Question
Present average age of 4 friends: βAβ, βBβ,
βCβ and βDβ is 36 years. If their ages are in the arithmetic progression in the same order and βDβ is 4 years older than βBβ, then find the sum of present ages of βAβ and βCβ.Solution
Common difference between ages = (4/2) = 2 years. So, let present ages of βAβ, βBβ, βCβ and βDβ be βxβ, (x + 2) years, (x + 4) years and (x + 6) years, respectively According to question: x + x + 2 + x + 4 + x + 6 = 144 Or, 4x = 132 Or, x = 33 years (Age of A) Present age of βCβ = x + 4 = 33 + 4 = 37 years Required sum = 33 + 37 = 70 years
(29.97%) of 9840 + ? + (45.17% of 1240) = (31.955% of 11750)
15.1 + 3.97 β 9.07 Γ 1.96 = β?Β
(?)2 + 9.113 = 31.92 β 39.03Β
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