Latest TCS Aptitude Question SOLUTION: 3 friends A,B,c went for week end party ti Mc donald resturent and there they measure there weights in some order IN 7 rounds. A,B,C,AB,BC,AC,ABC. final round measure is 155kgthen

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3 friends A,B,c went for week end party ti Mc donald resturent and there they measure there weights in some order IN 7 rounds.
A,B,C,AB,BC,AC,ABC. final round measure is 155kgthen find average weight of all the 7 rounds.

Asked In TCS bhanupriya gupta (11 years ago)

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Garima  (11 years ago)

A+B+C= 155 kg

Average of 7 rounds = (A+B+C+A+B+A+C+B+C+A+B+C)/7 = 4*(A+B+C)/7 = 4*155/7 = 620/7 kg

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onkar suresh salvi  (11 years ago)

They Measure their weights 1 by 1....
then they measure it together....
So we can for equation like this
=A+B+C+(A+B)+(B+C)+(A+C)+(A+B+C)
=(A+B+C)+(A+B+C)+(A+B+C)+(A+B+C)
=4(A+B+C)
And final measure is 155Kg. ie (A+B+C)=155kg
So, 4x155=620
Hence the avg is 620/7=88.571428 Kg.
Ans is 88.571428kg

Is this solution Helpfull? Yes (1) | No   

Divyakrishna  (11 years ago)

88.57
if last measure is 155 tht means ABC=155
so A+B+C= 155 (1st 3 measures)
nxt cmg to AB,BC,AC = 155*2 (since A,B,C occurs 2 tyms)
finally avg=(155+155+155*2)/7 = (155*4)/7 =88.57

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vikas kale  (11 years ago)

the ans is
155*4/7

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venkatrao  (11 years ago)

A+B+C+(A+B)+(B+C)+(A+C)+(A+B+C)
=(A+B+C)+(A+B+C)+(A+B+C)+(A+B+C)
=4(A+B+C)
And final measure is 155Kg. ie (A+B+C)=155kg
So, 4x155=620
Hence the avg is 620/7=88.571428 Kg.
Ans is 88.571428kg

Is this solution Helpfull? Yes | No   

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