Question
P can complete a piece of work in 15 days. Q in 20 days
and R in 30 days. P and R worked together for three days and then P was replaced by Q. In how many days, altogether, was the work completed (in days)?Solution
Work done by ( P + R ) in 3 days = 3(1/15+1/30) = 3((2+1)/30)=3/10 Remaining work = 1-3/10=7/10 ( Q + R )’s 1 day’s work = 1/20+1/30 = (3+2)/60 = 5/60 = 1/12 Time taken by ( Q + R ) to finish 7/10 part of the work = 12 × (7/10) = 42/5 days = 8(2/5) days Total time = 3 + 8  = 11`2/5`  days. Alternate Method:    P      Q      R   15     20      30 LCM = 60 units Eff 4  :   3  :   2 (P+R)'s 3 days work = (4+2)×3 = 18 units So remaining is done by P & R in = (60-18)/(3+2) = 42/5 days So total days = 3 + (42/5) = 57/5 days
35.25 71 146 296 624 1376
...22   39    58    81   ?      141
1 9 64 255 755 1835
...Find the odd one out: 11, 13, 17, 19, 21
32 63 121 ? 453 889
...1,2, 5, ‘?’, 41, 122, 365
7Â Â Â Â Â Â 15Â Â Â Â Â Â Â 31Â Â Â Â Â Â 63Â Â Â Â Â Â 127Â Â Â Â Â Â ?
...122, 116, 125, 119, 128, ?
13, 14, 18, 27, 43, ?
38    40    83    254    ?     5126
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