Algebra
- If p : q = r : s = t : u = 2 : 3, then (mp + nr + ot) : (mq + ns + ou) equals :
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p = r = t = 2 q s u 3 ⇒ p = q = k 2 3
⇒ = 2k, q = 3k
Similarly, r = 2k, s = 3k, t = 2k, u = 3kNow mp + nr + ot = m.2k + n.2k + o.2k mq + ns + ou m.3k + n.3k + o.3k = 2k(m + n + o) = 2 or 2 : 3 3k(m + n + o) 3 Correct Option: B
p = r = t = 2 q s u 3 ⇒ p = q = k 2 3
⇒ = 2k, q = 3k
Similarly, r = 2k, s = 3k, t = 2k, u = 3kNow mp + nr + ot = m.2k + n.2k + o.2k mq + ns + ou m.3k + n.3k + o.3k = 2k(m + n + o) = 2 or 2 : 3 3k(m + n + o) 3
- If x : y = 3 : 4 , then (7x +3y) : (7x – 3y) is equal to :
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= x = 3 ⇒ 7x = 7 × 3 = 7 y 4 3y 3 4 4
By componendo and dividendo,7x + 3y = 4 + 7 = 11 or 11 : 3 7x - 3y 7 - 4 3 Correct Option: C
= x = 3 ⇒ 7x = 7 × 3 = 7 y 4 3y 3 4 4
By componendo and dividendo,7x + 3y = 4 + 7 = 11 or 11 : 3 7x - 3y 7 - 4 3
- If a : b : c = (y – z) : (z – x) : (x – y) then the value of ax+ by+cz is
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= a = b = c = k y - z z - x x - y
⇒ a = k (y–z); b = k(z–x); c = k (x–y)
∴ ax + by + cz = k (xy – xz + yz – xy + xz – yz) = 0Correct Option: C
= a = b = c = k y - z z - x x - y
⇒ a = k (y–z); b = k(z–x); c = k (x–y)
∴ ax + by + cz = k (xy – xz + yz – xy + xz – yz) = 0
- If 50% of (p – q) =30% of (p + q), then p : q is equal to
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= 50 (p - q) = 30 (p + q) 100 100
⇒ 5(p – q) = 3 (p + q)
⇒ 5p – 5q = 3p + 3q
⇒ 2p = 8q
⇒ p = 4q
∴ p : q = 4 : 1Correct Option: B
= 50 (p - q) = 30 (p + q) 100 100
⇒ 5(p – q) = 3 (p + q)
⇒ 5p – 5q = 3p + 3q
⇒ 2p = 8q
⇒ p = 4q
∴ p : q = 4 : 1
- If x : y = 2 : 1, then (5x² – 13xy + 6y²) is equal to
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x = 2 ⇒ x = 2y y
⇒ x – 2y = 0 ...(i)
∴ 5x² – 13xy + 6y²
= 5x² – 10xy – 3xy + 6y²
= 5x (x – 2y) – 3y (x – 2y)
= (x – 2y) (5x – 3y)
= 0 × (5x – 3y) = 0 [Using (i)]Correct Option: C
x = 2 ⇒ x = 2y y
⇒ x – 2y = 0 ...(i)
∴ 5x² – 13xy + 6y²
= 5x² – 10xy – 3xy + 6y²
= 5x (x – 2y) – 3y (x – 2y)
= (x – 2y) (5x – 3y)
= 0 × (5x – 3y) = 0 [Using (i)]
