maths important question for SSLC board paper
Answers
Question 1: Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}.
Find A′, B′, A′ ∩ B′, A ∪ B and hence show that ( A ∪ B )′ = A′∩ B′.
Solution:
U = {1, 2, 3, 4, 5, 6} , A = {2, 3} and B = {3, 4, 5}
A’ = {1, 4, 5, 6}
B’ = {1, 2, 6}
A′ ∩ B′ = {1, 6}
(A U B) = {2, 3, 4, 5}
(A ∪ B )′ = {1, 6}
A′ ∩ B′ = {1, 6}
(A ∪ B)′ = A′ ∩ B′
Question 2: Prove that 2 + √3 is an irrational number.
Solution:
If (2 + √3) is rational, then,
(2 + √3) = a / b (say) where a and b are integers and b ≠ 0
⇒ √3 = a / b − (2)
⇒ √3 = [a − 2b] / b …(1)
∵ a and b are integers
∴ a – 2b is also an integer.
⇒ (a − 2b) / b is rational.
The L.H.S. of equation (1) is the square root of a prime number.
So, it is irrational and R.H.S. is rational.
It is a contradiction because a rational number and an irrational number can never be equal.
So, our assumption of 2 + √3 being rational is wrong.
Hence, 2 + √3 is an irrational number.
Question 3: Find the sum of all 2 digit natural numbers that are divisible by 5.
Solution:
The two-digit natural numbers divisible by 5 are 10, 20 …… 90, 95
The series forms an arithmetic progression.
The last term is an = a + (n – 1) d
95 = 10 + (n – 1) 5
95 – 10 = (n – 1) 5
85 / 5 = n – 1
17 = n – 1
17 + 1 = 18 = n
n = 18
Sum of n numbers in a series is given by,
Sn = (n / 2) (a + an)
= (18 / 2) (10 + 95)
= 9 * 105
= 945
Question 4: If 2 (nP2) + 50 = 2nP2, then find the value of n.
Solution:
nPr = n! / (n – r)!
2 (nP2) + 50 = 2nP2
2 * (n! / (n – 2)!) + 50 = 2n! / (2n – 2)!
2 * [n (n – 1) (n – 2)! / (n – 2)!] + 50 = 2n (2n – 1) (2n – 2)! / (2n – 2)!
2n (n – 1) + 50 = 2n (2n – 1)
2n2 – 2n + 50 = 4n2 – 2n
2n2 = 50
n2 = 50 / 2
= 25
n = √25
n = 5
Question 5: Rationalise the denominator and simplify [3√2] / [√5 − √2].
Solution:
[3√2] / [√5 − √2]
= {[3√2] / [√5 − √2]} * {[√5 + √2] / [√5 + √2]}
= [3√2 √5 + 3√2 √2] / [(√5)2 – (√2)2]
= [3√10 + 6] / 3
= 3 (√10 + 2) / 3
= √10 + 2
Question 6: A box has 4 red and 3 black marbles. Four marbles are picked up randomly. Find the probability that two marbles are red.
Solution:
There are 7 marbles, out of these 4 marbles can be drawn in 7C4 = 35 ways.
∴ n (S) = 35
Two marbles out of 4 red marbles can be drawn in 4C2 = 6 ways.
The remaining 2 marbles must be black and they can be drawn in 3C2 = 3 ways.
∴ n (A) = 4C2 3C2 = 6 * 3 = 18
P (A) = n (A) / n (S)
= 18 / 35
Question 7: Calculate the standard deviation for the following scores: 5, 6, 7, 8, 9.
Solution:
x
x2
5
25
6
36
7
49
8
64
9
81
∑x = 35
∑x2 = 255
Standard deviation = √(∑x2 / N) – (∑x / N)2
= √(255 / 5) – (35 / 5)2
= √51 – 49
= √2
= 1.414
Question 8: Find the radius of a circle whose centre is (– 5, 4) and which passes through the point (– 7, 1).
Solution:
(x1, y1) = (– 5, 4)
(x2, y2) = (– 7, 1)
∴ d = √(x2 − x1)2 + (y2 − y1)2
Radius of the circle = √[−7 − (−5)]2 + (1 − 4) 2
= √(− 7 + 5)2 + (– 3)2
= √(−2)2 + (−3)2
= √4 + 9
r = √13
Question 9: In Δ ABC, DE | |BC, if AD = 2 cm, DB = 5 cm and AE = 4 cm, find AC.
KSEEB class 10 important questions Q9
Solution:
In Δ ABC, DE || BC
∴ AD / DB = AE / EC [BPT]
2 / 5 = 4 / EC
EC = (4 * 5) / 2
= 10 cm
∴ AC = AE + EC
= 4 + 10
= 14 cm
Question 10: If cos θ = 5 / 13, then find the value of (sin θ + cos θ) / (sin θ – cos θ).
Solution:
KSEEB class 10 important questions Q10
cos θ = 5 / 13 = AB / AC
In triangle ABC, B = 90o.
BC2 = AC2 – AB2
BC2 = 132 – 52
BC = √169 – 25
= √144
= 12
sin θ = 12 / 13
(sin θ + cos θ) / (sin θ – cos θ)
= [(12 / 13) + (5 / 13)] / [(12 / 13) – (5 / 13)]
= (17 / 13) * (13 / 7)
= (17 / 7)
Question 11: From the top of a building 20 m high, the angle of elevation of the top of a vertical pole is 30° and the angle of depression of the foot of the same pole is 60°. Find the height of the pole.
Solution:
KSEEB class 10 important questions Q11
In △BED, DBE = 30o
tan 30o = DE / BE
1 / √3 = (x – 20) / BE
BE = √3 (x – 20)
In △ABC, ACB = 60o
tan 60o = AB / AC
√3 = 20 / [√3 (x – 20)]
3 (x – 20) = 20
3x – 60 = 20
3x = 80
x = 80 / 3 = 26.6 cm
Height of the pole = 26.6 cm