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07 Oct 2025, 00:52
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I went the algebraic way......
Range is 2 ..Total is 28....Keep these constraints......
With Range 2...
0 to 2 Possible? No..Coz maximum will be 20 then
1 to 3 Possible? Yes....
2 to 4 Possible? Yes...
3 to 5 Possible? No..Coz minimum will be 30 then...
Now...Either 1,2,3 (Case 1)
or 2,3,4 ( Case 2)
Looking at it ...0 is not possible in 1,2,3....because then maximum will be 20(If all give 2 each)
Coming to Case 2...
2,3,4..if 0 gave 3...means all the other 10 gave either 2 or 4.... Let those who gave 2 be = x ...so those who gave 4 should be 10-x
Equation....2x + 4(10-x) = 28 ...2x + 40 -4x = 28.....-2x = -12..x = 6...Its possible.....
II. 5....
Case 1...1,2,3....
if 5 Gave 3....means 15.....remaining is 13....(The other 5 will give a total of 13..) Is it possible? No...because the maximum they can give is 2 each....so max it can go to 10....
Case 2...2,3,4
5 gave 3..means 15...remaining 13....(The other 5 will give a total of 13...)....so 2(x) +4(5-x) = 13...Need not to solve...Something x 2 + Something x 4 should give us even number whereas we are getting odd number.....
So not possible......
III.... 9...
Means 9 x 3 is 27...the remaining 1 will give 1...hence 28 total..possible....
So I and III possible not II....answer D
Range is 2 ..Total is 28....Keep these constraints......
With Range 2...
0 to 2 Possible? No..Coz maximum will be 20 then
1 to 3 Possible? Yes....
2 to 4 Possible? Yes...
3 to 5 Possible? No..Coz minimum will be 30 then...
Now...Either 1,2,3 (Case 1)
or 2,3,4 ( Case 2)
Looking at it ...0 is not possible in 1,2,3....because then maximum will be 20(If all give 2 each)
Coming to Case 2...
2,3,4..if 0 gave 3...means all the other 10 gave either 2 or 4.... Let those who gave 2 be = x ...so those who gave 4 should be 10-x
Equation....2x + 4(10-x) = 28 ...2x + 40 -4x = 28.....-2x = -12..x = 6...Its possible.....
II. 5....
Case 1...1,2,3....
if 5 Gave 3....means 15.....remaining is 13....(The other 5 will give a total of 13..) Is it possible? No...because the maximum they can give is 2 each....so max it can go to 10....
Case 2...2,3,4
5 gave 3..means 15...remaining 13....(The other 5 will give a total of 13...)....so 2(x) +4(5-x) = 13...Need not to solve...Something x 2 + Something x 4 should give us even number whereas we are getting odd number.....
So not possible......
III.... 9...
Means 9 x 3 is 27...the remaining 1 will give 1...hence 28 total..possible....
So I and III possible not II....answer D
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