Selina Concise Solutions for ICSE Class 8 Mathematics Chapter 3 Squares and Square Roots

Exercise 3(A)

Question 1. Find the square of :
(i) 59
(ii) 6.3
(iii) 15
Answer:
(i) To find the square of 59, we multiply the number by itself: \( 59 \times 59 = 3481 \)
(ii) To calculate the square of 6.3, we multiply it by itself: \( 6.3 \times 6.3 = 39.69 \)
(iii) To obtain the square of 15, we multiply 15 by 15: \( 15 \times 15 = 225 \)
In simple words: To find the square of any number, just multiply that number by itself.

Exam Tip: Always remember that the square of a decimal with one decimal place will have exactly two decimal places.

Question 2. By splitting into prime factors, find the square root of :
(i) 11025
(ii) 396900
(iii) 194481
Answer:
(i) Let us find the prime factors of 11025:
\( 11025 = 3 \times 3 \times 5 \times 5 \times 7 \times 7 \)
Grouping the prime factors into identical pairs:
\( \sqrt{11025} = \sqrt{(3 \times 3) \times (5 \times 5) \times (7 \times 7)} \)
Taking one factor from each pair:
\( \sqrt{11025} = 3 \times 5 \times 7 = 105 \)

(ii) Let us find the prime factors of 396900:
\( 396900 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 7 \times 7 \)
Grouping the prime factors into identical pairs:
\( \sqrt{396900} = \sqrt{(2 \times 2) \times (3 \times 3) \times (3 \times 3) \times (5 \times 5) \times (7 \times 7)} \)
Taking one factor from each pair:
\( \sqrt{396900} = 2 \times 3 \times 3 \times 5 \times 7 = 630 \)

(iii) Let us find the prime factors of 194481:
\( 194481 = 3 \times 3 \times 3 \times 3 \times 7 \times 7 \times 7 \times 7 \)
Grouping the prime factors into identical pairs:
\( \sqrt{194481} = \sqrt{(3 \times 3) \times (3 \times 3) \times (7 \times 7) \times (7 \times 7)} \)
Taking one factor from each pair:
\( \sqrt{194481} = 3 \times 3 \times 7 \times 7 = 441 \)
In simple words: Write the number as a product of prime numbers. Put these prime numbers in matching pairs, choose one number from each pair, and multiply them to get the final square root.

Exam Tip: When using prime factorization, double-check that every prime factor has a matching partner. If any factor is left unpaired, the number is not a perfect square.

Question 3.
(i) Find the smallest number by which 2592 be multiplied so that the product is a perfect square.
(ii) Find the smallest number by which 12748 be multiplied so that the product is a perfect square?
Answer:
(i) Let us find the prime factorization of 2592:
\( 2592 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \)
Grouping identical factors into pairs:
\( 2592 = (2 \times 2) \times (2 \times 2) \times 2 \times (3 \times 3) \times (3 \times 3) \)
We find that one factor of 2 is left without a partner.
Therefore, we must multiply 2592 by 2 so that all prime factors exist in complete pairs. The smallest multiplying factor is 2.

(ii) Let us find the prime factorization of 12748:
\( 12748 = 2 \times 2 \times 3187 \)
Grouping identical factors into pairs:
\( 12748 = (2 \times 2) \times 3187 \)
Here, the prime factor 3187 is left without a partner.
To make the number a perfect square, we must multiply 12748 by 3187 to form a complete pair.
Therefore, the smallest multiplying factor is 3187.
In simple words: Break down the number into its prime factors and group them in pairs. Any factor left without a partner is the number you need to multiply by to make it perfect.

Exam Tip: Be meticulous with division when you encounter large prime factors like 3187 to verify they cannot be divided any further.

Question 4. Find the smallest number by which 10368 be divided, so that the result is a perfect square. Also, find the square root of the resulting numbers.
Answer:
Let us write down the prime factors of 10368:
\( 10368 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \)
Now, group these factors into identical pairs:
\( 10368 = (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times 2 \times (3 \times 3) \times (3 \times 3) \)
Here, one factor of 2 is left without a partner.
To convert this number into a perfect square, we must divide 10368 by this unpaired factor of 2. Thus, the smallest number to divide by is 2.
The resulting perfect square is:
\( \frac{10368}{2} = 5184 \)
The square root of this new number is:
\( \sqrt{5184} = 2 \times 2 \times 2 \times 3 \times 3 = 72 \)
In simple words: Find the prime factors and group them in pairs. Divide the number by the leftover prime factor that has no partner. The remaining paired factors will give you the square root of the new number.

Exam Tip: In division questions, once you divide the number by the unpaired factor, the new square root is simply the product of one factor from each of the remaining complete pairs.

Question 5. Find the square root of :
(i) 0.1764
(ii) \( 96\frac{1}{25} \)
(iii) 0.0169
Answer:
(i) First, convert the decimal 0.1764 into a fraction:
\( 0.1764 = \frac{1764}{10000} \)
Now, find the square root of both numerator and denominator:
\( \sqrt{0.1764} = \sqrt{\frac{1764}{10000}} = \frac{42}{100} = 0.42 \)

(ii) First, convert the mixed fraction \( 96\frac{1}{25} \) into an improper fraction:
\( 96\frac{1}{25} = \frac{96 \times 25 + 1}{25} = \frac{2400 + 1}{25} = \frac{2401}{25} \)
Now, calculate the square root of both top and bottom parts:
\( \sqrt{96\frac{1}{25}} = \sqrt{\frac{2401}{25}} = \frac{49}{5} = 9.8 \)

(iii) Convert the decimal 0.0169 into a fraction:
\( 0.0169 = \frac{169}{10000} \)
Now, find the square root of the numerator and denominator:
\( \sqrt{0.0169} = \sqrt{\frac{169}{10000}} = \frac{13}{100} = 0.13 \)
In simple words: Turn decimals or mixed fractions into normal fractions first. Then, take the square root of the top and bottom numbers separately to get your final decimal or fractional answer.

Exam Tip: Always change mixed numbers into improper fractions before you start finding square roots to prevent simple calculation errors.

Question 6. Evaluate
(i) \( \sqrt{\frac{14.4}{22.5}} \)
(ii) \( \sqrt{\frac{0.225}{28.9}} \)
(iii) \( \sqrt{\frac{25}{32} \times 2\frac{13}{18} \times 0.25} \)
(iv) \( \sqrt{1\frac{4}{5} \times 14\frac{21}{44} \times 2\frac{7}{55}} \)
Answer:
(i) Multiply both numerator and denominator by 10 to clear the decimals:
\( \sqrt{\frac{14.4}{22.5}} = \sqrt{\frac{144}{225}} = \frac{12}{15} = \frac{4}{5} = 0.8 \)

(ii) Multiply both parts by 1000 to remove the decimal points:
\( \sqrt{\frac{0.225}{28.9}} = \sqrt{\frac{225}{28900}} = \frac{15}{170} = \frac{3}{34} \)

(iii) Let us express all values as simple fractions first:
\( 2\frac{13}{18} = \frac{49}{18} \)
\( 0.25 = \frac{25}{100} = \frac{1}{4} \)
Substitute these fractions into the square root:
\( \sqrt{\frac{25}{32} \times \frac{49}{18} \times \frac{1}{4}} = \sqrt{\frac{25 \times 49 \times 1}{32 \times 18 \times 4}} = \sqrt{\frac{5^2 \times 7^2}{2^5 \times (2 \times 3^2) \times 2^2}} = \sqrt{\frac{5^2 \times 7^2}{2^8 \times 3^2}} \)
Taking the square root of the numerator and denominator:
\( = \frac{5 \times 7}{2^4 \times 3} = \frac{35}{16 \times 3} = \frac{35}{48} \)

(iv) Convert each mixed fraction to an improper fraction first:
\( 1\frac{4}{5} = \frac{9}{5} \)
\( 14\frac{21}{44} = \frac{14 \times 44 + 21}{44} = \frac{637}{44} \)
\( 2\frac{7}{55} = \frac{2 \times 55 + 7}{55} = \frac{117}{55} \)
Now, rewrite the full product expression:
\( \sqrt{\frac{9}{5} \times \frac{637}{44} \times \frac{117}{55}} \)
Let us break each component down to its prime factors:
\( 637 = 7 \times 7 \times 13 \)
\( 117 = 9 \times 13 = 3 \times 3 \times 13 \)
Substituting these factors back into the equation:
\( = \sqrt{\frac{3^2 \times (7 \times 7 \times 13) \times (3^2 \times 13)}{5 \times (4 \times 11) \times (5 \times 11)}} = \sqrt{\frac{3^4 \times 7^2 \times 13^2}{2^2 \times 5^2 \times 11^2}} \)
Extracting the square root:
\( = \frac{3^2 \times 7 \times 13}{2 \times 5 \times 11} = \frac{9 \times 91}{110} = \frac{819}{110} = 7\frac{49}{110} \)
In simple words: Convert decimals and mixed fractions into regular fractions first. Prime factorize all the terms to cancel out common elements, and then take the square root of the remaining terms.

Exam Tip: Simplify the fraction values inside the square root before attempting to calculate their square roots; this keeps the math simple and easy to track.

Question 7. Evaluate :
(i) \( \sqrt{3^2 \times 6^3 \times 24} \)
(ii) \( \sqrt{(0.5)^3 \times 6 \times 3^5} \)
(iii) \( \sqrt{\left(5 + 2\frac{21}{25}\right) \times \frac{0.169}{1.6}} \)
(iv) \( \sqrt{5\left(2\frac{3}{4} - \frac{3}{10}\right)} \)
(v) \( \sqrt{248 + \sqrt{52 + \sqrt{144}}} \)
Answer:
(i) Group the base factors into even powers to make taking the square root straightforward:
\( \sqrt{3^2 \times 6^3 \times 24} = \sqrt{3^2 \times 6^3 \times (6 \times 4)} = \sqrt{3^2 \times 6^4 \times 2^2} \)
Now, find the square root of each term:
\( = 3 \times 6^2 \times 2 = 3 \times 36 \times 2 = 216 \)

(ii) Group the numbers using exponent laws:
\( \sqrt{(0.5)^3 \times 6 \times 3^5} = \sqrt{(0.5)^3 \times (2 \times 3) \times 3^5} = \sqrt{(0.5)^3 \times 2 \times 3^6} \)
Since \( 0.5 \times 2 = 1 \), rewrite the expression as:
\( = \sqrt{(0.5)^2 \times (0.5 \times 2) \times 3^6} = \sqrt{(0.5)^2 \times 1 \times 3^6} \)
Taking the square root:
\( = 0.5 \times 3^3 = 0.5 \times 27 = 13.5 \)

(iii) First, evaluate the terms within the parentheses:
\( 5 + 2\frac{21}{25} = 5 + \frac{71}{25} = \frac{125 + 71}{25} = \frac{196}{25} \)
Next, simplify the decimal ratio:
\( \frac{0.169}{1.6} = \frac{169}{1600} \)
Substitute these values back into the product:
\( \sqrt{\frac{196}{25} \times \frac{169}{1600}} = \frac{\sqrt{196} \times \sqrt{169}}{\sqrt{25} \times \sqrt{1600}} = \frac{14 \times 13}{5 \times 40} = \frac{182}{200} = \frac{91}{100} = 0.91 \)

(iv) Find the difference within the parentheses first:
\( 2\frac{3}{4} - \frac{3}{10} = \frac{11}{4} - \frac{3}{10} = \frac{55 - 6}{20} = \frac{49}{20} \)
Now multiply the result by 5:
\( 5 \times \frac{49}{20} = \frac{49}{4} \)
Take the square root of the simplified fraction:
\( \sqrt{\frac{49}{4}} = \frac{7}{2} = 3\frac{1}{2} \)

(v) Work your way from the innermost radical to the outer one:
\( \text{Innermost radical: } \sqrt{144} = 12 \)
Substitute 12 back into the expression:
\( \sqrt{248 + \sqrt{52 + 12}} = \sqrt{248 + \sqrt{64}} \)
Find the square root of 64:
\( \sqrt{64} = 8 \)
Substitute 8 back:
\( \sqrt{248 + 8} = \sqrt{256} \)
Finally, find the square root of 256:
\( \sqrt{256} = 16 \)
In simple words: Break down each math problem into smaller pieces. Group powers into pairs, simplify decimal terms, and for nested square roots, solve the deepest square root first and then move outwards.

Exam Tip: For nested square root questions, show each step of your simplification systematically to avoid losing easy marks.

Question 8. A man, after a tour, finds that he had spent every day as many rupees as the number of days he had been on tour. How long did his tour last, if he had spent in all Rs. 1,296
Answer:
Let the duration of the man's tour be \( x \) days.
According to the given condition, the amount of money spent each day is also Rs. \( x \).
Therefore, the total money spent on the entire tour is:
Total Expenditure = \( x \times x = x^2 \) rupees.
We are given that the total spent amount is Rs. 1,296:
\( x^2 = 1296 \)
To find \( x \), we take the square root of 1296:
\( x = \sqrt{1296} \)
Using prime factorization or division:
\( 1296 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \)
Grouping into pairs:
\( x = \sqrt{(2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (3 \times 3)} \)
\( x = 2 \times 2 \times 3 \times 3 = 36 \)
Thus, the man's tour lasted for 36 days.
In simple words: Since the number of days equals the daily cost, multiplying them gives the total cost. Taking the square root of Rs. 1,296 tells us the tour lasted 36 days.

Exam Tip: In word problems, write down your assumptions clearly (e.g., let the number of days be \( x \)) before showing your calculations.

Question 9. Out of 745 students, maximum are to be arranged in the school field for a P.T. display, such that the number of rows is equal to the number of columns. Find the number of rows if 16 students were left out after the arrangement.
Answer:
Total strength of students = 745
Number of students left out after the arrangement = 16
Therefore, the number of students who are arranged in rows and columns = \( 745 - 16 = 729 \)
Let the number of rows in the arrangement be \( y \).
Since the number of rows equals the number of columns, the total number of students arranged is:
\( y \times y = y^2 = 729 \)
To find the number of rows, we calculate the square root of 729:
\( y = \sqrt{729} \)
Finding the prime factors of 729:
\( 729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \)
Grouping them in pairs:
\( y = \sqrt{(3 \times 3) \times (3 \times 3) \times (3 \times 3)} \)
\( y = 3 \times 3 \times 3 = 27 \)
Hence, there are 27 rows of students in the arrangement.
In simple words: First subtract the 16 left-out students from the total. This leaves 729 students in a perfect square shape, and taking the square root of 729 gives 27 rows.

Exam Tip: Always read word problems carefully to identify whether you need to subtract any leftover quantities before calculating the square root.

Question 10. 13 and 31 is a strange pair of numbers such that their squares 169 and 961 are also mirror images of each other. Find two more such pairs.
Answer:
We are looking for pairs of numbers whose squares are mirror images of each other.
Let's examine the pair 12 and 21:
\( (12)^2 = 144 \)
\( (21)^2 = 441 \)
Since 144 and 441 are mirror images, 12 and 21 form one such pair.

Next, let's examine the pair 102 and 201:
\( (102)^2 = 10404 \)
\( (201)^2 = 40401 \)
Since 10404 and 40401 are mirror images, 102 and 201 form another such pair.
In simple words: We find pairs of numbers where reversing the digits of the original number also reverses the digits of its square. Two examples of this are the pairs (12, 21) and (102, 201).

Exam Tip: Such mirror-image properties generally happen with numbers containing small digits like 0, 1, and 2, because multiplying them does not generate carry-overs that alter the order of the digits.

Question 11. Find the smallest perfect square divisible by 3, 4, 5 and 6.
Answer:
First, find the least common multiple (LCM) of 3, 4, 5, and 6:
\( \text{LCM}(3, 4, 5, 6) = 2 \times 2 \times 3 \times 5 = 60 \)
Now, express 60 in terms of its prime factors:
\( 60 = 2 \times 2 \times 3 \times 5 \)
We can see that the prime factors 3 and 5 do not exist in pairs.
To convert 60 into a perfect square, we must multiply it by 3 and 5 to complete their pairs:
Required perfect square = \( 60 \times 3 \times 5 = 60 \times 15 = 900 \)
In simple words: Find the LCM of the given numbers, which is 60. Then multiply 60 by the unpaired prime factors (3 and 5) to turn it into a perfect square, which gives 900.

Exam Tip: Remember that any perfect square divisible by a set of numbers must be a multiple of their LCM. Find the LCM first, and then multiply by the missing factor pairs.

Question 12. If \( \sqrt{784} = 28 \), find the value of:
(i) \( \sqrt{7.84} + \sqrt{78400} \)
(ii) \( \sqrt{0.0784} + \sqrt{0.000784} \)
Answer:
Given that \( \sqrt{784} = 28 \).

(i) To find the value of \( \sqrt{7.84} + \sqrt{78400} \):
\( \sqrt{7.84} = \sqrt{\frac{784}{100}} = \frac{28}{10} = 2.8 \)
\( \sqrt{78400} = \sqrt{784 \times 100} = 28 \times 10 = 280 \)
Adding these two values:
\( \sqrt{7.84} + \sqrt{78400} = 2.8 + 280 = 282.8 \)

(ii) To find the value of \( \sqrt{0.0784} + \sqrt{0.000784} \):
\( \sqrt{0.0784} = \sqrt{\frac{784}{10000}} = \frac{28}{100} = 0.28 \)
\( \sqrt{0.000784} = \sqrt{\frac{784}{1000000}} = \frac{28}{1000} = 0.028 \)
Adding these two values:
\( \sqrt{0.0784} + \sqrt{0.000784} = 0.28 + 0.028 = 0.308 \)
In simple words: Use the given square root of 784, which is 28. Determine the correct decimal point position for each term, and then add them up.

Exam Tip: Pay close attention to the number of decimal digits. When you find the square root of a decimal, the number of decimal places in the answer is exactly half of those in the original number.

Exercise 3(B)

Question 1. Find the square root of:
(i) 4761
(ii) 7744
(iii) 15129
(iv) 0.2916
(v) 0.001225
(vi) 0.023104
(vii) 27.3529
Answer:
(i) Using the division method for 4761:
Group into pairs: \(\overline{47}\,\overline{61}\). The largest square less than 47 is 36 (\(6 \times 6\)). Subtracting 36 gives 11. Bringing down 61 gives 1161. Double the quotient 6 to get 12. Find a digit \(d\) such that \(12d \times d = 1161\). With \(d = 9\), \(129 \times 9 = 1161\).
Therefore, \( \sqrt{4761} = 69 \).

(ii) Using the division method for 7744:
Group into pairs: \(\overline{77}\,\overline{44}\). The largest square less than 77 is 64 (\(8 \times 8\)). Subtracting 64 gives 13. Bringing down 44 gives 1344. Double the quotient 8 to get 16. Find a digit \(d\) such that \(16d \times d = 1344\). With \(d = 8\), \(168 \times 8 = 1344\).
Therefore, \( \sqrt{7744} = 88 \).

(iii) Using the division method for 15129:
Group into pairs: \(\overline{1}\,\overline{51}\,\overline{29}\). The largest square for 1 is 1 (\(1 \times 1\)), leaving 0. Bringing down 51 gives 51. Double 1 to get 2. Since \(22 \times 2 = 44\), write 2 in the quotient and subtract 44 from 51, leaving 7. Bringing down 29 gives 729. Double 12 to get 24. Since \(243 \times 3 = 729\), write 3 in the quotient.
Therefore, \( \sqrt{15129} = 123 \).

(iv) Using the division method for 0.2916:
Group into pairs: \(\overline{29}\,\overline{16}\). Since the integer part is 0, place the decimal point in the quotient. For 29, the largest square is 25 (\(5 \times 5\)), leaving 4. Bringing down 16 gives 416. Double 5 to get 10. Since \(104 \times 4 = 416\), write 4 in the quotient.
Therefore, \( \sqrt{0.2916} = 0.54 \).

(v) Using the division method for 0.001225:
Group into pairs: \(\overline{00}\,\overline{12}\,\overline{25}\). The quotient starts with 0.0. For 12, the largest square is 9 (\(3 \times 3\)), leaving 3. Bringing down 25 gives 325. Double 3 to get 6. Since \(65 \times 5 = 325\), write 5 in the quotient.
Therefore, \( \sqrt{0.001225} = 0.035 \).

(vi) Using the division method for 0.023104:
Group into pairs: \(\overline{02}\,\overline{31}\,\overline{04}\). Quotient starts with 0.1, since \(1 \times 1 = 1\), leaving 1. Bringing down 31 gives 131. Double 1 to get 2. Since \(25 \times 5 = 125\), subtract 125 from 131, leaving 6. Bringing down 04 gives 604. Double 15 to get 30. Since \(302 \times 2 = 604\), write 2 in the quotient.
Therefore, \( \sqrt{0.023104} = 0.152 \).

(vii) Using the division method for 27.3529:
Group into pairs: \(\overline{27}.\overline{35}\,\overline{29}\). For 27, the largest square is 25 (\(5 \times 5\)), leaving 2. Bringing down 35 gives 235. Place a decimal point after 5. Double 5 to get 10. Since \(102 \times 2 = 204\), subtract 204 from 235, leaving 31. Bringing down 29 gives 3129. Double 52 to get 104. Since \(1043 \times 3 = 3129\), write 3 in the quotient.
Therefore, \( \sqrt{27.3529} = 5.23 \).
In simple words: Group the numbers in pairs starting from the decimal point. Find the nearest square value for each pair, subtract, and double the current result to find the next divisor digit.

Exam Tip: Keep your columns aligned when executing long division of square roots to prevent simple mistakes when bringing down digits.

Question 2. Find the square root of:
(i) 4.2025
(ii) 531.7636
(iii) 0.007225
Answer:
(i) To find the square root of 4.2025 using long division:
Group the numbers: \(\overline{4}.\overline{20}\,\overline{25}\). For 4, the root is 2, leaving 0. Put the decimal point in the quotient. Bring down 20. Double 2 to get 4. Since \(41 \times 1 = 41 > 20\), write 0 in the quotient and bring down 25 to get 2025. Double 20 to get 40. Since \(405 \times 5 = 2025\), write 5 in the quotient.
Thus, \( \sqrt{4.2025} = 2.05 \).

(ii) To find the square root of 531.7636 using long division:
Group the numbers: \(\overline{5}\,\overline{31}.\overline{76}\,\overline{36}\). For 5, the largest square is 4 (\(2 \times 2\)), leaving 1. Bring down 31 to make 131. Double 2 to get 4. Since \(43 \times 3 = 129\), subtract 129 from 131, leaving 2. Bring down 76 to get 276 and place a decimal point in the quotient. Double 23 to get 46. Since \(461 \times 1 = 461 > 276\), write 0 in the quotient and bring down 36 to get 27636. Double 230 to get 460. Since \(4606 \times 6 = 27636\), write 6 in the quotient.
Thus, \( \sqrt{531.7636} = 23.06 \).

(iii) To find the square root of 0.007225 using long division:
Group the numbers: \(\overline{00}\,\overline{72}\,\overline{25}\). The quotient starts with 0.0. For 72, the largest square is 64 (\(8 \times 8\)), leaving 8. Bring down 25 to make 825. Double 8 to get 16. Since \(165 \times 5 = 825\), write 5 in the quotient.
Thus, \( \sqrt{0.007225} = 0.085 \).
In simple words: Perform standard long division. If a divisor is too large for the remainder, write a zero in the quotient and bring down the next pair of numbers.

Exam Tip: Remember to put a decimal point in the quotient as soon as you start working with digits to the right of the decimal point.

Question 3. Find the square root of:
(i) 245 correct to two places of decimal.
(ii) 496 correct to three places of decimal.
(iii) 82.6 correct to two places of decimal.
(iv) 0.065 correct to three places of decimal.
(v) 5.2005 correct to two places of decimal.
(vi) 0.602 correct to two places of decimal
Answer:
(i) For 245 correct to two decimal places:
Calculate the square root up to three decimal places by adding three pairs of zeros: \(245.000000\).
Using the division method, we get \( \sqrt{245} \approx 15.652 \).
Since the digit in the third decimal place is 2 (which is less than 5), we round down.
Therefore, the square root correct to two decimal places is 15.65.

(ii) For 496 correct to three decimal places:
Calculate up to four decimal places by adding four pairs of zeros: \(496.00000000\).
Using the division method, we get \( \sqrt{496} \approx 22.2710 \).
Rounding to three decimal places gives 22.271.

(iii) For 82.6 correct to two decimal places:
Write the number as \(82.600000\).
Using long division, we find \( \sqrt{82.6} \approx 9.088 \).
Since the third decimal digit is 8 (which is 5 or more), we round up by adding 1 to the second decimal digit.
Therefore, the square root correct to two decimal places is 9.09.

(iv) For 0.065 correct to three decimal places:
Write the number as \(0.06500000\).
Using long division, we find \( \sqrt{0.065} \approx 0.2549 \).
Since the fourth decimal digit is 9, we round up the third decimal digit from 4 to 5.
Therefore, the square root correct to three decimal places is 0.255.

(v) For 5.2005 correct to two decimal places:
Write the number as \(5.200500\).
Using long division, we find \( \sqrt{5.2005} \approx 2.280 \).
Rounding to two decimal places gives 2.28.

(vi) For 0.602 correct to two decimal places:
Write the number as \(0.602000\).
Using long division, we find \( \sqrt{0.602} \approx 0.775 \).
Since the third decimal digit is 5, we round up the second decimal digit from 7 to 8.
Therefore, the square root correct to two decimal places is 0.78.
In simple words: When a question asks for "correct to n decimal places," find the answer up to n + 1 decimal places. Then, look at that last digit: if it is 5 or more, round the previous digit up; if it is less than 5, round down.

Exam Tip: Always state both the unrounded value (e.g., 9.088) and the rounded value (e.g., 9.09) in your final answer to show your rounding logic clearly.

Question 4. Find the square root of each of the following correct to two decimal places:
(i) \( 3\frac{4}{5} \)
(ii) \( 6\frac{7}{8} \)
Answer:
(i) First, convert the mixed fraction \( 3\frac{4}{5} \) into a decimal, which is \( 3.8 \). To find the square root up to two decimal places, we add zeros to make it \( 3.800000 \). Using the long division method:

The quotient is \( 1.949 \). Rounding this value to two decimal places gives \( 1.95 \).

(ii) Next, convert \( 6\frac{7}{8} \) into a decimal, which is \( 6.875 \). We add a zero at the end to make it \( 6.8750 \) for our calculations. Using the long division method:

The quotient is \( 2.62 \). Thus, the square root correct to two decimal places is \( 2.62 \).
In simple words: To find the square root of a mixed fraction, first change it to a decimal. Then, use long division to find the root. Finally, round it to two decimal places.
Exam Tip: When finding square roots up to two decimal places, always calculate up to three decimal places so you can round off the last digit correctly.

Question 5. For each of the following, find the least number that must be subtracted so that the resulting number is a perfect square.
(i) 796
(ii) 1886
(iii) 23497
Answer:
(i) Finding the square root of 796 with long division:

Since there is a remainder of 12, the smallest number we need to subtract from 796 is 12.

(ii) Finding the square root of 1886 with long division:

Since there is a remainder of 37, the smallest number we need to subtract from 1886 is 37.

(iii) Finding the square root of 23497 with long division:

Since there is a remainder of 88, the smallest number we need to subtract from 23497 is 88.
In simple words: To find what number to subtract to make a perfect square, find the square root by long division. The leftover remainder is the number you need to subtract.
Exam Tip: Always double-check your subtraction: if you subtract the remainder from the original number, the resulting number should be exactly equal to the square of the quotient.

Question 6. For each of the following, find the least number that must be added so that the resulting number is a perfect square.
(i) 511
(ii) 7172
(iii) 55078
Answer:
(i) First, perform long division on 511:

We find a remainder of 27. This shows that \( 22^2 < 511 \). The next perfect square is \( 23^2 = 529 \).
Thus, the number to add is \( 529 - 511 = 18 \).

(ii) Next, perform long division on 7172:

We find a remainder of 116, which means \( 84^2 < 7172 \). The next perfect square is \( 85^2 = 7225 \).
The required number to add is \( 7225 - 7172 = 53 \).

(iii) Finally, perform long division on 55078:

This gives a remainder of 322, indicating \( 234^2 < 55078 \). The next perfect square is \( 235^2 = 55225 \).
So, the number to add is \( 55225 - 55078 = 147 \).
In simple words: To find what to add, first find the square root by division. The quotient tells you the current square. Find the square of the next number up, and subtract your starting number from it.
Exam Tip: For "what to add" questions, the answer is always \( (q+1)^2 - N \), where \( q \) is the quotient from the long division and \( N \) is the given number.

Question 7. Find the square root of 7 correct to two decimal places; then use it to find the value of \( \sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}} \) correct to three significant digits.
Answer:
First, we find \( \sqrt{7} \) by using long division up to three decimal places:

Rounding \( 2.645 \) to two decimal places gives \( 2.65 \).

Now, simplify the expression \( \sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}} \) by rationalizing the denominator:
\( \sqrt{\frac{(4+\sqrt{7})(4+\sqrt{7})}{(4-\sqrt{7})(4+\sqrt{7})}} \)
\( = \sqrt{\frac{(4+\sqrt{7})^2}{(4)^2 - (\sqrt{7})^2}} \)
\( = \sqrt{\frac{(4+\sqrt{7})^2}{16 - 7}} \)
\( = \sqrt{\frac{(4+\sqrt{7})^2}{9}} \)
\( = \frac{4+\sqrt{7}}{3} \)

Substitute \( \sqrt{7} \approx 2.65 \) in the expression:
\( \frac{4 + 2.65}{3} = \frac{6.65}{3} \approx 2.22 \) (correct to three significant digits).
In simple words: First, find the square root of 7. Then, simplify the complex fraction by multiplying the top and bottom by \( 4+\sqrt{7} \). Finally, plug in the value of \( \sqrt{7} \) to find the final answer.
Exam Tip: When rationalizing the denominator, multiply by the conjugate (opposite sign) of the denominator to eliminate the square root at the bottom.

Question 8. Find the value of \( \sqrt{5} \) correct to 2 decimal places; then use it to find the square root of \( \frac{3-\sqrt{5}}{3+\sqrt{5}} \) correct to 2 significant digits.
Answer:
First, find \( \sqrt{5} \) using long division up to three decimal places:

So, \( \sqrt{5} \approx 2.236 \approx 2.24 \) correct to 2 decimal places.

Now, simplify the expression \( \sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}} \) by rationalizing the denominator:
\( \sqrt{\frac{(3-\sqrt{5})(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}} \)
\( = \sqrt{\frac{(3-\sqrt{5})^2}{(3)^2 - (\sqrt{5})^2}} \)
\( = \sqrt{\frac{(3-\sqrt{5})^2}{9 - 5}} \)
\( = \sqrt{\frac{(3-\sqrt{5})^2}{4}} \)
\( = \frac{3-\sqrt{5}}{2} \)

Substitute \( \sqrt{5} \approx 2.24 \) in the simplified form:
\( \frac{3 - 2.24}{2} = \frac{0.76}{2} = 0.38 \).
This is already correct to 2 significant digits.
In simple words: Find \( \sqrt{5} \) up to 2 decimal places. Rationalize the fraction inside the square root to make it simpler, then plug in the value of \( \sqrt{5} \) to calculate the result.
Exam Tip: Remember that 'two significant digits' for \( 0.38 \) starts from the first non-zero digit, which are \( 3 \) and \( 8 \).

Question 9. Find the square root of:
(i) \( \frac{1764}{2809} \)
(ii) \( \frac{507}{4107} \)
(iii) \( \sqrt{108 \times 2028} \)
(iv) \( 0.01 + \sqrt{0.0064} \)
Answer:
(i) Using long division, find the square roots of 1764 and 2809:

And for 2809:

Thus, \( \sqrt{\frac{1764}{2809}} = \frac{42}{53} \).

(ii) First, reduce the fraction \( \frac{507}{4107} \) by dividing the numerator and denominator by 3:
\( \frac{507 \div 3}{4107 \div 3} = \frac{169}{1369} \)
Now find the square roots of 169 and 1369 using long division:

And for 1369:

Thus, \( \sqrt{\frac{169}{1369}} = \frac{13}{37} \).

(iii) Express the given multiplication as prime factors:
\( 108 = 2 \times 2 \times 3 \times 3 \times 3 \)
\( 2028 = 2 \times 2 \times 3 \times 13 \times 13 \)

Multiply them inside the square root:
\( \sqrt{108 \times 2028} = \sqrt{(2 \times 2 \times 3 \times 3 \times 3) \times (2 \times 2 \times 3 \times 13 \times 13)} \)
\( = \sqrt{2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 13 \times 13} \)
\( = 2 \times 2 \times 3 \times 3 \times 13 \)
\( = 4 \times 9 \times 13 = 468 \).

(iv) First find the square of \( 0.0064 \):

So, \( \sqrt{0.0064} = 0.08 \).
Now, perform the addition:
\( 0.01 + 0.08 = 0.09 \).
In simple words: To find the square root of a fraction, find the square roots of the top and bottom separately. For products, you can either multiply first or find their prime factors and pair them up.
Exam Tip: Simplifying a fraction first by dividing the numerator and denominator by their common factor can make it much easier to find the square root.

Question 10. Find the square root of 7.832 correct to :
(i) 2 decimal places
(ii) 2 significant digits.
Answer:
Let us find the square root of 7.832 using the long division method up to 4 decimal places by pairing digits:
\( \sqrt{7.83200000} \):

We get \( \sqrt{7.832} \approx 2.7985 \).

(i) Rounding to two decimal places gives \( 2.80 \).
(ii) Rounding to two significant digits gives \( 2.8 \).
In simple words: Calculate the square root using division up to three decimal places. For two decimal places, round 2.798 to 2.80. For two significant digits, round it to 2.8.
Exam Tip: Be careful with rounding: \( 2.80 \) has three significant digits (the trailing zero counts after a decimal point), whereas \( 2.8 \) has only two significant digits.

Question 11. Find the least number which must be subtracted from 1205 so that the resulting number is a perfect square.
Answer:
Calculate the square root of 1205 using the long division method:

Since there is a remainder of 49 left over, 49 must be subtracted from 1205 to make it a perfect square.
\( 1205 - 49 = 1156 \), which is \( 34^2 \).
In simple words: Find the square root of 1205. The leftover remainder is 49, so subtracting 49 from 1205 will give you a perfect square (1156).
Exam Tip: Whenever you subtract the remainder, the resulting number should be the square of the quotient.

Question 12. Find the least number which must be added to 1205 so that the resulting number is a perfect square.
Answer:
From the long division of 1205, we see that \( 34^2 < 1205 \).
The next higher perfect square is \( 35^2 = 1225 \).
So, the least number to add is \( 1225 - 1205 = 20 \).
In simple words: We know 1205 is between the square of 34 and 35. The square of 35 is 1225. Subtract 1205 from 1225 to find that we need to add 20.

Exam Tip: Use the quotient from the subtraction problem (Question 11) to quickly solve this. Since the quotient is 34, the next perfect square is \( 35^2 \).

Question 13. Find the least number which must be subtracted from 2037 so that the resulting number is a perfect square.
Answer:
Find the square root of 2037 using long division:

The remainder is 12. If we subtract 12 from 2037, we get a perfect square.
\( 2037 - 12 = 2025 \), which is \( 45^2 \).
In simple words: Find the square root of 2037. The remainder is 12, so subtract 12 to make it a perfect square.
Exam Tip: After subtracting, write down the resulting perfect square and its root to show a complete solution.

Question 14. Find the least number which must be added to 5483 so that the resulting number is a perfect square.
Answer:
Find the square root of 5483 using long division:

This shows \( 74^2 < 5483 \). The next perfect square is \( 75^2 = 5625 \).
So, the least number to add is \( 5625 - 5483 = 142 \).
In simple words: The long division shows that 5483 is a bit larger than \( 74^2 \). The next square is \( 75^2 = 5625 \). Subtracting 5483 from 5625 tells us we must add 142.
Exam Tip: For addition problems, find the square of \( q + 1 \) (where \( q \) is the quotient) and then subtract the original number.

Exercise 3(C)

Question 1. Seeing the value of the digit at unit’s place, state which of the following can be square of a number :
(i) 3051
(ii) 2332
(iii) 5684
(iv) 6908
(v) 50699
Answer:
Perfect squares can only end with the digits 0, 1, 4, 5, 6, or 9 at their unit's place. They can never end with 2, 3, 7, or 8.
Let's check the last digits of the given numbers:
- (i) 3051 ends in 1 (can be a square)
- (ii) 2332 ends in 2 (cannot be a square)
- (iii) 5684 ends in 4 (can be a square)
- (iv) 6908 ends in 8 (cannot be a square)
- (v) 50699 ends in 9 (can be a square)
Therefore, the numbers that can be perfect squares are 3051, 5684, and 50699, which correspond to subparts (i), (iii), and (v).
In simple words: Perfect squares can only end with the numbers 0, 1, 4, 5, 6, or 9. Looking at the last digit of each number helps us see which ones might be square numbers.

Exam Tip: Memorize that no perfect square ever ends in 2, 3, 7, or 8. This is a very quick way to eliminate options in exams.

Question 2. Squares of which of the following numbers will have 1 (one) at their unit’s place :
(i) 57
(ii) 81
(iii) 139
(iv) 73
(v) 64
Answer:
The square of a number has 1 at its unit's place if the number ends in either 1 or 9, because \( 1^2 = 1 \) and \( 9^2 = 81 \).
Let's look at the given options:
- (i) 57 ends in 7 (\( 7^2 = 49 \) - ends in 9)
- (ii) 81 ends in 1 (\( 1^2 = 1 \) - ends in 1)
- (iii) 139 ends in 9 (\( 9^2 = 81 \) - ends in 1)
- (iv) 73 ends in 3 (\( 3^2 = 9 \) - ends in 9)
- (v) 64 ends in 4 (\( 4^2 = 16 \) - ends in 6)
Thus, the squares of 81 and 139 will have 1 at their unit's place, which correspond to options (ii) and (iii).
In simple words: To get a 1 at the end of a squared number, the original number must end in 1 or 9. Here, 81 and 139 fit this rule.

Exam Tip: Always square just the last digit of the given number to find the units digit of its square.

Question 3. Which of the following numbers will not have 1 (one) at their unit’s place :
(i) 322
(ii) 572
(iii) 692
(iv) 3212
(v) 2652
Answer:
A squared number ends in 1 only if the original number ends in 1 or 9.
Since all the given options end in the digit 2 (whose square ends in 4), none of their squares will have 1 at their unit's place.
Thus, none of the options (i), (ii), (iii), (iv), and (v) will end in 1 when squared.
In simple words: None of these numbers end in 1 or 9. They all end in 2, so their squares will end in 4. None of them will have 1 at the unit's place.

Exam Tip: A number's square ends in 1 only if the number ends in 1 or 9. Since all options here end in 2, none of their squares will end in 1.

Question 4. Square of which of the following numbers will not have 6 at their unit’s place :
(i) 35
(ii) 23
(iii) 64
(iv) 76
(v) 98
Answer:
A perfect square has 6 at its unit's place only when the original number ends in 4 or 6, because \( 4^2 = 16 \) and \( 6^2 = 36 \).
Let's check each number:
- (i) 35 ends in 5 (\( 5^2 = 25 \) - ends in 5)
- (ii) 23 ends in 3 (\( 3^2 = 9 \) - ends in 9)
- (iii) 64 ends in 4 (\( 4^2 = 16 \) - ends in 6)
- (iv) 76 ends in 6 (\( 6^2 = 36 \) - ends in 6)
- (v) 98 ends in 8 (\( 8^2 = 64 \) - ends in 4)
Therefore, the squares of 35, 23, and 98 will not have 6 at their unit's place, which correspond to options (i), (ii), and (v).
In simple words: Only numbers ending in 4 or 6 have squares that end in 6. The numbers 35, 23, and 98 do not end in 4 or 6, so their squares will not end in 6.

Exam Tip: Quickly check the square of the unit digit: \( 5^2=25 \) (ends in 5), \( 3^2=9 \) (ends in 9), and \( 8^2=64 \) (ends in 4).

Question 5. Which of the following numbers will have 6 at their unit’s place :
(i) 264
(ii) 492
(iii) 346
(iv) 432
(v) 2444
Answer:
The square of a number ends in 6 if the number itself ends in 4 or 6. Let's look at the given values:
- (i) 264 ends in 4 (\( 4^2 = 16 \) - ends in 6)
- (ii) 492 ends in 2 (\( 2^2 = 4 \) - ends in 4)
- (iii) 346 ends in 6 (\( 6^2 = 36 \) - ends in 6)
- (iv) 432 ends in 2 (\( 2^2 = 4 \) - ends in 4)
- (v) 2444 ends in 4 (\( 4^2 = 16 \) - ends in 6)
Thus, the numbers 264, 346, and 2444 will have 6 at their unit's place, which are options (i), (iii), and (v).
In simple words: To have a square that ends in 6, the number itself must end in 4 or 6. Thus, 264, 346, and 2444 will have squares ending in 6.

Exam Tip: Only numbers ending in 4 or 6 produce squares that end in 6.

Question 6. If a number ends with 3 zeroes, how many zeroes will its square have ?
Answer:
If a number ends with \( n \) zeros, then its square will always end with \( 2n \) zeros.
Since the given number has 3 zeros, its square will have \( 3 \times 2 = 6 \) zeros.
In simple words: When you square a number, the number of zeros at the end gets doubled. So, 3 zeros become 6 zeros.

Exam Tip: Always multiply the number of ending zeros by 2 to find the number of zeros in its square.

Question 7. If the square of a number ends with 10 zeroes, how many zeroes will the number have ?
Answer:
We know that the square of a number with \( n \) zeros will have \( 2n \) zeros.
So, if the square ends with 10 zeros, the original number must have:
\( \frac{10}{2} = 5 \) zeros.
In simple words: Since squaring doubles the zeros, we divide the square's zeros by 2 to find how many zeros the original number had. So, 10 zeros become 5.

Exam Tip: Conversely, if you are given the number of zeros in a perfect square, divide by 2 to find the zeros of the original number.

Question 8. Is it possible for the square of a number to end with 5 zeroes ? Give reason.
Answer:
No, it is not possible. A perfect square must end with an even number of zeros (such as 2, 4, 6, etc.), which is represented by \( 2n \) zeros. Since 5 is an odd number, no perfect square can end with exactly 5 zeros.
In simple words: No, a square number must always have an even number of zeros at the end. Since 5 is an odd number, it is impossible.

Exam Tip: Perfect squares always have an even number of trailing zeros (like 2, 4, 6, etc.), never an odd number.

Question 9. Give reason to show that none of the numbers, given below, is a perfect square.
(i) 2162
(ii) 6843
(iii) 9637
(iv) 6598
Answer:
A perfect square never ends with the digits 2, 3, 7, or 8 at its unit's place.
Let's look at the ending digits of the given numbers:
- (i) 2162 ends in 2
- (ii) 6843 ends in 3
- (iii) 9637 ends in 7
- (iv) 6598 ends in 8
Since all of these numbers end with one of these forbidden digits, none of them can be a perfect square.
In simple words: A perfect square never ends in the digits 2, 3, 7, or 8. Since all these numbers end in one of these digits, none of them can be a perfect square.

Exam Tip: Just look at the last digit: if it is 2, 3, 7, or 8, the number is definitely not a perfect square.

Question 10. State, whether the square of the following numbers is even or odd?
(i) 23
(ii) 54
(iii) 76
(iv) 75
Answer:
The square of an odd number is always odd, and the square of an even number is always even.
- (i) 23 is an odd number, so its square is odd.
- (ii) 54 is an even number, so its square is even.
- (iii) 76 is an even number, so its square is even.
- (iv) 75 is an odd number, so its square is odd.
In simple words: Odd numbers have odd squares, and even numbers have even squares. So, 23 and 75 have odd squares, while 54 and 76 have even squares.

Exam Tip: This is a basic property of numbers: \( \text{even}^2 = \text{even} \), and \( \text{odd}^2 = \text{odd} \).

Question 11. Give reason to show that none of the numbers 640, 81000 and 3600000 is a perfect square.
Answer:
A number can only be a perfect square if it has an even number of zeros at the end. Let's check the number of ending zeros for each:
- 640 has 1 zero (odd number of zeros)
- 81000 has 3 zeros (odd number of zeros)
- 3600000 has 5 zeros (odd number of zeros)
Since all these numbers have an odd number of zeros at the end, none of them is a perfect square.
In simple words: A perfect square must have an even number of zeros at the end. Since these numbers have 1, 3, and 5 zeros, they cannot be perfect squares.

Exam Tip: Even if the non-zero part of the number is a perfect square (like 64, 81, 36), the number itself is not a perfect square if it has an odd number of ending zeros.

Question 12. Evaluate:
(i) \( 37^2 - 36^2 \)
(ii) \( 85^2 - 84^2 \)
(iii) \( 101^2 - 100^2 \)
Answer:
We can use the algebraic identity for any natural number \( n \):
\( (n + 1)^2 - n^2 = (n + 1) + n \)

(i) For \( 37^2 - 36^2 \):
\( 37^2 - 36^2 = (36 + 1)^2 - 36^2 = 37 + 36 = 73 \)

(ii) For \( 85^2 - 84^2 \):
\( 85^2 - 84^2 = (84 + 1)^2 - 84^2 = 85 + 84 = 169 \)

(iii) For \( 101^2 - 100^2 \):
\( 101^2 - 100^2 = (100 + 1)^2 - 100^2 = 101 + 100 = 201 \)
In simple words: To find the difference between the squares of two consecutive numbers, just add the two numbers together.

Exam Tip: Using the formula \( a^2 - b^2 = (a-b)(a+b) \) makes this very easy because the difference \( a-b \) is always 1 for consecutive numbers.

Question 13. Without doing the actual addition, find the sum of:
(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
(ii) 1 + 3 + 5 + 7 + 9 + ................ + 39 + 41
(iii) 1 + 3 + 5 + 7 + 9 + ................. + 51 + 53
Answer:
The sum of the first \( n \) odd natural numbers is always equal to \( n^2 \).

(i) In \( 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 \), there are 12 odd numbers.
Sum = \( 12^2 = 144 \).

(ii) In \( 1 + 3 + 5 + 7 + 9 + \dots + 39 + 41 \), the number of terms is \( \frac{41 + 1}{2} = 21 \).
Sum = \( 21^2 = 441 \).

(iii) In \( 1 + 3 + 5 + 7 + 9 + \dots + 51 + 53 \), the number of terms is \( \frac{53 + 1}{2} = 27 \).
Sum = \( 27^2 = 729 \).
In simple words: The sum of consecutive odd numbers starting from 1 is always equal to the square of how many numbers there are.

Exam Tip: To find how many odd numbers are there up to \( L \), use the formula \( n = \frac{L + 1}{2} \).

Question 14. Write three sets of Pythagorean triplets such that each set has numbers less than 30.
Answer:
A Pythagorean triplet is a set of three natural numbers \( a, b, c \) that satisfies \( a^2 + b^2 = c^2 \).
Three sets where all the values are smaller than 30 are:
1. \( 3, 4, \text{ and } 5 \) (since \( 3^2 + 4^2 = 9 + 16 = 25 = 5^2 \))
2. \( 6, 8, \text{ and } 10 \) (since \( 6^2 + 8^2 = 36 + 64 = 100 = 10^2 \))
3. \( 5, 12, \text{ and } 13 \) (since \( 5^2 + 12^2 = 25 + 144 = 169 = 13^2 \))
In simple words: Pythagorean triplets are sets of three numbers where the sum of the squares of the two smaller numbers equals the square of the largest number.

Exam Tip: You can find more triplets by multiplying any basic triplet (like 3, 4, 5) by any natural number (like 2, to get 6, 8, 10).