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There are 5 boxes of different weights... [#permalink]
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08 Jun 2016, 06:08
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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
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Re: There are 5 boxes of different weights... [#permalink]
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08 Jun 2016, 06:30
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



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There are 5 boxes of different weights... [#permalink]
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08 Jun 2016, 06:52
aayushagrawal wrote: 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 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 106 = 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
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There are 5 boxes of different weights... [#permalink]
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08 Jun 2016, 09:40
chetan2u wrote: aayushagrawal wrote: 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 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..\(\frac{110+112+113+114+115+116+117+118+120+121}{4}\)... 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 Thank you for the explanation chetan2u! I was just looking to understand exactly, why is it that the number is 4 times the total weight of all five boxes combined. Would it be correct to say that every number is counted 5 times instead of one, which makes it 4 times too much. And since it applies to every single number, we divide by 4 to cancel repetition? Thank you!
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Re: There are 5 boxes of different weights... [#permalink]
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08 Jun 2016, 09:48
fantaisie wrote: chetan2u wrote: aayushagrawal wrote: 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 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..\(\frac{110+112+113+114+115+116+117+118+120+121}{4}\)... 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 Thank you for the explanation chetan2u! I was just looking to understand exactly, why is it that the number is 4 times the total weight of all five boxes combined. Would it be correct to say that every number is counted 5 times instead of one, which makes it 4 times too much. And since it applies to every single number, we divide by 4 to cancel repetition? Thank you! Hi, It is 4 times the SUM of all five boxes because
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 106 = 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
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Re: There are 5 boxes of different weights... [#permalink]
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09 Oct 2017, 17:25
aayushagrawal wrote: 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 When the boxes are weighted 2 at a time, the weight of each box is counted 4 times. Thus, the sum 110 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 120 + 121 = 1,156 must be 4 times the sum of the weights of the 5 boxes. Therefore, the sum of the weights of the 5 boxes is 1,156/4 = 289. Note: If you can’t see why the weight of each box is counted 4 times, we can label the boxes as A, B, C, D, and E. Thus, when we pair them up, we have: AB AC AD AE BC BD BE CD CE DE We see that each of the letters appears 4 times in the list. Answer: B
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