Question
A child can do 1/3 rd. of work in same time as his
father can do the whole work and efficiency of child’s mother is equal to half the sum of father and child efficiency. If they complete 60% of work in 28 days, when they work alternatively, starting with father and child together on first day followed by father and mother together on second day. Find in how many days mother can complete 80% of work alone?ÂSolution
Let efficiency of child and father be x unit/day and 3x unit/day   Efficiency of mother = (x + 3x)/2 = 2x unit/day  ATQ, They start work alternatively— First day efficiency = (3x + x) = 4x Second day efficiency= (3x + 2x) = 5x    Total work = {[4x × 14 + 5x × 14]/60} × 100 = {126x/60} ×100 = 210x  Mother complete 80% of work in = [210x × (80/100)]/2x = (168x/2x) = 84 days
Evaluate: 360 ÷ [ {18 − (6×2)} × 5 ] + 72 − 33
(43)² - (28)² + (32)² = ?% of 2500
Evaluate:
√729 + √49 - √16 + 1/√64
What will come in place of (?) in the given expression.
12.5 + 7.75 - 3.6 = ?62 of 8 - 320 ÷ 4 = ?3 + 200
2(1/3) + 2(5/6) – 1(1/2) = ? – 6(1/6)
What will come in the place of question mark (?) in the given expression?
30% of 520 + 16% of 1500 = ? + 244
60% of 120 – ?% of 64 = 20% of 200
35% of 840 + 162 = ? – 25% × 300
20% of 240 + 18% of 200 = ?
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