Mixture and Alligation

Alligation: It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of the desired price.

or

A Process or rule for the solution of problems concerning the compounding or mixing of ingredients differing in price or quality.

Mean Price: The cost price No of a unit quantity of the mixture is called the mean Price.

Formula: If two ingredients A and B of price X and Y respectively are mixed and the price of the resultant mixture is M (Mean price) then the ratio (R) in which ingredients are mixed is given by, the rule of allegation.

R = (M - Y)/(X - M)

The above formula can be represented in diagram below, which in turn is more intuitive to grasp.

Thus the required ratio is,

R = (M - Y)/(X - M) or (Y - M)/(M - X)

Replacement of part of Solution Formula

Suppose a container contains a solution form which some quantity of solution is taken out and replaced with one of the ingredients. This process if repeated n times then,

Final Amount of ingredient that is not replaced = Initial Amount x (Vol. After removal/Vol. after replacing)ⁿ

Above Formula is not only true for absolute amounts but for ratios as well. So following formula is also valid:

Final ratio of ingredient not replaced to total = Initial Amount X (Vol. After removal/Vol. After replacing)ⁿ

Mixture of More Than Two Elements

These questions may seem a little tricky at first, but it is similar concept applied repeteadly.

In order to calculate final ratio of ingredients when mixture contains more than two ingredients,

• Take two ingredients such that 1st ingredient is Lower than the mean value and the other one is Higher than the mean value.
• Calculate the ratio of ingredients
• Repeat for all possible pairs
• Final ratio is the ratio obtained from step 2 (if an ingredient is common in the ratios, add values for this particular ingredient)

Mixture Containing 3 Ingredients

ex: In What ratio must a person mix three kind of tea each of which has a price of 70,80 and 120 rupees per kg, in such a way that the mixture costs him 100 rupees per kg?

Solution: Here the prices of tea are 70, 80 and 120 And Mean Price is 100.

Here, prices lower than the mean are 70 and 80. And prices higher than the mean is 120.

Thus possible pairs which can give mean value of 100 is : {70, 120} and {80, 120}

Let us denote tea of Rs. 70 with t70, tea of Rs. 80 with t80 and tea of Rs. 120 with t120

We apply the old alligation rule to All (two in this case) the pairs

For the 1st pair (t70, t120)

= t20: t120 = 20:30

Similarly for second pair, (t80, t120)

= t80: t120 = 20:20

so final ratio,

t70 : t80: t120 = 20:20: (30+20)

= t70 : t80: t120 = 2:2:5

Note: It's best to simplify the intermediate ratios at the end, else you may get wrong answer.

Here, if we had simplified the ratio to be, t70:t120 = 2:3

and t80 : t120 = 1:1

We could have easily gone to calculate ratio to be t70 : t80: t120 = 2:1 : (3+1)

t70: t80: t120 = 2:1:4 (wrong)

Exercise: Mixture Containing 4 Ingredients

How must a shop owner mix 4 types of rice worth Rs 95, Rs 60, Rs 90 and Rs 50 per kg so that he can make the mixture of these rice worth Rs 80 per kg?

Ans: 3:2:4:6

Related Article:-

• Speed Calculations

• Mixture and Alligation Shortcuts

• trigonometry tricks

• Learn Divisibility Rules

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Speed Calculations

Average Speed = Total Distance/Total Time

No matter which formula you choose to use, it will always boil down to this one. Keeping this in mind, let's discuss the various formulas we come across:

1. Average Speed = (a + b)/2

Applicable when one travels at speed a for half the time and speed b for other half of the time. In this case, average speed is the arithmetic mean of the two speeds.


2. Average Speed = 2ab/(a + b)

Applicable when one travels at speed a for half the distance and speed b for other half of the distance. In this case, average speed is the harmonic mean of the two speeds. On similar lines, you can modigy this formula for one-third distance.


3. Average Speed = 3abc/(ab + bc + ca)

Applicable when one travels at speed a for one-third of the distance, at speed b for another one-third of the distance and speed c for rest of the one-third of the distance.

Note that the generic Harmonic mean formula for n numbers is

Harmonic Mean = n/(1/a + 1/b + 1/c +.....)


4. You can also use weighted averages. Note that in case of average speed, the weight is always 'time'. So in case you are given the average speed, you can find the ratio of time as

t1/t2 = (a - Avg)/(Avg - b)

As you already know, this is just our weighted average formula.

Now, let's look at some simple questions where you can use these formulas

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Easy way to learn Divisibility Rules

Divisibility Rules

Divisible by means when you divide one number by another number the result should be the whole number with zero remainders.

Example:-

6/3 = 2; 6 is divisible by 3 because result 2 is the whole number and the remainder is 0.

7/3 = 2.33; 7 is not divisible by 3 because result 2.33 is not the whole number and the remainder is 1.

Divisibility by 2: Any number, the last digit of which is rather even or zero, is divisible by 2.

Example:- 

1875494: The last digit is 4 and it is an even number so this fat number 1875494 is divisible by 2.

Divisibility by 3: If the sum of the digits of a number is divisible by 3, the number is also divisible by 3.

Example:-

• 3789 is divisible by 3

Sum 3+7+8+9 = 27 is divisible by 3.

• 43266737 is not divisible by 3

Sum 4+3+2+6+6+7+3+7 = 38 is not divisible by 3.

Divisibility by 4: A number is divisible by 4 if the number's last two digits are divisible by 4.

Example:-

• 112 since the last two digits, 12, are divisible by 4, the number 112 satisfies this rule and is also divisible by 4.
• 10,948 the last two digits, 48, are divisible by 4. Therefore, the whole number is also.

Divisibility by 5: If a number ends in 5 or 0, the number is divisible by 5.

Example:-

• 1345: As its last digit is 5, it is divisible by 5.
• 1340: as its last digit is 0, it is divisible by 5.

Divisibility by 6: A number is divisible by 6 if it is even and if the sum of its digits is divisible by 3.

The rule for checking whether a number is divisible by 6 is quite tricky.

Since 6=2*3

We must apply the rules of 2 and 3 on a number to check if it is divisible by 6. So rules are 

• The last digit should be even number (Divisibility trick for 2)
• Add up all the digits and it should be multiple of 3 (Divisible by 3)

Example:

2742204: The last digit in 2742204 is 4 and it is the even number.

Add up all the digits so 2+7+4+2+2+0+4 = 24. 24 is divisible by 3.

This number 2742204 satisfy both the rules (2&3) so it is divisible by 6.

Divisibility by 7: Twice the last digit and subtract it from remaining number in given number, the result must be divisible by 7.

Example:

• 343 is divisible 7

34 - (2*3) = 28, 28 is divisible by 7.

Divisibility by 8: If the last three digits of a number are divisible by 8, the number is also divisible by 8, the number is also divisible by 8. Also, if the last three digits of a number are zeros, the number is divisible by 8.

Example:

• 1256: As 256 is divisible by 8, the number is also divisible by 8.
• 135923120: as 120 is divisible by 8, the number is also divisible by 8.

Divisibility by 9: A number is divisible by 9 if the sum of  the digits is an evenly divisible 9

Example:

• 39681: 3+9+6+8+1 = 27 is divisible by 9, hence the number is also divisible by 9.

Example: What least whole number should be added to 532869 to make it divisible by 9?

Solution: If a number is divisible by 9, the sum of its digits must be a multiple of 9.

Here, 5+3+2+8+6+9 = 33, the next multiple of 9 is 36.

3 must be added to 532869 to make it divisible by 9.

Divisibility by 10: The last digit must be 0

Example: 

• 456780 is divisible by 10. The last digit is 0.
• 78521 is not divisible by 10.  The last digit is 1.

Divisibility by 11: If the sums of digits at odd even places are equal or differ by a number divisible by 11, then the number is also divisible by 11.

Example:

3245682

A1 = 3+4+6+2 = 15   and A2 = 2+5+8 = 15

As A1 = A2, the number is divisible by 11.

Example: What is the least value of x such that 7648x is divisible by

Solution: A number is divisible by 11 When difference between the sum of digits at even places and at odd places is 0 or multiple of 11

The given number is 4876x.

(Sum of digits at Even places) - (Sum of digits at ODD places) = 0

(6+8) - (x + 7 + 6) = 0

14 - (X+13) = 0

Here the value of x must be 1.

Divisibility by 12: 12 = 3*4 so if the number follows the divisibility rules of 3 and  4 is divisible by 12.

• Add up all the digits and it should be multiple of 3 (divisible by 3)
• The last 2 digit (ones and tens should be divisible by 4)

Example: 

1. 834864: Add up all the digits: 8+3+4+8+6+4 = 33 and it is multiple of (3*11)/3

The last two digits are 64 and it is divisible by (4*16)/3 so this number 834864 is divisible by 12.

Divisibility by 13: Multiply last digit with 4 and add it to remaining number in given number, the result must be divisible by 13.

Example:

• 4568 is not divisible by 13

456 + (4*8) = 488

48 + (4*8) = 80, 80 is not divisible by 13.

Divisibility by 14: The number must be divisible by 2 and 7. Because 2 and 7 are prime factors of 14.

Divisibility by 15: The number should be divisible by 3 and 5. Because 3 and 5 are prime factors of 15.

Divisibility by 16: The number formed by last four digits in given number must be divisible by 16.

Example: 

7852176 is divisible be 16

2176 is divisible by 16

Divisibility by 17: Multiply last digit with 5 and subtract it from remaining number in given number, result must be divisible by 17. (You can again apply this to check for divisibility by 17)

Follow the similar examples given divisibility by 7 and divisibility by 13.

Divisibility by 18: Any number which is divisible by 9 and has its last digit even or zero, is divisible by 18.

Example:

926565: digit - sum is a multiple of nine (i.e. divisible by 9) and unit digit (8) is even, hence the number is divisible by 18.

Divisibility by 19: 

• Double the last digit.
• Add to the rest of the numbers until you get two or three digit number that should be divisible by 19

Example:

6688 8*2 = 16, Add 16 to the rest of the number so 668 + 16 = 684

Repeat it again: 4*2 = 8 and add to 68+8 = 76 and 76 is divisible by 19 so this number 6688 is divisible by 19 as well.

Divisibility by 20: 

• The last digit should be 0 (divisible by 10)
• The second last digit should be even

Example: 

4490

• The last digit is 0 but the last 2nd digit 9 is not an even number so 4490 is not divisible by 20.
• The last 2 digit should be divisible by 20.

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Mixture and Alligation Shortcuts

Alligation or Mixture's formula

Alligation and Mixture Solved Questions

Friends, today we shall discuss about an important concept for Banking and Other Competitive Exams - Mixture and Alligation. Usually people leave this section because its little confusing. But trust us friends, we will make it easier for you. Lets start with the basics.....


What is mixture?

A Mixture is a solution containing two or more products (we can simply call these products as Elements of the mixture) which bear a certain proportion to each other by quantity. This may be expressed in terms of

• Percentage (ex: 25% of salt in water)
• Fractions (ex: A solution of water and milk such that: water: milk = 5:7)

When you add two mixtures, the respective quantities of these elements of the mixtures get added, thereby affecting the overall contents of the combination. So, the resultant mixture would have " a weighted proportion" of the two elements.

It is presumed that no Chemical Reaction takes place while the tow mixtures are combined. So these are physical mixtures rather than chemical mixtures. (This clarification is for those chemical geniuses who thinks about Evaporation;)

Rule of Alligation

When two or more quantities of different products are mixed together to produce a mixture of a mean value, the ratios of their amounts  are inversely proportional to the differences of their values form the mean value. Please read the above statement. Its the heart of our concept.

Confused? lets read this in simple words....

If the gradients are mixed in a ratio, then 

Quantity of Cheaper/Quantity of Dearer  =  C.P of Dearer - Mean Price/ Mean Price - C.P of Cheaper

We can represent the above Statement as

Then, 

(Cheaper Quantity) : (Dearer Quantity) = (d - m) : (m - c)

• Where can we apply the rule of Mixture and Alligation?

  
  

  You can apply this when you have to find the proportion in which different ingredients of known value are to be mixed to produce a mixture of a given mean value. You can apply this when you have to find the mean value of a mixture when the proportion and value of its ingredients are known.

• What types of questions can I expect from Mixture and Allegation for Exams?

  
  

  Ex. 500 ml solution of water contains 10% of milk. How much pure water needs to be added to this solution in order to get the percentage of milk as 8% ?

  
  

  
  

  Sol: We can solve this problem using two methods. First we shall discuss solving this question using conventional method. Later we shall discuss the shortcut method.

  
  

Conventional Method

Let "X" ml of water to be added to the solution 

Water (ml) + Milk (ml) = Total Solution (ml)
	Water	Milk	Solution
Original	(100 – 10)% of 500 = 90% of 500 = 450	10% of 500 = 50	500
Add	X		X
Total	(450 + x)	50	(500 + x)

Now,

50/(500 + X) = 8/100

(500 + X) = 625

so, X = 125 ml

Now lets have a look at the shortcut method

Shortcut Methhod

This method is applicable only when the impurity level of one of the components of the mixture id reduced in the mixture is reduced in the resultant solution. This method makes use of only 3 values, as given in the question viz

1. Quantity of Original Total Solution = 500 ml
2. Original Level of 'Impurity' = 10%
3. The desired level of 'Impurity' = 8%

Now, the application proceeds as follows:

Original level of Impurity (10%)

                                                       ⇙                                                ⇘

                    Desired Level of Impurity(8%)                    Differenece (10% - 8%) = 2%

Quantity of Pure water to be added

(Difference in level of impurity / Required level of impurity)       

X (Quantity of Original Solution)

⇒ (2% / 8%) * 500 = 125ml

So, 125ml water should be added to get 8% milk.....

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trigonometry tricks

Quick Guide

Study notes

Tips & Tricks

Prev exam question

Definition of Trigonometry:- Trigonometry is the branch of mathematics that is concerned with calculating the angles of triangles or the lengths of their sides.

In this article we are provided the basic of trigonometry. Which are main topic in SSC Exams. 4-5 question are based on this topic in SSC Exam and in the mains of CGL 10-14 Question are asked. So, it is the important on for preparation.

First of all you have to learn that following table.

Ɵ	SinƟ	CosƟ	TanƟ	CosecƟ	SecƟ	CotƟ
0°	0	1	0	Not defined	1	Not  defined
30°	1 / 2	Ö3 / 2	1 / Ö3	2	2 / Ö3	Ö3
45°	1 / Ö2	1 / Ö2	1	Ö2	Ö2	1
60°	Ö3 / 2	1 / 2	Ö3	2 / Ö3	2	1 / Ö3
90°	1	0	Not defined	1	Not defined	0

 In trigonometry there is no trick based question asked but if you learn the basic concept of trigonometry then you are easily solve within few seconds. Candidates find this is very tough topic to prepare for SSC some candidates leaves these chapter question in exam but this chapter is so easily prepare if you practice hard with this chapter by learn table and identities.

Tricky Question of Trigonometry

Q.1  Sin 1.Sin2.Sin3......................Sin180 = 0

Q.2  Cos1.Cos2.Cos3....................Cos90 = 0

Q.3  Tan1.Tan2.Tan3.....................Tan89 = 1

Q.4  Cot1.Cot2.Cot3.....................Cot89 = 1

if X + Y = 90 so

Q.5 Sec²x- Cot²y =1

Q.6 Sin²x- Sin²y =1

Q.7 Cosec²x- Tan²y =1

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simple interest and compound interest shortcut tricks

CI Formula

SI Formula

Tips and Tricks

Previous Question

Important info

maths short tricks on simple interest formula and compounded interest calculator are one of the most important topics in exams. competitive exam preparation are all about time. If you manage your time then you can do well in those exams. Most of us skip that part. Here in this page we give few examples on simple interest and compound interest shortcut maths. We try to provide all types of maths shortcut tricks on simple interest and compound interest here. We request all visitors to read all examples carefully. These examples will help you to understand maths short tricks simple interest and compound interest.

Before starting anything just do a math practice set. Write down twenty math word problems related to this topic on a page. Do first ten maths using basic formula of this math topic. You alos need to keep track of Timing. Write down the time taken by you to solve those questions. Now practice our mathematics shortcuts tricks on Simple interest and compound interest and read examples carefully. After finishing this do remaining questions using simple interest and compound interest shortcut maths. Again keep track of the time. You will surely see the improvement in you timing this time. But this is not enough. You need to practice more to improve you timing more.

You all know that math portion is very much important in competition exams. It doesn't mean that other topics are not so important. But only math portion can leads you to a good score. A good score comes with practice and practice. You should do your math problems within time with correctness, and this can be achieved only by using maths short tricks. But it doesn't mean that without using maths short tricks you can't do any math problems. you may have that potential that you may do maths within time without using any maths shortcut tricks. But other peoples may not do the same. So Simple interest and Compound interest shortcut tricks here for those people. We always try to put all shortcut methods of the given topic. But it possible we miss any.

• What is Interest
• Compound Interest Example

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Time, Speed and Distance Shortcuts Tricks

Concepts

1. There is a relationship between speed, distance and time:

Speed = Distance / Time

             or

Distance = Speed * Time

2 . Average speed = 2xy / x+y

Where x km/hr is a speed for certain distance and y km/hr is a speed at for same distance covered.

Rmemeber that average speed is not just an average of two

 speeds i.e. x+y/2. It is equal to 2xy / x+y.

3. Always remember that during solving questions units must be same. Units can be km/hr, m/sec etc.

conversion of km/hr to m/sec and m/sec to km/hr

x km/hr = (x*5/18) m/sec i.e. u just need to multiply 5/18

Similarly, x m/sec = (x*18/5) km/sec

4. As we know, Speed = Distance/Time

Now, if in Questions Distance is constant then speed will be inversely proportional to time i.e. if speed increases, time taken will decrease and vice versa. 

• Time and Distance problems 

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