\).

Exam Tip: Direct textbook formula.

Question 17. How many numbers are there between \(5^2\) and \(6^2\)
(A) 11
(B) 12
(C) 10
(D) 13
Answer: (C) 10
Using formula \(2n\) where \(n = 5\):
\(2 \times 5 = 10\)
In simple words: Substitute \(n = 5\) into the formula \(2n\).

Answer: (C) 2000
Using formula \(2n\) where \(n = 1000\):
\(2 \times 1000 = 2000\)
In simple words: Multiply 1000 by 2.

Exam Tip: Fast application of the \(2n\) rule.

Question 19. Sum of successive odd numbers 1, 3, 5, 7, 9 and 11 is
(A) 36
(B) 25
(C) 30
(D) 40
Answer: (A) 36
There are 6 odd numbers. Sum = \(6^2 = 36\)
In simple words: Count the odd numbers (6) and square it (\(6^2 = 36\)).

Question 20. Sum of first 'n' odd natural numbers is
(A) (n + 1)
(B) \(n^2\)
(C) (n - 1)
(D) 2n
Answer: (B) \(n^2\)
Standard definition of sum of first \(n\) odd numbers.
In simple words: The sum of the first \(n\) odd numbers is always \(n^2\).

Question 21. Express \(21^2\) as the sum of two consecutive integers
(A) 220, 199
(B) 220 and 221
(C) 221, 222
(D) None
Answer: (B) 220 and 221
\(21^2 = 441\).
First integer = \(\frac{441 - 1}{2} = 220\)
Second integer = \(\frac{441 + 1}{2} = 221\)
In simple words: Split 441 into two consecutive numbers: 220 and 221.

Question 22. Product of two consecutive even or odd natural numbers \((a + 1) (a - 1)\) is
(A) \((a^2 - 1)\)
(B) (a + 1)
(C) (a - 1)
(D) None
Answer: (A) \((a^2 - 1)\)
Using algebraic identity \((a+1)(a-1) = a^2 - 1^2 = a^2 - 1\).
In simple words: Apply the difference of squares algebraic identity.

Question 23. Product of two consecutive even number 44 and 46 is
(A) 2025
(B) 2024
(C) 2020
(D) 2525
Answer: (B) 2024
Let numbers be \(45 - 1\) and \(45 + 1\).
Product = \(45^2 - 1 = 2025 - 1 = 2024\)
In simple words: Use the identity \((x-1)(x+1) = x^2 - 1\) with \(x = 45\).

Question 24. Pythagorean triplet whose one number 6 is
(A) 6, 8, 10
(B) 6, 9, 12
(C) 3, 4, 6
(D) 6, 5, 4
Answer: (A) 6, 8, 10
Check options using Pythagoras theorem: \(6^2 + 8^2 = 36 + 64 = 100 = 10^2\).
In simple words: Verify which set satisfies \(a^2 + b^2 = c^2\).

Question 25. Square of what number is 121
(A) Only 11
(B) Only (-11)
(C) Both 11 & (-11)
(D) None of these
Answer: (C) Both 11 & (-11)
Both \(11^2 = 121\) and \((-11)^2 = 121\).
In simple words: Squaring either a positive or negative number gives a positive result.

Question 26. Square root of 121 is
(A) -11
(B) 11
(C) Both
(D) None
Answer: (B) 11
By definition, the radical symbol denotes the principal (positive) square root.
In simple words: The standard square root symbol always represents the positive value.

Question 27. Square root of 6400 is
(A) 80
(B) 64
(C) 32
(D) 3200
Answer: (A) 80
\(\sqrt{6400} = \sqrt{64 \times 100} = 8 \times 10 = 80\).
In simple words: Find the square root of 64 (which is 8) and halve the number of zeros (from 2 to 1).

Question 28. What could be the possible "one's" digits of the square root of 99856.
(A) 6
(B) 4
(C) Both 4 & 6
(D) None
Answer: (C) Both 4 & 6
The number ends in 6. Both \(4^2 = 16\) and \(6^2 = 36\) end in 6.
In simple words: Numbers ending in 4 or 6 when squared yield 6 in the ones place.

Question 29. Number of digits in the square root of 9801 (without any calculation).
(A) 1
(B) 2
(C) 3
(D) 4
Answer: (B) 2
Number of digits in 9801 is 4 (even, \(n=4\)). Number of digits in square root = \(\frac{n}{2} = \frac{4}{2} = 2\).
In simple words: If the number of digits is even, divide by 2.

Question 30. Number of digits in the square root of 529 (without any calculation)
(A) 1
(B) 2
(C) 3
(D) 4
Answer: (B) 2
Number of digits in 529 is 3 (odd, \(n=3\)). Number of digits in square root = \(\frac{n+1}{2} = \frac{3+1}{2} = 2\).
In simple words: If odd, add 1 and divide by 2.

Question 31. The length of the side of a square whose area is 441 \(\text{m}^2\) is
(A) 29 m
(B) 21 m
(C) 31 m
(D) 41 m
Answer: (B) 21 m
Side = \(\sqrt{\text{Area}} = \sqrt{441} = 21 \text{ m}\).
In simple words: Take the square root of the given area.

Question 32. Square root is the inverse operation of
(A) 21
(B) square
(C) Cube
(D) None
Answer: (B) square
Squaring and taking square root are inverse mathematical operations.
In simple words: Square root undoes squaring.

Question 33. Positive square root of a number is denoted by the symbol
(A) \(\sqrt[3]{}\)
(B) \(\sqrt{}\)
(C) \(\sqrt[n]{}\)
(D) None of these
Answer: (B) \(\sqrt{}\)
The radical sign \(\sqrt{}\) denotes the principal square root.
In simple words: The standard checkmark-like radical symbol represents the square root.

Question 34. Square root of decimal number 12.25 is
(A) 6.05
(B) 2.5
(C) 3.5
(D) 0.25
Answer: (C) 3.5
\(3.5 \times 3.5 = 12.25\).
In simple words: Since \(35^2 = 1225\), \(3.5^2 = 12.25\).

Question 35. Number of digits in the square root of 27225 (without any calculation)
(A) 2
(B) 3
(C) 4
(D) 5
Answer: (B) 3
Number of digits = 5 (odd). Number of digits in square root = \(\frac{5 + 1}{2} = 3\).
In simple words: Apply the odd digits formula \((n+1)/2\).

Question 36. In a right triangle ABC, \(\angle B = 90^\circ\), If AC = 13cm, BC = 5cm then AB is equal to:
(A) 10cm
(B) 12cm
(C) 8cm
(D) 6cm
Answer: (B) 12cm
Using Pythagoras theorem: \(AB^2 = AC^2 - BC^2 = 13^2 - 5^2 = 169 - 25 = 144\).
\(AB = \sqrt{144} = 12 \text{ cm}\).
In simple words: Subtract the square of the shorter side from the hypotenuse square and take the square root.

Question 37. Square root of even square number is
(A) Odd
(B) Even
(C) Both
(D) None
Answer: (B) Even
The square of an even number is always even, and its square root is also even.
In simple words: Even numbers yield even square roots.

Question 38. Square of 39 is (without actual multiplication)
(A) 1501
(B) 1529
(C) 1521
(D) 1527
Answer: (C) 1521
\(39^2 = (40 - 1)^2 = 40^2 - 2(40)(1) + 1^2 = 1600 - 80 + 1 = 1521\).
In simple words: Use the algebraic identity \((a-b)^2 = a^2 - 2ab + b^2\).

Question 39. 49 as sum of 7 odd numbers in expressed as
(A) \(1 + 3 + 5 + 7 + 9 + 11 + 13\)
(B) \(1 + 3 + 5 + 7 + 11 + 13 + 15\)
(C) \(11 + 7 + 5 + 3 + 1 + 9 + 15\)
(D) None
Answer: (A) \(1 + 3 + 5 + 7 + 9 + 11 + 13\)
Sum of first 7 consecutive odd numbers is \(7^2 = 49\).
In simple words: List the first 7 consecutive odd numbers starting from 1.

Question 40. What will be the unit digit of the square of 555.
(A) 2
(B) 3
(C) 4
(D) 5
Answer: (D) 5
Units digit is 5. \(5^2 = 25\), which ends in 5.
In simple words: Squaring a number ending in 5 always results in 5.

Question 41. Number of digits in the square root of a perfect square number 14400.
(A) 3
(B) 5
(C) 2
(D) 4
Answer: (A) 3
Number of digits in 14400 is 5 (odd). Square root digits = \(\frac{5 + 1}{2} = 3\).
In simple words: Count the digits (5) and apply \((n+1)/2\).

Question 42. Find the number of digits in the square root of 390625.
Answer: Number of digits in 390625 is 6 (even).
Number of digits in square root = \(\frac{6}{2} = 3\).
In simple words: Divide the total number of digits by 2 since 6 is even.

Exam Tip: Quick formula \(n/2\) for even digit numbers.

Question 43. Find the Square root of 1764.
Answer: Prime factorization of 1764:
\(1764 = 2 \times 2 \times 3 \times 3 \times 7 \times 7 = 2^2 \times 3^2 \times 7^2\)
\(\sqrt{1764} = 2 \times 3 \times 7 = 42\)
In simple words: Factorize into prime pairs and take one factor from each pair.

Exam Tip: Prime factorization is reliable for medium-sized numbers.

Question 44. Find the square root of 3136 by division method.
Answer: Group digits in pairs from right: 31 36.
1. Greatest square \(\le 31\) is 25 (\(5 \times 5 = 25\)). Remainder = 6. Bring down 36 to get 636.
2. Double the quotient 5 to get 10. Find \(10x \times x \le 636\). For \(x = 6\), \(106 \times 6 = 636\).
\(\sqrt{3136} = 56\)
In simple words: Use long division grouping digits in pairs from the right.

Exam Tip: Long division method prevents errors on larger numbers.

Question 45. Area of a square plot is \(4489 \text{ m}^2\). Find the side of the square plot.
Answer: Side = \(\sqrt{\text{Area}} = \sqrt{4489}\)
Using long division or factorization, \(\sqrt{4489} = 67 \text{ m}\).
In simple words: Take the square root of the given area to find the side length.

Exam Tip: Include units (\(\text{m}\)) in the final answer.

Question 46. Find the square root of 7.29.
Answer: \(\sqrt{7.29} = \sqrt{\frac{729}{100}} = \frac{\sqrt{729}}{\sqrt{100}} = \frac{27}{10} = 2.7\)
In simple words: Convert the decimal to a fraction over 100 and take the square root of numerator and denominator.

Exam Tip: Converting to a fraction simplifies decimal square roots.

Question 47. Find the greatest number of 4 digits which is a perfect square.
Answer: The greatest 4-digit number is 9999.
Using long division to find \(\sqrt{9999}\), we get quotient 99 with a remainder of 198.
Greatest 4-digit square = \(9999 - 198 = 9801\)
Alternatively, \(99^2 = 9801\).
In simple words: Subtract the remainder from 9999 after performing long division.

Exam Tip: Always subtract the remainder from the original number to find the greatest square.

Question 48. Find the side of a square whose area is \(1024 \text{ m}^2\).
Answer: Side = \(\sqrt{1024} = 32 \text{ m}\)
In simple words: Take the square root of 1024 to get 32.

Exam Tip: Memorizing powers of 2 helps recognize 1024 as \(2^{10}\).

Question 49. For what value of x the following statement is correct. (2 marks)
\(\sqrt{8x} \times \sqrt{2x} = 144\)
Answer: Combine the square roots:
\(\sqrt{8x \times 2x} = 144\)
\(\sqrt{16x^2} = 144\)
\(4x = 144\)
\(x = \frac{144}{4} = 36\)
In simple words: Multiply terms inside the radical, simplify, and solve for \(x\).

Exam Tip: Use laws of radicals \(\sqrt{a}\sqrt{b} = \sqrt{ab}\).

Question 50. Find \(37^2\) using identity \((a + b)^2 = a^2 + 2ab + b^2\).
Answer: Express 37 as \(30 + 7\):
\(37^2 = (30 + 7)^2\)
\(= 30^2 + 2(30)(7) + 7^2\)
\(= 900 + 420 + 49\)
\(= 1369\)
In simple words: Split the number into tens and units and apply the binomial square identity.

Exam Tip: Splitting into round tens makes mental arithmetic easier.

Question 51. Find the square root of \(\frac{225}{3136}\).
Answer: \(\sqrt{\frac{225}{3136}} = \frac{\sqrt{225}}{\sqrt{3136}}\)
\(\sqrt{225} = 15\)
\(\sqrt{3136} = 56\)
Result = \(\frac{15}{56}\)
In simple words: Take the square roots of the numerator and denominator separately.

Exam Tip: \(\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}\).

Question 52. Find the Square root of 36 by Successive subtraction.
Answer: Subtract successive odd numbers starting from 1 until zero is reached:
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, \(\sqrt{36} = 6\).
In simple words: Subtract consecutive odd numbers sequentially; the count of steps is the square root.

Exam Tip: This method demonstrates the geometric property of square numbers.

Question 53. Find the smallest number that is divisible by each of the numbers 4, 9 and 10.
Answer: First, find the LCM of 4, 9, and 10.
Prime factorizations:
\(4 = 2^2\)
\(9 = 3^2\)
\(10 = 2 \times 5\)
LCM = \(2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180\)
To make 180 a perfect square, we examine its prime factors: \(2^2 \times 3^2 \times 5^1\).
Multiply by 5 to pair the factor 5:
Smallest square number = \(180 \times 5 = 900\)
In simple words: Find the LCM, then multiply by the missing factor to make all exponents even.

Exam Tip: Ensure all prime factors in a perfect square have even exponents.

Question 54. 2025 students are to be 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 = \(x \times x = x^2\)
\(x^2 = 2025\)
\(x = \sqrt{2025} = 45\)
Number of rows = 45, Students per row = 45.
In simple words: Take the square root of the total number of students.

Exam Tip: Equal rows and columns always translate to taking a square root.

Question 55. Find the square root of 0.9 up to two places of decimal.
Answer: Write as \(0.9000\). Using long division for \(\sqrt{0.9000}\):
1. Group decimal pairs: 0.90 00
2. \(0.9 \times 0.9 = 0.81\), remainder = 0.09. Bring down 00 to get 0.0900.
3. Double 0.9 to get 1.8. Test \(1.84 \times 4 = 0.736\). Remainder = 0.164.
4. Bring down another 00 to get 0.16400. Test \(1.889 \times 9 = 0.17001\) (approx).
\(\sqrt{0.9} \approx 0.948 \approx 0.95\)
In simple words: Append zeros to decimal places and perform long division.

Exam Tip: Calculate up to three decimal places and round to two.

Question 56. Find the square root of 2 correct to two places of decimal.
Answer: Using long division for \(\sqrt{2.0000}\):
1. \(1 \times 1 = 1\), remainder = 1. Bring down 00 to get 100.
2. Double 1 to get 2. Test \(24 \times 4 = 96\), remainder = 4. Bring down 00 to get 400.
3. Double 14 to get 28. Test \(281 \times 1 = 281\), remainder = 119.
\(\sqrt{2} = 1.414 \dots \approx 1.41\)
In simple words: Use long division on 2.0000 to find the decimal approximation.

Exam Tip: Memorize \(\sqrt{2} \approx 1.414\) and \(\sqrt{3} \approx 1.732\) for speed.

Question 57. Find the square root of 363609.
Answer: Using long division method for 36 36 09:
1. \(6 \times 6 = 36\), remainder = 0. Bring down 36.
2. Double 6 to get 12. Test \(120 \times 0 = 0\), remainder = 36. Bring down 09 to get 3609.
3. Double 60 to get 120. Test \(1203 \times 3 = 3609\), remainder = 0.
\(\sqrt{363609} = 603\)
In simple words: Group in pairs from right and apply long division step-by-step.

Exam Tip: Watch out for placing zeros in the quotient when pairs are smaller than the divisor.

Question 58. A general wishing to draw up his 64019 men in the form of a solid square, found that he had 10 men even. Find the number of men in the front row?
Answer: Total men = 64019. Since 10 men were left over (even/extra), subtract them to form the solid square:
Men in square = \(64019 - 10 = 64009\)
Number of men in the front row = \(\sqrt{64009}\)
Using long division for 64 00 9:
\(\sqrt{64009} = 253\)
In simple words: Subtract the extra men from the total and take the square root.

Exam Tip: Subtract remainder/extra quantities before taking the square root for solid square arrangements.

Question 59. Fill in the blanks
(i) \(\sqrt{49 \times 16} = 7 \times \square\)
(ii) \(\sqrt{9 \times \square} = 6 \times 6\)
(iii) \(\sqrt{28} = 2 \times \square\)
(iv) \(\sqrt{\square} = \sqrt{5} \times \sqrt{6} \times \sqrt{11}\)

Answer:
(i) \(\sqrt{49 \times 16} = \sqrt{49} \times \sqrt{16} = 7 \times 4\). Box = 4.
(ii) \(\sqrt{9 \times \square} = 6 \times 6 = 36\). Squaring both sides: \(9 \times \square = 36^2 = 1296\). Box = \(\frac{1296}{9} = 144\).
(iii) \(\sqrt{28} = \sqrt{4 \times 7} = \sqrt{4} \times \sqrt{7} = 2 \times \sqrt{7}\). Box = \(\sqrt{7}\).
(iv) \(\sqrt{5} \times \sqrt{6} \times \sqrt{11} = \sqrt{5 \times 6 \times 11} = \sqrt{330}\). Box = 330.
In simple words: Use radical properties to equate both sides and fill in the missing terms.

Exam Tip: Break down surds into simpler factors to compare coefficients.

Question 60. A society collected Rs 92.16. Each member collected as many pairs as there were members: How many members were there and how much did each contribute.
Answer: Let the number of members be \(x\).
Each member collected \(x\) pairs (which equals \(x\) items/amount units as per context).
Total collection = \(x \times x = x^2 = 92.16\)
\(x = \sqrt{92.16} = 9.6\)
Since membership must be a whole integer, interpret total in paise or re-evaluate: Rs 92.16 = 9216 paise. If each member contributed equal to number of members in paise or units, \(\sqrt{92.16} = 9.6\). Interpreting standard textbook framing: 96 members, each contributed Rs 0.96 or total units adjustment giving 96 members.
In simple words: Take the square root of the total amount to find the number of members.

Exam Tip: Pay attention to decimal placement when extracting square roots of money amounts.

Question 61. Find the least number which must be added to 893304 to obtain a perfect square.
Answer: Using long division for \(\sqrt{893304}\) (pairs: 89 33 04):
1. \(9 \times 9 = 81\), remainder = 8. Bring down 33 to get 833.
2. Double 9 to get 18. Test \(184 \times 4 = 736\), remainder = 97. Bring down 04 to get 9704.
3. Double 94 to get 188. Test \(1885 \times 5 = 9425\).
Next number after 945 is 946.
\(946^2 = 894916\)
Number to be added = \(894916 - 893304 = 1612\)
In simple words: Find the next integer's square that exceeds the number and subtract the original number.

Exam Tip: For "what must be added", find the square of (\(\text{quotient} + 1\) ) and subtract the original number.