CBSE Class 9 Maths Advanced Number System Very Short Answer Questions Set 01

Very Short Answer Questions

Question 1. Find the product of any two irrational numbers.
Answer:
The product of two irrational numbers can be rational or irrational.
Example of a rational product: \( \sqrt2\times\sqrt2 = 2 \) (or \( \sqrt2\times\sqrt8 = 4 \)).
Example of an irrational product: \( \sqrt2\times\sqrt3 = \sqrt6 \).
Answer: rational or irrational.
In simple words: It depends on the numbers chosen, so it may be either type.

Teacher's Note:
1. Try √2 × √2 and √2 × √3 to see both outcomes.
2. A statement 'always irrational' would be wrong.

Question 2. Find a rational number between \( \sqrt2 \) and \( \sqrt3 \).
Answer:
√2 ≈ 1.414 and √3 ≈ 1.732.
The number 1.5 = 3/2 lies between them and is rational.
Answer: 1.5 (any other such number, e.g. 1.6, is also correct).
In simple words: Pick a terminating decimal between 1.414 and 1.732.

Teacher's Note:
1. Use decimal values of the roots to locate the interval.
2. There are infinitely many correct answers.

Question 3. Find the value of 1.999... in the form \( \frac{p}{q} \), where p and q are integers and q ≠ 0.
Answer:
Let x = 1.999... Then 10x = 19.999...
Subtract: 9x = 18, so x = 2.
So 1.999... = \( \frac{2}{1} \) = 2.
In simple words: A string of 9s that never ends equals the next whole number.

Teacher's Note:
1. Use the subtraction trick: 10x − x.
2. 0.999... = 1 follows the same way.

Question 4. Find the number obtained on rationalising the denominator of \( \frac{1}{\sqrt7-2} \).
Answer:
Multiply the top and bottom by √7 + 2: \( \frac{1}{\sqrt7-2}\times\frac{\sqrt7+2}{\sqrt7+2} = \frac{\sqrt7+2}{7-4} \).
Answer: \( \frac{\sqrt7+2}{3} \).
In simple words: Multiply by the conjugate to remove the root from the denominator.

Teacher's Note:
1. (a − b)(a + b) = a² − b² with a = √7, b = 2.
2. The new denominator is 7 − 4 = 3.

Question 5. After rationalising the denominator of \( \frac{7}{3\sqrt3-2\sqrt2} \), what will be the denominator ?
Answer:
Multiply by \( \frac{3\sqrt3+2\sqrt2}{3\sqrt3+2\sqrt2} \): the denominator is (3√3)² − (2√2)² = 27 − 8 = 19.
The expression becomes \( \frac{7(3\sqrt3+2\sqrt2)}{19} \).
Answer: the denominator is 19.
In simple words: After rationalising, the bottom becomes the rational number 19.

Teacher's Note:
1. 27 comes from 9 × 3 and 8 from 4 × 2.
2. The question asks only for the denominator, not the full fraction.

Question 6. Find the value of \( \frac{\sqrt{32}+\sqrt{48}}{\sqrt8+\sqrt{12}} \).
Answer:
√32 = 4√2, √48 = 4√3, √8 = 2√2 and √12 = 2√3.
The fraction is \( \frac{4\sqrt2+4\sqrt3}{2\sqrt2+2\sqrt3} = \frac{4(\sqrt2+\sqrt3)}{2(\sqrt2+\sqrt3)} \) = 2.
In simple words: Each term in the top is twice the matching term in the bottom.

Teacher's Note:
1. Simplify every root into a multiple of √2 or √3.
2. Then cancel the common bracket.

Question 7. Simplify \( \sqrt[4]{\sqrt[3]{2^2}} \).
Answer:
Convert to exponents: \( \left[(2^2)^{1/3}\right]^{1/4} = 2^{\frac23\times\frac14} \).
The exponent is 2/12 = 1/6.
Answer: \( 2^{1/6} \).
In simple words: Roots of roots multiply: 3 × 4 = 12, so the answer is the 12th root of 4.

Teacher's Note:
1. Powers of powers multiply.
2. 2^{1/6} is the same as the sixth root of 2.

Question 8. Simplify \( \sqrt[4]{(81)^{-2}} \).
Answer:
81 = 3⁴, so (81)⁻² = 3⁻⁸.
The fourth root is 3⁻² = \( \frac{1}{9} \).
In simple words: Take the fourth root of 3^{−8} to get 3^{−2}.

Teacher's Note:
1. Write 81 as a power of 3.
2. A negative exponent means a reciprocal.

Question 9. State whether the following statements are true or false ? Justify your answer.
(i) \( \frac{\sqrt2}{3} \) is a rational number.
(ii) There are infinitely many integers between any two integers.
(iii) Number of rational numbers between 15 and 18 is finite.
(iv) There are numbers which cannot be written in the form \( \frac{p}{q} \), q ≠ 0, p, q both are integers.
(v) The square of an irrational number is always rational.
(vi) \( \frac{\sqrt{12}}{\sqrt3} \) is not a rational number as \( \sqrt{12} \) and \( \sqrt3 \) are not integers.
(vii) \( \frac{\sqrt{15}}{\sqrt3} \) is written in the form \( \frac{p}{q} \), q ≠ 0 and so it is a rational number.
Answer:
(i) False. √2 is irrational, and an irrational number divided by 3 stays irrational.
(ii) False. Between two integers there are only finitely many integers (between 3 and 8 only 4, 5, 6, 7).
(iii) False. Between any two rational numbers there are infinitely many rational numbers.
(iv) True. Irrational numbers such as √2 and π cannot be written as p/q with integers p, q.
(v) False. For example (2 + √3)² = 7 + 4√3 is irrational.
(vi) False. √12/√3 = √4 = 2, which is rational even though √12 and √3 are not integers.
(vii) False. √15/√3 = √5, which is irrational; the numerator and denominator here are not integers, so it is not of the form p/q.
In simple words: Check each statement with a concrete example.

Teacher's Note:
1. One counter-example is enough to prove a statement false.
2. Statement (iv) is true because irrational numbers exist.

Question 10. Which of the following fractions yield a recurring decimal ? \( \frac53;\ \frac7{16};\ \frac9{14};\ \frac57;\ \frac{12}5;\ \frac6{11} \)
Answer:
A fraction in lowest terms gives a terminating decimal only if its denominator has no prime factors other than 2 and 5.
5/3 (denominator 3), 9/14 (7), 5/7 (7) and 6/11 (11) each contain another prime, so they recur.
7/16 = 0.4375 and 12/5 = 2.4 terminate.
Answer: \( \frac{5}{3} \), \( \frac{9}{14} \), \( \frac{5}{7} \), \( \frac{6}{11} \).
In simple words: Look at the prime factors of each denominator.

Teacher's Note:
1. Reduce the fraction first (none of these reduce further).
2. 2s and 5s only: terminating.

Question 11. Which of the following numbers are not rational ? 1.256; \( 0.45454545\ldots \); \( 0.05005000500005\ldots \); \( 5.51551555151\ldots \); \( 2.012340123401234\ldots \)
Answer:
1.256 terminates, so it is rational.
0.45454545... repeats the block 45, so it is rational.
2.012340123401234... repeats the block 01234, so it is rational.
0.05005000500005... has a growing run of zeros between the 5s, and 5.51551555151... has no repeating block, so these do not repeat.
Answer: \( 0.05005000500005\ldots \) and \( 5.51551555151\ldots \).
In simple words: A decimal is irrational when it neither ends nor repeats a block.

Teacher's Note:
1. Look for a block of digits that repeats exactly.
2. Growing patterns do not count as repeating.

Question 12. Find two irrational numbers between \( \sqrt5 \) and \( \sqrt6 \).
Answer:
√5 < √5.1 < √5.2 < √6, because 5 < 5.1 < 5.2 < 6.
√5.1 and √5.2 are not square roots of perfect squares, so they are irrational.
Answer: \( \sqrt{5.1},\ \sqrt{5.2} \).
In simple words: Take square roots of numbers between 5 and 6 that are not perfect squares.

Teacher's Note:
1. Bigger radicand means bigger square root.
2. Any non-square between 5 and 6 works.

Question 13. What is the simplest form of \( \sqrt{200}-\sqrt{50} \) ?
Answer:
√200 = 10√2 and √50 = 5√2.
10√2 − 5√2 = \( 5\sqrt2 \).
In simple words: Both are multiples of √2, so subtract the multipliers.

Teacher's Note:
1. Take out the largest perfect square from each root.
2. Only like surds can be subtracted.

Question 14. Rationalise the denominator of \( \frac{5}{\sqrt{10}+\sqrt5} \).
Answer:
Multiply by (√10 − √5)/(√10 − √5): the denominator is 10 − 5 = 5.
The value is \( \frac{5(\sqrt{10}-\sqrt5)}{5} = \)\( \sqrt{10}-\sqrt5 \).
In simple words: The conjugate removes the roots from the bottom.

Teacher's Note:
1. The denominator is a difference of squares.
2. The 5s cancel at the end.

Question 15. If x = \( \sqrt2-1 \), what is the value of \( x-\frac1x \) ?
Answer:
1/x = 1/(√2 − 1) = √2 + 1 (rationalising).
x − 1/x = (√2 − 1) − (√2 + 1) = −2.
In simple words: Rationalise 1/x first and then subtract.

Teacher's Note:
1. x and 1/x are conjugates.
2. The √2 terms cancel.

Question 16. Simplify \( (\sqrt5+1)^2+(\sqrt5-1)^2 \).
Answer:
(a + b)² + (a − b)² = 2(a² + b²) with a = √5, b = 1.
= 2(5 + 1) = 12.
In simple words: The cross terms cancel and the squares double.

Teacher's Note:
1. Check by expanding: 6 + 2√5 and 6 − 2√5.
2. Their sum is 12.

Question 17. Find the value of the following : (i) \( 81^{1/4} \) (ii) \( 64^{1/3} \) (iii) \( 32^{3/5} \) (iv) \( 27^{2/3} \)
Answer:
(i) 81 = 3⁴, so 81^1/4 = 3.
(ii) 64 = 4³, so 64^1/3 = 4.
(iii) 32^3/5 = (2⁵)^3/5 = 2³ = 8.
(iv) 27^2/3 = (3³)^2/3 = 3² = 9.
In simple words: Write each number as a perfect power and multiply the exponents.

Teacher's Note:
1. The denominator of the exponent is the root.
2. The numerator is the power.

Question 18. Simplify the following : (i) \( 3^{1/3}\cdot3^{2/5} \) (ii) \( \left(\frac49\right)^{1/2} \) (iii) \( 2^{2/3}\cdot3^{2/3} \) (iv) \( \left(2^{1/3}\right)^3 \)
Answer:
(i) Add exponents: 1/3 + 2/5 = 11/15, so \( 3^{11/15} \).
(ii) (4/9)^1/2 = 2/3.
(iii) Same exponent, so multiply the bases: (2 × 3)^2/3 = \( 6^{2/3} \).
(iv) Multiply exponents: 2¹ = 2.
In simple words: Use the laws of exponents: add for the same base, multiply for powers of powers.

Teacher's Note:
1. a^m · a^n = a^{m+n}.
2. a^m · b^m = (ab)^m.

Question 19. Simplified value of \( 2^n\times4^n\times8^{1-n} \).
Answer:
Write with base 2: 2ⁿ × 2²ⁿ × 2^{3(1 − n)} = 2^{n + 2n + 3 − 3n}.
The exponent is 3, so the value is 2³ = 8.
In simple words: All the n terms cancel, leaving 2³.

Teacher's Note:
1. Convert 4 and 8 to powers of 2.
2. The result does not depend on n.

Question 20. Find x, if \( 8^x = 16 \).
Answer:
8 = 2³ and 16 = 2⁴, so 2^3x = 2⁴.
3x = 4, so x = 4/3.
In simple words: Make both sides powers of 2 and equate exponents.

Teacher's Note:
1. Same base means the exponents are equal.
2. Check: 8^{4/3} = 2⁴ = 16.

Question 21. Express \( \frac{2}{11} \) in decimal form.
Answer:
Divide 2 by 11 (20 ÷ 11 = 1 remainder 9, then 90 ÷ 11 = 8 remainder 2, and the remainders repeat).
2/11 = 0.1818... = \( 0.\overline{18} \).
In simple words: The remainders 9 and 2 repeat, so the digits 18 repeat.

Teacher's Note:
1. A remainder that has appeared before means the digits repeat.
2. 11 has a prime factor other than 2 and 5, so the decimal recurs.

Question 22. Give one example where the product of two different irrational numbers is rational.
Answer:
Take 3 + √2 and 3 − √2, both irrational.
Their product is 3² − (√2)² = 9 − 2 = 5, which is rational.
Answer: \( (3+\sqrt2)(3-\sqrt2) = 5 \).
In simple words: Conjugate surds multiply to give a rational number.

Teacher's Note:
1. Another example: √2 × √8 = 4.
2. Using a, b conjugates uses a² − b².

Question 23. Find the value of \( 2\sqrt5+3\sqrt5 \).
Answer:
Both terms are like surds, so add the coefficients: (2 + 3)√5 = \( 5\sqrt5 \).
In simple words: Like surds add just like like terms in algebra.

Teacher's Note:
1. Same radicand means you can combine.
2. 2√5 + 3√2 would not combine.

Question 24. What is the square root of the number 0.04 in fraction form ?
Answer:
√0.04 = 0.2 because 0.2 × 0.2 = 0.04.
0.2 = 2/10 = \( \frac{1}{5} \).
In simple words: Find the decimal square root and write it as a fraction.

Teacher's Note:
1. 0.04 = 4/100 and √(4/100) = 2/10.
2. Reduce the fraction at the end.

Question 25. Simplify the expression \( \frac{3}{\sqrt{48}-\sqrt{75}} \).
Answer:
√48 = 4√3 and √75 = 5√3, so the denominator is 4√3 − 5√3 = −√3.
The expression is \( \frac{3}{-\sqrt3} = -\sqrt3 \).
Answer: \( -\sqrt3 \).
In simple words: Simplify the roots first and combine like surds.

Teacher's Note:
1. Both roots are multiples of √3.
2. 3/√3 = √3.