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

Short Answer Questions

Question 1. If \( \sqrt2=1.4142 \), then find the value of \( \sqrt{\frac{\sqrt2-1}{\sqrt2+1}} \).
Answer:
Rationalise the fraction inside: \( \frac{\sqrt2-1}{\sqrt2+1}\times\frac{\sqrt2-1}{\sqrt2-1} = \frac{(\sqrt2-1)^2}{2-1} = (\sqrt2-1)^2 \).
So the square root is √2 − 1 = 1.4142 − 1 = 0.4142.
In simple words: The fraction is a perfect square once rationalised.

Teacher's Note:
1. Multiply top and bottom by the conjugate of the denominator.
2. The denominator becomes 1.

Question 2. Find the product \( \sqrt[3]{2}\cdot\sqrt[4]{2}\cdot\sqrt[12]{32} \).
Answer:
Write as powers of 2: 2^1/3 × 2^1/4 × 2^5/12 (since 32 = 2⁵).
Add the exponents: 4/12 + 3/12 + 5/12 = 12/12 = 1.
The product is 2¹ = 2.
In simple words: The three exponents add up to exactly 1.

Teacher's Note:
1. Use 12 as the common denominator.
2. 32 = 2⁵ so the twelfth root of 32 is 2^{5/12}.

Question 3. Find the value of \( (256)^{0.16}\times(256)^{0.09} \).
Answer:
Same base, so add the exponents: 0.16 + 0.09 = 0.25.
256^0.25 = 256^1/4 = 4 (since 4⁴ = 256).
In simple words: Adding the exponents gives the fourth root of 256.

Teacher's Note:
1. 0.25 is 1/4.
2. 256 = 4⁴ = 2⁸.

Question 4. Locate \( \sqrt{13} \) on the number line.
Answer:
Write 13 as a sum of two squares: 13 = 9 + 4 = 3² + 2².
On the number line take OA = 3 units. Draw AB = 2 units perpendicular to OA and join OB.
By Pythagoras, OB = √(3² + 2²) = √13.
With O as centre and OB as radius draw an arc cutting the number line at C. Then C represents \( \sqrt{13} \) (the remark: OA = 2 and AB = 3 also works).

In simple words: Build a right triangle with legs 3 and 2 and swing its hypotenuse down to the line.

Teacher's Note:
1. Look for two squares that add up to 13.
2. The arc is drawn with the compass centred at O.

Question 5. Express \( 0.12\overline{3} \) in the form \( \frac{p}{q} \), where p and q are integers and q ≠ 0.
Answer:
Let x = 0.12333... Then 10x = 1.2333...
Multiply again by 10: 100x = 12.333... and 1000x = 123.333...
Subtract: 1000x − 100x = 123.333... − 12.333... = 111, so 900x = 111.
x = 111/900 = \( \frac{37}{300} \).
In simple words: Shift the decimal so the repeating 3s line up, then subtract.

Teacher's Note:
1. Direct check: (123 − 12)/900 = 111/900.
2. Reduce by 3 at the end.

Question 6. Find the value of a in the following : \( \frac{6}{3\sqrt2-2\sqrt3} = 3\sqrt2-a\sqrt3 \).
Answer:
Multiply the top and bottom by 3√2 + 2√3. The denominator is (3√2)² − (2√3)² = 18 − 12 = 6.
The fraction becomes 6(3√2 + 2√3)/6 = 3√2 + 2√3.
So 3√2 + 2√3 = 3√2 − a√3, giving −a = 2 and a = −2.
In simple words: Rationalise and compare the √3 terms on both sides.

Teacher's Note:
1. The rational and irrational parts must match separately.
2. Sign of a: −a√3 = 2√3.

Question 7. Simplify : \( \left[5\left(8^{1/3}+27^{1/3}\right)^3\right]^{1/4} \).
Answer:
8^1/3 = 2 and 27^1/3 = 3, so the bracket is 5(2 + 3)³ = 5 × 5³ = 5⁴.
(5⁴)^1/4 = 5.
In simple words: Everything collapses to the fourth root of 5⁴.

Teacher's Note:
1. Work from the inside out.
2. 5 × 5³ = 5⁴.

Question 8. Represent the following numbers on the number line : \( 7,\ 7.2,\ \frac{-3}{2},\ \frac{-12}{5} \)
Answer:
7 is the integer 7 on the line.
7.2 lies two tenths to the right of 7.
−3/2 = −1.5 lies half-way between −2 and −1.
−12/5 = −2.4 lies between −3 and −2, four tenths to the left of −2.

In simple words: Convert each number to a decimal and place it at that distance from 0.

Teacher's Note:
1. Negative numbers go to the left of 0.
2. Convert fractions to decimals to place them.

Question 9. Represent the number \( \frac{3}{5} \) on the number line.
Answer:
Draw a line and mark O (0) and A (1) on it.
Divide OA into 5 equal parts (the denominator is 5). The third division point from O is P, which represents 3/5.

In simple words: Split the unit length into 5 parts and count 3 of them.

Teacher's Note:
1. The denominator tells the number of parts.
2. The numerator tells how many parts to count.

Question 10. Find a fraction between \( \frac{3}{8} \) and \( \frac{2}{5} \).
Answer:
Take the mean: \( \frac12\left(\frac38+\frac25\right) = \frac12\cdot\frac{15+16}{40} = \frac{31}{80} \).
Check: 3/8 = 0.375, 31/80 = 0.3875 and 2/5 = 0.4.
Answer: \( \frac{31}{80} \).
In simple words: The average of two numbers lies between them.

Teacher's Note:
1. Make the denominators equal before adding.
2. Several other fractions also work.

Question 11. Insert 5 rational numbers between 3 and 4.
Answer:
Multiply the top and bottom of 3 and 4 by (5 + 1) = 6: 3 = 18/6 and 4 = 24/6.
Five numbers between them are \( \frac{19}{6} \), \( \frac{20}{6} \), \( \frac{21}{6} \), \( \frac{22}{6} \), \( \frac{23}{6} \).
In simple words: Write both numbers with denominator 6 and list the numerators 19 to 23.

Teacher's Note:
1. Using n + 1 gives exactly n choices.
2. 20/6 and 21/6 can also be reduced.

Question 12. Represent \( 1.\overline{129} \) (1.129129129...) as a fraction.
Answer:
Let x = 1.129129... Three digits repeat, so 1000x = 1129.129129...
Subtract: 999x = 1128.
x = \( \frac{1128}{999} \) = \( \frac{376}{333} \) in lowest terms.
In simple words: Shift by three places and subtract.

Teacher's Note:
1. The 1 before the decimal point stays in the whole number part.
2. 1128 and 999 have the common factor 3.

Question 13. Represent \( \sqrt6 \) on the number line.
Answer:
Take x = 6. Mark A and O so that AO = 6 units, and put B on the line so that OB = 1 unit (beyond O).
Draw a semicircle on AB as diameter (centre C). At O draw a perpendicular meeting the semicircle at D.
OD² = AO × OB = 6 × 1, so OD = √6. With centre O and radius OD draw an arc meeting the line at P: OP = √6.

In simple words: Use the semicircle construction with AO = 6 and OB = 1.

Teacher's Note:
1. The perpendicular height in a semicircle is the geometric mean of the two parts.
2. Here that mean is √(6 × 1).

Question 14. Represent \( \sqrt{7.3} \) on the number line.
Answer:
Mark AO = 7.3 units and OB = 1 unit. Draw a semicircle on AB (diameter 8.3 units).
The perpendicular OD at O meets the semicircle at D, and OD² = 7.3 × 1, so OD = √7.3.
With O as centre and OD as radius cut the line at P. Then OP = √7.3 ≈ 2.70.

In simple words: The same construction works for any positive number x.

Teacher's Note:
1. Check: 2.7² = 7.29.
2. Use a ruler or compass for the measurements.

Question 15. If \( x=\frac{\sqrt5-2}{\sqrt5+2} \), find the value of \( x+\frac1x \).
Answer:
x = (√5 − 2)²/(5 − 4) = 9 − 4√5 and 1/x = (√5 + 2)/(√5 − 2) = 9 + 4√5.
x + 1/x = (9 − 4√5) + (9 + 4√5) = 18.
In simple words: x and 1/x are conjugates, so the √5 parts cancel.

Teacher's Note:
1. Rationalise each fraction.
2. The denominator is 5 − 4 = 1.

Question 16. Find x, y, z if \( 15^3\times12^2\times16^4 = 2^x\times3^y\times5^z \).
Answer:
15³ = 3³ × 5³, 12² = (2² × 3)² = 2⁴ × 3², and 16⁴ = 2¹⁶.
Multiply: 2^{4 + 16} × 3^{3 + 2} × 5³ = 2²⁰ × 3⁵ × 5³.
Comparing: x = 20, y = 5, z = 3.
In simple words: Break every number into primes and add the powers.

Teacher's Note:
1. Prime factorisation is unique.
2. Group the powers of 2, 3 and 5 separately.

Question 17. Give three rational numbers between −2 and −1.
Answer:
−2 = −8/4 and −1 = −4/4.
Three numbers between them are \( \frac{-7}{4} \), \( \frac{-6}{4} \) (= \( \frac{-3}{2} \)), \( \frac{-5}{4} \).
In simple words: Write both with denominator 4 and pick numerators −7, −6, −5.

Teacher's Note:
1. Negative numbers: −7/4 is closer to −2.
2. Any other three numbers in between are fine.

Question 18. Find five rational numbers between \( \frac{3}{5} \) and \( \frac{4}{5} \).
Answer:
Multiply top and bottom by 6: 3/5 = 18/30 and 4/5 = 24/30.
Five numbers between them: \( \frac{19}{30} \), \( \frac{20}{30} \), \( \frac{21}{30} \), \( \frac{22}{30} \), \( \frac{23}{30} \) (these can be reduced, e.g. 20/30 = 2/3).
In simple words: Use a bigger common denominator to make room for five fractions.

Teacher's Note:
1. Multiplying by (n + 1)/(n + 1) with n = 5 gives 6.
2. Reduce the answers if you wish.

Question 19. Write three irrational numbers between \( \sqrt3 \) and \( \sqrt5 \).
Answer:
√3 = 1.732... and √5 = 2.236...
Numbers such as 1.801001..., 1.901001... and 2.01001... have non-repeating, non-terminating decimals and lie between 1.732 and 2.236.
Answer: 1.801001..., 1.901001..., 2.01001....
In simple words: Build decimals with a changing pattern between 1.732 and 2.236.

Teacher's Note:
1. A repeating block would make the number rational.
2. The runs of zeros grow in each number.

Question 20. Find three different irrational numbers between \( \frac{5}{7} \) and \( \frac{9}{11} \).
Answer:
5/7 = 0.714285... and 9/11 = 0.818181...
Choose non-repeating decimals in between: 0.7507500750007..., 0.7670767007670007... and 0.808008000800008...
Answer: 0.7507500750007..., 0.7670767007670007..., 0.808008000800008....
In simple words: Find the decimal range first, then make irrational decimals inside it.

Teacher's Note:
1. 0.714 < each number < 0.818.
2. Patterns with growing runs never repeat.

Question 21. Multiply \( 3\sqrt{28} \) by \( 2\sqrt7 \).
Answer:
√28 = 2√7, so 3√28 = 6√7.
6√7 × 2√7 = 12 × 7 = 84.
In simple words: Simplify √28 first; then √7 × √7 = 7.

Teacher's Note:
1. Simplifying the radical makes the multiplication easy.
2. Multiply the coefficients 6 and 2.

Question 22. Find the value of \( \frac{6}{\sqrt5-\sqrt3} \), if \( \sqrt3=1.732 \) and \( \sqrt5=2.236 \).
Answer:
Rationalise: \( \frac{6}{\sqrt5-\sqrt3}\times\frac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3} = \frac{6(\sqrt5+\sqrt3)}{2} = 3(\sqrt5+\sqrt3) \).
= 3(2.236 + 1.732) = 3 × 3.968 = 11.904.
In simple words: Rationalise first so that you only need to add the given decimals.

Teacher's Note:
1. The denominator becomes 5 − 3 = 2.
2. Do not divide by the difference 0.504 directly.

Question 23. Multiply \( \sqrt{27a^3b^2c^4}\times\sqrt[3]{128a^7b^9c^2}\times\sqrt[6]{729ab^{12}c^2} \).
Answer:
Convert all to the sixth root: the three radicands become (27a³b²c⁴)³, (128a⁷b⁹c²)² and 729ab¹²c², that is 3⁹a⁹b⁶c¹², 2¹⁴a¹⁴b¹⁸c⁴, 3⁶ab¹²c².
Product: 2¹⁴ 3¹⁵ a²⁴ b³⁶ c¹⁸ under the sixth root.
Take out sixth powers: 2² · 3² · a⁴ · b⁶ · c³ and what is left is ⁶√(2² · 3³).
Answer: \( 36\,a^4b^6c^3\sqrt[6]{108} \).
In simple words: Bring everything to the sixth root and collect the powers.

Teacher's Note:
1. L.C.M. of 2, 3, 6 is 6.
2. Each exponent is divided by 6 and the remainder stays inside.

Question 24. Find the values of a and b : (i) \( \frac{\sqrt{11}-\sqrt7}{\sqrt{11}+\sqrt7} = a-b\sqrt{77} \) (ii) \( \frac{3+\sqrt6}{\sqrt3+\sqrt2} = a+b\sqrt3 \).
Answer:
(i) Multiply by (√11 − √7): numerator (√11 − √7)² = 18 − 2√77 and denominator 11 − 7 = 4.
The value is 9/2 − (1/2)√77, so a = 9/2, b = 1/2.
(ii) Multiply by (√3 − √2): numerator 3√3 − 3√2 + √18 − √12 = 3√3 − 3√2 + 3√2 − 2√3 = √3; the denominator is 3 − 2 = 1.
So the value is √3 = 0 + 1·√3, giving a = 0, b = 1.
In simple words: Rationalise and match the rational and irrational parts.

Teacher's Note:
1. Note √18 = 3√2 and √12 = 2√3.
2. Compare coefficients on both sides.

Question 25. Find the value of x : \( 5^{x-2}\times3^{2x-3}=135 \).
Answer:
135 = 5 × 27 = 5¹ × 3³.
Compare the powers of 5 and 3: x − 2 = 1 and 2x − 3 = 3.
Both give x = 3.
In simple words: Write 135 as a product of prime powers and match the exponents.

Teacher's Note:
1. Both equations must agree on x.
2. Check: 5¹ × 3³ = 135.