number set qs
Fri, 2008-03-07 00:00
The set of Y consist of following numbers Y = {1, 31/2 , 3, 33/2. . . . . .319/2, 310}
In how many ways can a pair of distinct numbers be selected from the set Y such that their product is greater than or equal to 310 ?
a. 110 b. 210 c. 105 d. 100
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The set of Y consist of following numbers Y = {1, 31/2 , 3, 33/2. . . . . .319/2, 310}
In how many ways can a pair of distinct numbers be selected from the set Y such that their product is greater than or equal to 310 ?
a. 110 b. 210 c. 105 d. 100
Y = {1, 31/2 , 3, 33/2. . . . . .319/2, 310}
so Y = {3^0, 31/2 , 3, 33/2. . . . . .319/2, 3^20/2}
total 21 number
if we take 3^20/2 then possible number from 3^0 to 3^19/2=20 number
if we take 3^19/2 then possible number from 3^1/2 to 3^18/2=18 number
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so total=20+18+16+14+12+10+8+6+4+2+0=110
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