CBSE Class 9 Maths Advanced Number System Long Answer Questions Set 01

Long Answer Questions

Question 1. Find three rational numbers between
(i) −1 and −2   (ii) 0.1 and 0.11   (iii) \( \frac{5}{7} \) and \( \frac{6}{7} \)   (iv) \( \frac{1}{4} \) and \( \frac{1}{5} \)
Answer:
(i) −2 = −8/4 and −1 = −4/4. Three numbers between them: \( \frac{-7}{4} \), \( \frac{-3}{2} \), \( \frac{-5}{4} \).
(ii) 0.1 = 0.100 and 0.11 = 0.110. Three numbers between them: 0.101, 0.105, 0.109.
(iii) 5/7 = 20/28 and 6/7 = 24/28. Three numbers: \( \frac{21}{28} \) (= \( \frac{3}{4} \)), \( \frac{22}{28} \) (= \( \frac{11}{14} \)), \( \frac{23}{28} \).
(iv) 1/5 = 0.20 and 1/4 = 0.25. Three numbers: 0.21, 0.22, 0.23 (that is \( \frac{21}{100} \), \( \frac{22}{100} \), \( \frac{23}{100} \)).
In simple words: Give the two numbers a common denominator with enough room, then list fractions in between.

Teacher's Note:
1. Use (n + 1)/(n + 1) with n = 3 to get 4.
2. Decimals are easiest for (ii) and (iv).

Question 2. Insert a rational number and an irrational number between the following :
(i) 2 and 3   (ii) \( \frac{1}{3} \) and \( \frac{1}{2} \)   (iii) \( \frac{-2}{5} \) and \( \frac{1}{2} \)   (iv) \( \sqrt2 \) and \( \sqrt3 \)   (v) 0.0001 and 0.001
Answer:
(i) Rational: 2.5. Irrational: \( \sqrt6 \) (≈ 2.449).
(ii) 1/3 = 0.333... and 1/2 = 0.5. Rational: 5/12 (≈ 0.4167). Irrational: 0.4101001000100001... .
(iii) −2/5 = −0.4 and 1/2 = 0.5. Rational: 0. Irrational: 0.1010010001... .
(iv) √2 ≈ 1.414 and √3 ≈ 1.732. Rational: 1.5. Irrational: \( \sqrt[4]{6} \) (≈ 1.565).
(v) Rational: 0.0005. Irrational: 0.0005010010001... .
In simple words: Rational numbers are terminating or repeating decimals; irrational ones never end or repeat.

Teacher's Note:
1. For irrational numbers, change the pattern as you go (more zeros each time).
2. √(ab) lies between a and b when a, b are positive.

Question 3. Locate \( \sqrt5,\ \sqrt{10} \) and \( \sqrt{17} \) on the number line.
Answer:
5 = 2² + 1², 10 = 3² + 1², 17 = 4² + 1².
For √5: on the number line take OA = 2 and draw AB = 1 perpendicular to OA. Then OB = √5; the arc with centre O and radius OB cuts the line at the point √5.
For √10: take OA = 3 and AB = 1; OB = √10; the arc meets the line at √10.
For √17: take OA = 4 and AB = 1; OB = √17; the arc meets the line at √17.

In simple words: Each root is the hypotenuse of a right triangle with legs (n, 1).

Teacher's Note:
1. Choose n so that n² + 1 equals the number.
2. The arc is centred at O in each case.

Question 4. Represent geometrically the following numbers on the number line : (i) \( \sqrt{4.5} \) (ii) \( \sqrt{2.3} \)
Answer:
(i) √4.5. Take AO = 4.5 units and OB = 1 unit. Draw a semicircle on AB (diameter 5.5 units). The perpendicular at O meets it at D, with OD² = 4.5 × 1, so OD = √4.5. Cut the line with an arc of radius OD centred at O to get P, the point √4.5 ≈ 2.12.

(ii) √2.3. Take AO = 2.3 units and OB = 1 unit; draw a semicircle on AB (3.3 units). The perpendicular at O meets it at D, with OD = √2.3 ≈ 1.52, and the arc centred at O with radius OD meets the line at P.

In simple words: Build a semicircle on a line of length x + 1 and use the perpendicular at the join.

Teacher's Note:
1. The height of the perpendicular is √(x × 1).
2. Draw both semicircles carefully with a compass.

Question 5. Simplify the following :
(i) \( \sqrt{45}-3\sqrt{20}+4\sqrt5 \) (ii) \( \frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9} \) (iii) \( \sqrt[4]{12}\times\sqrt[7]{6} \)
(iv) \( 4\sqrt{28}\div3\sqrt7\div\sqrt[3]{7} \) (v) \( 3\sqrt3+2\sqrt{27}+\frac{7}{\sqrt3} \) (vi) \( (\sqrt3-\sqrt2)^2 \)
(vii) \( \sqrt[4]{81}-8\sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225} \) (viii) \( \frac{3}{\sqrt8}+\frac{1}{\sqrt2} \) (ix) \( \frac{2\sqrt3}{3}-\frac{\sqrt3}{6} \)
Answer:
(i) 3√5 − 3(2√5) + 4√5 = 3√5 − 6√5 + 4√5 = \( \sqrt5 \).
(ii) √24 = 2√6 and √54 = 3√6, so 2√6/8 + 3√6/9 = √6/4 + √6/3 = \( \frac{7\sqrt6}{12} \).
(iii) L.C.M. of 4 and 7 is 28: \( \sqrt[28]{12^7}\times\sqrt[28]{6^4} = \sqrt[28]{2^{14}3^7\cdot2^43^4} = \sqrt[28]{2^{18}\cdot3^{11}} \).
(iv) 4√28 = 8√7, so (8√7)/(3√7) ÷ ∛7 = (8/3) ÷ 7^1/3 = \( \frac{8}{3\sqrt[3]{7}} \).
(v) 3√3 + 6√3 + 7√3/3 = (27 + 7)√3/3 = \( \frac{34}{\sqrt3} \) (= 34√3/3).
(vi) 3 + 2 − 2√6 = \( 5-2\sqrt6 \).
(vii) 3 − 8(6) + 15(2) + 15 = 3 − 48 + 30 + 15 = 0.
(viii) 3/(2√2) + 1/√2 = (3 + 2)/(2√2) = 5/(2√2) = \( \frac{5\sqrt2}{4} \).
(ix) (4√3 − √3)/6 = 3√3/6 = \( \frac{\sqrt3}{2} \).
In simple words: Simplify each root to a multiple of the simplest surd, then combine like surds.

Teacher's Note:
1. Bring different orders to a common order for products.
2. Only like surds can be added.

Question 6. Rationalise the denominator of the following :
(i) \( \frac{2}{3\sqrt3} \) (ii) \( \frac{\sqrt{40}}{\sqrt3} \) (iii) \( \frac{3+\sqrt2}{4\sqrt2} \) (iv) \( \frac{16}{\sqrt{41}-5} \) (v) \( \frac{2+\sqrt3}{2-\sqrt3} \)
(vi) \( \frac{\sqrt6}{2+\sqrt3} \) (vii) \( \frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2} \) (viii) \( \frac{3\sqrt5+\sqrt3}{\sqrt5-\sqrt3} \) (ix) \( \frac{4\sqrt3+5\sqrt2}{\sqrt{48}+\sqrt{18}} \)
Answer:
(i) Multiply by √3/√3: \( \frac{2\sqrt3}{9} \).
(ii) Multiply by √3/√3: \( \frac{\sqrt{120}}{3} = \frac{2\sqrt{30}}{3} \).
(iii) Multiply by √2/√2: \( \frac{3\sqrt2+2}{8} \).
(iv) Multiply by (√41 + 5): \( \frac{16(\sqrt{41}+5)}{41-25} = \sqrt{41}+5 \).
(v) Multiply by (2 + √3): \( \frac{(2+\sqrt3)^2}{4-3} = 7+4\sqrt3 \).
(vi) Multiply by (2 − √3): \( \frac{\sqrt6(2-\sqrt3)}{1} = 2\sqrt6-3\sqrt2 \).
(vii) Multiply by (√3 + √2): \( \frac{(\sqrt3+\sqrt2)^2}{1} = 5+2\sqrt6 \).
(viii) Multiply by (√5 + √3): \( \frac{15+3\sqrt{15}+\sqrt{15}+3}{2} = 9+2\sqrt{15} \).
(ix) √48 + √18 = 4√3 + 3√2. Multiply by (4√3 − 3√2): numerator 48 − 12√6 + 20√6 − 30 = 18 + 8√6, denominator 48 − 18 = 30. Result \( \frac{9+4\sqrt6}{15} \).
In simple words: Multiply top and bottom by the conjugate (or by the root itself) so the bottom becomes rational.

Teacher's Note:
1. The product (a + b)(a − b) = a² − b².
2. Simplify the roots in the denominator first, as in part (ix).

Question 7. Find the values of a and b in each of the following :
(i) \( \frac{5+2\sqrt3}{7+4\sqrt3} = a-6\sqrt3 \) (ii) \( \frac{3-\sqrt5}{3+2\sqrt5} = a\sqrt5-\frac{19}{11} \)
(iii) \( \frac{\sqrt2+\sqrt3}{3\sqrt2-2\sqrt3} = 2-b\sqrt6 \) (iv) \( \frac{7+\sqrt5}{7-\sqrt5}-\frac{7-\sqrt5}{7+\sqrt5} = a+\frac{7}{11}\sqrt5\,b \)
Answer:
(i) Multiply by (7 − 4√3): numerator 35 − 20√3 + 14√3 − 24 = 11 − 6√3; denominator 49 − 48 = 1. So the value is 11 − 6√3 and a = 11.
(ii) Multiply by (3 − 2√5): numerator 9 − 6√5 − 3√5 + 10 = 19 − 9√5; denominator 9 − 20 = −11. The value is −19/11 + (9/11)√5, so a = 9/11.
(iii) Multiply by (3√2 + 2√3): numerator 6 + 2√6 + 3√6 + 6 = 12 + 5√6; denominator 18 − 12 = 6. The value is 2 + (5/6)√6 = 2 − b√6, so b = −5/6.
(iv) The difference is [(7 + √5)² − (7 − √5)²]/(49 − 5) = 28√5/44 = (7/11)√5. So a + (7/11)√5 b = (7/11)√5, giving a = 0 and b = 1.
In simple words: Rationalise the left side and compare the rational and irrational parts.

Teacher's Note:
1. Equal numbers of the form p + q√r have equal rational parts and equal irrational parts.
2. Take care with the minus signs in (ii) and (iii).

Question 8. If a = 2 + √3, then find the value of \( a-\frac1a \).
Answer:
1/a = 1/(2 + √3) = (2 − √3)/(4 − 3) = 2 − √3.
a − 1/a = (2 + √3) − (2 − √3) = \( 2\sqrt3 \).
In simple words: Rationalise 1/a, then subtract.

Teacher's Note:
1. The rational parts cancel.
2. a + 1/a would be 4 here.

Question 9. If a = \( 5+2\sqrt6 \) and b = \( \frac1a \), then what will be the value of \( a^2+b^2 \) ?
Answer:
b = 1/(5 + 2√6) = (5 − 2√6)/(25 − 24) = 5 − 2√6.
a + b = 10 and ab = 25 − 24 = 1.
a² + b² = (a + b)² − 2ab = 100 − 2 = 98.
In simple words: Use the identity a² + b² = (a + b)² − 2ab.

Teacher's Note:
1. First find b by rationalising.
2. ab = 1 because b is the reciprocal of a.

Question 10. If \( \sqrt2=1.414 \) and \( \sqrt3=1.732 \), then find the value of \( \frac{4}{3\sqrt3-2\sqrt2}+\frac{3}{3\sqrt3+2\sqrt2} \).
Answer:
Rationalise: 4/(3√3 − 2√2) = 4(3√3 + 2√2)/19 and 3/(3√3 + 2√2) = 3(3√3 − 2√2)/19.
Sum = [12√3 + 8√2 + 9√3 − 6√2]/19 = (21√3 + 2√2)/19.
= (21 × 1.732 + 2 × 1.414)/19 = (36.372 + 2.828)/19 = 39.2/19 ≈ 2.063.
In simple words: The common denominator is (3√3)² − (2√2)² = 19.

Teacher's Note:
1. Substitute the decimals only at the end.
2. Answer is approximate because the roots are approximate.

Question 11. Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236, up to three places of decimal.
(i) \( \frac{4}{\sqrt3} \) (ii) \( \frac{6}{\sqrt6} \) (iii) \( \frac{\sqrt{10}-\sqrt5}{2} \) (iv) \( \frac{\sqrt2}{2+\sqrt2} \) (v) \( \frac{1}{\sqrt3+\sqrt2} \)
Answer:
(i) 4/√3 = 4√3/3 = 4 × 1.732/3 ≈ 2.309.
(ii) 6/√6 = √6 = √2 × √3 = 1.414 × 1.732 ≈ 2.449.
(iii) √10 = √2 × √5 = 1.414 × 2.236 = 3.162, so (3.162 − 2.236)/2 = 0.463.
(iv) √2/(2 + √2) = √2(2 − √2)/(4 − 2) = √2 − 1 = 0.414.
(v) 1/(√3 + √2) = √3 − √2 = 1.732 − 1.414 = 0.318.
In simple words: Rationalising first means you only add or multiply the given decimals.

Teacher's Note:
1. Dividing by a decimal root directly is harder and less accurate.
2. Products such as √2 × √5 give √10.

Question 12. Simplify :
(i) \( \left(1^3+2^3+3^3\right)^{1/2} \) (ii) \( \left(\frac35\right)^4\left(\frac85\right)^{-12}\left(\frac{32}5\right)^6 \) (iii) \( \left(\frac1{27}\right)^{-2/3} \) (iv) \( \left[\left(625^{-1/2}\right)^{-1/4}\right]^2 \)
(v) \( \frac{9^{1/3}\times27^{-1/2}}{3^{1/6}\times3^{-2/3}} \) (vi) \( 64^{-1/3}\left[64^{1/3}-64^{2/3}\right] \) (vii) \( \frac{8^{1/3}\times16^{1/3}}{32^{-1/3}} \)
Answer:
(i) 1 + 8 + 27 = 36 and 36^1/2 = 6.
(ii) (3/5)⁴ = 81/625; (8/5)⁻¹² = 5¹²/2³⁶; (32/5)⁶ = 2³⁰/5⁶. The product = 81 × 5^{12 − 4 − 6} × 2^{30 − 36} = 81 × 25/64 = \( \frac{2025}{64} \).
(iii) (1/27)^−2/3 = 27^2/3 = 3² = 9.
(iv) 625^−1/2 = 1/25, raised to −1/4 gives 25^1/4 = √5; squared gives 5.
(v) Numerator 3^2/3 × 3^−3/2 = 3^−5/6; denominator 3^{1/6 − 2/3} = 3^−1/2. The ratio is 3^{−5/6 + 1/2} = \( 3^{-1/3} \).
(vi) 64^1/3 = 4 and 64^2/3 = 16, so (1/4)(4 − 16) = −3.
(vii) 8^1/3 = 2, 16^1/3 = 2^4/3 and 32^−1/3 = 2^−5/3. The value is 2 × 2^4/3 × 2^5/3 = 2^{1 + 3} = 16.
In simple words: Rewrite each base as a prime power and use the laws of exponents.

Teacher's Note:
1. Be careful with negative exponents.
2. Check part (v) by writing 9 = 3² and 27 = 3³.

Question 13. If \( a=\frac{3+\sqrt5}{2} \), then find the value of \( a^2+\frac{1}{a^2} \).
Answer:
1/a = 2/(3 + √5) = 2(3 − √5)/(9 − 5) = (3 − √5)/2.
a + 1/a = (3 + √5)/2 + (3 − √5)/2 = 3.
a² + 1/a² = (a + 1/a)² − 2 = 9 − 2 = 7.
In simple words: First find a + 1/a; then square and subtract 2.

Teacher's Note:
1. Rationalise 1/a.
2. The √5 terms cancel in the sum.

Question 14. If \( x=\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2} \) and \( y=\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2} \), then find the value of \( x^2+y^2 \).
Answer:
x = (√3 + √2)²/1 = 5 + 2√6 and y = 5 − 2√6.
x + y = 10 and xy = 1.
x² + y² = 100 − 2 = 98.
In simple words: x and y are reciprocals; use (x + y)² − 2xy.

Teacher's Note:
1. Rationalise each to find the sum and product.
2. xy = 1.

Question 15. Express \( 0.6+0.\overline7+0.4\overline7 \) in the form \( \frac{p}{q} \), where p and q are integers and q ≠ 0.
Answer:
0.6 = 3/5.
0.7 repeating = 7/9.
0.4777... = (47 − 4)/90 = 43/90.
Sum = 3/5 + 7/9 + 43/90 = 54/90 + 70/90 + 43/90 = \( \frac{167}{90} \).
In simple words: Convert each decimal to a fraction and add with denominator 90.

Teacher's Note:
1. Use the direct method for recurring decimals.
2. The L.C.M. of 5, 9 and 90 is 90.

Question 16. Simplify : \( \frac{7\sqrt3}{\sqrt{10}+\sqrt3}-\frac{2\sqrt5}{\sqrt6+\sqrt5}-\frac{3\sqrt2}{\sqrt{15}+3\sqrt2} \).
Answer:
First term: 7√3(√10 − √3)/(10 − 3) = √30 − 3.
Second term: 2√5(√6 − √5)/(6 − 5) = 2√30 − 10.
Third term: 3√2(√15 − 3√2)/(15 − 18) = −√2(√15 − 3√2) = −√30 + 6.
Expression = (√30 − 3) − (2√30 − 10) − (−√30 + 6) = √30 − 3 − 2√30 + 10 + √30 − 6 = 1.
In simple words: Rationalise each term; the √30 parts cancel.

Teacher's Note:
1. Keep track of the minus signs between terms.
2. The final answer is rational.

Question 17. Find the value of \( \frac{4}{(216)^{-2/3}}+\frac{1}{(256)^{-3/4}}+\frac{2}{(243)^{-1/5}} \).
Answer:
(216)^−2/3 = 6⁻² = 1/36, so 4 ÷ (1/36) = 144.
(256)^−3/4 = 4⁻³ = 1/64, so 1 ÷ (1/64) = 64.
(243)^−1/5 = 3⁻¹ = 1/3, so 2 ÷ (1/3) = 6.
Total = 144 + 64 + 6 = 214.
In simple words: A negative exponent in the denominator moves the power to the numerator.

Teacher's Note:
1. Write 216 = 6³, 256 = 4⁴, 243 = 3⁵.
2. Add the three results at the end.