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Let, the initial no. of male students be X . X × 130% + (13,200 – X ) × 120% = 16,440 ⇒ 1.30 X + 15,840 – 1.20 X = 16,440 ⇒ 0.10 X = 600 ⇒ X = 6,000 No. of female students = 13,200 – 6,000 = 7,200 Required difference = 7,200 – 6,000 = 1,200 Or Use Alligation Net increase in Population =`(16440-13200)/(13200)xx 100 = 3240/13200 xx 100 = 270/11%` Male Female 20% 30% Net Change% = 270/11% So New Values, Male Female 220 330 Net Change = 270
If x = √7 + √6 and y = √7 - √6 , then the value of is (x2 + y2)/(x3 + y3).
...If (x + y + z) = 68, (x/z) = (3/4) and (z/y) = (2/5), then find the value of ‘y’.
(123×123×123 + 130×130×130)/(123×123 - 123×130 + 130×130) = ?
If (x2 + y2 + z2 - 4x + 6y + 13) = 0, then find the value of (x + y + z).
(288 ÷ 8)² × (144 ÷ 24)³ = 24 × ? × (51840 ÷ 20)
If (a3+1)/(a+1) = (a3-1)/(a-1) and a ≠ 1, -1. Find the value of 'a'
What is the highest common factor of (x³ - x² - x - 15) and (x³ - 3x² - 3x + 9)?
If (a + b) = 7 and ab = 9, then find the value of (a2 + b2).