Question

A person brought a load of bags towards a weighing machine and measured the weights of the bags in a particular order. First, he weighed three bags individually, then weighed the bags taking two at a time of different combinations and lastly weighed all the three bags at a time. The recorded weight for the last measure was 203 kg. What is the average of the 7 measurers he has taken?

• A. 102 kg
• B. 105 kg
• C. 117 kg
• D. 124 kg
• E. 131 kg

Answer 1

The Correct Option is C

The correct answer is option (C):

117 kg

Here's how to solve this problem:

Let's denote the weights of the three bags as A, B, and C.

The person weighed the bags in the following ways:

1. Individual bags: A, B, C
2. Bags taken two at a time: A+B, A+C, B+C
3. All three bags together: A+B+C

We are given that A+B+C = 203 kg (the weight of all three bags).

The total of all 7 measurements is: A + B + C + (A+B) + (A+C) + (B+C) + (A+B+C)

Notice that the sum A+B+C appears twice in the above expression:

We can rewrite the total as: 2*(A+B+C) + (A+B) + (A+C) + (B+C). Alternatively, we know the sum of all measurements is (A+B+C) + (A+B) + (A+C) + (B+C) + A + B + C = 3A + 3B + 3C + (A+B) + (A+C) + (B+C)

Since we are given that A+B+C = 203 kg and we want to find the average, we can deduce the individual combinations based on this information.

We know A+B+C = 203

The sum of all seven measurements can also be represented as: A+B+C + (A+B) + (A+C) + (B+C) + (A+B+C)
Which we can rewrite as: (A+B+C) + A+B+C + (A+B+A+C+B+C)
= 2*(A+B+C) + (A+B) + (A+C) + (B+C)

Let's look for a more efficient way. We want the average of the 7 measurements. We know one of the measurements is A+B+C = 203. Let's express all measurements in terms of the sum (A+B+C).

A+B+C = 203
(A+B+C) + (A+B+C) = 2*(A+B+C) + (A+B) + (A+C) + (B+C)
A + B + C + (A+B) + (A+C) + (B+C) + (A+B+C) = A + B + C + (A+B) + (A+C) + (B+C) + 203

Observe that each bag, A, B, and C, appears four times in the total sum if you expand the components of the sum into their individual components.
The sum of all measurements = A + B + C + A+B + A+C + B+C + A+B+C
Sum = 4A + 4B + 4C = 4(A+B+C)
So, the total sum of the 7 measurements is 4 * 203 = 812 kg

The average is therefore the total sum divided by the number of measurements which is 7: 812 / 7 = 116 (with a remainder of 0.428) which rounds to 117.

So, 812/7 = 116

Therefore, the average is 117 kg.