CBSE Class 8 Mathematics Revision Assignment Set 02

Q1: The product of two rational numbers is (- 15/7) .If one of them is (-10/21), then find the number?
Answer: Let the required rational number be \(x\).
According to the question,
\(x \times \left(\frac{-10}{21}\right) = \frac{-15}{7}\)
\(x = \frac{-15}{7} \div \left(\frac{-10}{21}\right)\)
\(x = \frac{-15}{7} \times \left(\frac{21}{-10}\right)\)
\(x = \frac{-15 \times 21}{7 \times -10}\)
\(x = \frac{-3 \times 3}{1 \times -2} = \frac{-9}{-2} = \frac{9}{2}\)
In simple words: Divide the product by the given number to find the unknown number.

Exam Tip: Remember to reciprocal the divisor fraction when changing division to multiplication.

Q2: Write five rational numbers between 25/10 & 27/10?
Answer: The given rational numbers are \(\frac{25}{10}\) and \(\frac{27}{10}\).
To find 5 rational numbers, multiply the numerator and denominator of both numbers by \(5 + 1 = 6\) (or a larger power of 10 for easier decimals).
Let us convert them with denominator 60:
\(\frac{25}{10} = \frac{25 \times 6}{10 \times 6} = \frac{150}{60}\)
\(\frac{27}{10} = \frac{27 \times 6}{10 \times 6} = \frac{162}{60}\)
Five rational numbers between \(\frac{150}{60}\) and \(\frac{162}{60}\) are:
\(\frac{151}{60}, \frac{152}{60}, \frac{153}{60}, \frac{154}{60}, \frac{155}{60}\)
In simple words: Make denominators equal and expand the gap between numerators to insert required numbers.

Exam Tip: There are infinitely many rational numbers between any two rational numbers; any valid set within the range is correct.

Q3: A train is at a uniform speed covers a distance of 350Km in 3 ½ hours. Find the speed of the train & the distance covered by the train in 5Hrs?
Answer: Distance = 350 km
Time = \(3\frac{1}{2}\text{ hours} = \frac{7}{2}\text{ hours}\)
Speed = \(\frac{\text{Distance}}{\text{Time}}\)
Speed = \(350 \div \frac{7}{2} = 350 \times \frac{2}{7} = 50 \times 2 = 100\text{ km/h}\)
Now, distance covered in 5 hours at uniform speed:
Distance = Speed \(\times\) Time
Distance = \(100\text{ km/h} \times 5\text{ hours} = 500\text{ km}\)
In simple words: First find the speed by dividing distance by time, then multiply speed by 5 to get the new distance.

Exam Tip: Always convert mixed fractions into improper fractions before performing division or multiplication.

Q4: Naina had 11 ¼ long ribbon. She cuts it into 9 equal pieces. Find the length of each piece?
Answer: Total length of ribbon = \(11\frac{1}{4} = \frac{45}{4}\text{ units}\)
Number of equal pieces = 9
Length of each piece = \(\frac{\text{Total length}}{\text{Number of pieces}}\)
Length of each piece = \(\frac{45}{4} \div 9 = \frac{45}{4} \times \frac{1}{9} = \frac{5}{4} = 1\frac{1}{4}\text{ units}\)
In simple words: Divide the total length of the ribbon by 9 to get the length of one piece.

Q5: Find the value of x:
a) (4/5)³ ÷ (4/5)⁻³ = (4/5)³ˣ
Answer: Using law of exponents \(a^m \div a^n = a^{m-n}\):
\(\left(\frac{4}{5}\right)^{3 - (-3)} = \left(\frac{4}{5}\right)^{3x}\)
\(\left(\frac{4}{5}\right)^{3+3} = \left(\frac{4}{5}\right)^{3x}\)
\(\left(\frac{4}{5}\right)^{6} = \left(\frac{4}{5}\right)^{3x}\)
Comparing the exponents on both sides:
\(3x = 6\)
\(x = \frac{6}{3} = 2\)
In simple words: Subtract the powers during division, then equate the exponents since bases are equal.
Exam Tip: When bases on both sides of an equation are identical, you can directly equate their powers.

b) (4/15)³ × (4/15)⁻⁶ = (4/15)²ˣ⁺¹
Answer: Using law of exponents \(a^m \times a^n = a^{m+n}\):
\(\left(\frac{4}{15}\right)^{3 + (-6)} = \left(\frac{4}{15}\right)^{2x+1}\)
\(\left(\frac{4}{15}\right)^{-3} = \left(\frac{4}{15}\right)^{2x+1}\)
Comparing the exponents on both sides:
\(2x + 1 = -3\)
\(2x = -3 - 1\)
\(2x = -4\)
\(x = \frac{-4}{2} = -2\)
In simple words: Add the powers during multiplication, then solve the linear equation for \(x\).

Exam Tip: Watch out for signs when adding negative exponents.

Q6: Find the squares of the following numbers using the column method:
a) 99
Answer: Let \(a = 9\) and \(b = 9\).
Using algebraic identity \((a + b)^2 = a^2 + 2ab + b^2\), we set up three columns:
Column I: \(a^2 = 9^2 = 81\)
Column II: \(2ab = 2 \times 9 \times 9 = 162\)
Column III: \(b^2 = 9^2 = 81\)
 

From Column III, keep 1 and add 8 to Column II (\(162 + 8 = 170\)).
From Column II, keep 0 and add 17 to Column I (\(81 + 17 = 98\)).
So, \(99^2 = 9801\).
In simple words: Break the number into tens and units, apply the column formula, and carry over digits from right to left.

Exam Tip: Always carry over the tens digit from the right column to the middle column, and from the middle to the left column.

b) 89
Answer: Let \(a = 8\) and \(b = 9\).
Column I: \(a^2 = 8^2 = 64\)
Column II: \(2ab = 2 \times 8 \times 9 = 144\)
Column III: \(b^2 = 9^2 = 81\)
 

From Column III, keep 1 and add 8 to Column II (\(144 + 8 = 152\)).
From Column II, keep 2 and add 15 to Column I (\(64 + 15 = 79\)).
So, \(89^2 = 7921\).
In simple words: Use the column method table to calculate squares step-by-step using carries.

Exam Tip: Double check your multiplication in Column II (\(2ab\)) to avoid calculation errors.

Q7: Find the square root of 0.9 correct up to three decimal places. (Using long division method)?
Answer: To find the square root of 0.9 up to three decimal places, write it as \(0.900000\).
Grouping digits in pairs from the decimal point: \(0.90\) 00\ 00 \)
1. Trial for first pair 90: \(9 \times 9 = 81\). Remainder = \(9\). Quotient = \(0.9\).
2. Bring down pair 00 to make 900. Double the quotient (9) to get 18. Trial: \(184 \times 4 = 736\). Remainder = \(164\). Quotient = \(0.94\).
3. Bring down pair 00 to make 16400. Double the quotient (94) to get 188. Trial: \(1888 \times 8 = 15104\). Remainder = \(1296\). Quotient = \(0.948\).
4. Bring down pair 00 to make 129600. Double the quotient (948) to get 1896. Trial: \(18966 \times 6 = 113796\). Remainder = \(15804\). Quotient = \(0.9486 \rightarrow 0.949\).
Therefore, \(\sqrt{0.9} \approx 0.949\).
In simple words: Append zeros to make pairs after the decimal point and continue long division until four decimal places to round off to three.

Exam Tip: Always calculate one extra decimal place than required so you can correctly round off the final answer.

Q8: Find the cube root of the following numbers by finding their ones & tens digit:
A) 300763
Answer: 1. Group the digits in sets of three from right to left: \(300\) and \(763\).
2. Consider the first group \(763\): The units digit is 3. Since \(7^3 = 343\), the units digit of the cube root is 7.
3. Consider the second group \(300\): The cube numbers are \(6^3 = 216\) and \(7^3 = 343\). Since \(216 < 300 < 343\), the tens digit is 6.
Therefore, \(\sqrt[3]{300763} = 67\).
In simple words: Split the number into groups of three, use the units digit to find the ones place, and range comparison for the tens place.

Exam Tip: Memorize cubes of numbers from 1 to 10 to quickly identify the tens digit range.

b) 636056
Answer: 1. Group the digits in sets of three from right to left: \(636\) and \(056\).
2. Consider the first group \(056\): The units digit is 6. Since \(6^3 = 216\), the units digit of the cube root is 6.
3. Consider the second group \(636\): The cube numbers are \(8^3 = 512\) and \(9^3 = 729\). Since \(512 < 636 < 729\), the tens digit is 8.
Therefore, \(\sqrt[3]{636056} = 86\).
In simple words: Split into two groups of three, find the matching cube ending for ones digit, and lower bound cube for tens digit.

Exam Tip: Numbers ending in 0, 1, 4, 5, 6, and 9 have the exact same digits at the units place of their cubes.

Q9: Divide by long division method:
a) \(12x^3 - 20x^2 - 9x + 15\) by \((3x-5)\)
Answer: Dividend = \(12x^3 - 20x^2 - 9x + 15\), Divisor = \(3x - 5\)
1. Divide first term: \(\frac{12x^3}{3x} = 4x^2\)
Multiply \(4x^2\) by \((3x - 5)\): \(12x^3 - 20x^2\)
Subtract: \((12x^3 - 20x^2) - (12x^3 - 20x^2) = 0\). Bring down \(-9x + 15\).
2. Divide new first term: \(\frac{-9x}{3x} = -3\)
Multiply \(-3\) by \((3x - 5)\): \(-9x + 15\)
Subtract: \((-9x + 15) - (-9x + 15) = 0\).
Quotient = \(4x^2 - 3\), Remainder = 0.
In simple words: Divide the leading term of the dividend by the leading term of the divisor at each step, multiply, and subtract.

Exam Tip: Always arrange the polynomial in descending order of powers before starting long division.

b) \(6y^5 + 4y^4 - 3y^3 - 1\) by \((3y^2 - y + 1)\)
Answer: Dividend = \(6y^5 + 4y^4 - 3y^3 + 0y^2 - 1\), Divisor = \(3y^2 - y + 1\)
1. First term quotient: \(\frac{6y^5}{3y^2} = 2y^3\)
Multiply \(2y^3(3y^2 - y + 1) = 6y^5 - 2y^4 + 2y^3\)
Subtract from dividend: \((6y^4 + 4y^4 - 3y^3) - (6y^5 - 2y^4 + 2y^3) = 6y^4 - 5y^3\). Bring down \(0y^2\).
2. Second term quotient: \(\frac{6y^4}{3y^2} = 2y^2\)
Multiply \(2y^2(3y^2 - y + 1) = 6y^4 - 2y^3 + 2y^2\)
Subtract: \((-5y^3) - (-2y^3) - 2y^2 = -3y^3 - 2y^2\). Bring down \(-1\).
3. Third term quotient: \(\frac{-3y^3}{3y^2} = -y\)
Multiply \(-y(3y^2 - y + 1) = -3y^3 + y^2 - y\)
Subtract: \((-2y^2) - (y^2) - (-y) = -3y^2 + y - 1\).
4. Fourth term quotient: \(\frac{-3y^2}{3y^2} = -1\)
Multiply \(-1(3y^2 - y + 1) = -3y^2 + y - 1\)
Subtract: remainder = 0.
Quotient = \(2y^3 + 2y^2 - y - 1\), Remainder = 0.
In simple words: Include placeholder terms with zero coefficients for missing powers to keep columns aligned during polynomial division.

Exam Tip: Be extremely careful with sign changes when subtracting polynomial rows.

Q10: Factorize the following expressions:
a) \(16x^4 - (z-x)^4\)
Answer: Rewrite the expression as a difference of squares: \((4x^2)^2 - ((z-x)^2)^2\)
Using identity \(A^2 - B^2 = (A - B)(A + B)\), where \(A = 4x^2\) and \(B = (z-x)^2\):
\(= [4x^2 - (z-x)^2][4x^2 + (z-x)^2]\)
Now expand terms inside the first bracket using difference of squares again: \(4x^2 = (2x)^2\)
\(= [(2x)^2 - (z-x)^2][4x^2 + (z-x)^2]\)
\(= [(2x - (z - x))(2x + (z - x))][4x^2 + z^2 - 2zx + x^2]\)
\(= [(2x - z + x)(2x + z - x)][5x^2 - 2zx + z^2]\)
\(= (3x - z)(x + z)(5x^2 - 2zx + z^2)\)
In simple words: Apply the difference of squares formula repeatedly until no further factorization is possible.

Exam Tip: Always check if any of the resulting factors can be factored further using algebraic identities.

b) \(x^8 - y^8 + x^4 - y^4\)
Answer: Group terms: \((x^8 - y^8) + (x^4 - y^4)\)
Factor \(x^8 - y^8\) using difference of squares: \((x^4)^2 - (y^4)^2 = (x^4 - y^4)(x^4 + y^4)\)
Substitute back into the expression:
\(= (x^4 - y^4)(x^4 + y^4) + (x^4 - y^4)\)
Factor out the common binomial term \((x^4 - y^4)\):
\(= (x^4 - y^4)[(x^4 + y^4) + 1]\)
Now factor \(x^4 - y^4\) further as \((x^2)^2 - (y^2)^2 = (x^2 - y^2)(x^2 + y^2)\):
\(= (x^2 - y^2)(x^2 + y^2)(x^4 + y^4 + 1)\)
Finally, factor \(x^2 - y^2 = (x - y)(x + y)\):
\(= (x - y)(x + y)(x^2 + y^2)(x^4 + y^4 + 1)\)
In simple words: Group terms, factor out common expressions, and apply the difference of squares identity to each polynomial factor.

Exam Tip: Taking common factors out first simplifies higher-degree polynomial factorizations significantly.