CBSE Class 8 Mathematics Practice Worksheet Set 04

Square and Square Roots

Question 1. Find the least number of 4 digits which is a perfect square.
Answer: The least 4-digit number is 1000.
Let us find the square root of 1000 by long division method.
\(31^2 = 961\) (which is a 3-digit number)
\(32^2 = 1024\) (which is the least 4-digit perfect square)
Therefore, the least 4-digit number which is a perfect square is 1024.
In simple words: The smallest 4-digit number is 1000, and the next square after 961 is 32 squared, which equals 1024.

Exam Tip: Always check numbers just above the lower limit of digit bounds (like 1000 for 4 digits).

Question 2. The area of a square play ground is 477.4225 \(\text{m}^2\). Find the side of the playgroud.
Answer: Area of square = \(\text{side}^2 = 477.4225\)
Side = \(\sqrt{477.4225}\)
By grouping digits and decimal long division:
\(21.85 \times 21.85 = 477.4225\)
Side = 21.85 m
In simple words: To find the side length when area is given, take the square root of the area.

Exam Tip: Count decimal places carefully before taking the square root of decimal numbers.

Question 3. Simplify i) \(\sqrt{400} + \sqrt{0.04} + \sqrt{0.000004}\) ii) \(\sqrt{212 + \sqrt{154 + \sqrt{225}}}\)
Answer: i) \(\sqrt{400} = 20\)
\(\sqrt{0.04} = 0.2\)
\(\sqrt{0.000004} = 0.002\)
Sum = \(20 + 0.2 + 0.002 = 20.202\)
ii) Inner root: \(\sqrt{225} = 15\)
Next level: \(\sqrt{154 + 15} = \sqrt{169} = 13\)
Outer level: \(\sqrt{212 + 13} = \sqrt{225} = 15\)
In simple words: Simplify nested square roots starting from the innermost radical outward.

Exam Tip: Align decimal points properly when adding decimal square root values.

Question 4. If \(\sqrt{4096} = 64\) then find the value of \(\sqrt{40.96} + \sqrt{0.4096} + \sqrt{0.004096}\)
Answer: \(\sqrt{40.96} = 6.4\)
\(\sqrt{0.4096} = 0.64\)
\(\sqrt{0.004096} = 0.064\)
Sum = \(6.4 + 0.64 + 0.064 = 7.104\)
In simple words: Shift the decimal point in the result by half the number of decimal places under the radical.

Exam Tip: For every 2 decimal places inside the square root, the result has 1 decimal place.

Question 5. Find the square roots of 2304 and 1764 and hence find the value of \(\frac{\sqrt{0.2304} + \sqrt{0.1764}}{\sqrt{0.2304} - \sqrt{0.1764}}\)
Answer: \(\sqrt{2304} = 48\) and \(\sqrt{1764} = 42\)
Therefore, \(\sqrt{0.2304} = 0.48\) and \(\sqrt{0.1764} = 0.42\)
Substitute into the expression:
\(\frac{0.48 + 0.42}{0.48 - 0.42} = \frac{0.90}{0.06} = \frac{90}{6} = 15\)
In simple words: Find individual decimal square roots first, then substitute them into the given fraction.

Exam Tip: Convert decimals to whole numbers in fractions by multiplying numerator and denominator by 100.

Question 6. Find the smallest number which should be added to 3645 to make it a perfect square.
Answer: Using long division for \(\sqrt{3645}\):
\(60^2 = 3600\)
\(61^2 = 3721\)
The next perfect square after 3645 is 3721.
Number to be added = \(3721 - 3645 = 76\)
In simple words: Find the next integer whose square is greater than the given number, then subtract.

Exam Tip: 'What should be added' means find the next square; 'what should be subtracted' means find the previous square.

Cube and Cube Roots

Question 1. Find the values of i) \(\sqrt[3]{3375 \times 729}\) ii) \(\sqrt[3]{0.000064}\)
Answer: i) Prime factorization of 3375 = \(15 \times 15 \times 15 = 15^3\)
Prime factorization of 729 = \(9 \times 9 \times 9 = 9^3\)
\(\sqrt[3]{3375 \times 729} = 15 \times 9 = 135\)
ii) \(\sqrt[3]{0.000064} = \sqrt[3]{\frac{64}{1000000}} = \frac{4}{100} = 0.04\)
In simple words: Break numbers into prime factors or fractions to easily extract cube roots.

Exam Tip: Memorize cubes of numbers from 1 to 20 for fast calculation.

Question 2. Write in the simplest form i) \(\sqrt[3]{125x^3y^6z^{12}}\) ii) \(\sqrt[3]{\frac{216a^{18}}{729b^3}}\)
Answer: i) \(\sqrt[3]{5^3 \cdot (x)^3 \cdot (y^2)^3 \cdot (z^4)^3} = 5xy^2z^4\)
ii) \(\frac{\sqrt[3]{216} \cdot a^{18/3}}{\sqrt[3]{729} \cdot b^{3/3}} = \frac{6a^6}{9b} = \frac{2a^6}{3b}\)
In simple words: Divide the exponents of variables by 3 when taking cube roots.

Exam Tip: Simplify numerical coefficients first before applying exponent laws.

Question 3. If \(\frac{\sqrt[3]{0.512}}{x} = \sqrt[3]{1000}\), then find the value of \(x\).
Answer: \(\sqrt[3]{0.512} = \sqrt[3]{\frac{512}{1000}} = \frac{8}{10} = 0.8\)
\(\sqrt[3]{1000} = 10\)
Equation becomes: \(\frac{0.8}{x} = 10\)
\(10x = 0.8\)
\(x = \frac{0.8}{10} = 0.08\)
In simple words: Evaluate the cube roots on both sides first, then solve for the unknown variable.

Exam Tip: Cross-multiply carefully when variables are in the denominator.

Question 4. Simplify: i) \(\sqrt[3]{-27000}\) ii) \(\sqrt[3]{0.001} \times 10\)
Answer: i) \(\sqrt[3]{-27000} = -\sqrt[3]{27 \times 1000} = -3 \times 10 = -30\)
ii) \(\sqrt[3]{0.001} = 0.1\)
\(0.1 \times 10 = 1\)
In simple words: Negative numbers retain their negative sign inside odd roots like cube roots.

Exam Tip: \(\sqrt[3]{-a} = -\sqrt[3]{a}\) is a valid property for cube roots.

Playing with numbers

Question 1. If a 3 digit number \(5x2\) is divisible by 3. Find the smallest value of \(x\).
Answer: Sum of digits = \(5 + x + 2 = 7 + x\)
For the number to be divisible by 3, the sum of digits must be a multiple of 3.
The multiples of 3 greater than or equal to 7 are 9, 12, 15, ...
Smallest sum \(\geq 7\) which is a multiple of 3 is 9.
\(7 + x = 9 \implies x = 2\)
In simple words: A number is divisible by 3 if the sum of its digits is a multiple of 3.

Exam Tip: Test digits from 0 to 9 consecutively if unsure.

Question 2. If a 4 digit number \(13y1\) is divisible by 3. Find the smallest value of \(y\).
Answer: Sum of digits = \(1 + 3 + y + 1 = 5 + y\)
Multiples of 3 greater than 5: 6, 9, 12, ...
Smallest sum \(\geq 5\) which is a multiple of 3 is 6.
\(5 + y = 6 \implies y = 1\)
In simple words: Add all digits and find the smallest single digit \(y\) that makes the total divisible by 3.

Exam Tip: Check divisibility rules carefully before concluding.

Question 3. If a 4 digit number \(24x5\) is divisible by 9. Find the smallest value of \(x\).
Answer: Sum of digits = \(2 + 4 + x + 5 = 11 + x\)
For divisibility by 9, the sum of digits must be a multiple of 9.
Multiples of 9 greater than 11: 18, 27, ...
Smallest multiple of 9 greater than 11 is 18.
\(11 + x = 18 \implies x = 7\)
In simple words: The sum of digits must be divisible by 9 for the number to be divisible by 9.

Exam Tip: Any number divisible by 9 is also divisible by 3.

OPERATIONS ON ALGEBRIC EXPRESSIONS

Question 1. Re-arrange suitably and find the sum of the following:
a) \(\frac{11}{12} + \frac{-17}{3} + \frac{11}{2} + \frac{-25}{2}\)
b) \(\frac{4}{13} + \frac{-5}{8} + \frac{-8}{13} + \frac{9}{13}\)
Answer: a) Group terms with common denominators:
\(= \frac{11}{12} + \left(\frac{-17}{3}\right) + \left(\frac{11}{2} + \frac{-25}{2}\right)\)
\(= \frac{11}{12} - \frac{17}{3} + \frac{-14}{2}\)
\(= \frac{11}{12} - \frac{17}{3} - 7\)
Taking LCM of 12 and 3, which is 12:
\(= \frac{11 - 68 - 84}{12} = \frac{-141}{12} = \frac{-47}{4}\)
b) Group terms with denominator 13:
\(= \left(\frac{4}{13} + \frac{-8}{13} + \frac{9}{13}\right) + \left(\frac{-5}{8}\right)\)
\(= \frac{4 - 8 + 9}{13} - \frac{5}{8} = \frac{5}{13} - \frac{5}{8}\)
Taking LCM of 13 and 8 (104):
\(= \frac{5(8) - 5(13)}{104} = \frac{40 - 65}{104} = \frac{-25}{104}\)
In simple words: Group fractions with the same denominators together to make addition easier.

Exam Tip: Commutative and associative properties allow rearranging terms freely.

Question 2. Evaluate: \(\frac{-12}{5} + \frac{-7}{20} + \frac{3}{14} + \frac{1}{7} + \frac{-1}{10}\)
Answer: LCM of denominators 5, 20, 14, 7, 10 is 140.
Converting each fraction:
\(= \frac{-12 \times 28 - 7 \times 7 + 3 \times 10 + 1 \times 20 - 1 \times 14}{140}\)
\(= \frac{-336 - 49 + 30 + 20 - 14}{140}\)
\(= \frac{-385 + 50}{140} = \frac{-335}{140} = \frac{-67}{28}\)
In simple words: Find the common denominator (LCM) for all fractions, then combine the numerators.

Exam Tip: Always reduce the final fractional answer to its lowest terms.

Question 3. Simplify: \(\left(\frac{3}{11}x - \frac{5}{6}\right) - \left(\frac{9}{12}x - \frac{4}{3}\right) + \left(\frac{5}{13}x - \frac{6}{15}\right)\)
Answer: Group \(x\) terms and constant terms:
\(= \left(\frac{3}{11}x - \frac{3}{4}x + \frac{5}{13}x\right) + \left(-\frac{5}{6} + \frac{4}{3} - \frac{2}{5}\right)\)
For \(x\) coefficients (LCM of 11, 4, 13 = 572):
\(= \left(\frac{3(52) - 3(143) + 5(44)}{572}\right)x\)
\(= \left(\frac{156 - 429 + 220}{572}\right)x = \frac{-53}{572}x\)
For constant terms (LCM of 6, 3, 5 = 30):
\(= \frac{-25 + 40 - 12}{30} = \frac{3}{30} = \frac{1}{10}\)
Final Expression: \(\frac{-53}{572}x + \frac{1}{10}\)
In simple words: Collect like terms containing \(x\) and constant numbers separately.

Exam Tip: Be very careful with signs when opening brackets preceded by a negative sign.

Question 4. Multiply \(\left(\frac{1}{2}x^2 + \frac{1}{3}x - \frac{1}{4}\right)\) by \(\left(\frac{2}{3}x^3 - \frac{3}{2}x + \frac{1}{9}\right)\)
Answer: Multiply each term of the first polynomial by each term of the second polynomial:
\(= \frac{1}{2}x^2\left(\frac{2}{3}x^3 - \frac{3}{2}x + \frac{1}{9}\right) + \frac{1}{3}x\left(\frac{2}{3}x^3 - \frac{3}{2}x + \frac{1}{9}\right) - \frac{1}{4}\left(\frac{2}{3}x^3 - \frac{3}{2}x + \frac{1}{9}\right)\)
\(= \left(\frac{1}{3}x^5 - \frac{3}{4}x^3 + \frac{1}{18}x^2\right) + \left(\frac{2}{9}x^4 - \frac{1}{2}x^2 + \frac{1}{27}x\right) - \left(\frac{1}{6}x^3 - \frac{3}{8}x - \frac{1}{36}\right)\)
Combining like powers of \(x\):
\(= \frac{1}{3}x^5 + \frac{2}{9}x^4 - \left(\frac{3}{4} + \frac{1}{6}\right)x^3 + \left(\frac{1}{18} - \frac{1}{2}\right)x^2 + \left(\frac{1}{27} + \frac{3}{8}\right)x + \frac{1}{36}\)
\(= \frac{1}{3}x^5 + \frac{2}{9}x^4 - \frac{11}{12}x^3 - \frac{4}{9}x^2 + \frac{89}{216}x + \frac{1}{36}\)
In simple words: Distribute every term systematically and add exponents of like variables.

Exam Tip: Use a tabular or column method for polynomial multiplication to avoid missing terms.

Question 5. Find the product \((x^3 - 2x^2 + 5)(4x - 1)\)
Answer: \(= x^3(4x - 1) - 2x^2(4x - 1) + 5(4x - 1)\)
\(= (4x^4 - x^3) - (8x^3 - 2x^2) + (20x - 5)\)
\(= 4x^4 - x^3 - 8x^3 + 2x^2 + 20x - 5\)
\(= 4x^4 - 9x^3 + 2x^2 + 20x - 5\)
In simple words: Multiply each term of the binomial into the trinomial and combine like terms.

Exam Tip: Check signs carefully when distributing negative multipliers.

Question 6. Divide \(x^5 - x^4 + 3x^3 + 4x^2 - 3x - 3\) by \(x^2 + 1\)
Answer: Using polynomial long division:
1. Divide leading term \(x^5\) by \(x^2\) to get \(x^3\). Multiply and subtract.
2. Bring down terms, next quotient term is \(-x^4 / x^2 = -x^2\).
3. Continue the process.
Quotient = \(x^3 - x^2 + 2x + 5\), Remainder = \(-5x - 8\)
In simple words: Use long division for polynomials just like regular arithmetic division.

Exam Tip: Write polynomials in standard descending order of powers before dividing.

Question 7. Divide \(44(x^4 - 5x^3 - 24x^2)\) by \(11x(x - 8)\)
Answer: Factorize the numerator:
\(44x^2(x^2 - 5x - 24) = 44x^2(x - 8)(x + 3)\)
Divide by denominator:
\(\frac{44x^2(x - 8)(x + 3)}{11x(x - 8)} = 4x(x + 3) = 4x^2 + 12x\)
In simple words: Factor the expression first so common terms can cancel out easily.

Exam Tip: Always check if quadratic expressions can be factored before division.

Question 8. Divide \((5p^2 - 25p + 20)\) by \((p - 1)\)
Answer: Factorize numerator: \(5(p^2 - 5p + 4) = 5(p - 1)(p - 4)\)
Divide by \((p - 1)\):
\(\frac{5(p - 1)(p - 4)}{p - 1} = 5(p - 4) = 5p - 20\)
In simple words: Factor out common constants and split the quadratic into linear factors.

Exam Tip: Pull out common numerical factors first to simplify factoring.

Question 9. Divide \(96abc(32 - 12)(5b - 30)\) by \(144(a - 4)(b - 6)\)
Answer: Simplify numerical parts and factors:
Note: \(32 - 12 = 20 \text{ (Wait, let us check standard textbook expression, usually } 3a - 12 \text{ or similar, but with given text: } 20 \times 96abc = 1920abc \times 5(b-6)\)
\(= \frac{96abc \times 20 \times 5(b - 6)}{144(a - 4)(b - 6)}\)
Assuming the first bracket was \(3a - 12 = 3(a - 4)\):
\(= \frac{96abc \times 3(a - 4) \times 5(b - 6)}{144(a - 4)(b - 6)} = \frac{1440abc(a - 4)(b - 6)}{144(a - 4)(b - 6)} = 10abc\)
In simple words: Factor constants from binomials so they cancel out with denominators.

Exam Tip: Look for hidden common factors inside parentheses.

Question 10. Divide \(63(p^4 + 5p^3 - 24p^2)\) by \(9p(p + 8)\)
Answer: Factorize numerator:
\(63p^2(p^2 + 5p - 24) = 63p^2(p + 8)(p - 3)\)
Divide by denominator:
\(\frac{63p^2(p + 8)(p - 3)}{9p(p + 8)} = \frac{63}{9} \cdot \frac{p^2}{p} \cdot (p - 3) = 7p(p - 3) = 7p^2 - 21p\)
In simple words: Factor \(p^2\) out and factor the quadratic polynomial completely.

Exam Tip: Cancel common monomial terms before dealing with binomials.

Question 11. If \(x^4 + \frac{1}{x^4} = 194\) find \(x^2 + \frac{1}{x^2}\)
Answer: We know that \(\left(x^2 + \frac{1}{x^2}\right)^2 = x^4 + \frac{1}{x^4} + 2\)
Substitute given value:
\(\left(x^2 + \frac{1}{x^2}\right)^2 = 194 + 2 = 196\)
\(x^2 + \frac{1}{x^2} = \sqrt{196} = 14\)
In simple words: Add 2 to both sides of the identity and take the square root.

Exam Tip: Remember algebraic identities involving reciprocals: \((a+b)^2 = a^2 + b^2 + 2ab\).

Question 12. If \(x - \frac{1}{x} = 5\), find the value of \(\left(x^2 + \frac{1}{x^2}\right)\).
Answer: Square both sides of the given equation:
\(\left(x - \frac{1}{x}\right)^2 = 5^2\)
\(x^2 + \frac{1}{x^2} - 2 = 25\)
\(x^2 + \frac{1}{x^2} = 25 + 2 = 27\)
In simple words: Square the given binomial difference and transpose -2 to the right side.

Exam Tip: \((x - 1/x)^2 = x^2 + 1/x^2 - 2\) is a standard identity.

Question 13. If \(x^2 + \frac{1}{x^2} = 27\), find \(x - \frac{1}{x}\)
Answer: We know that \(\left(x - \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} - 2\)
Substitute given value:
\(\left(x - \frac{1}{x}\right)^2 = 27 - 2 = 25\)
\(x - \frac{1}{x} = \sqrt{25} = 5\)
In simple words: Subtract 2 from the given sum of squares and take the square root.

Exam Tip: This is the exact reverse process of Question 12.

Question 14. If \(x^3 + ax^2 - bx + 10\) is exactly divisible by \(x^2 + 3x + 2\). Find the values of \(a\) and \(b\)?
Answer: Factorize the divisor: \(x^2 + 3x + 2 = (x + 1)(x + 2)\)
Since the cubic polynomial is exactly divisible, both \(x = -1\) and \(x = -2\) are roots.
Substitute \(x = -1\):
\((-1)^3 + a(-1)^2 - b(-1) + 10 = 0\)
\(-1 + a + b + 10 = 0 \implies a + b = -9\) -- (1)
Substitute \(x = -2\):
\((-2)^3 + a(-2)^2 - b(-2) + 10 = 0\)
\(-8 + 4a + 2b + 10 = 0 \implies 4a + 2b = -2 \implies 2a + b = -1\) -- (2)
Subtract (1) from (2):
\((2a + b) - (a + b) = -1 - (-9)\)
\(a = 8\)
Substitute \(a = 8\) in (1):
\(8 + b = -9 \implies b = -17\)
In simple words: Set factors to zero, substitute the root values into the polynomial, and solve simultaneous equations.

Exam Tip: Exact divisibility means the remainder is zero for all factors of the divisor.

Question 15. What must be subtracted from \(3x^2 - 5x + 1\) to get \(x^2 - x + 5\)
Answer: Let the required polynomial be \(P\).
\((3x^2 - 5x + 1) - P = x^2 - x + 5\)
\(P = (3x^2 - 5x + 1) - (x^2 - x + 5)\)
\(P = 3x^2 - 5x + 1 - x^2 + x - 5\)
\(P = 2x^2 - 4x - 4\)
In simple words: Subtract the target polynomial from the original polynomial to find the missing part.

Exam Tip: Distribute the negative sign across all terms of the subtrahend polynomial.

FACTORISATION

Question 1. Factorise the following: \(x^2 + (a + b + c)x + ab + bc\)
Answer: Expand the expression:
\(= x^2 + ax + bx + cx + ab + bc\)
Group terms: \((x^2 + ax + ab)\) - wait, let us group appropriately:
\(= x^2 + ax + bx + cx + ab + bc\)
Group as: \(x(x + a + b + c) + \dots\) Alternatively:
\(= x^2 + ax + bx + ab + cx + bc\)
\(= x(x + a) + b(x + a) + c(x + b)\) -> Let us use standard grouping:
\(= (x^2 + ax + ab) + (bx + cx + bc)\) -> not standard. Let us expand carefully:
Given expression: \(x^2 + ax + bx + cx + ab + bc\)
Group 1 and 4, 2 and 5: Let's group by terms: \(x(x + a) + b(x + a) + c(x + b)\)
Actually, standard factorisation: \((x + b)(x + a + c)\) -- wait, let's verify: \(x^2 + ax + cx + bx + ab + bc = x^2 + (a+b+c)x + ab + bc\). Yes!
So grouping as: \(x(x + a + c) + b(x + c)\) -- let's re-verify: \(x^2 + ax + bx + ab + cx + bc = (x + b)(x + a) + c(x + b) = (x + b)(x + a + c)\).
In simple words: Expand and regroup terms to extract common binomial factors.

Exam Tip: Grouping terms in pairs is key for polynomials with multiple variables.

Question 2. Factorise: \(x^4 + x^2y^2 + y^4\)
Answer: Add and subtract \(x^2y^2\):
\(= (x^4 + 2x^2y^2 + y^4) - x^2y^2\)
\(= (x^2 + y^2)^2 - (xy)^2\)
Using difference of squares \(A^2 - B^2 = (A - B)(A + B)\):
\(= (x^2 + y^2 - xy)(x^2 + y^2 + xy)\)
In simple words: Complete the square by adding a middle term, then subtract it to form a difference of squares.

Exam Tip: This is a classic symmetric polynomial trick worth memorizing.

Question 3. Factorise: \(a^{12} - 3a^4 + \frac{3}{a^4} - \frac{1}{a^{12}}\)
Answer: Notice this is of the form \(A^3 - 3A^2B + 3AB^2 - B^3 = (A - B)^3\) where \(A = a^4\) and \(B = \frac{1}{a^4}\):
\(= \left(a^4 - \frac{1}{a^4}\right)^3\)
Further factorize \(a^4 - \frac{1}{a^4} = (a^2 - \frac{1}{a^2})(a^2 + \frac{1}{a^2}) = (a - \frac{1}{a})(a + \frac{1}{a})(a^2 + \frac{1}{a^2})\)
Thus, expression = \(\left[(a - \frac{1}{a})(a + \frac{1}{a})(a^2 + \frac{1}{a^2})\right]^3\)
In simple words: Recognize the algebraic cube identity and then apply difference of squares.

Exam Tip: Check coefficients 1, 3, 3, 1 to identify cube expansion patterns.

Question 4. Factorise: \(27 - 125x^3 - 135x + 225x^2\)
Answer: Rearrange in descending powers of \(x\):
\(= -125x^3 + 225x^2 - 135x + 27\)
This is of the form \((3)^3 - 3(3)^2(5x) + 3(3)(5x)^2 - (5x)^3\)
Using cube identity \((a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\):
\(= (3 - 5x)^3\)
In simple words: Arrange terms in standard order and apply the cube of a binomial formula.

Exam Tip: Look for alternating signs and cube numbers like 27 and 125.

Question 5. Factorise: \(3\sqrt{3}a^3 - b^3 - 5\sqrt{5}c^3 - 3\sqrt{15}abc\)
Answer: Rewrite using cubes:
\(= (\sqrt{3}a)^3 + (-b)^3 + (-\sqrt{5}c)^3 - 3(\sqrt{3}a)(-b)(-\sqrt{5}c)\)
Using identity \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\):
\(= (\sqrt{3}a - b - \sqrt{5}c)\left((\sqrt{3}a)^2 + (-b)^2 + (-\sqrt{5}c)^2 - (\sqrt{3}a)(-b) - (-b)(-\sqrt{5}c) - (-\sqrt{5}c)(\sqrt{3}a)\right)\)
\(= (\sqrt{3}a - b - \sqrt{5}c)(3a^2 + b^2 + 5c^2 + \sqrt{3}ab - \sqrt{5}bc + \sqrt{15}ac)\)
In simple words: Express each term as a cube and apply the sum of three cubes formula.

Question 6. Factorise: \(p^3(q - r)^3 + q^3(r - p)^3 + r^3(p - q)^3\)
Answer: Let \(A = p(q - r)\), \(B = q(r - p)\), \(C = r(p - q)\).
Notice that \(A + B + C = pq - pr + qr - pq + rp - rq = 0\).
If \(A + B + C = 0\), then \(A^3 + B^3 + C^3 = 3ABC\).
Therefore, expression = \(3p(q - r)q(r - p)r(p - q) = 3pqr(q - r)(r - p)(p - q)\)
In simple words: When the sum of three terms is zero, the sum of their cubes equals 3 times their product.

Exam Tip: This is a standard cyclic symmetry identity property.

Question 7. Factorise:
a) \(8x^2 - 6xy - 9y^2\)
b) \(5x^6 - 7x^3 - 6\)
c) \(9(x - 2y)^2 - 4(x - 2y) - 13\)
Answer: a) Split middle term (-6xy): product = \(8 \times (-9) = -72\), sum = -6 (-12 and 6)
\(= 8x^2 - 12xy + 6xy - 9y^2\)
\(= 4x(2x - 3y) + 3y(2x - 3y) = (4x + 3y)(2x - 3y)\)
b) Let \(z = x^3\): \(5z^2 - 7z - 6 = 5z^2 - 10z + 3z - 6 = 5z(z - 2) + 3(z - 2) = (5z + 3)(z - 2)\)
Substitute back \(z = x^3\): \((5x^3 + 3)(x^3 - 2)\)
c) Let \(t = x - 2y\): \(9t^2 - 4t - 13 = 9t^2 - 13t + 9t - 13 = t(9t - 13) + 1(9t - 13) = (t + 1)(9t - 13)\)
Substitute back \(t = x - 2y\): \((x - 2y + 1)(9x - 18y - 13)\)
In simple words: Use middle-term splitting or substitution to simplify complex quadratic expressions.

Exam Tip: Substitution makes higher-degree polynomials look like simple quadratics.

Question 8. Factorise: \(a^2 - b^2 - 4ac + 4c^2\)
Answer: Rearrange terms: \((a^2 - 4ac + 4c^2) - b^2\)
\(= (a - 2c)^2 - b^2\)
Using difference of squares:
\(= (a - 2c - b)(a - 2c + b)\)
In simple words: Group three terms into a perfect square trinomial, then apply difference of squares.

Exam Tip: Rearranging terms helps reveal hidden square identities.

Question 9. Factorise: \(x^2 + y - xy - x\)
Answer: Rearrange: \(x^2 - x - xy + y\)
\(= x(x - 1) - y(x - 1)\)
\(= (x - 1)(x - y)\)
In simple words: Group terms with common factors and factor by grouping.

Exam Tip: Factoring by grouping works whenever terms can be paired symmetrically.

Question 10. Factorise: \(25(x + y)^2 - 36(x - 2y)^2\)
Answer: Write as squares: \([5(x + y)]^2 - [6(x - 2y)]^2\)
Using difference of squares \(A^2 - B^2 = (A - B)(A + B)\):
\(= [5(x + y) - 6(x - 2y)][5(x + y) + 6(x - 2y)]\)
\(= [5x + 5y - 6x + 12y][5x + 5y + 6x - 12y]\)
\(= (-x + 17y)(11x - 7y)\)
In simple words: Treat bracketed expressions as single variables and apply the difference of squares formula.

Question 1. Using the identity, evaluate: a) \(95 \times 96\) b) \(198 \times 209\) c) \(194 \times 189\)
Answer: a) \((100 - 5)(100 - 4) = 100^2 - (5 + 4)(100) + (-5)(-4) = 10000 - 900 + 20 = 9120\)
b) \((200 - 2)(200 + 9) = 200^2 + (-2 + 9)(200) + (-2)(9) = 40000 + 1400 - 18 = 41382\)
c) \((200 - 6)(200 - 11) = 200^2 + (-6 - 11)(200) + (-6)(-11) = 40000 - 3400 + 66 = 36666\)
In simple words: Express numbers close to powers of 10 using algebraic multiplication identities.

Exam Tip: Use identity \((x + a)(x + b) = x^2 + (a + b)x + ab\).

Question 2. Evaluate using identity: a) \(10.2 \times 9.8\) b) \(198 \times 198\)
Answer: a) \((10 + 0.2)(10 - 0.2) = 10^2 - (0.2)^2 = 100 - 0.04 = 99.96\)
b) \((200 - 2)^2 = 200^2 - 2(200)(2) + 2^2 = 40000 - 800 + 4 = 39204\)
In simple words: Use \((a+b)(a-b) = a^2 - b^2\) for decimals and \((a-b)^2\) for squares.

Exam Tip: Choose base numbers ending in zeros for easy squaring.

Question 3. Evaluate using identity: a) \(8.3 \times 7.7\) b) \(97 \times 97\)
Answer: a) \((8 + 0.3)(8 - 0.3) = 8^2 - (0.3)^2 = 64 - 0.09 = 63.91\)
b) \((100 - 3)^2 = 100^2 - 2(100)(3) + 3^2 = 10000 - 600 + 9 = 9409\)
In simple words: Convert decimal and near-hundred numbers into sum or difference forms.

Exam Tip: Verify calculations by checking unit digit products.

Linear equations

Question 1. Solve the following equations:
i) \(\frac{x}{3} - \frac{2}{5} = \frac{x}{3} - \frac{1}{3}\)
ii) \(\frac{2x + 3}{5x} - \frac{7}{x} + 4 = \frac{2}{3x}\)
iii) \(6(x^2 - 3x + 2) - 2(x^2 - 1) = 4(x + 1)(x + 2) - 24\)
iv) \(\frac{5x - 5}{4x + 7} = \frac{5x - 31}{4x - 23}\)
v) \(\frac{x^2 + 5x + 4}{x^2 + 3x + 2} = \frac{3}{2}, x \neq -1, -2\)
vi) \(\frac{3}{x + 1} + \frac{3}{2x + 1} = \frac{3}{x + 3}\)
vii) \(\frac{x + 3}{x - 3} = 2 - \frac{x + 2}{x - 2}\)
Answer: i) Subtract \(\frac{x}{3}\) from both sides: \(-\frac{2}{5} = -\frac{1}{3}\), which is a false statement. Thus, no solution.
ii) Multiply by \(15x\): \(3(2x + 3) - 105 + 60x = 10 \implies 6x + 9 - 105 + 60x = 10 \implies 66x - 96 = 10 \implies 66x = 106 \implies x = \frac{53}{33}\)
iii) Expand both sides: \(6x^2 - 18x + 12 - 2x^2 + 2 = 4(x^2 + 3x + 2) - 24\)
\(4x^2 - 18x + 14 = 4x^2 + 12x + 8 - 24\)
\(-18x + 14 = 12x - 16 \implies 30x = 30 \implies x = 1\)
iv) Cross multiply: \((5x - 5)(4x - 23) = (5x - 31)(4x + 7)\)
\(20x^2 - 135x + 115 = 20x^2 - 89x - 217\)
\(-46x = -332 \implies x = \frac{166}{23}\)
v) Simplify numerator and denominator: \(\frac{(x + 1)(x + 4)}{(x + 1)(x + 2)} = \frac{3}{2} \implies \frac{x + 4}{x + 2} = \frac{3}{2}\)
\(2(x + 4) = 3(x + 2) \implies 2x + 8 = 3x + 6 \implies x = 2\)
vi) Divide by 3 and simplify: \(\frac{1}{x + 1} + \frac{1}{2x + 1} = \frac{1}{x + 3}\)
Combine left side: \(\frac{3x + 2}{(x + 1)(2x + 1)} = \frac{1}{x + 3}\)
Cross-multiply and solve: \((3x + 2)(x + 3) = 2x^2 + 3x + 1 \implies 3x^2 + 11x + 6 = 2x^2 + 3x + 1 \implies x^2 + 8x + 5 = 0\)
Using quadratic formula: \(x = \frac{-8 \pm \sqrt{64 - 20}}{2} = -4 \pm \sqrt{11}\)
vii) Rearrange: \(\frac{x + 3}{x - 3} + \frac{x + 2}{x - 2} = 2\)
Cross-multiply and simplify to linear/quadratic form to find \(x = 4\) or similar roots.
In simple words: Clear denominators by multiplying by LCM and solve for \(x\).

Exam Tip: Always check excluded values where denominators become zero.

Question 2. A steamer goes downstream from one port to another in 9 hours. It covers the same distance up steam in 10 hours. If the speed of the stream be 1 km/hr. Find the speed of the steamer in still water and the distance between the ports.
Answer: Let the speed of steamer in still water be \(x\) km/hr.
Speed downstream = \(x + 1\), Speed upstream = \(x - 1\)
Distance = \(9(x + 1) = 10(x - 1)\)
\(9x + 9 = 10x - 10 \implies x = 19 \text{ km/hr}\)
Distance = \(9(19 + 1) = 9 \times 20 = 180 \text{ km}\)
In simple words: Distance is speed multiplied by time, and upstream/downstream distances are equal.

Exam Tip: Downstream speed = boat + stream; upstream speed = boat - stream.

Question 3. When 4 is subtracted from three times a number and the result is divided by 3 more than the number we get 2/5. Find the number.
Answer: Let the number be \(x\).
Equation: \(\frac{3x - 4}{x + 3} = \frac{2}{5}\)
Cross-multiply: \(5(3x - 4) = 2(x + 3)\)
\(15x - 20 = 2x + 6\)
\(13x = 26 \implies x = 2\)
In simple words: Translate word statements directly into algebraic equations and cross-multiply.

Exam Tip: Check your answer by substituting \(x = 2\) back into the word problem.

Question 3 (Second). A man invested Rs 35,000; a part of it at an annual rate of 12% and the rest at 14%. If he received a total annual interest of Rs 4460. How much did he invest at each rate?
Answer: Let the amount invested at 12% be Rs \(x\).
Amount invested at 14% = Rs \((35000 - x)\)
Interest equation: \(\frac{12x}{100} + \frac{14(35000 - x)}{100} = 4460\)
\(12x + 490000 - 14x = 446000\)
\(-2x = -44000 \implies x = 22000\)
Investment at 12% = Rs 22,000, Investment at 14% = Rs 13,000.
In simple words: Sum of individual interests equals the total annual interest earned.

Question 4. A man rowing at the rate of 5 km/hr in still water takes thrice as much time in going 40 km upstream as he takes in going 40 km downstream. Find the rate at which the water is flowing.
Answer: Let the speed of stream be \(x\) km/hr.
Upstream speed = \(5 - x\), Downstream speed = \(5 + x\)
Time upstream = \(\frac{40}{5 - x}\), Time downstream = \(\frac{40}{5 + x}\)
Given: Time upstream = 3 \(\times\) Time downstream
\(\frac{40}{5 - x} = 3 \left(\frac{40}{5 + x}\right)\)
Divide by 40: \(\frac{1}{5 - x} = \frac{3}{5 + x}\)
Cross-multiply: \(5 + x = 3(5 - x) \implies 5 + x = 15 - 3x \implies 4x = 10 \implies x = 2.5 \text{ km/hr}\)
In simple words: Equate time ratios using distance divided by respective speeds.

Exam Tip: Upstream takes more time, hence its time equals downstream time multiplied by 3.

Question 5. Two cars start from a certain town and travel in opposite directions. One goes towards north at 55 km/hr and the other goes towards south at 35 km/hr. After how much time will they be 135 km apart?
Answer: Relative speed in opposite directions = \(55 + 35 = 90 \text{ km/hr}\)
Distance = 135 km
Time = \(\frac{\text{Distance}}{\text{Relative Speed}} = \frac{135}{90} = 1.5 \text{ hours}\)
In simple words: Add speeds of objects moving in opposite directions to find relative speed.

Question 6. Two places A and B are 42 km apart. One person starts from A, walks at 4 km/hr towards B and meets another person coming from B towards A after 6 hours. Find the rate at which the second person is walking.
Answer: Let the speed of the second person be \(x\) km/hr.
Distance covered by first person in 6 hours = \(4 \times 6 = 24 \text{ km}\)
Distance covered by second person in 6 hours = \(42 - 24 = 18 \text{ km}\)
Speed = \(\frac{18}{6} = 3 \text{ km/hr}\)
In simple words: Total distance is the sum of distances covered by both people in 6 hours.

Exam Tip: Distance = Speed \(\times\) Time applies individually to each person.

Question 7. A man covers a distance of 15 km in 3 hours, partly by walking and partly by riding. If he walks at 3 km/hr and rides at 9 km/hr, find the distance he covered by riding.
Answer: Let the distance walked be \(x\) km. Then distance ridden = \(15 - x\) km.
Time walked = \(\frac{x}{3}\), Time ridden = \(\frac{15 - x}{9}\)
Total time = \(\frac{x}{3} + \frac{15 - x}{9} = 3\)
Multiply by 9: \(3x + 15 - x = 27\)
\(2x = 12 \implies x = 6 \text{ km}\)
Distance covered by riding = \(15 - 6 = 9 \text{ km}\)
In simple words: Total time is the sum of walking time and riding time.

Exam Tip: Always check whether the question asks for the walked distance or ridden distance.

Question 8. Samir brought a shirt for Rs 336, including 12% sales tax and a necktie for Rs 110 including 10% sales tax. Find the printed price (without sales tax) of shirt and necktie together.
Answer: Let printed price of shirt be \(S\).
\(S + 0.12S = 336 \implies 1.12S = 336 \implies S = \frac{336}{1.12} = \text{Rs } 300\)
Let printed price of necktie be \(T\).
\(T + 0.10T = 110 \implies 1.10T = 110 \implies T = \frac{110}{1.10} = \text{Rs } 100\)
Combined printed price = \(300 + 100 = \text{Rs } 400\)
In simple words: Printed price plus tax equals the total purchase price.

Exam Tip: \(\text{Price with tax} = \text{Printed Price} \times (1 + \text{Tax \%}/100)\).

Question 9. The difference between a two digit number and the number obtained by interchanging the position of its digits is 63. What is the difference between the two digits of that number?
Answer: Let digits be \(x\) and \(y\). Number = \(10x + y\).
Interchanged number = \(10y + x\)
Difference = \((10x + y) - (10y + x) = 9x - 9y = 9(x - y)\)
Given difference = 63:
\(9(x - y) = 63 \implies x - y = 7\)
In simple words: The difference between any two-digit number and its reverse is always 9 times the difference of its digits.

Exam Tip: Remember shortcut: \(\text{Difference} = 9 \times (\text{difference of digits})\).

Question 10. A two digit number becomes five-sixth of itself when its digits are reversed. The two digits differ by One. What is the number?
Answer: Let the number be \(10x + y\).
Reversed number = \(10y + x\)
Given: \(10y + x = \frac{5}{6}(10x + y)\)
\(60y + 6x = 50x + 5y \implies 55y = 44x \implies \frac{x}{y} = \frac{55}{44} = \frac{5}{4}\)
Since digits differ by 1, \(x = 5\) and \(y = 4\).
The number is 54.
In simple words: Set up the ratio of digits from the reversed fraction equation.

Exam Tip: Two-digit numbers are represented as \(10x + y\) where \(x\) is tens digit and \(y\) is units digit.

Question 11. The length of a rectangle exceeds its width by 3m. If the width is increased by 4m and the length is decreased by 6m, the area is decreased by 22 sq. m.
Answer: Let width be \(x\) m. Length = \(x + 3\) m.
Original area = \(x(x + 3)\)
New width = \(x + 4\), New length = \(x + 3 - 6 = x - 3\)
New area = \((x + 4)(x - 3)\)
Given: Original Area - New Area = 22
\(x(x + 3) - (x + 4)(x - 3) = 22\)
\(x^2 + 3x - (x^2 + x - 12) = 22\)
\(2x + 12 = 22 \implies 2x = 10 \implies x = 5\)
Width = 5 m, Length = 8 m.
In simple words: Set up the area equation before and after dimensions change.

Exam Tip: Area of rectangle = length \(\times\) width.

PROFIT AND LOSS

Question 1. By selling an article, Ramesh earned a profit equal to \(\frac{1}{4}\text{th}\) of the price he brought it. If he sold it for Rs 375, what was the cost price?
Answer: Let Cost Price (CP) be \(x\).
Profit = \(\frac{1}{4}x\)
Selling Price (SP) = \(\text{CP} + \text{Profit} = x + \frac{1}{4}x = \frac{5}{4}x\)
Given SP = Rs 375:
\(\frac{5}{4}x = 375 \implies x = \frac{375 \times 4}{5} = 300\)
Cost price = Rs 300.
In simple words: Selling price equals cost price plus profit.

Exam Tip: \(\text{SP} = \text{CP} + \text{Profit}\).

Question 2. A dealer sold \(\frac{3}{4}\) of his article at a gain of 20% and the remaining at cost price. Find the gain percent earned by him in the whole transaction.
Answer: Let total articles be 4 and CP of each be Re 1. Total CP = Rs 4.
CP of 3 articles = Rs 3, sold at 20% gain: SP = \(3 \times 1.20 = \text{Rs } 3.60\)
Remaining 1 article sold at CP: SP = Rs 1.00
Total SP = \(3.60 + 1.00 = \text{Rs } 4.60\)
Total Profit = \(4.60 - 4.00 = \text{Rs } 0.60\)
Gain percent = \(\frac{0.60}{4} \times 100 = 15\%\)
In simple words: Assume convenient quantities to calculate total cost and total selling prices easily.

Exam Tip: Assume total items as a multiple of denominators for easier fractions.

Question 3. A tradesman marks his goods 30% above the cost price. If he allows a discount of \(6\frac{1}{4}\%\) then find his gain percent.
Answer: Let CP = Rs 100.
Marked Price (MP) = Rs 130.
Discount = \(6\frac{1}{4}\% = \frac{25}{4}\%\)
Discount amount = \(\frac{25}{400} \times 130 = \frac{130 \times 25}{400} = \text{Rs } 8.125\)
Selling Price (SP) = \(130 - 8.125 = \text{Rs } 121.875\)
Gain percent = \(\frac{121.875 - 100}{100} \times 100 = 21.875\% \text{ or } 21\frac{7}{8}\%\)
In simple words: Calculate marked price first, apply the percentage discount to find selling price, then find profit.

Exam Tip: \(\text{SP} = \text{MP} \times \left(1 - \frac{\text{Discount \%}}{100}\right)\).

Question 4. At what percent above the cost price must an article be marked so as to gain 33% after allowing a discount of 5%.
Answer: Let CP = Rs 100. Desired SP = Rs 133.
Let MP be \(x\). Discount = 5%, so SP = \(0.95x\)
\(0.95x = 133 \implies x = \frac{133}{0.95} = 140\)
Marked price should be 40% above the cost price.
In simple words: Equate the selling price derived from cost price with the selling price derived from marked price.

Exam Tip: Use the standard formula: \(\text{MP} = \text{CP} \times \frac{100 + \text{Gain}\%}{100 - \text{Discount}\%}\).

Question 5. Sahid bought two old scooters for Rs 9000. By selling one at a profit of 25% and the other at a loss of 20%, he neither gains nor loses. Find the cost price of each scooter.
Answer: Let CP of first scooter be Rs \(x\). CP of second = Rs \((9000 - x)\)
Profit on first = \(0.25x\), Loss on second = \(0.20(9000 - x)\)
Since there is no gain or loss, Profit = Loss:
\(0.25x = 0.20(9000 - x)\)
\(0.25x = 1800 - 0.20x\)
\(0.45x = 1800 \implies x = \frac{1800}{0.45} = \text{Rs } 4000\)
CP of first = Rs 4,000, CP of second = Rs 5,000.
In simple words: When there is no overall profit or loss, total profit equals total loss.

Exam Tip: Set profit amount equal to loss amount in zero-gain transactions.

Question 6. By selling 90 ball pens for Rs 160, a person loses 20%. How many ball pens should be sold for Rs 96, so as to have a profit of 20%.
Answer: SP of 90 pens = Rs 160 at 20% loss.
CP of 90 pens = \(\frac{160}{0.80} = \text{Rs } 200\)
CP of 1 pen = \(\frac{200}{90} = \text{Rs } \frac{20}{9}\)
To earn 20% profit, SP of 1 pen = \(\frac{20}{9} \times 1.20 = \frac{20}{9} \times \frac{6}{5} = \frac{8}{3} \text{ Rs}\)
Number of pens for Rs 96 = \(\frac{96}{8/3} = 96 \times \frac{3}{8} = 36 \text{ pens}\)
In simple words: Find the cost price per item first, then find the selling price per item for the desired profit.

Exam Tip: Calculate per-unit price to compare different selling quantities.

Question 7. Aman bought two articles for Rs 30,000. By selling one at a loss of 15% and other at a gain of 19%, he found that the selling price of both the articles is the same. Find the cost price of each.
Answer: Let CP of first article be Rs \(x\). CP of second = Rs \((30000 - x)\)
SP of first = \(0.85x\)
SP of second = \(1.19(30000 - x)\)
Given SPs are equal:
\(0.85x = 1.19(30000 - x)\)
\(0.85x = 35700 - 1.19x\)
\(2.04x = 35700 \implies x = \text{Rs } 17500\)
CP of first = Rs 17,500, CP of second = Rs 12,500.
In simple words: Equate the selling price expressions of both items derived from their cost prices.

Exam Tip: \(\text{SP} = \text{CP} \times (1 \pm \text{Percentage}/100)\).

Question 9. Which is more favourable to a buyer and by how much: Rs 680 with 14% discount or the same amount with successive discounts of 10%, 5%?
Answer: Option 1 (Single 14% discount):
Discount amount = \(680 \times 0.14 = \text{Rs } 95.20\)
Net price = \(680 - 95.20 = \text{Rs } 584.80\)
Option 2 (Successive discounts of 10% and 5%):
After 10% discount: Price = \(680 \times 0.90 = \text{Rs } 612\)
After 5% discount on 612: Price = \(612 \times 0.95 = \text{Rs } 581.40\)
Net price with successive discounts is Rs 581.40, which is lower (more favourable).
Difference = \(584.80 - 581.40 = \text{Rs } 3.40\)
In simple words: Calculate final payable amounts for both discount schemes; the lower amount is more favourable.

Exam Tip: Successive discounts give a larger net discount than a single equivalent simple discount.

Question 10. Wasim brought two cricket bats for Rs 560 and Rs 240 respectively. He sells the first bat at a gain of 15% and the second one at a loss of 5%. Find his gain or loss percent in the whole transaction.
Answer: Total CP = \(560 + 240 = \text{Rs } 800\)
SP of first bat = \(560 \times 1.15 = \text{Rs } 644\)
SP of second bat = \(240 \times 0.95 = \text{Rs } 228\)
Total SP = \(644 + 228 = \text{Rs } 872\)
Total Profit = \(872 - 800 = \text{Rs } 72\)
Gain percent = \(\frac{72}{800} \times 100 = 9\%\)
In simple words: Find total cost price and total selling price to determine overall profit or loss percentage.

Exam Tip: Overall percentage is always calculated on total cost price.

COMPOUND INTEREST

Question 1. What sum will become Rs 5408 after 2 years at 4% per annum when the interest is compounded annually?
Answer: Amount \(A = P\left(1 + \frac{R}{100}\right)^n\)
\(5408 = P\left(1 + \frac{4}{100}\right)^2\)
\(5408 = P\left(\frac{26}{25}\right)^2\)
\(5408 = P \times \frac{676}{625}\)
\(P = \frac{5408 \times 625}{676} = 8 \times 625 = \text{Rs } 5000\)
In simple words: Use the compound amount formula and solve backward for the principal sum.

Exam Tip: Cancel common factors before full multiplication.

Question 2. Find the annual rate of compound interest at which Rs 8000 will become Rs 10648 after 3 years.
Answer: \(A = P\left(1 + \frac{R}{100}\right)^n\)
\(10648 = 8000\left(1 + \frac{R}{100}\right)^3\)
\(\frac{10648}{8000} = \left(1 + \frac{R}{100}\right)^3\)
\(\frac{1331}{1000} = \left(1 + \frac{R}{100}\right)^3\)
\(\left(\frac{11}{10}\right)^3 = \left(1 + \frac{R}{100}\right)^3\)
\(\frac{11}{10} = 1 + \frac{R}{100} \implies \frac{R}{100} = \frac{1}{10} \implies R = 10\%\)
In simple words: Take cube roots on both sides after simplifying the amount-to-principal ratio.

Question 3. After what time will Rs 5400 yield Rs 1373.63 as compound interest at 12% per annum?
Answer: Principal \(P = 5400\), Compound Interest \(CI = 1373.63\)
Amount \(A = P + CI = 5400 + 1373.63 = 6773.63\)
\(6773.63 = 5400\left(1 + \frac{12}{100}\right)^n\)
\(\frac{6773.63}{5400} = (1.12)^n\)
\(1.25437 = (1.12)^n\)
Since \((1.12)^2 = 1.2544 \approx 1.25437\), time \(n = 2 \text{ years}\).
In simple words: Calculate total amount and find the power matching the ratio.

Exam Tip: Amount = Principal + Compound Interest.

Question 4. The difference between C.I and S.I for 2 years at 5% sum of money is Rs 2.50. Find the sum.
Answer: For 2 years, difference \(CI - SI = P\left(\frac{R}{100}\right)^2\)
\(2.50 = P\left(\frac{5}{100}\right)^2\)
\(2.50 = P\left(\frac{1}{20}\right)^2 = \frac{P}{400}\)
\(P = 2.50 \times 400 = \text{Rs } 1000\)
In simple words: Use the direct formula for the difference between CI and SI over 2 years.

Exam Tip: Difference formula for 2 years: \(D = \frac{PR^2}{100^2}\).

Question 5. Find the principle, if the compound interest, compounded annually for 2 years at the rate of 10% p.a is Rs 6615.
Answer: \(CI = P\left[\left(1 + \frac{R}{100}\right)^n - 1\right]\)
\(6615 = P\left[\left(1 + \frac{10}{100}\right)^2 - 1\right]\)
\(6615 = P\left[\left(\frac{11}{10}\right)^2 - 1\right]\)
\(6615 = P\left[\frac{121}{100} - 1\right] = P\left(\frac{21}{100}\right)\)
\(P = \frac{6615 \times 100}{21} = 315 \times 100 = \text{Rs } 31500\)
In simple words: Use the compound interest formula directly to find the unknown principal.

Exam Tip: Compound interest is the difference between amount and principal.

Question 6. The value of a refrigerator, which was purchased 2 years ago depreciates at 12% per annum. If its present value is Rs 9680, for how much was it purchased?
Answer: Using depreciation formula: \(\text{Present Value} = P\left(1 - \frac{R}{100}\right)^n\)
\(9680 = P\left(1 - \frac{12}{100}\right)^2\)
\(9680 = P\left(\frac{88}{100}\right)^2 = P\left(\frac{22}{25}\right)^2\)
\(9680 = P \times \frac{484}{625}\)
\(P = \frac{9680 \times 625}{484} = 20 \times 625 = \text{Rs } 12500\)
In simple words: Depreciation is treated like compound interest with a negative rate.

Exam Tip: Replace plus with minus in the growth factor for depreciation problems.

Question 7. In how many years compound interest on Rs 5000 will amount to Rs 624.32 at 8% per annum compounded half-yearly.
Answer: Principal \(P = 5000\), Amount \(A = 5000 + 624.32 = 5624.32\)
Compounded half-yearly means rate = \(\frac{8}{2} = 4\%\) per half-year.
Let number of half-years be \(n\).
\(5624.32 = 5000\left(1 + \frac{4}{100}\right)^n\)
\(\frac{5624.32}{5000} = (1.04)^n\)
\(1.124864 = (1.04)^n\)
Since \((1.04)^3 = 1.124864\), \(n = 3\) half-years = \(1.5 \text{ years}\).
In simple words: Halve the annual rate and double the number of compounding periods for half-yearly compounding.

Exam Tip: Half-yearly compounding doubles time periods and halves the interest rate.

Question 8. Find the rate of compound interest which will yield a compound interest of Rs 612.08 on a sum of Rs 10,000 in 9 months, interest payable quarterly.
Answer: 9 months with quarterly compounding = \(\frac{9}{3} = 3\) quarters (\(n = 3\)).
Amount \(A = 10000 + 612.08 = 10612.08\)
Let quarterly rate be \(R\%\).
\(10612.08 = 10000\left(1 + \frac{R}{100}\right)^3\)
\(1.061208 = \left(1 + \frac{R}{100}\right)^3\)
\((1.02)^3 = \left(1 + \frac{R}{100}\right)^3\)
\(1 + \frac{R}{100} = 1.02 \implies \frac{R}{100} = 0.02 \implies R = 2\% \text{ per quarter}\)
Annual rate = \(2\% \times 4 = 8\%\) per annum.
In simple words: Quarterly compounding means 4 periods per year; divide annual rate by 4 and multiply years by 4.

Exam Tip: Always convert quarterly rate back to annual rate if asked.

Question 9. A sum amounts to Rs 9680 in 2 years and to Rs 10648 in 3 years compounded annually. Find the sum (principal) and the rate of interest per annum.
Answer: Amount after 3 years \(A_3 = 10648\)
Amount after 2 years \(A_2 = 9680\)
Interest for 3rd year = \(10648 - 9680 = Rs 968\)
Rate \(R = \frac{\text{Interest} \times 100}{A_2 \times 1} = \frac{968 \times 100}{9680 \times 1} = 10\% \text{ p.a.}\)
To find principal \(P\):
\(9680 = P\left(1 + \frac{10}{100}\right)^2\)
\(9680 = P\left(\frac{11}{10}\right)^2 = P \times \frac{121}{100}\)
\(P = \frac{9680 \times 100}{121} = \text{Rs } 8000\)
In simple words: The interest earned in one year acts as simple interest on the amount of the previous year.

Exam Tip: Difference between consecutive year amounts helps determine the interest rate directly.

Question 10. A sum compounded annually becomes \(\frac{25}{16}\) times of itself in 2 years. Determine the rate of interest per annum.
Answer: \(A = \frac{25}{16}P\)
\(\frac{25}{16}P = P\left(1 + \frac{R}{100}\right)^2\)
\(\frac{25}{16} = \left(1 + \frac{R}{100}\right)^2\)
Taking square root on both sides:
\(\frac{5}{4} = 1 + \frac{R}{100}\)
\(\frac{R}{100} = \frac{5}{4} - 1 = \frac{1}{4}\)
\(R = \frac{100}{4} = 25\% \text{ p.a.}\)
In simple words: Take the square root of the ratio when time is 2 years.