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
50 62 ? 96 126 138
...89Â Â Â Â Â 78Â Â Â Â Â 74Â Â Â Â Â 67Â Â Â Â Â 63Â Â Â Â ?
4 6 ? 31 119 601
...Find the maximum number of trees which can be planted, 25 meters apart, on the two sides of a straight road 2125 meters long
55 46 60 41 65 36 ?
8 6 2 ? -1 9.5
...9999 6666 ? 2962.66 1975.11 1316. 74
...2, 17, 147, 1167, 8157, 48957
3,   2,   3,  8,  31,  ?
Study the given pattern carefully and select the number that can replace the question mark (?) in it.
4 7 6
15 ? 21
44 68 60
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