Data Relations are a subgroup of the arithmetic relations. This section, in the exam paper, has substantial weight. The questions in data relations are based on PEMDAS or BODMAS rules. You will be given a couple of relations and asked to find some variable based on the relationship that you develop from the given data. This is how data relations work. Let us begin!

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## Data Relations

Let us start with a question. The sum of your age and that of your friend is 70 years. What was the sum of your ages, five years ago? Well, that is easy you say! Five years ago, the sum of the ages will be 70 – 5 = 65. You would be wrong! Let us figure out the answer and develop a method that we can trust to solve similar problems. The first step in such questions is to figure out the “Relation”.

This is the most important step. The relation could be a rule or an algebraic expression or even an equation. For example in the above case, we can say that let ‘a’ represent your age and ‘b’ be the age of your friend. Then as per the question, the relation is a + b = 70. Now five years ago, your age would be a – 5 and that of your friend will also be b – 5 years. So the new sum of your ages will be [(a – 5) + (b – 5)].

This can be written as [a + b] – 10. Step two, use the relation: From above, we have a + b = 70. Using that above, we have the sum of the ages five years ago will have been [70] – 10 = 60.

### Shortcut

Take the number of years that have to subtract (5 in the above example) and multiply it by the number of people (2 in the above example). The difference between this number and the given sum is the answer. Let us solve other types of problems.

### Fraction Based Problems – Data Insufficiency

In a hospital, one-fifth of the staff and one-third of the patients admitted to smoking inside the hospital. What fraction of the total people present in the hospital, considering that at any point a person in the hospital is either a patient or a member of the staff, smoke in the hospital premises?

- 25
- 35
- Data inadequate
- None of the above

Answer: Well, if you don’t quantify this, you are bound to make mistakes. By quantifying, we mean that you should try and form an algebraic expression from the given data. In these questions since the number of medical staff or the patients is not given, you might be tempted to take some arbitrary number but that is not right, as we will see. Here the data is in the form of fractions.

#### Let us solve it now

Let ‘m’ denote the total number of the people of the medical staff and ‘p’ denote the total number of patients in the hospital. Since we have been asked to find a fraction of the total number of people of the hospital that would be (m + p) = N {say}………(1)

Step 1: From the given condition, we have: ⅕(m) + ⅓(p) = Smokers

Therefore, we may write: [3m + 5p]/15 = Smokers …………(2)

Since in equation (1) and (2) we have two equations and four variables, it is impossible to determine the solution with the help of algebra. To solve four variables, we need at least four equations. Hence the answer is C. The data is not sufficient or the data is inadequate.

### Fraction Based Problems – Data Sufficiency

Three friends live together in a shared flat. The rent of the flat is 9999 rupees. One of the friends pays ⅕ of his salary in rent. The other two pay ⅓ of their salary in rent. If all the friends contribute equally to the rent, what is the salary of each friend?

Answer: This is a straightforward question. Let the salary of the three friends be A, B, and C respectively. Since each friend contributes equally towards the rent, friend one will contribute ⅕(A) to the rent. Similarly, friend two and friend three will each contribute ⅓(B) and ⅓(C) towards the rent. Now the rent splits three ways. In other words, ⅕(A) = ⅓(B) = ⅓(C) = (9999)/3 = 1111 rupees.

So the salary of the first friend is 5(1111) = 5555 rupees. Similarly, the salary of the second and the third friend are 3333 rupees each.

### Miscellaneous Type

A person goes to the market and purchases some apples and oranges. An apple costs 8 rupees per piece and an orange costs 4 rupees per piece. If the person spends 64 rupees and buys both apples and oranges, how many apples has he bought?

- 1
- 7
- Data insufficient
- 6

Answer: Let a be the number of apples and o be the number of oranges. The from the given condition, we have 8a + 4o = 64. This gives, 8a = 64 – 4o or a = [64 – 4o]/8.

Now you might say that this can’t be solved and the option is C) data insufficient nut that is not right. There is a hidden condition here that the number of apples and oranges can’t be a fraction. They must be whole numbers. This happens when o = 2. In that case, a = 7. So the man has bought 7 apples and 2 oranges.

## Practice Questions

Q 1: A person started counting on his fingers of the left hand. The thumb was 1, the index finger 2, middle finger 3, and so on up to the little finger. He then reversed direction calling the ring finger 6, middle finger 7 and so on. Then he counts up to 1024. His counting ended on which finger?

- Ring finger
- Middle finger
- Index finger
- Thumb

Ans: C. Index finger

Q 2: A telephone has all the numbers except the number 7. What is the product of all the numbers that can be dialed with this telephone?

- 0
- Infinity
- Data insufficient
- 99999999999999999

Ans: A) 0

Q 3: There are three friends A, B, and C. A is three years younger than C and three years older than B. B has a twin X. What is the age difference between C and X?

- 3
- 6
- 9
- 12

Ans: B) 6

The answer to the question on apples and oranges is wrong. Apart from this there are a lot of mistakes in the whole syllabus. Please correct them