Given, A and B be any two sets such that A - B = A ⇒ B = Φ
Now, A ∩ B = A ∩ Φ = Φ
Now, A ∩ B = A ∩ Φ = Φ
CTET Solved Paper MCQ - UPTET October 2017 Paper2
Given, equation x2 - bx + c =0
Let roots of the given equation α and α + 1, then sum of roots,
α + (α + 1) = -(-b)/1 = b
⇒ 2α + 1 = b and product of roots, α(α + 1) = c
Now, according to the question,
b2 - 4c = (2α + 1 )2 - 4α(α + 1)
⇒ (4α2 + 1 + 4α) - (4α2 + 4α) = 1
Let roots of the given equation α and α + 1, then sum of roots,
α + (α + 1) = -(-b)/1 = b
⇒ 2α + 1 = b and product of roots, α(α + 1) = c
Now, according to the question,
b2 - 4c = (2α + 1 )2 - 4α(α + 1)
⇒ (4α2 + 1 + 4α) - (4α2 + 4α) = 1
We know that
LCM(a,b) × HCF(a,b) = a × b
Let other number is x.
Then, according to the question,
1989 × 13 = 117 × x
⇒ x = (1989 × 13)/117 ⇒ x = 221
LCM(a,b) × HCF(a,b) = a × b
Let other number is x.
Then, according to the question,
1989 × 13 = 117 × x
⇒ x = (1989 × 13)/117 ⇒ x = 221
Let value of an article = 100
Since, Jatin buys an article for 10% less than its vlue
∴ CP = 90
and sells it for 10% more than its value ∴ SP = 110
∴ His percentage gain
Since, Jatin buys an article for 10% less than its vlue
∴ CP = 90
and sells it for 10% more than its value ∴ SP = 110
∴ His percentage gain
Given, equation x2 - px + 54 = 0
Let roots of the given equation are 2α and 3α
Then, 2α + 3α = p ⇒ 5α = p ⇒ α = p / 5
and also, (2α)(3α) = 54
⇒ 6(α)2 = 54
⇒ 6(p/5)2 = 54 (∴ α = p/5)
⇒ 6 × p2/25 = 54 ⇒ p2 = 9 × 25
⇒ p = 3 × 5 = 15
Let roots of the given equation are 2α and 3α
Then, 2α + 3α = p ⇒ 5α = p ⇒ α = p / 5
and also, (2α)(3α) = 54
⇒ 6(α)2 = 54
⇒ 6(p/5)2 = 54 (∴ α = p/5)
⇒ 6 × p2/25 = 54 ⇒ p2 = 9 × 25
⇒ p = 3 × 5 = 15
