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Question

Once my peon went to the office of SKYLINE COURIER with 4 different envelopes. The clerk in the office measured the weights in all possible pairs. The weights obtained are 59 gm, 61 gm, 62 gm, 63 gm, 64 gm and 66 gm. The weight of the heaviest envelope is :

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Solution

The correct option is **C** 34 gm

If the highest weight be 35 gm, then the second highest weight will be 31 gm. Again if the second highest will be 31, then the third highest will be 33 which is inadmissible, since then 35 + 33 = 68 which is not the greatest possible combination. Hence wrong.

Similarly 36 (1. e. , option b) is also invalid

HighestSec.HighestThird Highest363034

Thus 36 + 34 = 70 > 66, hence wrong.

The greatest possible combination can not be greater than 66.

Now, consider option (c)

HighestSec.HighestThird Highest 343232χ(since weights are different) 3231χ(since 65 is not a combination) 3230✓

So, the highest weight 34

Sec. highest weight 32

Third highest weight 30

Lowest weight 29

Since, all the weights obtained give all the 6 different combinations, hence 34 is the highest possible weight of an envelope.

If the highest weight be 35 gm, then the second highest weight will be 31 gm. Again if the second highest will be 31, then the third highest will be 33 which is inadmissible, since then 35 + 33 = 68 which is not the greatest possible combination. Hence wrong.

Similarly 36 (1. e. , option b) is also invalid

HighestSec.HighestThird Highest363034

Thus 36 + 34 = 70 > 66, hence wrong.

The greatest possible combination can not be greater than 66.

Now, consider option (c)

HighestSec.HighestThird Highest 343232χ(since weights are different) 3231χ(since 65 is not a combination) 3230✓

So, the highest weight 34

Sec. highest weight 32

Third highest weight 30

Lowest weight 29

Since, all the weights obtained give all the 6 different combinations, hence 34 is the highest possible weight of an envelope.

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