CBSE Class 8 Mathematics Practice Worksheet Set 12

LINEAR EQUATIONS IN ONE VARIABLE

Question 1. Solve the following equations
a. \((b - 2) \times 4 = 8\)
Answer: Divide both sides by 4:
\(b - 2 = \frac{8}{4}\)
\(b - 2 = 2\)
Add 2 to both sides:
\(b = 2 + 2 = 4\)
In simple words: Isolate the variable by performing inverse operations step by step.

Exam Tip: Always check your answer by substituting the value back into the original equation.

Question 1. b. \(2y + 1 = 1\)
Answer: Subtract 1 from both sides:
\(2y = 1 - 1\)
\(2y = 0\)
Divide by 2:
\(y = 0\)
In simple words: Subtracting 1 leaves 0 on the right side, making \(y = 0\).

Exam Tip: Zero divided by any non-zero number is always zero.

Question 1. c. \(5x - 5 = 15\)
Answer: Add 5 to both sides:
\(5x = 15 + 5\)
\(5x = 20\)
Divide by 5:
\(x = \frac{20}{5} = 4\)
In simple words: Move constant terms to one side and divide by the coefficient of \(x\).

Exam Tip: Transposing a term changes its sign from negative to positive.

Question 1. d. \(6n = 24\)
Answer: Divide both sides by 6:
\(n = \frac{24}{6} = 4\)
In simple words: Divide 24 by 6 to find the value of \(n\).

Question 1. e. \(4(m - 9) = 12\)
Answer: Divide both sides by 4 first:
\(m - 9 = \frac{12}{4}\)
\(m - 9 = 3\)
Add 9 to both sides:
\(m = 3 + 9 = 12\)
In simple words: Either expand the bracket or divide both sides by 4 to simplify quickly.

Exam Tip: Dividing first avoids large numbers and makes expansion unnecessary.

Question 1. f. \((5x - 6) / 2 = 7\)
Answer: Multiply both sides by 2:
\(5x - 6 = 7 \times 2\)
\(5x - 6 = 14\)
Add 6 to both sides:
\(5x = 14 + 6\)
\(5x = 20\)
Divide by 5:
\(x = 4\)
In simple words: Clear the denominator by multiplying by 2, then solve normally.

Exam Tip: Always eliminate fractions first by cross multiplication or multiplying by the LCM.

Question 1. g. \(m + 2 = -7\)
Answer: Subtract 2 from both sides:
\(m = -7 - 2\)
\(m = -9\)
In simple words: Move 2 to the right side to get \(-7 - 2\).

Exam Tip: Combining two negative numbers results in a larger negative sum.

Question 1. h. \(7 + (3m / 4) = -2\)
Answer: Subtract 7 from both sides:
\(\frac{3m}{4} = -2 - 7\)
\(\frac{3m}{4} = -9\)
Multiply by 4:
\(3m = -9 \times 4\)
\(3m = -36\)
Divide by 3:
\(m = \frac{-36}{3} = -12\)
In simple words: Isolate the fraction term first, then clear the denominator and coefficient.

Exam Tip: Maintain proper negative signs throughout each transposition step.

Question 1. i. \(0.5x + 7.5 = 11.5\)
Answer: Subtract 7.5 from both sides:
\(0.5x = 11.5 - 7.5\)
\(0.5x = 4.0\)
Divide by 0.5:
\(x = \frac{4}{0.5} = 8\)
In simple words: Subtract decimals first, then divide 4 by 0.5 to get 8.

Exam Tip: Dividing by 0.5 is equivalent to multiplying by 2.

Question 1. j. \(3a - 3.25 = 5.75\)
Answer: Add 3.25 to both sides:
\(3a = 5.75 + 3.25\)
\(3a = 9.00\)
Divide by 3:
\(a = \frac{9}{3} = 3\)
In simple words: Add decimals to get 9, then divide by 3.

Exam Tip: Align decimal points carefully during addition.

Question 1. k. \((10y + 1.5) / 6 = 1.5\)
Answer: Multiply both sides by 6:
\(10y + 1.5 = 1.5 \times 6\)
\(10y + 1.5 = 9.0\)
Subtract 1.5 from both sides:
\(10y = 9.0 - 1.5\)
\(10y = 7.5\)
Divide by 10:
\(y = \frac{7.5}{10} = 0.75\)
In simple words: Multiply by 6, subtract 1.5, and divide by 10.

Exam Tip: Dividing by 10 shifts the decimal point one place to the left.

Question 1. l. \((x / 1.6) + 2.8 = 8.8\)
Answer: Subtract 2.8 from both sides:
\(\frac{x}{1.6} = 8.8 - 2.8\)
\(\frac{x}{1.6} = 6.0\)
Multiply by 1.6:
\(x = 6.0 \times 1.6\)
\(x = 9.6\)
In simple words: Subtract 2.8 to get 6, then multiply by 1.6.

Exam Tip: \(6 \times 1.6 = 9.6\).

Question 1. m. \(5(y - 2) + (y - 3) = 2(2y + 1) - 9\)
Answer: Expand all parentheses:
\(5y - 10 + y - 3 = 4y + 2 - 9\)
Combine like terms on both sides:
\(6y - 13 = 4y - 7\)
Transpose \(4y\) to the left and \(-13\) to the right:
\(6y - 4y = -7 + 13\)
\(2y = 6\)
Divide by 2:
\(y = 3\)
In simple words: Expand brackets, combine like terms, and shift variables to one side.

Exam Tip: Watch out for signs when transposing terms across the equals sign.

Question 1. n. \((m - 1) / 2 - (m - 2) / 3 = (m - 4) / 7\)
Answer: Find the LCM of denominators 2, 3, and 7, which is 42. Multiply the entire equation by 42:
\(21(m - 1) - 14(m - 2) = 6(m - 4)\)
Expand the brackets:
\(21m - 21 - 14m + 28 = 6m - 24\)
Simplify the left side:
\(7m + 7 = 6m - 24\)
Transpose \(6m\) to the left and 7 to the right:
\(7m - 6m = -24 - 7\)
\(m = -31\)
In simple words: Multiply by the LCM to remove all fractions, then solve normally.

Exam Tip: Be extremely careful with negative signs distributed inside parentheses.

Question 1. o. \((3x + 4) / (2 - 6x) = -2 / 5\)
Answer: Cross-multiply:
\(5(3x + 4) = -2(2 - 6x)\)
Expand both sides:
\(15x + 20 = -4 + 12x\)
Transpose \(12x\) to the left and 20 to the right:
\(15x - 12x = -4 - 20\)
\(3x = -24\)
Divide by 3:
\(x = -8\)
In simple words: Cross-multiply to eliminate fractions, then simplify and solve.

Exam Tip: \(-2 \times -6x = +12x\).

Question 1. p. \((5a - 7) / (3a + 1) = 4 / 5\)
Answer: Cross-multiply:
\(5(5a - 7) = 4(3a + 1)\)
Expand both sides:
\(25a - 35 = 12a + 4\)
Transpose \(12a\) to the left and \(-35\) to the right:
\(25a - 12a = 4 + 35\)
\(13a = 39\)
Divide by 13:
\(a = 3\)
In simple words: Cross-multiply, group variables, and solve for \(a\).

Exam Tip: Check your solution by substituting \(a = 3\) back into the original ratio.

Question 1. q. \(6 / (2m - (3 - 4m)) = 2 / 3\)
Answer: Simplify the denominator first:
\(2m - 3 + 4m = 6m - 3\)
So the equation is:
\(\frac{6}{6m - 3} = \frac{2}{3}\)
Cross-multiply:
\(6 \times 3 = 2(6m - 3)\)
\(18 = 12m - 6\)
Add 6 to both sides:
\(24 = 12m\)
Divide by 12:
\(m = 2\)
In simple words: Simplify the inner brackets first, then cross-multiply and solve.

Exam Tip: Always simplify complex denominators before cross-multiplication.

Question 1. r. \((9z - 8) / (4z + 5) = 5 / 3\)
Answer: Cross-multiply:
\(3(9z - 8) = 5(4z + 5)\)
Expand both sides:
\(27z - 24 = 20z + 25\)
Transpose \(20z\) to the left and \(-24\) to the right:
\(27z - 20z = 25 + 24\)
\(7z = 49\)
Divide by 7:
\(z = 7\)
In simple words: Cross-multiply to clear denominators, then isolate \(z\).

Question 1. s. \((7a - 6) / 3 = (3a + 1) / 2\)
Answer: Cross-multiply:
\(2(7a - 6) = 3(3a + 1)\)
Expand both sides:
\(14a - 12 = 9a + 3\)
Transpose \(9a\) to the left and \(-12\) to the right:
\(14a - 9a = 3 + 12\)
\(5a = 15\)
Divide by 5:
\(a = 3\)
In simple words: Cross-multiply and solve for \(a\).

Exam Tip: Cross-multiplication is the fastest method when fraction equals fraction.

Question 1. t. \(2(4x - 1) / 3 - 2x - (1 + x) / 3 = (x + 4) / 3\)
Answer: Multiply the entire equation by 3 to eliminate denominators:
\(2(4x - 1) - 3(2x) - (1 + x) = x + 4\)
Expand all terms:
\(8x - 2 - 6x - 1 - x = x + 4\)
Combine like terms on the left:
\((8x - 6x - x) + (-2 - 1) = x + 4\)
\(x - 3 = x + 4\)
Subtract \(x\) from both sides:
\(-3 = 4\)
Since \(-3 = 4\) is a false statement, there is no solution for this equation.
In simple words: Clear denominators, simplify both sides, and observe that variables cancel out into a false statement.

Exam Tip: When variables cancel out leaving an impossible equality, the equation has no solution.

Question 1. u. \((x - 2) / 8 - (x - 4) / 7 = (2x - 3) / 12 - 2 \frac{3}{4}\)
Answer: Convert mixed fraction \(2 \frac{3}{4}\) to improper fraction \(\frac{11}{4}\):
\(\frac{x - 2}{8} - \frac{x - 4}{7} = \frac{2x - 3}{12} - \frac{11}{4}\)
LCM of denominators 8, 7, 12, and 4 is 168. Multiply the entire equation by 168:
\(21(x - 2) - 24(x - 4) = 14(2x - 3) - 168 \times \frac{11}{4}\)
\(21(x - 2) - 24(x - 4) = 14(2x - 3) - 462\)
Expand brackets:
\(21x - 42 - 24x + 96 = 28x - 42 - 462\)
\(-3x + 54 = 28x - 504\)
Transpose \(-3x\) to the right and \(-504\) to the left:
\(54 + 504 = 28x + 3x\)
\(31x = 558\)
Divide by 31:
\(x = \frac{558}{31} = 18\)
In simple words: Convert mixed numbers, multiply by the LCM of denominators, and solve for \(x\).

Exam Tip: Finding the LCM of all denominators is crucial for simplifying complex multi-fraction equations.

Question 2. The sum of three consecutive multiples of 7 is 84. Find the three multiples. (Ans : 21, 28, 35)
Answer: Let the three consecutive multiples of 7 be \(7x\), \(7(x + 1)\), and \(7(x + 2)\).
According to the question, their sum is 84:
\(7x + 7(x + 1) + 7(x + 2) = 84\)
Divide the entire equation by 7:
\(x + (x + 1) + (x + 2) = \frac{84}{7}\)
\(3x + 3 = 12\)
\(3x = 12 - 3\)
\(3x = 9\)
\(x = 3\)
Therefore, the three multiples are:
\(7(3) = 21\)
\(7(3 + 1) = 28\)
\(7(3 + 2) = 35\)
In simple words: Represent consecutive multiples as \(7x\), \(7x+7\), \(7x+14\), add them to 84, and solve for \(x\).

Exam Tip: Dividing by the common factor 7 at the beginning simplifies the calculation significantly.

Question 3. The perimeter of a rectangle is 62 cm. If the length is 5 cm more than the breadth, find the length and breadth. (Ans : l = 18 cm, b = 13 cm)
Answer: Let the breadth of the rectangle be \(b = x\) cm.
Then the length \(l = (x + 5)\) cm.
Perimeter formula: \(2(l + b) = 62\)
Substitute \(l\) and \(b\):
\(2((x + 5) + x) = 62\)
\(2(2x + 5) = 62\)
Divide by 2:
\(2x + 5 = 31\)
\(2x = 31 - 5\)
\(2x = 26\)
\(x = 13\)
Breadth \(b = 13\) cm.
Length \(l = 13 + 5 = 18\) cm.
In simple words: Use the perimeter formula \(2(\text{length} + \text{breadth}) = 62\) with length expressed in terms of breadth.

Exam Tip: Always include proper units (cm) in the final answer for geometry word problems.

Question 4. The denominator of a fraction is 7 more than the numerator. If 1 is added to the numerator and 6 is added to the denominator, the value of the fraction will be \(\frac{1}{2}\). Find the numerator and the denominator. (Ans : 11/18)
Answer: Let the numerator be \(x\). Then the denominator is \(x + 7\).
Original fraction = \(\frac{x}{x + 7}\)
According to the condition, when 1 is added to numerator and 6 to denominator, the fraction becomes \(\frac{1}{2}\):
\(\frac{x + 1}{(x + 7) + 6} = \frac{1}{2}\)
\(\frac{x + 1}{x + 13} = \frac{1}{2}\)
Cross-multiply:
\(2(x + 1) = 1(x + 13)\)
\(2x + 2 = x + 13\)
\(2x - x = 13 - 2\)
\(x = 11\)
Numerator = 11.
Denominator = \(11 + 7 = 18\).
Fraction = \(\frac{11}{18}\).
In simple words: Set up the fraction with numerator \(x\) and denominator \(x+7\), apply the changes, and cross-multiply.

Exam Tip: Clearly state what \(x\) represents before framing the equation.

Question 5. The ratio of two numbers is 2:3. If both the numbers are increased by 8, their ratio becomes 10:13. Find the numbers. (Ans : 12 and 18)
Answer: Let the two numbers be \(2x\) and \(3x\).
If both are increased by 8, they become \(2x + 8\) and \(3x + 8\).
According to the question, their new ratio is 10:13:
\(\frac{2x + 8}{3x + 8} = \frac{10}{13}\)
Cross-multiply:
\(13(2x + 8) = 10(3x + 8)\)
Expand both sides:
\(26x + 104 = 30x + 80\)
Transpose \(26x\) to the right and 80 to the left:
\(104 - 80 = 30x - 26x\)
\(24 = 4x\)
\(x = 6\)
Therefore, the numbers are:
\(2(6) = 12\)
\(3(6) = 18\)
In simple words: Use \(2x\) and \(3x\) for ratio quantities, add 8 to each, equate to the new ratio, and solve.

Exam Tip: Always multiply \(x\) back by the ratio multipliers to get the final numbers.

Question 6. The length of a rectangle is 7 cm more than its breadth. If the perimeter is 62 cm, find the length and breadth. (Ans : 19 cm and 12 cm)
Answer: Let breadth \(b = x\) cm.
Length \(l = (x + 7)\) cm.
Perimeter = \(2(l + b) = 62\)
\(2((x + 7) + x) = 62\)
\(2(2x + 7) = 62\)
Divide by 2:
\(2x + 7 = 31\)
\(2x = 31 - 7\)
\(2x = 24\)
\(x = 12\)
Breadth \(b = 12\) cm.
Length \(l = 12 + 7 = 19\) cm.
In simple words: Use the perimeter formula with length as \(x+7\) and breadth as \(x\).

Exam Tip: This is a variation of Question 3 with different numerical values.

Question 7. Ameen's age is one fifth of his father's age. After 6 years his age will be one third of his father's age. How old are they now? (Ans : 6 years and 30 years)
Answer: Let Ameen's present age be \(x\) years.
Then his father's present age is \(5x\) years.
After 6 years:
Ameen's age = \(x + 6$
Father's age = \(5x + 6$
According to the question, after 6 years Ameen's age will be \(\frac{1}{3}\) of his father's age:
\(x + 6 = \frac{1}{3}(5x + 6)\)
Multiply both sides by 3:
\(3(x + 6) = 5x + 6\)
\(3x + 18 = 5x + 6\)
Transpose \(3x$ to the right and 6 to the left:
\(18 - 6 = 5x - 3x\)
\(12 = 2x\)
\(x = 6\)
Ameen's present age = 6 years.
Father's present age = \(5 \times 6 = 30\) years.
In simple words: Set Ameen's age as \(x\) and father as \(5x\), set up the future age equation, and solve.

Exam Tip: Always add the same number of years (6) to both persons' present ages.

Question 8. Mother's age is 6 times Amal's age and grandmother's age is twice mother's age. After 5 years the sum of their ages will be 120 years. Find the present age.
Answer: Let Amal's present age be \(x\) years.
Mother's present age = \(6x$ years.
Grandmother's present age = \(2 \times (\text{Mother's age}) = 2(6x) = 12x$ years.
After 5 years:
Amal's age = \(x + 5$
Mother's age = \(6x + 5$
Grandmother's age = \(12x + 5$
According to the question, the sum of their ages after 5 years is 120:
\((x + 5) + (6x + 5) + (12x + 5) = 120\)
Combine like terms:
\(19x + 15 = 120\)
\(19x = 120 - 15\)
\(19x = 105\)
\(x = \frac{105}{19} \approx 5.53\) years (or check source data if integer expected).
In simple words: Express all ages in terms of Amal's age \(x\), add 5 to each for future age, and sum them to 120.

Exam Tip: Remember to add 5 years to each of the three persons when calculating the future sum.

Question 9. The sum of 5 consecutive odd numbers is 135. Find the numbers. (Ans : 23, 25, 27, 29, 31)
Answer: Let the 5 consecutive odd numbers be \(x\(, \(x + 2\), \(x + 4\(, \(x + 6\), and \(x + 8\).
Their sum is 135:
\(x + (x + 2) + (x + 4) + (x + 6) + (x + 8) = 135\)
\(5x + 20 = 135\)
\(5x = 135 - 20\)
\(5x = 115\)
\(x = \frac{115}{5} = 23\)
The five consecutive odd numbers are:
\(23, 25, 27, 29, 31\)
In simple words: Consecutive odd numbers differ by 2, so use \(x, x+2, x+4, x+6, x+8\).

Exam Tip: Consecutive odd numbers increase by 2, just like consecutive even numbers.

Question 10. The ages of Reena and Neema are in the ratio of 5:6. Five years from now their ages will be in the ratio of 6:7. Find their present age.
Answer: Let Reena's present age be \(5x\(and Neema's present age be \(6x\).
Five years from now:
Reena's age = \(5x + 5$
Neema's age = \(6x + 5$
Ratio after 5 years is 6:7:
\(\frac{5x + 5}{6x + 5} = \frac{6}{7}\)
Cross-multiply:
\(7(5x + 5) = 6(6x + 5)\)
\(35x + 35 = 36x + 30\)
Transpose \(35x$ to the right and 30 to the left:
\(35 - 30 = 36x - 35x\)
\(x = 5\)
Reena's present age = \(5(5) = 25$ years.
Neema's present age = \(6(5) = 30$ years.
In simple words: Use \(5x\) and \(6x\) for present ages, add 5 to both, and equate the ratio to \(\frac{6}{7}\).

Exam Tip: Cross-multiplication is the standard technique for solving ratio word problems.

Question 11. Femina's mother is 22 years younger than Femina's grandmother and 27 years older than Femina. The sum of their ages is 121 years. Find their present ages. (Ans : 15, 42, 64)
Answer: Let Femina's age be \(x$ years.
Mother's age = \((x + 27)\) years.
Grandmother's age = \((x + 27) + 22 = x + 49\) years.
The sum of their ages is 121:
\(x + (x + 27) + (x + 49) = 121\)
\(3x + 76 = 121\)
\(3x = 121 - 76\)
\(3x = 45\)
\(x = 15\)
Femina's age = 15 years.
Mother's age = \(15 + 27 = 42$ years.
Grandmother's age = \(42 + 22 = 64$ years.
In simple words: Express all three ages in terms of Femina's age \(x\), sum them to 121, and solve.

Question 12. Two numbers are in the ratio of 7:5. If they differ by 16, what are the numbers? (Ans : 56, 40)
Answer: Let the two numbers be \(7x\(and \(5x\).
Given that they differ by 16:
\(7x - 5x = 16\)
\(2x = 16\)
\(x = 8\)
The numbers are:
\(7(8) = 56\)
\(5(8) = 40\)
In simple words: Use \(7x\) and \(5x\), subtract them to equal 16, and solve for \(x\).

Exam Tip: "Differ by" means the difference between the two quantities.

Question 13. Three consecutive integers are such that when they are taken in increasing order and multiplied by 3, 4, and 5 respectively, they add up to 386. Find these numbers. (Ans : 31, 32, 33)
Answer: Let the three consecutive integers be \(x\(, \(x + 1\), and \(x + 2\).
Multiply them by 3, 4, and 5 respectively and add them:
\(3(x) + 4(x + 1) + 5(x + 2) = 386\)
Expand brackets:
\(3x + 4x + 4 + 5x + 10 = 386\)
Combine like terms:
\(12x + 14 = 386\)
\(12x = 386 - 14\)
\(12x = 372\)
\(x = \frac{372}{12} = 31\)
The three numbers are 31, 32, and 33.
In simple words: Multiply consecutive integers \(x, x+1, x+2\) by 3, 4, 5, sum them to 386, and solve.

Exam Tip: Consecutive integers increase by 1 each time.

Question 14. The angles of a triangle are \(3x\), \((2x + 20)\), and \((5x - 40)\). Find the angles.
Answer: The sum of all angles in a triangle is \(180^\circ\):
\(3x + (2x + 20) + (5x - 40) = 180\)
Combine like terms:
\(10x - 20 = 180\)
\(10x = 180 + 20\)
\(10x = 200\)
\(x = 20\)
Substitute \(x = 20\) into each angle expression:
First angle = \(3(20) = 60^\circ\)
Second angle = \(2(20) + 20 = 40 + 20 = 60^\circ\)
Third angle = \(5(20) - 40 = 100 - 40 = 60^\circ\)
In simple words: Use the angle sum property of triangles (\(180^\circ\)) to set up the equation.

Exam Tip: Always verify that the sum of the calculated angles equals \(180^\circ\).

Question 15. One angle of a triangle is equal to the sum of other two. If the ratio of other two angles is 7:8. Find the angles of the triangle. (Ans : \(90^\circ, 42^\circ, 48^\circ\))
Answer: Let the two angles in ratio 7:8 be \(7x$ and \(8x\).
The third angle is equal to their sum: \(7x + 8x = 15x\).
The sum of all angles in a triangle is \(180^\circ\):
\(7x + 8x + 15x = 180\)
\(30x = 180\)
\(x = 6\)
The angles are:
First angle = \(7(6) = 42^\circ$
Second angle = \(8(6) = 48^\circ$
Third angle = \(15(6) = 90^\circ$
In simple words: Let the two angles be \(7x\) and \(8x\), the third is \(15x\), and their total sum is \(180^\circ\).

Exam Tip: A triangle with an angle of \(90^\circ\) is a right-angled triangle.

Rational Numbers

Question 1) Using appropriate properties find
a) \(-\frac{3}{4} \times \frac{2}{3} + \frac{2}{7} \times -\frac{14}{8} - \frac{5}{6} \times \frac{3}{4}\)
Answer: Rearrange and group terms using commutativity and distributivity:
\(= \left(-\frac{3}{4} \times \frac{2}{3}\right) - \left(\frac{5}{6} \times \frac{3}{4}\right) + \left(\frac{2}{7} \times -\frac{14}{8}\right)\)
Factor out common \(\frac{3}{4}\) from the first two terms:
\(= \frac{3}{4} \left(-\frac{2}{3} - \frac{5}{6}\right) + \left(\frac{2}{7} \times -\frac{14}{8}\right)\)
Simplify inside the bracket (\(-\frac{4}{6} - \frac{5}{6} = -\frac{9}{6} = -\frac{3}{2}\)):
\(= \frac{3}{4} \left(-\frac{3}{2}\right) + \left(-\frac{28}{56}\right)\)
\(= -\frac{9}{8} - \frac{1}{2}\)
Convert to common denominator 8:
\(= -\frac{9}{8} - \frac{4}{8} = -\frac{13}{8}\)
In simple words: Group terms with common factors, simplify using distributive property, and compute the final fraction.

Exam Tip: Look for common multipliers across terms to apply distributive property.

Question 1. b) \(-\frac{7}{8} \times \frac{2}{3} - \frac{1}{4} \times \frac{3}{5} + \frac{4}{5} \times \frac{7}{8}\)
Answer: Rearrange terms containing \(\frac{7}{8}\):
\(= \left(-\frac{7}{8} \times \frac{2}{3}\right) + \left(\frac{4}{5} \times \frac{7}{8}\right) - \left(\frac{1}{4} \times \frac{3}{5}\right)\)
Factor out \(\frac{7}{8}$ from the first two terms:
\(= \frac{7}{8} \left(-\frac{2}{3} + \frac{4}{5}\right) - \frac{3}{20}\)
Simplify inside the bracket (LCM of 3 and 5 is 15):
\(-\frac{2}{3} + \frac{4}{5} = \frac{-10 + 12}{15} = \frac{2}{15}\)
Multiply by \(\frac{7}{8}\):
\(= \frac{7}{8} \times \frac{2}{15} - \frac{3}{20}\)
\(= \frac{14}{120} - \frac{3}{20} = \frac{7}{60} - \frac{9}{60} = -\frac{2}{60} = -\frac{1}{30}\)
In simple words: Rearrange terms to group common factors, apply distributive property, and simplify.

Exam Tip: Always reduce fractions to their simplest form at the end.

Question 2) Fill in the blanks and state the property.
a) \(-\frac{2}{9} \div \left(-\frac{2}{9}\right) = 1\)
Answer: \(1\)
Property: Inverse property / Division of a number by itself.
In simple words: Any non-zero number divided by itself equals 1.

Exam Tip: Memorize standard algebraic properties for rational numbers.

Question 2. b) \(\frac{11}{7} \div \left(-\frac{11}{7}\right) = -1\)
Answer: \(-\frac{11}{7}\)
Property: Division of a number by its negative counterpart.
In simple words: Dividing a number by its negative yields \(-1\).

Exam Tip: Pay close attention to negative signs in division.

Question 2. c) \(-\frac{3}{4} \times \left(\frac{9}{5}\right) = -\frac{9}{5} \times -\frac{3}{4}\)
Answer: \(\frac{9}{5}$\br /> Property: Commutative property of multiplication (\(a \times b = b \times a\)).
In simple words: Changing the order of multiplication does not change the product.

Exam Tip: Commutativity applies to both addition and multiplication.

Question 2. d) \(-\frac{3}{13} \times 0 = 0 \times -\frac{3}{13}\)
Answer: \(0$
Property: Multiplicative property of zero.
In simple words: Multiplying any rational number by zero gives zero.

Exam Tip: Zero is the additive identity, but multiplying by zero always yields zero.

Question 2. e) \(\frac{7}{15} \times 1 = \frac{7}{15}\)
Answer: \(1$
Property: Multiplicative identity (1 is the multiplicative identity).
In simple words: Multiplying by 1 leaves the rational number unchanged.

Exam Tip: 1 is the multiplicative identity; 0 is the additive identity.

Question 2. f) \(\frac{4}{3} + 0 = \frac{4}{3}\)
Answer: \(\frac{4}{3}$ (or 0 in blank depending on presentation format)
Property: Additive identity property.
In simple words: Adding zero to a number leaves it unchanged.

Question 2. g) \(\frac{2}{3} + \frac{5}{7} = \frac{5}{7} + \frac{2}{3}\)
Answer: \(\frac{2}{3}$
Property: Commutative property of addition.
In simple words: Changing the order of addition does not change the sum.

Exam Tip: Addition is commutative for all rational numbers.

Question 2. h) \(\left(-\frac{3}{4} + \frac{1}{2}\right) + \frac{1}{6} = -\frac{3}{4} + \left(\frac{1}{2} + \frac{1}{6}\right)\)
Answer: \(\frac{1}{6}$
Property: Associative property of addition.
In simple words: Grouping of numbers in addition does not affect the sum.

Exam Tip: Associative property applies to addition and multiplication, not subtraction or division.

Question 2. i) \(\left(-\frac{2}{9} \times \frac{3}{5}\right) + \left(-\frac{2}{9} \times \frac{7}{11}\right) = -\frac{2}{9} \times \left(\frac{3}{5} + \frac{7}{11}\right)\)
Answer: \(+\frac{7}{11}$ (with plus sign inside bracket)
Property: Distributive property of multiplication over addition.
In simple words: Multiplication distributes over addition.

Exam Tip: Distributivity helps simplify complex algebraic expressions.

Question 3) Find the additive inverse of
a) \(\frac{3}{7}\)
Answer: \(-\frac{3}{7}\)
In simple words: The additive inverse is the number with the opposite sign.

Exam Tip: A number plus its additive inverse always equals 0.

Question 3. b) \(-\frac{5}{-11}\)
Answer: First simplify \(-\frac{5}{-11} = \frac{5}{11}\).
Additive inverse = \(-\frac{5}{11}\).
In simple words: Simplify the double negative first, then change its sign.

Exam Tip: Always simplify signs before finding inverses.

Question 3. c) \(-\frac{3}{7} \times \frac{5}{-4}\)
Answer: First compute the product: \(-\frac{3}{7} \times -\frac{5}{4} = \frac{15}{28}\).
Additive inverse = \(-\frac{15}{28}\).
In simple words: Multiply first to get a single fraction, then find its additive inverse.

Exam Tip: Negative times negative is positive.

Question 3. d) \(\frac{8}{-9}\)
Answer: Rewrite as \(-\frac{8}{9}\).
Additive inverse = \(\frac{8}{9}\).
In simple words: The negative sign in the denominator makes the whole number negative, so its inverse is positive.

Exam Tip: Standardize position of negative signs in the numerator.

Question 4) Find \(x\)
a) \(\frac{2}{7} + x = \frac{5}{4}\)
Answer: \(x = \frac{5}{4} - \frac{2}{7}\)
LCM of 4 and 7 is 28:
\(x = \frac{35 - 8}{28} = \frac{27}{28}\)
In simple words: Transpose \(\frac{2}{7}\) to the right side and subtract using a common denominator.

Exam Tip: Find the LCM of denominators for fraction subtraction.

Question 4. b) \(-\frac{1}{8} + x = \frac{11}{72}\)
Answer: \(x = \frac{11}{72} + \frac{1}{8}\)
Since \(72 = 8 \times 9\), convert \(\frac{1}{8}\) to \(\frac{9}{72}\):
\(x = \frac{11 + 9}{72} = \frac{20}{72} = \frac{5}{18}\)
In simple words: Add \(\frac{1}{8}\) to both sides and simplify the resulting fraction.

Exam Tip: Always reduce fractions to lowest terms.

Question 4. c) \(-\frac{7}{8} \times x = -\frac{1}{2}\)
Answer: \(x = -\frac{1}{2} \div -\frac{7}{8}\)
\(x = -\frac{1}{2} \times -\frac{8}{7}\)
\(x = \frac{8}{14} = \frac{4}{7}\)
In simple words: Divide both sides by \(-\frac{7}{8}\) by multiplying by its reciprocal.

Exam Tip: Division by a fraction is multiplication by its reciprocal.

Question 5) Write the multiplicative inverse of the following rational numbers
a) \(-\frac{2}{5} \times -\frac{3}{8}\)
Answer: Compute the product: \(-\frac{2}{5} \times -\frac{3}{8} = \frac{6}{40} = \frac{3}{20}\).
Multiplicative inverse = \(\frac{20}{3}\).
In simple words: Multiply first, then flip the resulting fraction to find its reciprocal.

Exam Tip: Multiplicative inverse is also known as the reciprocal.

Question 5. b) \(5\)
Answer: \(\frac{1}{5}\)
In simple words: The reciprocal of 5 is \(\frac{1}{5}\).

Question 5. c) \(-\frac{19}{12}\)
Answer: \(-\frac{12}{19}\)
In simple words: Flip the numerator and denominator while keeping the negative sign.

Exam Tip: Reciprocal of a negative number is negative.

Question 5. d) \(-\frac{6}{-19}\)
Answer: Simplify \(-\frac{6}{-19} = \frac{6}{19}\).
Multiplicative inverse = \(\frac{19}{6}\).
In simple words: Simplify the double negative to \(\frac{6}{19}\), then flip it.

Exam Tip: Always simplify signs before determining inverses.

Question 6) Verify that \(-(-x) = x\) if
a) \(x = -\frac{7}{8}\)
Answer: Substitute \(x = -\frac{7}{8}\):
\(-\left(-\left(-\frac{7}{8}\right)\right) = -\frac{7}{8}\)
LHS: \(-(-(-\frac{7}{8})) = -(\frac{7}{8}) = -\frac{7}{8}\).
Since LHS = RHS, verified.
In simple words: Substitute the value and apply sign rules successively.

Exam Tip: An odd number of negative signs results in negative; an even number results in positive.

Question 6. b) \(x = \frac{8}{17}\)
Answer: Substitute \(x = \frac{8}{17}\):
\(-\left(-\frac{8}{17}\right) = \frac{8}{17}\)
LHS: \(-(-\frac{8}{17}) = \frac{8}{17}\).
Since LHS = RHS, verified.
In simple words: Two negative signs cancel each other out.

Exam Tip: This demonstrates the property of additive inverses.

Question 7. Is 0.21 the multiplicative inverse of \(4 \frac{16}{21}\)?
Answer: Convert \(4 \frac{16}{21}\) to an improper fraction:
\(4 \frac{16}{21} = \frac{4 \times 21 + 16}{21} = \frac{84 + 16}{21} = \frac{100}{21}\)
Convert 0.21 to fraction: \(\frac{21}{100}\).
Multiply them: \(\frac{21}{100} \times \frac{100}{21} = 1\).
Since the product is 1, yes, 0.21 is the multiplicative inverse of \(4 \frac{16}{21}\).
In simple words: Two numbers are multiplicative inverses if their product is 1. Here, \(\frac{21}{100} \times \frac{100}{21} = 1\).

Exam Tip: Always convert decimals and mixed fractions to improper fractions before checking reciprocals.

Question 8. Is \(33 \frac{1}{3}\) the multiplicative inverse of 0.03?
Answer: Convert \(33 \frac{1}{3}\) to improper fraction: \(\frac{100}{3}\).
Convert 0.03 to fraction: \(\frac{3}{100}\).
Multiply them: \(\frac{100}{3} \times \frac{3}{100} = 1\).
Since the product is 1, yes, \(33 \frac{1}{3}\) is the multiplicative inverse of 0.03.
In simple words: Multiply \(\frac{100}{3}\) by \(\frac{3}{100}\); since the product is 1, they are inverses.

Exam Tip: Product equal to 1 confirms multiplicative inverse relation.

Question 9. Draw a number line and represent the following:
a) \(-\frac{7}{8}\)
Answer: Draw a number line between 0 and \(-1$, divide the segment into 8 equal parts, and mark the 7th division to the left of 0 as \(-\frac{7}{8}\).
In simple words: Divide the unit interval between 0 and \(-1\) into 8 equal parts and count 7 parts to the left.

Exam Tip: The denominator indicates the number of equal subdivisions between integers.

Question 9. b) \(\frac{8}{3}\) and \(-\frac{8}{3}\)
Answer: \(\frac{8}{3} = 2 \frac{2}{3}\) (lies between 2 and 3).
\(-\frac{8}{3} = -2 \frac{2}{3}\) (lies between \(-2$ and \(-3\)).
Divide each integer interval into 3 equal parts and mark the respective points.
In simple words: Convert to mixed fractions to locate between integers on positive and negative sides of the number line.

Exam Tip: Improper fractions should be converted to mixed numbers for easy plotting.

Question 9. c) \(-\frac{11}{6}\)
Answer: \(-\frac{11}{6} = -1 \frac{5}{6}\) (lies between \(-1$ and \(-2\)).
Divide the segment between \(-1$ and \(-2\) into 6 equal parts and mark the 5th subdivision from \(-1\).
In simple words: Plot between \(-1\) and \(-2\) by subdividing into 6 parts.

Exam Tip: Negative fractions lie to the left of zero.

Question 9. d) \(\frac{10}{4}\)
Answer: \(\frac{10}{4} = \frac{5}{2} = 2.5\) (lies exactly halfway between 2 and 3).
In simple words: Simplify to \(2.5\) and mark midway between 2 and 3.

Exam Tip: Simplifying fractions first makes number line representation easier.

Question 10. Find 5 rational numbers between
a) \(\frac{1}{4}\) and \(\frac{7}{8}\)
Answer: Convert to a common denominator of 8: \(\frac{1}{4} = \frac{2}{8}\) and \(\frac{7}{8}\).
Multiply numerator and denominator by 6 to get enough space: \(\frac{2 \times 6}{8 \times 6} = \frac{12}{48}\) and \(\frac{7 \times 6}{8 \times 6} = \frac{42}{48}\).
Five rational numbers: \(\frac{13}{48}, \frac{14}{48}, \frac{15}{48}, \frac{16}{48}, \frac{17}{48}\).
In simple words: Make denominators equal, then scale up both fractions to insert required rational numbers.

Exam Tip: Scaling by a larger factor provides a wider range of numbers to choose from.

Question 10. b) \(0\) and \(2\)
Answer: Write 0 and 2 with denominator 6: \(\frac{0}{6}\) and \(\frac{12}{6}\).
Five rational numbers: \(\frac{1}{6}, \frac{2}{6}, \frac{3}{6}, \frac{4}{6}, \frac{5}{6}\).
In simple words: Express integers as fractions with the same denominator and pick numbers between them.

Exam Tip: Any integers can be expressed as rational numbers with denominator 1 or higher.

Question 10. c) \(-7\) and \(-8\)
Answer: Write with denominator 6: \(-7 = -\frac{42}{6}\) and \(-8 = -\frac{48}{6}\).
Five rational numbers: \(-\frac{43}{6}, -\frac{44}{6}, -\frac{45}{6}, -\frac{46}{6}, -\frac{47}{6}\).
In simple words: Scale integers to fractions with denominator 6 to find intermediate values.

Question 10. d) \(-\frac{7}{9}\) and \(\frac{1}{9}\)
Answer: Denominators are already equal. The numerators range from \(-7\) to \(1\).
Five rational numbers: \(-\frac{6}{9}, -\frac{5}{9}, -\frac{4}{9}, -\frac{3}{9}, -\frac{2}{9}\).
In simple words: Choose any five fractions between \(-\frac{7}{9}\) and \(\frac{1}{9}\).

Exam Tip: When denominators match, simply pick values between the numerators.

Question 11. Write any 10 rational numbers less than
a) \(+1\)
Answer: \(0, -1, -2, -3, -4, -5, -6, -7, -8, -9$
In simple words: Any 10 numbers smaller than 1, such as integers or fractions.

Exam Tip: Integers are also rational numbers.

Question 11. b) \(-\frac{3}{8}\)
Answer: \(-\frac{4}{8}, -\frac{5}{8}, -\frac{6}{8}, -\frac{7}{8}, -\frac{8}{8}, -\frac{9}{8}, -\frac{10}{8}, -\frac{11}{8}, -\frac{12}{8}, -\frac{13}{8}\)
In simple words: Choose fractions smaller (more negative) than \(-\frac{3}{8}\).

Exam Tip: On the number line, numbers to the left are smaller.

Squares and Square Roots

Question 1. How many natural numbers lie between:
a. \((13)^2\) and \((14)^2\)
Answer: For any natural number \(n\), the number of non-square natural numbers between \(n^2\) and \((n + 1)^2\) is \(2n$.
Here \(n = 13\).
Number of natural numbers = \(2 \times 13 = 26\).
In simple words: Use the formula \(2n\) where \(n\) is the smaller number.

Exam Tip: Shortcut formula: between \(n^2\) and \((n+1)^2\) there are exactly \(2n$ non-square numbers.

Question 1. b. \((2001)^2\) and \((2002)^2\)
Answer: Here \(n = 2001\).
Number of natural numbers = \(2 \times 2001 = 4002\).
In simple words: Multiply 2001 by 2 using the \(2n\) property.

Exam Tip: This property saves time compared to calculating actual squares.

Question 2. Without adding, find sum:
A) \(1 + 3 + 5 + 7 + 9 + 11\)
Answer: The sum of the first \(n\) odd natural numbers is \(n^2\).
Here there are 6 consecutive odd numbers (\(n = 6\)).
Sum = \(6^2 = 36\).
In simple words: Count the number of terms (\(n=6\)) and square it.

Exam Tip: Sum of first \(n$ odd numbers = \(n^2\).

Question 2. B) \(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21\)
Answer: There are 11 consecutive odd numbers (\(n = 11\)).
Sum = \(11^2 = 121\).
In simple words: Count the 11 terms and square 11 to get 121.

Exam Tip: Always count the exact number of terms before applying \(n^2\).

Question 3. Find the squares using the given pattern.
\(45^2 = (4 \times 5) \times 100 + 25$
a) \(65\)
Answer: \(65^2 = (6 \times 7) \times 100 + 25 = 42 \times 100 + 25 = 4200 + 25 = 4225\).
In simple words: Multiply the tens digit by the next consecutive integer, multiply by 100, and add 25.

Exam Tip: This pattern works for any two-digit number ending in 5: \((10a + 5)^2 = a(a+1) \times 100 + 25\).

Question 3. b) \(85\)
Answer: \(85^2 = (8 \times 9) \times 100 + 25 = 72 \times 100 + 25 = 7200 + 25 = 7225\).
In simple words: Multiply 8 by 9 to get 72, attach 00, and add 25.

Exam Tip: Very fast mental math trick for numbers ending in 5.

Question 4. Write Pythagorean triplet whose one member is:
a) \(20\)
Answer: For any integer \(m > 1\), a Pythagorean triplet is given by \((2m, m^2 - 1, m^2 + 1)\).
Let \(2m = 20 \implies m = 10\).
Second member = \(m^2 - 1 = 10^2 - 1 = 100 - 1 = 99\).
Third member = \(m^2 + 1 = 10^2 + 1 = 100 + 1 = 101\).
Triplet: \((20, 99, 101)\).
In simple words: Set \(2m\) equal to the given even number to find \(m$, then calculate \(m^2-1\) and \(m^2+1\).

Exam Tip: Pythagorean triplets satisfy \(a^2 + b^2 = c^2\).

Question 4. b) \(12\)
Answer: Let \(2m = 12 \implies m = 6\).
Second member = \(6^2 - 1 = 36 - 1 = 35\).
Third member = \(6^2 + 1 = 36 + 1 = 37\).
Triplet: \((12, 35, 37)\).
In simple words: Take \(m=6\), then find \(6^2-1\) and \(6^2+1\).

Exam Tip: Verify: \(12^2 + 35^2 = 144 + 1225 = 1369 = 37^2\).

Question 5. Find the square using \((a + b)^2 = (a + b)(a + b)\).
a) \(42\)
Answer: Express 42 as \((40 + 2)\):
\(42^2 = (40 + 2)^2 = (40 + 2)(40 + 2)\)
\(= 40(40 + 2) + 2(40 + 2)\)
\(= 1600 + 80 + 80 + 4\)
\(= 1600 + 160 + 4 = 1764\)
In simple words: Split 42 into \((40+2)\) and expand using algebraic multiplication.

Exam Tip: Splitting into tens and units makes mental squaring easy.

Question 5. b) \(87\)
Answer: Express 87 as \((80 + 7)\):
\(87^2 = (80 + 7)^2 = (80 + 7)(80 + 7)\)
\(= 80(80 + 7) + 7(80 + 7)\)
\(= 6400 + 560 + 560 + 49\)
\(= 6400 + 1120 + 49 = 7569\)
In simple words: Split 87 into \((80+7)\) and multiply out.

Question 6. Find the square root by prime factorization.
a) \(11025\)
Answer: Prime factorization of 11025:
\(11025 = 3 \times 3 \times 5 \times 5 \times 7 \times 7 = 3^2 \times 5^2 \times 7^2\)
Square root = \(\sqrt{11025} = 3 \times 5 \times 7 = 105\).
In simple words: Find prime factors, group them in pairs, and take one factor from each pair.

Exam Tip: Check divisibility by 5 and 3 to simplify prime factorization.

Question 6. b) \(8469\)
Answer: Prime factorization of 8469:
\(8469 = 3 \times 3 \times 31 \times 31 = 3^2 \times 31^2\)
Square root = \(\sqrt{8469} = 3 \times 31 = 93\).
In simple words: Factorize into prime pairs \((3 \times 31)\).

Exam Tip: Group identical prime factors into pairs for square roots.

Question 6. c) \(17956\)
Answer: Prime factorization of 17956:
\(17956 = 2 \times 2 \times 7 \times 7 \times 19 \times 19 = 2^2 \times 7^2 \times 19^2\)
Square root = \(\sqrt{17956} = 2 \times 7 \times 19 = 266\).
In simple words: Factorize into \(2^2 \times 7^2 \times 19^2\), then multiply base factors.

Exam Tip: Prime factorization is reliable for perfect squares.

Question 7. Find the value using the property of squares.
a) \(89^2 - 88^2\)
Answer: Use identity \(a^2 - b^2 = (a - b)(a + b)\):
\(89^2 - 88^2 = (89 - 88)(89 + 88)\)
\(= (1)(177) = 177\)
In simple words: Use the difference of squares identity \((a-b)(a+b)\) to compute mentally.

Exam Tip: For consecutive numbers, \(a - b = 1\), so \(a^2 - b^2 = a + b\).

Question 7. b) \(121^2 - 120^2\)
Answer: Use identity \(a^2 - b^2 = (a - b)(a + b)\):
\(121^2 - 120^2 = (121 - 120)(121 + 120)\)
\(= (1)(241) = 241\)
In simple words: Add 121 and 120 since their difference is 1.

Exam Tip: This algebraic identity avoids large multiplication steps.

Question 8. Find the square root by long division method.
a) \(961\)
Answer: Grouping digits from right: \(09 \,\, 61\).
1. Largest square \(\le 9\) is \(3^2 = 9\). Remainder 0.
2. Bring down 61. Trial divisor \(3 \times 2 = 6\).
3. Place 1 in quotient (\(61 \times 1 = 61\)). Remainder 0.
Square root = \(31\).
In simple words: Use long division in pairs of digits from right to left.

Exam Tip: Long division method is efficient for larger numbers.

Question 8. b) \(3364\)
Answer: Grouping digits: \(33 \,\, 64\).
1. Largest square \(\le 33$ is \(5^2 = 25\). Remainder \(33 - 25 = 8\).
2. Bring down 64 to get 864. Trial divisor \(5 \times 2 = 10\).
3. Testing \(108 \times 8 = 864\). Remainder 0.
Square root = \(58\).
In simple words: Group in pairs, find closest square, and continue long division steps.

Question 8. c) \(7225\)
Answer: Grouping digits: \(72 \,\, 25\).
1. Largest square \(\le 72$ is \(8^2 = 64\). Remainder \(72 - 64 = 8\).
2. Bring down 25 to get 825. Trial divisor \(8 \times 2 = 16\).
3. Testing \(165 \times 5 = 825\). Remainder 0.
Square root = \(85\).
In simple words: Apply long division pairing from right.

Exam Tip: Ending in 25 indicates 5 in the unit place of the square root.

Question 8. d) \(7396\)
Answer: Grouping digits: \(73 \,\, 96\).
1. Largest square \(\le 73$ is \(8^2 = 64\). Remainder \(73 - 64 = 9\).
2. Bring down 96 to get 996. Trial divisor \(8 \times 2 = 16\).
3. Testing \(166 \times 6 = 996\). Remainder 0.
Square root = \(86\).
In simple words: Pair digits and perform standard square root long division.

Exam Tip: Check division product carefully at each stage.

Question 9. Find the least number of 5 digits which is a perfect square.
Answer: Smallest 5-digit number is 10000.
Notice that \(100^2 = 10000\).
Therefore, 10000 is already the least 5-digit perfect square.
In simple words: Since 10000 is \(100^2$, it is the smallest 5-digit perfect square.

Exam Tip: Always check powers of 10 for boundary digit questions.

Question 10. Find the least square number exactly divisible by each of the numbers: 7, 16, and 21.
Answer: First find the LCM of 7, 16, and 21:
Prime factors: \(7 = 7\), \(16 = 2^4\), \(21 = 3 \times 7\).
\(\text{LCM} = 2^4 \times 3 \times 7 = 16 \times 3 \times 7 = 336\)
Prime factorization of 336: \(336 = 2^4 \times 3 \times 7 = 2^4 \times 3^1 \times 7^1\).
To make 336 a perfect square, multiply by missing pairs (3 and 7):
Least square number = \(336 \times 3 \times 7 = 336 \times 21 = 7056\).
In simple words: Find LCM, then multiply by missing prime factors to make all exponents even.

Exam Tip: A number is a perfect square only if all prime factors have even exponents.

Question 11. Find the smallest number that should be (a) multiplied (b) divided so that the following numbers become a perfect square. Also find the square root of the number obtained.
a) \(2475\)
Answer: Prime factorization of 2475:
\(2475 = 3 \times 3 \times 5 \times 5 \times 11 = 3^2 \times 5^2 \times 11^1\)
(a) To make it a perfect square, multiply by 11.
New number = \(2475 \times 11 = 27225\).
Square root = \(3 \times 5 \times 11 = 165\).
(b) To make it a perfect square, divide by 11.
New number = \(2475 / 11 = 225\).
Square root = \(3 \times 5 = 15\).
In simple words: Factorize into primes; multiply or divide by the unpaired factor (11).

Exam Tip: Unpaired prime factors determine the multiplier or divisor.

Question 11. b) \(720\)
Answer: Prime factorization of 720:
\(720 = 2^4 \times 3^2 \times 5^1\)
Unpaired factor is 5.
(a) Multiply by 5. New number = \(720 \times 5 = 3600\), square root = \(2^2 \times 3 \times 5 = 60\).
(b) Divide by 5. New number = \(720 / 5 = 144\), square root = \(2^2 \times 3 = 12\).
In simple words: Multiply or divide by 5 to pair up the single 5.

Exam Tip: Check prime factorization exponents.

Question 11. c) \(600\)
Answer: Prime factorization of 600:
\(600 = 2^3 \times 3^1 \times 5^2 = 2^2 \times 2 \times 3 \times 5^2\)
Unpaired factors are 2 and 3, so product is \(2 \times 3 = 6\).
(a) Multiply by 6. New number = \(600 \times 6 = 3600\), square root = 60.
(b) Divide by 6. New number = \(600 / 6 = 100\), square root = 10.
In simple words: Multiply or divide by the product of all unpaired prime factors (\(2 \times 3 = 6\)).

Exam Tip: All unpaired factors must be multiplied or divided together.

Question 11. d) \(2448\)
Answer: Prime factorization of 2448:
\(2448 = 2^4 \times 3^2 \times 17^1\)
Unpaired factor is 17.
(a) Multiply by 17. New number = \(2448 \times 17 = 41616\), square root = \(2^2 \times 3 \times 17 = 204\).
(b) Divide by 17. New number = \(2448 / 17 = 144\), square root = \(2^2 \times 3 = 12\).
In simple words: Multiply or divide by 17.

Exam Tip: Prime factorization is essential for these types of questions.

Question 12. Find the least number which must be (a) subtracted (b) added to the following numbers to make them a perfect square. Also find the square root of the number obtained.
a) \(7671\)
Answer: Using long division for \(\sqrt{7671}\):
Square root comes to 87 with a remainder of 42.
(a) Subtracted: Subtract the remainder 42. New number = \(7671 - 42 = 7629\), square root = 87.
(b) Added: Next square is \((88)^2 = 7744\). Amount to add = \(7744 - 7671 = 73\), square root = 88.
In simple words: Use long division to find remainder for subtraction, and compare with next integer square for addition.

Exam Tip: Remainder gives the amount to subtract; difference to the next square gives the amount to add.

Question 12. b) \(2361\)
Answer: Using long division for \(\sqrt{2361}\):
Square root comes to 48 with a remainder of 57.
(a) Subtracted: Subtract 57. New number = \(2361 - 57 = 2304\), square root = 48.
(b) Added: Next square is \((49)^2 = 2401\). Amount to add = \(2401 - 2361 = 40\), square root = 49.
In simple words: Subtract the remainder 57 or add 40 to reach \(49^2\).

Exam Tip: Long division remainder indicates what excess needs removal.

Question 12. c) \(5678\)
Answer: Using long division for \(\sqrt{5678}\):
Square root comes to 75 with a remainder of 53.
(a) Subtracted: Subtract 53. New number = \(5678 - 53 = 5625\), square root = 75.
(b) Added: Next square is \((76)^2 = 5776\). Amount to add = \(5776 - 5678 = 98\), square root = 76.
In simple words: Subtract 53 or add 98 to reach \(76^2\).

Exam Tip: Always verify perfect squares by squaring the obtained root.

Question 13. 9604 soldiers are arranged in an auditorium in such a way that there are as many soldiers in a row as there are in a column. How many rows and columns are there in the auditorium?
Answer: Let the number of rows be \(x\). Then the number of columns is also \(x$.
Total soldiers = \(x \times x = x^2 = 9604\).
\(x = \sqrt{9604}\)
Using prime factorization or long division:
\( 9604 = 2^2 \times 7^4 = 2 \times 7^2 = 2 \times 49 = 98 $
Therefore, there are 98 rows and 98 columns.
In simple words: Find the square root of 9604 to find the number of rows and columns.

Exam Tip: Equal rows and columns always form a square arrangement.

Question 14. In a right-angled triangle PQR, angle \(Q = 90^\circ\). If
a) \(PQ = 9\) cm; \(QR = 40\) cm, find \(PR\)
Answer: Using Pythagoras theorem in \(\triangle PQR\):
\(PR^2 = PQ^2 + QR^2\)
\(PR^2 = 9^2 + 40^2\)
\(PR^2 = 81 + 1600 = 1681\)
\(PR = \sqrt{1681} = 41\) cm.
In simple words: Apply Pythagoras theorem \(hypotenuse^2 = base^2 + perpendicular^2\).

Exam Tip: \((9, 40, 41)\) is a standard Pythagorean triplet.

Question 14. b) \(PR = 37\) cm; \(PQ = 12\) cm, find \(QR\)
Answer: Using Pythagoras theorem:
\(PR^2 = PQ^2 + QR^2\)
\(37^2 = 12^2 + QR^2\)
\(1369 = 144 + QR^2\)
\(QR^2 = 1369 - 144 = 1225\)
\(QR = \sqrt{1225} = 35\) cm.
In simple words: Subtract the square of PQ from the square of PR to find \(QR^2\).

Exam Tip: \((12, 35, 37)\) forms a well-known Pythagorean triplet.