Short Trick on Centres of Triangle in Geometry

In this article we will discuss about the various centers of triangle. In SSC CGL exams 1 or 2  question generally come in exam from this chapter.

There are four centres  in a triangle triangle:

• In-centre
• Circum-centre
• Centroid
• Ortho centre

In-centre:

The point of intersection of the all the three angle bisectors of a triangle is called as In-centre

The distance of the in-centre from the all the three sides is equal(ID=IE=IF=inradius “r”)
In-radius (r)= Area of triangle/Semiperimetre=A/S
∠BIC = 90 +∠A/2
∠AIC=90+∠B/2
∠AIB=90+∠C/2

Circumcentre:

The point of intersection of the perpendicular bisectors of the three sides of a triangle is called its circumcentre.

The distance between the circumcentre and the three vertices of a triangle is always equal.
OA=OB =OC=R(circumradius)=abc/4A
∠BOC=2∠A
∠AOC=2∠B
∠AOB=2∠C

Location of circumcentre in various types of triangle:

Acute angle triangle: Lies inside the triangle
Obtuse angle triangle: Lies outside the triangle
Right angle triangle: Lies at the midpoint of the hypotenuse.

Orthocentre:

It is the point of intersection of all the three altitudes of the triangle.

∠BHC=180-∠A
∠AHB=180-∠C
∠AHC=180-∠B

Position of orthocentre inside the triangle:

Acute angled triangle: lies inside the triangle.
Obtuse angle triangle: lies outside the triangle on the backside of the obtuse angle. Orthocentre and circumcentre lie opposite to each other in obtuse angle triangle.
Right angle triangle: lies on the right angle of the triangle.

Centriod:

• It is the point of the intersection of the three median of the triangle. It is denoted by G.
• A centroid divides the area of the triangle in exactly three parts.

Medians:

• A line segment joining the midpoint of the side with the opposite vertex is called median.
• Median bisects the opposite side as well as divide the area of the triangle in two equal parts.

Some important tricks are as follows:

(1)In a right angle triangle ABC,∠B=90° & AC is the hypotenuse of the triangle. The perpendicular BD is dropped on the hypotenuse AC from the right angle vertex B,

BD=AB*BC/AC
AD=AB²/AC
CD=BC²/AC
1/BD=1/AB² +1/BC²
(2)The ratio of the areas of the triangles with equal bases is equal to the ratio of their heights.
(3)The ratio of the areas of two triangles is equal to the ratio of products of base and its corresponding sides.

Area (Triangle ABC)/Area (Triangle PQR) =AC*BD/PR*QS
(4)If two triangles have the same base and lie between the same parallel lines then the area of the two triangles will be equal.


Area of Triangle ABC= Area of Triangle ADB
(5)In a triangle ABC, AE, CD  and BF are the medians then
3(AB²+BC²+AC²) = 4(CD²+BF²+AE²)
6)Sum of  any two sides of the triangle is always greater than the third sides.
(7)The difference of any two sides of a triangle is always less than the third sides.

Simple and Compound Interest Tricks - 1

Simple and Compound Interest Tricks - 1

 

After reading this series (Part-1 and Part-2), you will be able to solve all the questions that are asked by SSC from this topic.

(1) The basic concept of CI and SI
Let's say you have Rs. 30000 and you keep this money in three different banks for 2 years(Rs. 10000 each). The three banks have different policy :
a) Bank A keeps your money at simple interest and offers you 5% interest
b) Bank B keeps your money at compound interest and offers you 5% interest. The interest is compounded annually.
Bank C keeps your money at compound interest and offers you 5% interest. The interest is compounded half-yearly.
After 2 years, which bank will give you most interest?
Let us calculate
Case (A)
Simple interest is calculated simply as (P*R*T)/100
Here P= 10000, r = 5% and T = 2 years
T = 2 years. Let's divide this period in two equal intervals of 1 year each
Hence SI received for the period 0 to 1 = 10000*5*1/100 = Rs. 500
SI received for the period 1 to 2 = 10000*5*1/100 = Rs. 500
So after 2 years, you will get Rs. 10000 + 500 + 500 = Rs. 11000
Note : Simple Interest is proportional. The interest received is same each year. So in the above example where SI was Rs. 500 for 1 years, that will mean the SI for 3 years is Rs. 1500, the SI for 5 years is Rs. 2500 and so on.

Case (B)
Compounded annually means whatever interest you will earn on first year, that interest will be added to the principal to calculate the interest for 2nd year. Let us see how
We know the CI formula is, Amount = P(1 + r/100)^t   (where Amount = P + CI)
CI received for the period 0 to 1 = Amount - Principal = 10000(1 + 5/100)^1 - 10000= Rs. 500
Now the amount received after 1 year will act as the Principal for calculating the Amount for next year
For calculating the amount for second year, you won't take P as 10000, but as Rs. 10500. So unlike SI where the interest was same each year, in CI the interest increases every year (because the principal increases every year)
CI received for the period 1 to 2 = Amount - Principal = 10500(1 + 5/100)^1 - 10500 = Rs. 525
Total interest received after two years = Rs. 500 + Rs. 525 = Rs. 1025
Total amount received after two years = Rs. 11025
Note :  In Case (b), to calculate the amount received after 2 years, I had divided the calculation into 2 intervals. It was done just for the sake of explanation. You can calculate the amount received after 2 years directly by 10000(1 + 5/100)^2

Case (C)
Just like case (b), where Principal was getting updated every year, in case (c) we will update the Principal every 6 months (half-year)
Since I have given the explanation in case (b), so in this case I will directly apply the formula
Amount received after 2 years = 10000(1 + 2.5/100)^4 = Rs. 11038 approx.

So sum it up
Case A - amount received after two years= Rs. 11000
Case B - amount received after two years= Rs. 11025
Case C - amount received after two years= Rs. 11038

Case C is giving the maximum return and rightly so because in Case (C) principal is increasing every 6 months.

Important formulas for Compund Interest - 

(2) A sum of money becomes x times in T years. In how many years will it become y times?

The approach to solve such questions is different for SI and CI

For SI : Formula = [(y - 1)/(x - 1)] * T
Q. 1) A sum of money becomes three times in 5 years. In how many years will the same sum become 6 times at the same rate of simple interest?
Solution : [(6 - 1)/(3 - 1)] * 5 = 5/2 * 5 = 12.5 years
Answer : 12.5 years

For CI : Formula = (logy/logx) * T
Now dont worry, I wont be asking you to study logarithms :)
But just remember one property of logs and that is enough to solve the questions
log(x ^y) = y.log(x)
Hence log(8) = log(2³) = 3.log(2)
Q. 2) A sum of money kept at compound interest becomes three times in 3 years. In how many years will it be 9 times itself?
Solution : (log9/log3) * 3         ... (1)
log9 = log(3²) = 2.log(3)
Put this value in (1)
= 2.log(3)/log(3) * 3
= 2 * 3 = 6 years
Answer : 6 years

(3) Interest for a number of days

Q. 3)

Here P = 306.25
R = 15/4 %
T = Number of days/365
Number of days = Count  the days from March 3rd to July 27th but omit the first day, i.e., 3rd March
   = 28 days(March) + 30 days(April) + 31 days(May) + 30 days(June) + 27 days(July)
   = 146 days
We know SI = (P * r * t)/100

Answer : Rs. 4.59

(4) Annual Instalments
This is the most dreaded topic of CI-SI. Before giving you the direct formula, I would like to tell you what actually is the concept of annual instalments(if you only want the formula and not the explanation, you can skip this part. But I would like you to read it)
Suppose you want to purchase an iPhone and its price is Rs. 100000 but you dont have Rs. 1 lakh as of now. What would you do? You have two options - either you can sell your kidney (which most the iphone buyers do :D), or you can go for instalments. But if you want to buy the iPhone through this instalment route, the seller will incur a loss. How? Had you paid Rs. 1 lakh in one go, the seller would have kept that money in his savings account and earned some interest on it. But you will pay this Rs. 1 lakh in instalments and that means the seller will get his Rs. 1 lakh after several years. So the seller is incurring a loss. The seller will compensate for this loss and will charge interest from you.

Let the annual instalment be Rs. x. and you pay it for 4 years.

After 1 year you will pay Rs. x and the seller will immediately put this money in his savings account (or somewhere else) to earn interest. He will earn interest on this Rs. x at the rate of r% for 3 years (because the total duration is 4 years and 1 year has already passed)

Hence the amount which the seller will get from this Rs. x instalment = x(1 + r/100)^3

After 2nd year, you will again pay Rs. x and the seller will earn interest on this Rs. x for 2 years.

The amount which the seller will get from this Rs. x instalment = x(1 + r/100)^2

After 3rd year, you will again pay Rs. x and the seller will earn interest on this Rs. x for 1 year.

The amount which the seller will get from this Rs. x instalment = x(1 + r/100)^1

After 4th year, you will pay Rs. x and your debt would be paid in full (no interest on this Rs. x)

The amount which the seller will get from this Rs. x instalment = x

Now let's add all the above four amounts to get the total amount the seller would get from all the instalments =
x(1 + r/100)^3 + x(1 + r/100)^2 + x(1 + r/100)^1 + x     ... (1)

Now, had you paid Rs. 1 lakh in one go (without going for the instalment route), then the amount received by the seller after 4 years would have been = 100000(1+r/100)^4      ... (2)

Now (1) should be equal to (2) because only then the two routes (instalment route and direct payment route) will give the same return and seller would have no problem in giving you the iPhone in instalments.

100000(1+r/100)^4 = x(1 + r/100)^3 + x(1 + r/100)^2 + x(1 + r/100)^1 + x  
[Remember the above equation for solving questions of compound interest]

P + P*r*4/100 = (x + x*r*3/100) + (x + x*r*2/100) + (x + x*r*1/100) + x  
[Remember the above equation for solving questions of simple interest]

Although for Simple Interest, we have a direct formula-
The annual instalment value is given by-

Now coming to the questions. There are two types of questions and they are bit confusing. In one type, the Amount is given and in another type, Principal is given

Type 1(Amount is given):

Q. 4) What annual installment will discharge a debt of Rs.6450 due in 4 years at 5% simple interest?
When the language the question is like "what annual payment will discharge a debt of ...", it means the Amount is given in the question.
In this question, the Amount(A) is given, i.e., Rs. 6450. So we can apply the formula directly
Here A = 6450, r = 5%, t = 4 years
Solution : 100*6450/[100*4 + 5*4*3/4]
Answer : 1500

Type 2 (Principal is given) :

Q. 5) A sum of Rs. 6450 is borrowed at 5% simple interest and is paid back in 4 equal annual installments. What is amount of each installment?

Here the sum is given. Sum means Principal.
But our formula requires Amount(A)
So we will calculate Amount from this Principal
A = P + SI = 6450 + 6450*5*4/100 = Rs. 7740

Now put the values in the formula
A = 7740, r = 5%, t = 4
Annual instalment = 100*7740/(100*4 + 5*4*3/2)

Answer : Rs. 1800

Q. 6) 

"Sum borrowed" means Principal.
This question is of Compound Interest and hence we cant apply the direct formula. We will solve this question with the help of the equation we derived earlier.

P(1+r/100)^2 = x(1+r/100) + x

P(1 + 5/100)^2 = 17640(1 + 5/100) + 17640

Solve for P, you will get P = Rs.32800
Answer : (B)

Q. 7) What annual instalment will discharge a loan of Rs. 66000, due in 3 years at 10% Compound Interest?
Solution : Here again the question is of "Compound Interest" and hence we will solve it by equation :
Let each annual instalment be of Rs. x. Note that in this question, amount is given
Amount = x(1 + 10/100)^2 + x(1 + 10/100)^1 + x
66000 = x (1.21 + 1.1 + 1)
So x = Rs. 19939.58

Q. 8) What annual instalment will discharge a loan of Rs. 66000, due in 3 years at 10% Simple Interest?
I have just converted Q.7 into Simple Interest
Now we can either solve it by direct formula, or by equation
By Equation method :
66000 = (x + x*10*2/100) + (x + x*10*1/100) + x
66000 = x(3 + 0.2 + 0.1)
x = Rs. 20000
By Direct formula method :
A = 66000, t = 3, r = 10%
x = 100A/[100t + t(t-1)r/2]
x = 100*66000/[100*3 + 3*2*10/2]
x = 6600000/(300 + 30)
x = Rs. 20000
Don't forget to read Part - 2

•   Simple and Compound Interest Part-2

If you have any doubt Please Drop a comment...

Simple and Compound Interest Tricks - 2

Simple and Compound Interest Tricks - 2

 

Please Check Part-1 Before Reading Part-2

• Simple and Compound Interest Part- 1

In this post I will give some extremely important formulas that will save your time. Read this post carefully and note down all the formulas in a piece of paper for quick revision.

Q. 1) If a sum of money becomes 3 times itself in 20 years at simple interest. What is the rate of interest?
In such questions apply the direct formula-
Rate of interest = [100*(Multiple factor - 1)]/T
So R = 100*(3 - 1)/20
Answer : 10%
Note : With this formula you can find Rate if Time is given and Time if rate is given.

Q. 2) 

In such questions, just write this line :
1st part : 2nd part : 3rd part = 1/(100+T1 * r) : 1/(100+T2 * r) : 1/(100+T3 * r)
 = 1/(100+2 * 5) : 1/(100+3 * 5) : 1/(100+4*5)
 = 1/110 : 1/115 : 1/120
 = 23*24 : 22*24 : 23*22
Hence 1st part = (23*24)/ (23*24 + 22*24 + 23*22) * 2379
Answer : 828
Note : Surprisingly, such questions when asked mostly have this same data, i.e., R=5% and T1, T2, T3 = 2, 3, 4 years, respectively. Only the Principal is changed. So it would be wise if you can just mug this line :
1st part : 2nd part : 3rd part = 23*24 : 22*24 : 23*22
Based on the above line, you would be able to solve such questions in a jiffy. 
But note that it will only work if the question is on Simple Interest. Like the below question appeared in SSC CGL Tier 2-

Q. 4

Here the data is same. i.e., R=5% and T1, T2, T3 = 2, 3, 4 years, respectively. So we will write directly -
1st part : 2nd part : 3rd part = 23*24 : 22*24 : 23*22
A received = (23*24)/ (23*24 + 22*24 + 23*22) * 7930
Answer : Rs. 2760

Q. 5) If a certain sum of money P lent out for a certain time T amounts to P₁ at R₁% per annum and to P₂ at R₂% per annum, then

The above formula is for calculating the Time, if the question asks the rate, then just interchange the rate and time. Hence the formula will become

R = (P_{1 -} P_2)*100/P₂T_{1 -} P₁T₂

Apply the formula:
R = (650-600)*100/600*6 - 650*4
R = 5%

Alternative Method :

You can solve such questions quickly without mugging the above formula. How?

The sum amounts to Rs. 600 in 4 years and Rs. 650 in 6 years. This means the simple interest is Rs. 50 for 2 years (because the amount increased from Rs. 600 to Rs. 650 in 2 years)

So the SI for 4 years is Rs. 100 (we have seen earlier than SI is proportional. So if SI = 100 for 2 years, then SI = 150 for 3 years, SI = 250 for 5 years and so on)

Now SI = Rs. 100; P = 600-100 = Rs. 500; t = 4 years

R = 100*SI/(P*t) = 10000/2000

Answer: 5%

For CI, the formula is different

Difference between CI and SI

This topic is very important from examination point of view. Note the following things-

If t=1 year, then SI = CI

If t=2 years then difference between CI and SI can be given by two formulas-

 If t=3 years then difference between CI and SI can be given by two formulas-

In all the above formulas we have assumed that the interest is compounded annually

Let us solve some CGL questions

Q. 6

A = P(1+r/100)^t

Given, A=1.44P t = 2 years

1.44P = P(1 + r/100)^2

r = 20%

Answer : (D)

Q. 7

Here the interest is compounded half yearly, so the formulas we mugged earlier are of no use here. We will have to solve this question manually

SI = P*10*1.5/100 = 0.15P

CI = P(1 + 5/100)^3 - P = P(1.05^3 - 1)

Given CI - SI = 244

P(1.05^3 - 1) - 0.15P = 244
P = Rs. 32000

Answer : (C)

Q. 8

Time = 2 years

Hence apply the formula: Difference(D) = R*SI/200

CI - SI = R*SI/200

CI - SI = (12.5/200)*SI

510 = 1.0625*SI    [Since CI = Rs. 510]

SI = Rs. 480

Answer : (D)

       Q. 9

CI for 1st year = 10% of 1800 = Rs. 180

CI for 2nd years = 180 + 10% of 180 = Rs. 198

Total = 180+198 = Rs. 378

Hence time = 2 years

Or you can apply the formula

A = P(1+r/100)^t

Answer : (B)

Q. 10

2.5 = P*R*2/100 - P*r*2/100
2.5 = 10R - 10r
R - r = 0.25
Answer : (D)

       Q. 11

CI for 1st year = 5% of P = 0.05P

CI for 2nd year = 5% of P + 5% of (5% of P) = 0.05P + 0.0025P = 0.0525P

Total CI = 0.05P + 0.0525P = 0.1025P

Given, 0.1025P = 328

P = Rs. 3200

Answer : (C)

Note : You can solve this question by the formula A = P(1+r/100)^t as well

Q. 12

Note that in this question the CI for 2 years in not given, but the CI for the 2nd year is given.

CI for 2nd year = 10% of P + 10% of (10% of P) = 0.1P + 0.01P = 0.11P

Given, 0.11P = 132

P = Rs. 1200

Answer : (D)

Q. 13

Interest = Re. 1 per day = Rs. 365 for 1 year

SI = P*r*t/100

t=1, r=5%, SI = Rs. 365

So, P = 365*100/5 = Rs. 7300

Answer : (A)

Q. 14

We know 

Difference = P(r/100)^2(r/100 + 3)

P = Rs. 10000, r = 5%, t = 3 years

Hence D = Rs. 76.25

Answer : (C)

Q. 15

We know, R = [(y/x)^(1/T2 - T1) - 1]*100

 = [(1587/1200)^1/(3 - 1) - 1]*100

 = [(1587/1200)^1/2 - 1]*100

 = 3/20 * 100

 = 15%

Answer : (B)

So this is the end of CI and SI series. If you have any doubt in this topic, please drop a comment...

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- 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.