Question
Aβ and βBβ alone can do a certain work in 20 days
and 30 days, respectively. Both started the work together but after βxβ days βAβ left the work and the rest work was completed by B alone in β2xβ days. Find the total time taken to complete the whole work in this way.Solution
Total work = 60 units (LCM of 20 and 30) Efficiency of βAβ = 60/20 = 3 units/day Efficiency of βBβ = 60/30 = 2 units/day According to the question, βAβ worked for βxβ days and βBβ worked for β2x days. Therefore, 3x + 2 Γ 2x = 120 Or 3x + 4x = 60 Or x = 60/7days total time taken to complete the whole work = 2Γ60/7=120/7days
√3598 × √(230 ) ÷ √102= ?
7(1/2) – 3(5/6) = ? − 2(7/12)
Simplify the following expression.
(3-3 Γ 3 + 3 Γ· 3 + 3 Γ 5) Γ 2 of 5 + (2 + 2 Γ· 2 + 2 Γ 2 - 2)
What will come in the place of question mark (?) in the given expression?
(437 + ? - 167) x 2.5 = 875I. 8xΒ² - 74x + 165 = 0
II. 15yΒ² - 38y + 24 = 0
?2 = β20.25 Γ 10 + β16 + 32
15(2/9) + 11(2/9) + 17(1/9) + 13(4/9) = ?
- What will come in the place of question mark (?) in the given expression?
32% of 74% of ? = 16% of 37% of 180 15% of 695 – 12.5% of 250 =? – 1200