Most Asked Question of Number System Class 9th Himachal Board
Question 1. Represent √10 on the number line.
Solution:
Step 1: Draw a number line and mark point O at 0 and point A at 3 units from O.
Step 2: At A, draw a perpendicular line of length 1 unit. Mark its endpoint as B.
Step 3: Join O to B.
Using the Pythagorean theorem:
OB² = OA² + AB²
OB² = 3² + 1²
OB² = 9 + 1 = 10
Therefore, OB = √10.
Using a compass, transfer the length OB onto the number line. The marked point represents √10.
Question 2. Simplify: √48 − √27 + √75.
Solution:
√48 = √(16 × 3) = 4√3
√27 = √(9 × 3) = 3√3
√75 = √(25 × 3) = 5√3
Therefore,
√48 − √27 + √75
= 4√3 − 3√3 + 5√3
= (4 − 3 + 5)√3
= 6√3
Answer: 6√3
Question 3. Rationalise the denominator: 5/(√7 + √2).
Solution:
Multiply the numerator and denominator by the conjugate, √7 − √2.
= 5/(√7 + √2) × (√7 − √2)/(√7 − √2)
= 5(√7 − √2)/(7 − 2)
= 5(√7 − √2)/5
= √7 − √2
Answer: √7 − √2
Question 4. Express 0.123123123... in the form p/q.
Solution:
Let x = 0.123123123...
Since three digits repeat, multiply both sides by 1000.
1000x = 123.123123...
Subtract the first equation from the second:
1000x − x = 123.123123... − 0.123123...
999x = 123
x = 123/999
Simplifying by dividing by 9:
x = 41/333
Answer: 41/333
Question 5. Without performing long division, determine whether 13/160 has a terminating decimal expansion.
Solution:
A rational number p/q in its lowest form has a terminating decimal expansion if the prime factors of q are only 2 and/or 5.
Here,
13/160 is already in its lowest form.
160 = 2⁵ × 5
The denominator contains only the prime factors 2 and 5.
Therefore, 13/160 has a terminating decimal expansion.
Answer: Terminating decimal.
Question 6. Simplify: (5⁻² × 5³) ÷ 5⁻¹.
Solution:
Using the laws of exponents:
5⁻² × 5³ = 5⁻²⁺³ = 5¹
Now divide by 5⁻¹:
5¹ ÷ 5⁻¹ = 5¹⁻⁽⁻¹⁾
= 5²
= 25
Answer: 25
Question 7. Prove that √7 is an irrational number.
Solution:
Assume that √7 is rational.
Then √7 = p/q, where p and q are coprime integers and q ≠ 0.
Squaring both sides:
7 = p²/q²
p² = 7q²
Therefore, p² is divisible by 7, so p is divisible by 7.
Let p = 7k for some integer k.
Substituting:
49k² = 7q²
q² = 7k²
Therefore, q is also divisible by 7.
This contradicts the assumption that p and q are coprime.
Hence, √7 is irrational.
Question 8. Find two irrational numbers between 3 and 4.
Solution:
Squaring the given numbers:
3² = 9 and 4² = 16
Choose the integers 10 and 11, which lie between 9 and 16.
Taking square roots:
√10 and √11
Since 9 < 10 < 16 and 9 < 11 < 16,
3 < √10 < 4
3 < √11 < 4
Both √10 and √11 are irrational numbers.
Answer: √10 and √11.
Question 9. Simplify: (√5 + √3)² − (√5 − √3)².
Solution:
Using the identities:
(a + b)² = a² + 2ab + b²
(a − b)² = a² − 2ab + b²
Therefore,
(√5 + √3)² = 5 + 2√15 + 3
= 8 + 2√15
Similarly,
(√5 − √3)² = 5 − 2√15 + 3
= 8 − 2√15
Subtracting:
(8 + 2√15) − (8 − 2√15)
= 8 + 2√15 − 8 + 2√15
= 4√15
Answer: 4√15
Question 10. If x = 2 + √3, find the value of x + 1/x.
Solution:
Given:
x = 2 + √3
To find 1/x, rationalise the denominator:
1/x = 1/(2 + √3)
Multiply numerator and denominator by 2 − √3:
1/x = (2 − √3)/[(2 + √3)(2 − √3)]
= (2 − √3)/(4 − 3)
= 2 − √3
Now,
x + 1/x = (2 + √3) + (2 − √3)
= 2 + 2
= 4
Answer: 4