# CMAT & SNAP Quant: Weighing Scales and Measures-Based Problems

(Photo: Sixtybolts )

Problems based on weighing machines and pan balances are little more common in exams like CMAT and SNAP compared to the CAT because the nature of these questions is such that once you understand the basic concepts and formulae, there isn’t much thinking required to solve the problems. Having said that, many test-takers do not try to understand the logic behind the problems and just try to mug their way out of problems like these. It works on most occasions but not always.

Before we get started, we should first understand the two basic categories of weighing problems which get asked in MBA entrance exams,

1. Problems based on weighing machines

2. Problems based on a two-pan balance

Let us first try and understand how these devices measure and then we can look at a few problems based on them.

Weighing Scale or Weighing Machine is a measuring instrument for determining the weight of an object. It will display a reading of the amount of the weight that is put on it. A spring scale is a type of a weighing scale and looks like this,

Pan Balance, as described by Wikipedia, was “the first mass measuring instrument invented. In its traditional form, it consists of a pivoted horizontal lever of equal length arms, called the beam, with a weighing pan suspended from each arm. The unknown mass is placed in one pan, and standard weights are added to the other pan until the beam is as close to equilibrium as possible. It basically works on the concept of comparison of two weights.” A lot of profit and loss problems which talk about cheating shopkeepers are essentially about shopkeepers rigging the standard weights. Among other things, the pan balance made appearances in nearly every Bollywood film made in the 1970s and even played the title role in couple of films.

(Photo credit: Nikodem Nijaki)

Number of weighings required to find the faulty weight in a weighing machine

Out of 8 equal weights of 1 kg, one is faulty.

The weights can be divided in two parts, the ones on the weighing machine and the ones which are not on it. So, in such cases we will split the weights into equal parts. We will split the 8 weights into two parts of 4 and 4 each and measure one part. The following scenarios can occur.

Example 1: The possible readings are 4 kg or 3.9 kg. If it is 4 kg, the 4 weights which were not measured yet have the faulty weight. If it is 3.9 kg, the 4 weights which were measured contain the faulty weight.

Example 2: The possible readings are 4 kg or 4.1 kg. If it is 4 kg, the 4 weights which were not measured have the faulty weight. If it is 4.1 kg, the 4 weights which were measured have the faulty weight.

Example 3: The possible readings are 4 kg or something else. If it is 4 kg, the 4 weights which were not measured have the faulty weight. If it is something else, the 4 weights which were measured have the faulty weight.

With this one weighing, we have identified the set of 4 weights which has the faulty weight.

With the second weighing, we will be able to identify the set of 2 weights which has the faulty weight.

With the third weighing, we will be able to identify the faulty weight.

So, in any of the three cases, number of weighings required is 3.

Now, you might be thinking that this question was simple because I considered 8 weights initially and it was easy for me to split it into equal parts. What if I am not able to split it equally. How would the weighings behave in that case?

Let us assume, 25 weights initially, which contain 1 faulty weight which needs to be identified.

For the first weighing, we split 25 as 13 and 12.

Considering the worst case scenario, we get that the faulty weight is in the set of 13.

For the second weighing, we split 13 as 7 and 6.

Considering the worst case scenario, we get that the faulty weight is in the set of 7.

For the third weighing, we split 7 as 4 and 3.

Considering the worst case scenario, we get that the faulty weight is in the set of 4.

For the fourth weighing, we split 4 as 2 and 2. We will be able to identify the set of 2 weights which has the faulty weight.

With the fifth weighing, we will be able to identify the faulty weight.

Here are some key observations, which you can verify for yourself, that will give you the pattern required for the formula,

1 weighing is required to find out the heavier / lighter faulty weight from 2 weights.

2 weighings are required to find out the heavier / lighter faulty weight from [3,4] weights.

3 weighings are required to find out the heavier / lighter faulty weight from [5,8] weights.

4 weighings are required to find out the heavier / lighter faulty weight from [9,16] weights.

Or, ‘n’ weighings are required to find out the heavier / lighter faulty weight from [2n-1 + 1, 2n] weights in the case of a weighing machine or scale.

Number of weighings required to find the faulty weight in a two-pan balance

Out of 9 equal weights of 1 kg, one is faulty (900 g).

Out of 9 equal weights of 1 kg, one is faulty (1100 g).

The weights will now be divided into three parts, two would go on the pan balance and one would not be measured.

If the two which are measured come out to be equal, the set which is not measured has the faulty weight.

If the two which are measured come out to be unequal, the set which is lighter (in case 1) or the set which is heavier (in case 2) will have the faulty weight. We will not be able to predict the faulty set in case 3 as we do not know whether the faulty weight is heavier or lighter. So, to determine it we would need an extra weighing. In this extra weighing, we would compare the set which was not measured with one of the sets which was measured. The result of this extra weighing or comparison will not only give us the faulty set but also the fact that the faulty weight is heavier or lighter.

Case 1 and Case 2 will require two weighings.

In the first weighing, 9 weights would be split into sets of 3-3-3. A faulty set of 3 would be determined.

In the second weighing, 3 weights would be split into 1-1-1. The faulty weight would be determined.

Case 3 will require three weighings. The extra weighing would be for determining whether the faulty weight was heavier or lighter.

Here are some observations, which you can verify for youself, that will give you the pattern required for the formula,

1 weighing is required to find out the heavier / lighter faulty weight from 3 weights.

2 weighings are required to find out the heavier / lighter faulty weight from [4,9] weights.

3 weighings are required to find out the heavier / lighter faulty weight from [10,27] weights.

‘n’ weighings are required to find out the heavier / lighter faulty weight from [3n-1 + 1, 3n] weights.

1 extra weighing is required if it is not known whether the faulty weight is heavier / lighter.

A more complex version of this is the 12 coin problem. It might not be relevant for management entrance exams, but hey – it is a good read.

Ravi Handa has taught Quantitative Aptitude at various coaching institutes for seven years. An alumnus of IIT Kharagpur where he studied a dual-degree in computer science, he currently runs an online CAT 2013 coaching course and the website Handa Ka Funda.