# Averages Concepts from Arithmetic – How to apply in CAT Questions?

Wednesday, April 1st, 2020 Average is mainly defined as the sum of observations divided by the number of observations. Keeping in view the Quantitative Aptitude section of any competitive exam and especially CAT, 1-2 questions are asked from this chapter. It is not a difficulty chapter if you know the tricks to solve the question in very less time.

## Averages Concepts and Properties:

• Average= (Sum of observations)/(Number of observations)

Important Points: In past year Cat questions, a number of times concepts of age/height/weight and average has been mixed together. Here are some important points to remember,

1. When a person replaces another person then:
• If the average is increased, then
Age of new person= Age of person who left + (Increase in average * total number of persons)
• If the average is decreased, then
Age of new person= Age of person who left – (Decrease in average * total number of persons)
2. When a person joins the group:
• In case of an increase in average,
Age of new member= Previous average+ (Increase in average * Number of members including new member)
• In case of decrease in average,
Age of new member= Previous average- (Decrease in average * Number of members including new member)
3. In the Arithmetic Progression there are two cases:
• When the number of terms is odd – the average will be the middle term.
• When number of terms is even – the average will be the average of two middle terms.
1. Sum of 1st n consecutive natural numbers = [n(n+1)]/2

Average of 1st n consecutive natural numbers = (n+1)/2

To understand the concept properly, let’s try doing some questions.

## CAT Questions:

Example 1: The average weight of a class of 24 students is 36 years. When the weight of the teacher is also included, the average weight increases by 1kg. What is the weight of the teacher?

Solution: We know that, in case of increase in average,

Age of new member= Previous average+ (Increase in average X Number of members including new member)
Therefore,
Age of new member = 36 + (1 × 25)
Age of new member = 36 + 25 = 61

Example 2: Find the average of first 30 natural numbers.

Solution:         Average of first n natural numbers = [(n+1)/2]
Therefore, Average=((30+1))/(2)
= (31)/(2) = 15.5

Example 3:  A man buys spirit at Rs. 60 per liter, adds water to it and then sells at Rs. 75 per liter. What is the ratio of spirit to water if his profit in the deal is 37.5% ?

A. 9:1                      B. 10:1                    C. 11: 1                      D. None of the above

Solution: Selling price of the spirit and solution water is given as = Rs. 75/l

Profit on the solution = 37.5%

Therefore, CP of the mixture = 75/1.375 = Rs. 54.54/l

This should be actually the weighted average of the costs of spirit and water.

Now, Cost of water is 0.

Let in 1 liter of solution, x liters be spirit and (1-x) liter be water

Thus, 60x + (1-x)*0 = 54.54

Thus, x = Amount of spirit = 54.54/60

And, 1-x = Amount of water = 1 – (54.54/60) = 5.46/60

Required Ratio = x / (1-x)     = (54.54/60) * (60/ 5.46) = 54.54 / 5.46 ~ 10: 1

Example 4: Shyam went from Delhi to Shimla via Chandigarh by car. The distance from Delhi to Chandigarh is ¾ times the distance from Chandigarh to Shimla. The average speed from Delhi to Chandigarh was again half as much as from Chandigarh to Shimla. If the average sped for the entire journey was 49 kmph, what was the average speed from Chandigarh to Shimla?

a.) 2 kmph

b.) 63 kmph

c.) 42 kmph

d.) None of these

Solution:  It is clear from the question that the ratio of distances between (Delhi-Chandigarh): (Chandigarh-Shimla) = 3:4

Also, the ratio of the speeds between (Delhi-Chandigarh): (Chandigarh-Shimla) = 3:2.

Let the distances be 3x and 4x respectively and speeds be 3s and 2s.

So, the time taken from Delhi to Chandigarh will be = (x/s)

Also, time taken from Chandigarh to Shimla will be = (2x / s)

Average speed = Total distance / Total time taken. = (3x + 4x) / [(x/s) + (2x/s)] = 7s/3

Now, 7s / 3 = 49

Thus, s = 21

Average speed from Chandigarh to Shimla = 2s = 2*21 = 42 kmph

Example 5: A car rental agency has the following terms. If a car is rented for 5 hr. or less, then the charge is Rs. 60 per hour or Rs. 12 per kilometer whichever is more. On the other hand, if the car is rented for more than 5 hr., the charge is Rs. 50 per hour or Rs. 7.50 per kilometer whichever is more. Akil rented a car from this agency, drove it for 30 km and ended up paying Rs. 300. For how many hours, did he rent the car?

1. 4 hr.                   B. 5 hr.                        C. 6 hr.        D. None of these

Solution:  We can solve this question by using trial and error method.

Scenario 1: Let us consider that Akil rented the car for 5 hr. or less.

In this case, Cost could be either (12 * 30) or (60 * 5)   [Considering maximum of 5 hours]

Here, 12 * 30 = 360 > 300. So if Akil would have taken the car for 5 hours or less, he would have paid 360 for sure. But he has paid only Rs. 300. So, he has not taken the car for 5 hours or less.

Scenario 2: Let us consider that Akil rented the car for more than 5 hours.

Thus, Going by kilometer method, he would have paid = (30 * 7.50) = Rs. 225

But, Akil has paid INR 300, which means payment was done on per hour basis.

Thus, Number of hours for which Akil rented the car would be = Rs. 300 / Rs. 50 per hour = 6 hours.

Example 6:  Consider a sequence of seven consecutive integers. The average of first five integers is n. The average of all seven integers is:

a). n

b). n + 1

c). k x n, where k is a function of n

d). n +(2/7).

Solution:   Let’s solve this by assumption. Let the 7 consecutive integers be 1,2,3,4,5,6,7

Average of first five integers = 3

Average of first seven integers = 4.

Alternative Method:

Let a set of any 5 integers be a, a+1, a+2, a+3, a+4

Here, the average will be the middle term i.e. a+2.

It is given that a+2= n.

Now, similarly, a set of 7 integers can be expressed as a, a+1, a+2, a+3, a+4, a+5, a+6.

In this set, the average will be the middle term i.e. a+3.

The term a+3 can be written as; a+2+1.

As a+2= n,

∴a+3= n+1.

So, the answer is option (b).

Example 7: A train traveling at 72 kmph crosses a platform in 30 seconds and a man standing on the platform in 18 seconds. What is the length of the platform in meters?

1. 240 meters
2. 360 meters
3. 420 meters
4. 600 meters

Solution: When the train crosses a man standing on a platform, the distance covered by the train is equal to the length of the train.
However, when the same train crosses a platform, the distance covered by the train is equal to the length of the train plus the length of the platform.
The extra time that the train takes when crossing the platform is on account of the extra distance that it has to cover = length of the platform.
Therefore, length of the platform = speed of train * extra time taken to cross the platform
Length of platform = 72 kmph * (30-18) seconds = 72 kmph * 12 seconds
Converting 72 kmph into m/sec, we get 72 kmph =   (5/18) * 72 m/sec = 20 m/sec
Therefore, length of the platform = 20 * 12 = 240 meters.

Conclusion: The concept of averages has various applications while solving different quantitative aptitude question. Some of the important application is mentioned above along with examples. Concept of average is often combined with various other concepts such as age, speed and distance, natural numbers etc.   If we know the basics and have solved some of the CAT questions on these topics, it is not difficult to solve average related problems.

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### One response to “Averages Concepts from Arithmetic – How to apply in CAT Questions?”

1. MJ VIGNESH says:

classes are good sir

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