CBSE Class 8 Maths Advanced Rational Numbers Long Answer Questions Set 01

Long Answer Questions

Question 1. (i) Verify that | x + y | ≤ | x | + | y | by taking x = \( \frac{-5}{12} \), y = \( \frac{-7}{18} \)
(ii) Verify that | x × y | = | x | × | y | by taking x = \( \frac{-2}{3} \), y = \( \frac{-9}{8} \).
Answer:
(i) x + y = −5/12 − 7/18 = (−15 − 14)/36 = −29/36, so |x + y| = 29/36.
|x| + |y| = 5/12 + 7/18 = (15 + 14)/36 = 29/36.
So |x + y| = |x| + |y|, and the relation |x + y| ≤ |x| + |y| holds (with equality, since x and y have the same sign).
(ii) x × y = (−2/3) × (−9/8) = 18/24 = 3/4, so |x × y| = 3/4.
|x| × |y| = 2/3 × 9/8 = 18/24 = 3/4.
Hence |x × y| = |x| × |y|.
In simple words: Calculate both sides separately and compare them.

Teacher's Note:
1. The equality in (i) happens because both numbers are negative.
2. Absolute value of a product is the product of absolute values.

Question 2. (i) Which rational number is its own additive inverse ? (ii) Is the difference of two rational numbers a rational number ? (iii) Is addition commutative on rational numbers ? (iv) Is addition associative on rational numbers ? (v) Is subtraction commutative on rational numbers ? (vi) Is subtraction associative on rational numbers ? (vii) What is the negative of negative rational number ?
Answer:
(i) 0, because 0 + 0 = 0.
(ii) Yes. The difference of two rational numbers is always rational (closure under subtraction).
(iii) Yes: a + b = b + a.
(iv) Yes: (a + b) + c = a + (b + c).
(v) No: 5 − 3 = 2 but 3 − 5 = −2.
(vi) No: (8 − 4) − 2 = 2 but 8 − (4 − 2) = 6.
(vii) A positive rational number, since −(−a) = a.
In simple words: Addition obeys both laws, subtraction obeys neither, and the negative of a negative is positive.

Teacher's Note:
1. Use small numbers as counter-examples.
2. Only 0 is equal to its own additive inverse.

Question 3. The cost of \( 3\frac{7}{9} \) m cloth is Rs. \( 212\frac{4}{5} \). Find the cost of \( 7\frac{2}{3} \) m cloth.
Answer:
3 7/9 = 34/9 m costs 212 4/5 = 1064/5 rupees.
Cost of 1 m = (1064/5) ÷ (34/9) = 1064/5 × 9/34 = 9576/170 = 4788/85 rupees.
7 2/3 = 23/3 m, so the cost = 4788/85 × 23/3 = 1596 × 23/85 = 36708/85.
36708 ÷ 85 = 431 with remainder 73, so the cost is Rs. \( 431\frac{73}{85} \).
In simple words: Find the cost of one metre first, then multiply by the new length.

Teacher's Note:
1. Convert mixed numbers to improper fractions.
2. Cost of 1 m is Rs. 56 28/85, roughly Rs. 56.33.

Question 4. Simplify : (i) \( 2\frac{3}{4}\times 1\frac{2}{3}+9\frac{11}{12}-1\frac{5}{6} \) (ii) \( 5-\left[\frac{3}{4}+\left\{2\frac{1}{2}-\left(0.5+\overline{\frac{1}{6}-\frac{1}{7}}\right)\right\}\right] \)
Answer:
(i) 2 3/4 × 1 2/3 = 11/4 × 5/3 = 55/12.
9 11/12 = 119/12 and 1 5/6 = 11/6 = 22/12.
Value = 55/12 + 119/12 − 22/12 = 152/12 = 38/3 = \( 12\frac{2}{3} \).
(ii) The bar groups 1/6 − 1/7 = 1/42. Then 0.5 + 1/42 = 21/42 + 1/42 = 22/42 = 11/21.
2 1/2 − 11/21 = 5/2 − 11/21 = (105 − 22)/42 = 83/42.
3/4 + 83/42 = (63 + 166)/84 = 229/84.
5 − 229/84 = (420 − 229)/84 = 191/84 = \( 2\frac{23}{84} \).
In simple words: Do the multiplication first in (i); in (ii) work from the innermost bracket outwards.

Teacher's Note:
1. Convert mixed numbers into improper fractions.
2. BODMAS: brackets (innermost first), then of, division, multiplication, addition, subtraction.