Question
M, N and O can complete a task in βtβ, 100, and βt
+ 20β days respectively. All three started together, but after 15 days, N dropped out. Then, 30 days before the work finished, M also left. The work was finished in 60 days. Find the value of βtβ.Solution
ATQ,
M worked for 30 days, N worked for 15 days and O worked for 60 days So according to question: (30/t) + (15/100) + (60/(t + 20)) = 1 β (30/t) + (60/(t+20)) = 1 β (3/20) β (30t + 600 + 60t)/(tΒ² + 20t) = 17/20 β 17tΒ² + 340t = 600t + 12000 β 17tΒ² β 260t β 12000 = 0 β 17tΒ² β 340t + 80t β 12000 = 0 β 17t(t β 20) + 80(t β 20) = 0 β (17t + 80)(t β 20) = 0 β t = 20 (valid), t = -80/17 (not valid)
Evaluate:
β729 + β49 - β16 + 1/β64
Simplify:

(1/5)(40% of 800 β 120) = ? Γ 5
2/5 of 3/4 of 7/9 of 7200 = ?
`sqrt(5476)` + 40% of 1640 = ? `xx` 4 - 2020
? = (22% of 25% of 60% of 3000) + 21
Determine the simplified value of the given mathematical expression.
(342 β 20% of 5280) = ? Γ· 3
β157464 =?