Mixture and Allegation Tricks - 1

Mixture and Alligation Tricks - 1

 

"One Topic, to rule them all"

Yes...Mixture and Alligation has its own charm. A good hold of it can help you in solving a wide range of questions.

Alligation is a method of solving arithmetic problems related to mixtures of ingredients. Please note that alligation method is applied for percentage value, ratio, rate, prices, speed, etc. and not for absolute value. That is whenever per cent, per km, per hour, per kg, are being compared, we can use Alligation.

Common trick for Ratio-Proportion and Mixture Alligation : Almost 50% of the questions are solvable just by going through the options. Just go through the questions I have solved in this article and you will know the approach.

Rule of Alligation

Ingredient A : Ingredient B = M - Y : X - M

Here Mean Price is something which applies on the whole thing. If two varieties  of tea costing Rs. X and Rs. Y respectively are mixed and sold at Rs. Z, then Z is the mean price because it is price of the mixture.

Now I will take up some SSC CGL questions of Ratio-proportion and Mixture-Alligation.

Q.1 

Note that Rs. 180/kg and Rs. 280/kg are cost prices, while Rs. 320 is the selling price. To apply the alligation formula, all the three prices should be similar. So we will convert SP into CP
Given SP = Rs. 320/kg, Profit = 20%
Hence CP = 320/1.2 = Rs. 800/3
So the Mean price is Rs. 800/3 per kg
Now you can apply the formula-
Type 1 : Type 2 = 280 - 800/3 : 800/3 - 180 = 2 : 13
Answer : (B)

Q. 2)

Both the containers have equal capacity. Let us assume that both containers are of 28 litres. Why 28? Because 28 is the LCM of (3 + 1) and (5 + 2) or 4 and 7. So taking the capacity as 28 litres will make your calculations easier.
In Container 1, we have (3/4)*28 = 21 litres of milk and (1/4)*28 = 7 litres of water.
In Container 1, we have (5/7)*28 = 20 litres of milk and (2/7)*28 = 8 litres of water.
Total milk in both the containers = 21 + 20 = 41
Total water in both the containers = 7 + 8 = 15
Milk : Water = 41 : 15
Answer : (D)
Shortcut
Container 1 has 3 times more milk than water
Container 2 has 2.5 times more milk than water
When the contents of the two containers are mixed, the milk will still be more than water. How much more ? Somewhere between 2.5 and 3 times
(D) is the only option where the quantity of milk is around 2.7 times (i.e. between 2.5 and 3) that of the water.

Q. 3) 

Milk in vessel A = 4/7
Milk in vessel B = 2/5
Milk in vessel C = 1/2 (because in vessel C, milk and water are present in 1:1 ratio)
You have to mix 4/7 and 2/5, to produce 1/2. Hence 1/2 is the Mean Price.
A : B = (1/2 - 2/5)/(4/7 - 1/2) = 14 : 10 = 7 : 5

Q. 4)

Shortcut
Final ratio of the three varieties is 5 : 7 : 9
The question asks us the quantity of third variety of tea in the final mixture. From the above ratio, it is clear that the quantity of the third variety is a multiple of 9. So 45 is the only option possible.
Answer : (D)
Method
Let the three quantities be 4x, 5x and 8x
New quantities are 4x + 5, 5x + 10 and 8x + p
Now 4x + 5 : 5x + 10 : 8x + p = 5 : 7 : 9
(4x + 5)/(5x + 10) = 5/7 and (4x + 5)/(8x + p) = 5/9
Solving 1st equation, we get x = 5
Solving 2nd equation, we get p = 5
In the final mixture the quantity of the third variety is 8x + p = 8*5 + 5 = 45

Q. 5)

In this question we will use the below formula

So from the above formula
(Quantity of acid left)/(Quantity of acid in the original mixture) = (1 - 4/20)^2 = 16:25
Answer : (A)

Q. 6)

Let the original quantities of A and B be 4x and x
In 10 litres, quantity of A = 4/5 * 10 = 8 litres
In 10 litres, quantity of B = (10 - 8) = 2 litres
New quantities of A and B are 2x and 3x
(Original Quantity of A) - (New quantity of A) = 8 litres[Because after taking out 10 litres of the mixture, the quantity of liquid A reduced by 8 litres]
So, 4x - 2x = 8
or x = 4
Hence quantity of liquid A in original mixture = 4*4 = 16 litres
Answer : (C)
Note : In the above question, there were two different ratios 4:1 and 2:3, then too I took the same constant of proportionality for them, i.e. 'x' because the following two conditions were met:
1. The volume of mixture did not change (Like in this question 10 litres were replaced, not removed)
2. The two ratios had same no. of parts (4:1 and 2:3 both have 5 parts)
You can take different constant to solve the question, but that will make the calculations little lengthy.

Q. 7)

Since the ratio of alcohol and water is 1:4, hence quantities of alcohol and water in the mixture are 3 litres and 12 litres respectively.
Total volume will become 18 litres after adding 3 litres water
% of alcohol = 3/18 * 100 = 50/3%
Answer : (B)

Q. 8) 

Shortcut

Originally there are 1512 story books

Final ratio of Story books : Others = 15:4

That means the story books are a multiple of 15.

Just look at the options and see which number when added to 1512, will give a multiple of 15

Answer : (C)

Method

Let the no of story and other books be 7x and 2x respectively

Given 7x = 1512

x = 216

Now let the final quantity of story and other books be 15y and 4y respectively.

Since only story books are added to the collection, hence the quantity of other books has remained unchanged.

So 2x = 4y or y = x/2 = 108

We have to find 15y - 7x = 15y - 14y = y [Since x = 2y]

So answer is 108.

Don't Forget to check the Part-2 

• Mixture and allegation Part-2

Algebra Tricks- 3 [Approximation]

Algebra Tricks - 3 [Approximation]

 

This is another useful article for you all and requires no prior knowledge of any sort.

Approximation is a very important tool that can help you solve some complex and time taking questions. I will solve the below questions from SSC CGL with approximation technique to give you an idea of how it works. But before that, some basic rules of approximation:

1. Establish a limit within which the variable is falling.
2. Neglect the smaller terms of the expression (fractions with Denominator>Numerator)
3. Please use this technique only when the options have a significant difference between them. E.g. If in a question the 4 options are A. 4, B. 5, C. 6, D. 7, you can't use the approximation technique because the options are fairly close.

These rules will make sense once you go through the below CGL questions-

Q. 1

Now how will you approach this question if you dont know how to solve it?
Given, x^4 + 1/x^4 = 119
We can safely assume that 3<x<4 because 3^4 = 81 and 4^4 = 256  (119 lies between 81 and 256). Moreover x will be closer to 3 as 119 is more close to 81 than 256
We have established the limit of the variable.
Let us take our first value. Go with x = 3.2
3.2^4 = 104 (approx), which is still a little away from 119
Hence let us take x = 3.3 as our second value
3.3^4 = 118 (approx) [PERFECT]
Now we have to find x^3 - 1/x^3
Note that 1/x^3 is negligible and hence we can neglect it
So just find the value of 3.3^3
Answer : (C)
Note : You won't take much time in calculating 3.2^4 or 3.3^4 if you know a fast method to calculate squares. I have written an article about it. You can check it here. 

Q. 2

Here again no need to figure out how to solve the question

√3 = 1.73, √5 = 2.23

Hence √x = 1.73 - 2.23 or x = 0.25

Put the value of x

(0.25)^2 - 16*0.25 + 6

= 0.0625 - 4 + 6

= 2 (approx)

Answer : (C)

Q. 3

x = √5 + 2 = 4.23
(x^4 - 1)x^2 = x^2 - 1/x^2
Neglect 1/x^2
x^2 = 4.23^2 = 17 (approx)
Answer : (A)

Q. 4

Again if you dont know how to solve the above question, then observe the above equation
If a=1, then LHS = 5.33 (which is little more than RHS, i.e., 5). We need to decrease the value of 'a'.
Hence let's take a = 0.9
5a + 1/3a = 4.8
Now LHS is more than RHS. We need to increase the value of 'a' slightly
So lets lock the final value a = 0.95 (Now no need to check the value of LHS for a=0.95)
9a^2 + 1/25a^2
Neglect 1/25a^2
9a^2 = 8 (approx)
Answer : (D)

Q. 5

x = 2 + √3 = 2 + 1.73 = 3.73

Now you have to find the value of √x + 1/√x

√x = √3.73
You know that 19^2 = 361
Hence √3.73 = 1.9 (approx)

1/√x = 1/1.9 = 0.5

√x + 1/√x = 1.9 + 0.5 = 2.4 (which is close to √6)

Answer : (B)

Q. 6

3x - 1/4y = 6
Put x=1 and solve the equation for y
y = -1/12
Put x = 1 and y=-1/12 in the expression (4x - 1/3y)
You will get 8
Answer : (D)

Q. 7

Here again we can say that the approx value of x is 8, because 8^2 = 64
Put x = 8 in the expression
 = (64 - 1 + 16)/8
 = 10 (approx)
Answer : (A)

Q. 8)

Put x=2, the LHS becomes 6 and it is little more than RHS. So we need to decrease its value slightly
Let us take x = 1.8
LHS = (1.8)^2 + 1.8 = 5(approx)
LHS is almost equal to RHS, hence x=1.8 is a perfect value
Neglect 1/(x + 3)^3.
Now we only need to find the value of (x + 3)^3
(x + 3)^3 = (1.8 + 3)^3 = 110 (approx)
Answer : (A)


Below are some important formulas for Algebra, see if you can memorize them :)

Don't Forget to check Part-1,Part-2 and Part-4

• Algebra Tricks Part- 1
• Algebra Tricks Part- 2
• Algebra Tricks Part- 4

Keep Reading :)

Algebra Tricks- 2

Algebra Tricks - 2

 

Sometimes the equations are complex and you will find it difficult to assume values for the variables. Example -

Q . 1

Although the equation is symmetrical, we can't assume a=b=c, because that will make the LHS = 0. Such situations specially arise when the RHS is non-zero (here it is 1). Now what should we do? The trick is simple, there are three terms on the LHS, hence assume each term to be 1/3 (so that all the three terms will add up to give 1). Why have we taken 1/3, although it is obvious but still it comes from the formula - (value on RHS) / (No. of terms on LHS)
Here RHS = 1 and No. of terms on LHS = 3, hence we have assumed the value of each term as 1/3.

(a² - bc)/(a² + bc) = 1/3 ...      (1)

(b² - ca)/(b² + ca) = 1/3 ...      (2)

(b² - ca)/(b² - ca) = 1/3  ...      (3)

From (1), we get a² = 2bc ...       (4)

. Similarly from (2) and (3), we get b² = 2ca ...      (5)

and c² = 2ab      ...    (6)

Put the values of a², b² and  c² from (4), (5) and (6) in the expression whose value we have to find...

You will get 2 as the answer

Answer : (B)

Q. 2 

Here again, value on RHS = 3, No of terms = 3. Hence we assume each term to be 3/3 = 1

Therefore, (m – a²)/(b² + c²) = 1

or m = a² + b² + c²

Answer : (B)

Please note that this hack is also applicable only for symmetrical equations.

Now let us see some other questions where you can assume the values.

Q. 3

Put x = 1, then since A is the average of x and 1/x, it's value will also be A = 1

The average of x³ and 1/ x³ = 1

Put A = 1 in all the 4 options to check which option will give '1' as the output.
Answer : (D)

Q . 4

Here on putting a=1, you will find that both the options A and B will give the same result. Hence put a = 4, then x = 1.25
The value of the expression = 3/2
Answer : (A)
Here, I have straight away put a=4, instead of 2 or 3 because in the question we have to calculate the square root of a. So if you will take a perfect square(like 4), the calculations will be much easier.
Note : In this question we calculated the square of 1.25, which is 1.5625. For those who don't know the trick for calculating the square of numbers ending with '5' (like 15, 65, 135, 225, etc.), let me share it.

In such cases, the last two digits are always 25.
E.g. the square of 65
The last two digits = 25
First 2 digits = 6*(6+1) = 42
Hence square of 65 = 4225
Similarly square of 125
The last two digits = 25
First 3 digits = 12*(12+1) = 156
Square of 125 = 15625

Q. 5.

Put x = 0

1st term = 1/a^2

2nd term = 1/a^2

3rd term = 0

1st term - 2nd term + 3rd term = 0
Answer : (D)

You can choose any value for 'x' and you would get the same answer. For e.g. let us take x = 1

= 1/(a^2 + a + 1) - 1/(a^2 - a + 1) + 2a/(a^4 + a^2 + 1)

= -2a/(a^4 + a^2 + 1) + 2a/(a^4 + a^2 + 1)

= 0

Don't Forget to check Part-1, Part-3 and Part-4

•  Algebra Tricks Part- 1
•  Algebra Tricks Part- 3
•  Algebra Tricks Part- 4

If you have any doubt in any question that has been asked by SSC, please drop a comment.

Algebra Tricks - 1

Algebra Tricks - 1

 

Algebra is the easiest topic for SSC because you don't have to memorize any formula for it and all the questions can be solved within 10 seconds with jugaad.
First let me share with you the concept of symmetrical expressions(as I call it). A symmetrical expression is the one in which the weight of all the variables (a, b, c, etc.) is equal. Examples will make things clear.
Examples of symmetrical expressions -

• a³ + b³ + c³
• 3a + 3b + 3c
• a² + b² + c²
• a + b + c
• ab + bc + ca

Examples of non - symmetrical expressions -

• a - b + c
• 2a + 3b + 3c
• a³ + b² + c³
• a + b + c²

Hack - 1 : "Whenever you encounter a symmetrical equation in any question, you can safely assume : a = b = c (even if it is not given in the question)"

Let's solve previous year questions -

Q . 1.

Here you can see that the LHS as well as the RHS of the equation is symmetrical, hence a = b = c

Answer : (A)

Q .2. 

We put a = b = c, hence (a+c)/b becomes (a+a)/a, or 2

Answer : (B)

Q . 3.

In this question we have to find the value of x.
Here the equation is completely symmetrical, hence we assume a = b = c
Put b=a, c=a (so that the whole equation is in terms of 'a')
Now LHS becomes 3(x - a²)/2a
RHS = 12a
Solving this, you will get, x = 9a²
From here we get that the value of x is 9a²

Now put a = b = c in all the 4 options and check which option gives you the value 9a²

A) 9a²
B) 3a²
C) 3a²
D) 0

Answer : (A)

Q . 4. 

bc + ab + ca = abc is symmetrical and hence we can assume a=b=c
Now put b=a and c=a in this equation. We will get -
3a² = a³
So a = 3
Now put a=b=c=3 in the expression whose value we have to find. You will get the answer as 1.
Answer : (B)

Hack - 2 : "When only a single equation is given and based on that you have to find the value of an expression, you can assume the value of variables yourself. But make sure to assume only such values that will not make the denominator zero"

Examples :

Q . 5

In this question, only a single equation is given, i.e., x + y + z = 0, and based on this equation we have to find the value of an expression

We can assume x = -1, y = 1 and z = 0 (such that x + y + z = 0)

Now on putting these values in the expression, we get the answer as 2

Answer : (D)

           Q . 6. 

a + b = 1

Let's assume a = 1 and b = 0

Put the values in the expression, and you will get 0.

Answer : (A)

      Q . 7.

In this question the values of x and y both depend on a constant 'a'. We can assume any value for 'a' and this will give the values of x and y. Let us assume a=1

This will give x = 2 and y = 0

Put these values in the expression and you will get the answer as 16.

Answer : (A)

Q . 8

Pick values for a, b and c, such that their sum is 2s. Let us assume a = 2s, b = s and c = -s (here you should not assume a,b or c to be zero because that will make the elimination of options difficult)

Put these values in the expression and you will get 2s²

Now check all the four options to see which of them will give the value 2s² on putting a=2s, b=s and c=-s

Answer : (C)

Don't forget to read Algebra Tricks - 2, 3 and 4

•   Algebra Tricks Part - 2
•   Algebra Tricks Part - 3
•   Algebra Tricks Part - 4  

If you have any Doubt Please Comment Below:

Profit and Loss - Part-1

Profit and Loss - Part-1

 

Profit/Loss is another easy topic of SSC CGL. Most of the questions can be solved in less than 30 seconds. First let me introduce a formula that will be used in solving 50% of the questions.

Where,
SP = Selling price
CP = Cost Price
f = Profit/loss factor

What is this profit/loss factor ? It’s simple, ‘f’ depends on the profit/loss %
If profit% = 10, then f = 1.1
If profit% = 30, then f = 1.3
If profit% = 15, then f = 1.15
If loss% = 10, then f = 0.9
If loss% = 25, then f = 0.75
If loss% = 12.5, then f = 0.875
Please note that f depends on profit/loss percentage and not on the absolute value of Profit/Loss. So if in any question it is given that the profit is Rs. 30, then it doesn't mean that f = 1.3

Let us see some SSC CGL questions that can be solved with this formula

Q. 1

Let 's' be the SP of 1 article and 'c' be the CP of 1 article.
Given, 6c = 4s
Therefore s/c = 1.5
Gain % = 50
Answer : (B)

Q .2. 

Let 's' be the SP of 1 metre of cloth and 'c' be the CP of 1 metre of cloth
Total SP = 20s, Profit = 4s
CP = SP - Profit = 16s
The ratio s/c = 20s/16s = 1.25
Gain % = 25
Answer : (C)

Q . 3.

SP of 1 article(s) = Rs. 10/8 = 5/4
CP of 1 article(c) =Rs. 8/10 = 4/5
s/c = 25/16 = 1.5625
Gain % = 56.25
Answer : (A)

Q) 4. Kunal sold a shirt at a loss of 10%. Had he sold it for Rs 60 more, he would have gained 5% on it. Find the CP of the shirt.

In this question we have to find the CP of the article

c = s/f
From basic mathematics or elementary science we know that putting delta (∆) sign in numerator and denominator doesn't change the equation. ∆ stands for 'change'
c = ∆s/∆f
where ∆s = change in SP
∆f = change in factor
Therefore ∆s = New SP - Old SP = Rs. 60
∆f = New factor - Old factor
New factor is the factor when profit is 5%. Old factor is the one with loss = 10%
So ∆f = 1.05 - 0.9 = 0.15
c = ∆s/∆f = 60/0.15 = Rs. 400
Answer : Rs 400

Please Check the Link for the Part-2

• Profit and Loss Tricks Part-2

If you have any doubt in any question that has been asked by SSC, please drop a comment.

Book for mathamatics

Mensuration Tricks - 1

Mensuration Tricks - 1

 

Mensuration is a pure formula-based topic and tricks/shortcuts are seldom applied here. So in this series I will try to solve all the mensuration problems that have appeared in CGL lately and in the process I will share the important concepts/formulas.

Q. 1)

For Prism and Calendar (figures with uniform girth) -
Lateral Surface Area = Height * Perimeter of the Base
Volume = Height * Area of the Base

In this question the Total surface area is being asked
Total Surface Area of a Prism = Lateral Surface Area + Area of the two bases
Height of the prism = 10 cm
Perimeter of the base = 5 + 12 + 13 (Calculate the hypotenuse with Pythagoras Theorem) = 30 cm
So Lateral Surface Area = 10 * 30 = 300 cm
Area of the base = 1/2 * base * height = 1/2 * 5 * 12 = 30 cm
So Total Surface Area = 300 + 2*30 = 360 cm
Answer : (A)

Q. 2)

In such questions, remember one thing SIMILARITY

r/R = h/H     ...    (1)

where r = radius of small cone

R = radius of Big cone

h = height of small cone

H = height of big cone

Volume of cone = 1/3 * π* r² * h

Given, Volume of big cone = 27 * Volume of small cone

1/3 * π * R² * H = 27 * 1/3 * π * r² * h

27 * (r/R)^2=H/h
Put the value of r/R from equation (1)

27 * (h/H)^2 = H/h

27 * h³ = H³

Put H = 30 cm

So h = 10 cm

The question asks us the height above the base, which is (30 - h) = 30 - 10 = 20 cm

Answer : (B)

Q. 3)

The base of the prism looks like the figure above.

AD = 12 cm, AB = 9 cm

Hence BD = 15 cm (Pythagoras Theorem)

Area of base = Area of triangle ABD + Area of triangle BDC

Area of triangle ABD = 1/2 * 9 * 12 = 54 cm

Area of triangle BDC = 84 cm (Apply Heron's formula)
Area of base/quadrilateral = 84 + 54 = 138 cm
Volume = Height * Area of the Base
2070 = Height * 138
So, Height of the prism = 15 cm

Lateral Surface Area = Height * Perimeter of the Base
Perimeter of the base = AB + BC + CD + DA = 48 cm
Lateral Surface Area = 48 * 15 = 720 cm^2
Answer : (A)

Q. 4)

Area of the base = √3/4 * a^2, where a is the side of the equilateral triangle

Perimeter of the base = 3a
Volume of the prism = Area of the base * Height =√3/4 * a^2 * h ... (1)
Lateral surface Area of the prism = Perimeter of the base * Height = 3a * h ...(2)
Divide equation (1) by (2)
Volume/Area = (1/4√3) * a
40√3/120 = a/4√3 [Since Volume = 40√3 and Lateral surface Area = 120]
a = 160 * 3/120

a = 4 cm

Answer : (A)

Q. 5)

This question is about 'Pyramid'. So let me just give you some basic understanding of Pyramids. CGL can ask questions about two types of Pyramids - Pyramid with a Triangular Base and Pyramid with a square base. Both these pyramids have different formulas. Look at the below figure and understand the labellings, i.e., Slant edge and Slant Height

In the below image, I have written formulas for both types of Pyramids. The formula for Volume is same for both the Pyramids-

V = 1/3 * A * h

where A = Area of the base (calculation of A will be different for both)

h = Height of the Pyramid

When lateral surface area is asked, you will first calculate the 'Slant Height'. Then with the help of slant height  you will find the area of one lateral face (let's call this area M). If the pyramid is having a triangular base then multiply M with 3, to get the lateral surface area of the pyramid. And if the pyramid is having a square base, then multiply M with 4.

The area of the square is 324, hence its side is 18 cm

Volume of the pyramid = 1/3 * Area of the base * Height

1296 = 1/3 * 324 * Height

So Height = 12 cm

Slant Height of the pyramid with square base  = √ (h^2 + a^2/4) = √(12^2 + 18^2/4)

Slant Height = 15cm

Area of the lateral face = 1/2 * Base * Height = 1/2 * 18 * 15 = 135 cm^2

Pyramid with a square base has 4 lateral faces, so lateral surface area of the pyramid = 4 * 135 = 540 cm^2

Answer : (D)

Please check the link for Part-2

• Mensuration Tricks Part-2

If you have any  doubt please comment below: