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The quant. section includes many important topics like- percentage, average, simple and compound interest, time and work, time and distance, algebra etc. We provide you with a guide on how to solve percentage and average related problems.
A percentage is a number or ratio expressed as a fraction of 100.Percentage is considered a very important chapter.
When we say 100 percent in mathematical notation we write 100% so 35 % means 35 per 100 and 65% means 65 per 100 it is a proportion per hundred and it is used to find marks, profit percent or loss percent of a particular product.
Here are some different percent related problems given with shortcut tricks below.
Percentage is a fraction whose denominator is always 100. x percentage is represented by x%.
We know that x/y = (x/y× 100 )
[ R/(100+R)×100 ] %
[ R/(100-R)×100 ] %
Let the present value of a machine be P. Suppose depreciates at the rate of R% per annum Then :
1. Value of the machine after n Years
[ R/(100+R)×100 ]%
Example 1: The average mark of 70 students in a class is 80. Out of these 70 students, if the average mark of 40 students is 75, what is the average mark of the remaining 30 students?
Average of 70 students = 80.
Hence total marks of all the 70 students = 70×80 = 5600
Out of these 70, average of 40 students = 75
Hence total marks of these 40 students = 40×75 = 3000.
So the total marks of the remaining 30 students = 5600-3000 = 2600
Hence the average of the remaining 30 students = 2600/30 = 86.66
Now there will be a serious difference in the time taken if the numbers given here are not multiples of 10. Just look at example 2 below:
Example 2: “Average of 75 students is 82, out of which average of 42 students is 79. What is the average of the remaining 33 students?
Let’s go back to Example 1
Given that average of 70 students is 80. This class is split into two groups of 40 and 30. I am sure you will agree with me that if the average of each of these groups is 80, then the average of the total group i.e. 70 students will also be 80.
But the first group of 40 students, instead of 80 they had an average of only 75. So definitely the average of the remaining 40 students must be greater than 80.
The loss incurred because of the first group must be compensated by the second group.
Let us look at the loss incurred because of the first group
We want them to have an average of 80, but they managed only 75. So we lost an average of 5 upon 40 students. So the loss in the sum = 40×5.
Now this loss of 40×5 must be compensated by our second group i.e. 30 students.
So their average must be not only the initial 80, but also the average meant to compensate the loss incurred because of the first group.
Hence the average of the remaining 30 students = 80 + (40×5)/30 = 80 + 20/3 = 86.66
Now let us look at Example 2 where the numbers are not comfortable.
Example 3: Average of 75 students is 82, out of which average of 42 students is 79.
What is the average of the remaining 33 students?
Average of 75 members = 82.
Two groups of 42 and 33 and we want each group to have an average of 82.
But the first group i.e. 42 students they had an average of 79.
We fell short by 3 in the average.
So the loss in the sum= 3×42.
So the average of the second group i.e. 33 students must be 82 + (3×42)/33 = 82+ 42/11 = 85.82
So folks it’s time to get into some more different situations that arise when we calculate averages.
Let’s look at one more sum similar to the one discussed above.
The average height of 74 students in a class is 168 cm out of which 42 students had an average height of 170 cm. Find the average height of the remaining 32 students.
Average of 74 students = 168.
Here we can observe that, in the case of the first group we had a gain.
Since the average of 42 students is 170 cm, the gain is 2 cm in the average upon 42 students.
So the gain in the sum = 42×2
Hence average of the remaining 32 students = 168 – (42×2)/32= 168-(21/8) = 165.37
The average height of 40 students in a class is 172 cm. 30 students whose average height is 172.5 cm left the class and 40 students whose average height is 170.5 cm joined the class. Find the average height of the present class.
Going back to the ideal situation, we want the average of the students leaving the class as well as joining the class to be 172 so that the average remains the same.
But it is given that the average of the 40 students leaving the class is 172.5 (more than 172).
So we will incur a loss of 0.5 cm in the average upon 30 students.
Hence the loss in the sum = 0.5x 30 = 15 cm
Also, since the average of the 40 students joining the class is 170.5 (less than 172) we will incur a loss in this case as well. The loss in the average is 1.5 cm upon 40 students.
Hence the loss in the sum = 1.5×40 = 60 cm
Thus the total loss in the sum = 75 cm. This loss will be shared by 50 students which is the present strength of the class.
Hence the average of the present class = 172 – 75/50 = 170.5 cm.
Important Concepts and Formulas – Permutations and Combinations 1. Multiplication Theorem (Fundamental Principles of Counting) If an operation can be performed in m different ways and following which a second operation can be performed in n different ways, then the two operations in succession can be performed in m×n different ways.
If an operation can be performed in m different ways and a second independent operation can be performed in n different ways, either of the two operations can be performed in (m+n) ways.
3. Factorial Let n
be a positive integer. Then n factorial can be defined as
Examples 5!=5×4×3×2×1=120 3!=3×2×1=6
Special Cases 0!=1 1!=1
Suppose we want to select three out of three girls P, Q, R. Then, only possible combination is PQR
Sometimes, it will be clearly stated in the problem itself whether permutation or combination is to be used. However if it is not mentioned in the problem, we have to find out whether the question is related to permutation or combination.
Consider a situation where we need to find out the total number of possible samples of two objects which can be taken from three objects P, Q, R. To understand if the question is related to permutation or combination, we need to find out if the order is important or not.
If order is important, PQ will be different from QP, PR will be different from RP and QR will be different from RQ .If order is not important, PQ will be same as QP, PR will be same as RP and QR will be same as RQ. Hence,If the order is important, problem will be related to permutations. If the order is not important, problem will be related to combinations.
For permutations, the problems can be like “What is the number of permutations the can be made”, “What is the number of arrangements that can be made”, “What are the different number of ways in which something can be arranged”, etc.
For combinations, the problems can be like “What is the number of combinations the can be made”, “What is the number of selections the can be made”, “What are the different number of ways in which something can be selected”, etc. pq and qp are two different permutations, but they represent the same combination.
Mostly problems related to word formation, number formation etc will be related to permutations. Similarly most problems related to selection of persons, formation of geometrical figures, distribution of items (there are exceptions for this) etc will be related to combinations.
The term repetition is very important in permutations and combinations. Consider the same situation described above where we need to find out the total number of possible samples of two objects which can be taken from three objects P, Q, R.
If repetition is allowed, the same object can be taken more than once to make a sample. i.e., PP, QQ, RR can also be considered as possible samples.
If repetition is not allowed, then PP, QQ, RR cannot be considered as possible samples.
Normally repetition is not allowed unless mentioned specifically.
8. Number of permutations of n distinct things taking r at a time : Number of permutations of n distinct things taking r at a time can be given by
nPr = n!(n−r)!
=n(n−1)(n−2)...(n−r+1) where 0≤r≤n
Special Cases nP0 = 1 nPr = 0 for r>n
nPr is also denoted by P(n,r). nPr has importance outside combinatorics as well where it is known as the falling factorial and denoted by (n)r or nr
Examples 8P2 = 8 × 7 = 56
5P4= 5 × 4 × 3 × 2 = 120
9. Number of permutations of n distinct things taking all at a time
Number of permutations of n distinct things taking them all at a time
= nPn = n!
Number of combinations of n distinct things taking r at a time ( nCr) can be given by
nCr = n!(r!)(n−r)!
=n(n−1)(n−2)⋯(n−r+1)r! where 0≤r≤n
Special Cases nC0 = 1, nCr = 0 for r>n
nCr is also denoted by C(n,r). nCr occurs in many other mathematical contexts as well where it is known as binomial coefficient and denoted by (nr)
Examples 8C2 = 8×7/2×1= 28, 5C4= 5×4×3×2/4×3×2×1= 5
Here’s an easy way to remember: permutation sounds complicated, doesn’t it? And it is. With permutations, every little detail matters. Alice, Bob and Charlie is different from Charlie, Bob and Alice (insert your friends’ names here).
Combinations, on the other hand, are pretty easy going. The details don’t matter. Alice, Bob and Charlie is the same as Charlie, Bob and Alice.
Permutations are for lists (order matters) and combinations are for groups (order doesn’t matter).
A joke: A “combination lock” should really be called a “permutation lock”. The order you put the numbers in matters. (A true “combination lock” would accept both 10-17-23 and 23-17-10 as correct.)
Let’s start with permutations, or all possible ways of doing something. We’re using the fancy-pants term “permutation”, so we’re going to care about every last detail, including the order of each item. Let’s say we have 8 people:
1: Alice 2: Bob 3: Charlie 4: David 5: Eve 6: Frank 7: George 8: Horatio
How many ways can we award a 1st, 2nd and 3rd place prize among eight contestants? (Gold / Silver / Bronze)
We’re going to use permutations since the order we hand out these medals matters. Here’s how it breaks down:
We picked certain people to win, but the details don’t matter: we had 8 choices at first, then 7, then 6. The total number of options was 8 · 7 · 6 = 336.
Let’s look at the details. We had to order 3 people out of 8. To do this, we started with all options (8) then took them away one at a time (7, then 6) until we ran out of medals.
We know the factorial is:
Unfortunately, that does too much! We only want 8 · 7 · 6. How can we “stop” the factorial at 5?
This is where permutations get cool: notice how we want to get rid of 5 · 4 · 3 · 2 · 1. What’s another name for this? 5 factorial!
So, if we do 8!/5! we get:
And why did we use the number 5? Because it was left over after we picked 3 medals from 8. So, a better way to write this would be:
where 8!/(8-3)! is just a fancy way of saying “Use the first 3 numbers of 8!”. If we have n items total and want to pick k in a certain order, we get:
And this is the fancy permutation formula: You have n items and want to find the number of ways k items can be ordered:
Combinations are easy going. Order doesn’t matter. You can mix it up and it looks the same. Let’s say I’m a cheapskate and can’t afford separate Gold, Silver and Bronze medals. In fact, I can only afford empty tin cans.
How many ways can I give 3 tin cans to 8 people?
Well, in this case, the order we pick people doesn’t matter. If I give a can to Alice, Bob and then Charlie, it’s the same as giving to Charlie, Alice and then Bob. Either way, they’re equally disappointed.
This raises an interesting point — we’ve got some redundancies here. Alice Bob Charlie = Charlie Bob Alice. For a moment, let’s just figure out how many ways we can rearrange 3 people.
Well, we have 3 choices for the first person, 2 for the second, and only 1 for the last. So we have 3 · 2 · 1 ways to re-arrange 3 people.
Wait a minute… this is looking a bit like a permutation! You tricked me!
Indeed I did. If you have N people and you want to know how many arrangements there are for all of them, it’s just N factorial or N!
So, if we have 3 tin cans to give away, there are 3! or 6 variations for every choice we pick. If we want to figure out how many combinations we have, we just create all the permutations and divide by all the redundancies. In our case, we get 336 permutations (from above), and we divide by the 6 redundancies for each permutation and get 336/6 = 56.
The general formula is
which means “Find all the ways to pick k people from n, and divide by the k! variants”. Writing this out, we get our combination formula, or the number of ways to combine k items from a set of n:
Here’s a few examples of combinations (order doesn’t matter) from permutations (order matters).
Permutation: Picking a President, VP and Waterboy from a group of 10. P(10,3) = 10!/7! = 10 · 9 · 8 = 720.
Permutation: Listing your 3 favorite desserts, in order, from a menu of 10. P(10,3) = 720.
Don’t memorize the formulas, understand why they work. Combinations sound simpler than permutations, and they are. You have fewer combinations than permutations.
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