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08 Jun 2016, 06:08
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B
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Difficulty:
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Question Stats:
69% (02:49) correct
31%
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There are 5 boxes of different weights. They are weighed in pairs taken 2 at a time and the weights obtained are 110, 112, 113, 114, 115, 116, 117, 118, 120 and 121. What is the sum of the weights of the 5 boxes ?
A) 250
B) 289
C) 290
D) 295
E) 305
08 Jun 2016, 06:30
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Let the 5 boxes be a, b, c, d and e. The ten weights correspond to all the possible combinations of 2 weights out of 5.
On adding all the given weights, each boxes weight gets added 4 times:
a+b, a+c, a+d, a+e.
Similarly for b:
b+c, b+d, b+e and a+b (already included above)
Similarly for c, d and e.
Therefore, 110+112+113+114 .....+120+121 = 4* (a+b+c+d+e)
or, 4* (a+b+c+d+e) = 1156
or, (a+b+c+d+e) = 1156/4 = 289
On adding all the given weights, each boxes weight gets added 4 times:
a+b, a+c, a+d, a+e.
Similarly for b:
b+c, b+d, b+e and a+b (already included above)
Similarly for c, d and e.
Therefore, 110+112+113+114 .....+120+121 = 4* (a+b+c+d+e)
or, 4* (a+b+c+d+e) = 1156
or, (a+b+c+d+e) = 1156/4 = 289
08 Jun 2016, 06:52
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aayushagrawal
Hi,
since the weights given are of 2 boxes taken at one time..
so the SUM of all the weights measured taking two boxes at any time will be 4 times total weight of all five combined..
We choose 2 boxes out of 5 in 5C2 ways = 5!/3!2! = 10 ways....
How many ways 1 box say A will not be included = choosing 2 out of 4 boxes other than A = 4C2 = 4!/2!2! =6..
so NUMBER of times A has been counted is 10-6 = 4.....
similarily all other boxes also are counted 4 times...
eg say A,B,C,D,E - AB,AC,AD,AE has A in it
Now since these 10 weights of pair of boxes given has 4 times each box..
we divide the SUM of these 10 weights to get SUM of single boxes
\(\frac{110+112+113+114+115+116+117+118+120+121}{4}\)...
UNITS digit is 6 of the NUMERATOR..
so when you divide by 4, 0 and 5 cannot be the UNITS digit..
ONLY B has units digit other than 0 and 5..
ans B
