Extra Questions: Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply

Practice questions beyond the textbook exercises, testing deeper understanding of the distributive property and algebraic identities for Class 8 Maths Chapter 6: We Distribute, Yet Things Multiply. These Class 8 Mathematics Chapter 6 important questions are handy for last-minute exam practice.

Extra Questions: Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply

1. 1 (Assertion-Reason). Assertion: (x+7)² and (x−7)² differ by 28x. Reason: (a+b)²−(a−b)² = 4ab.
   Solution: Both true, and the reason correctly explains the assertion — (a+b)²−(a−b)²=4ab, so with a=x, b=7 the difference is 4×x×7=28x.
2. 2 (Numerical). Find the value of 998² using a suitable identity.
   Solution: 998² = (1000−2)² = 1000000−4000+4 = 996004.
3. 3 (Short Answer). Is (a+b)³ the same as a³+b³?
   Solution: No — (a+b)³ = a³+3a²b+3ab²+b³, which has two extra cross terms not present in a³+b³.
4. 4 (Applied). A square garden of side (x+5) m has a square flower bed of side x m cut from one corner. Find the remaining area.
   Solution: Remaining area = (x+5)²−x² = x²+10x+25−x² = (10x+25) m².
5. 5 (Assertion-Reason). Assertion: 205×195 can be found quickly using a difference-of-squares identity. Reason: 205×195 = (200+5)(200−5) = 200²−5².
   Solution: Both true, and the reason correctly explains the assertion — 200²−5² = 40000−25 = 39975.
6. 6 (Numerical). Two consecutive even numbers have squares differing by 84. Find the numbers.
   Solution: Let the numbers be 2n and 2n+2. (2n+2)²−(2n)² = 8n+4 = 84 ⇒ n=10. The numbers are 20 and 22 (22²−20²=484−400=84).
7. 7 (Synthesis). If a+b=10 and ab=21, find a²+b².
   Solution: a²+b² = (a+b)²−2ab = 100−42 = 58.
8. 8 (Applied). Simplify 104×96 using a suitable identity.
   Solution: 104×96 = (100+4)(100−4) = 10000−16 = 9984.
9. 9 (Short Answer). Why is (x−y)² always non-negative, even when x−y itself is negative?
   Solution: Squaring removes the sign of any number, so (x−y)² = (y−x)² ≥ 0 regardless of whether x−y is positive or negative.
10. 10 (Assertion-Reason). Assertion: For any integer n, (n+1)²−(n−1)² is always a multiple of 4. Reason: (n+1)²−(n−1)² = 4n.
    Solution: Both true, and the reason correctly explains the assertion — the expression simplifies exactly to 4n, which is always a multiple of 4.

Written by Satish

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