### How to solve questions of Algebra in SSC CGL (Pre & Mains)

KNOWLEDGE PHILIC PRESENTS.....

*Students preparing for SSC often face difficulties while solving the problems on*

**Algebra,Trigonometry and Number systems**. SSC is now putting stress on such streams of mathematics rather on conventional profit-loss, work and time etc.F*irstly,*

*I discuss Algebra and later take on Trigonometry and number system.*

*The low merit in mathematics section in recent exams conducted suggests that students are facing difficulties while solving such problems. They often loose hope when they come across such problems and could not find any approach to solve them.*

*But the fact is that such problems are only a bit tricky ones and need lesser time than those conventional problems. I am here to suggest how to approach these problems.*

Most of the questions are from these following formulas learn them

*1.*

*2.*

*3.*

*4.*

*5.*

*6.*

*7.*

*8.*

*when ,*

*( because RHS = 0)*

__NOTE THIS:__

__when ,__

__LHS can be equal to zero only when x = 0 , y = 0 and z = 0__

__because square of any number is always positive.__

*Now, lets solve most important questions with the view to crack tier-2 by knowledgePhilic.com*

**1. If , then the value of**

*How to approach this problems,*

*we need square of (a-2) and 1/(a-2)*

*direct square of both side will not give the result,*

*we need to get the desire result,*

*so,*

*and now square both sides,*

*or,*

**2. then the value of**

*Approach :*

Since,

*Since we have to get*

**which is in term of x^3 and Y^3**

**so, we have to take cube on both the sides**

or,

or,

or, (since

**)***so, (answer)*

**3. If then the value of is,**

*whenever you find cube of three variables and 3abc or 3xyz always think about the formula,*

*And,*

*when ,*

*then,*

*here*

*therefore,*

**4. If be a square number, then the value of 'x' is**

*Approach: knowledgePhilic.com*

*we know that,*

*1.*

*2.*

*The problem is,*

*we have to convert in a square, any sum is square of a number if we get it in the above two forms*

*here focus on 7954X7956 , they can become a square formula or ,*

*if, either you write, 7954 = 7956 - 2 or 7956 = 7954 + 2*

*lets take 7956 = 7954 + 2*

*then,*

*clearly if you put x = 1 , it will give a perfect square,*

*therefore , x = 1*

**5. If , then the value of**

*Approach*

*Here, no relation is apparent between and*

*start the working, why it is given or ?*

*or,*

*same variable put together,*

*or or can make square if we add 1,*

*write 3 = 1 + 1 + 1*

*then,*

*or,*

*since sum of the square of the numbers never be zero unless all are zero*

*therefore, , and*

*therefore,*

*and so, = 2.1-(-1)+(-1) = 2*

**6. If , then the value of**

*adjustment approach:*

*you have to find*

*and given,*

*you need x and x^2 , common denominator a should be on one side,*

*therefore,*

*or,*

*cross multiplication,*

*therefore,*

**7. If a,b,c are non-zero, a + 1/b =1 and b + 1/c =1, then the value of abc is:**

*Sol: ab + 1 = b and bc + 1 = c*

*Approach:*

*since we find the value of abc therefore multiply both side by c*

*since ab + 1 = b*

*abc + c = bc*

*or, abc + c = c - 1 ( since bc + 1 = c)*

*or, abc = -1*

**8. The Square root of is :**

Sol: To solve this type of problems, we have to find the number whose square is equal to

Sol: To solve this type of problems, we have to find the number whose square is equal to

*Approach:*

*we know that,*

*1.*

*2.*

*The problem is,*

*we have to convert it in a square. Any sum is square of a number if we get it in the above two forms*

here focus on , 35 is the product of two prime numbers , 35 = 7 X 5

here focus on , 35 is the product of two prime numbers , 35 = 7 X 5

or,

or,

, they can become a square formula ,

, they can become a square formula ,

*if, you write,*

**9. If a + b + c = 0, then the value of**

Sol:

Sol:

Approach:

Approach:

it is like an adjustment problem,

it is like an adjustment problem,

*simple approach,*

since a + b + c = 0 => a + b = -c, b + c = -a, c + a = -b

since a + b + c = 0 => a + b = -c, b + c = -a, c + a = -b

therefore, (-c/c + -a/a + -b/b)(a/(-a) + b/(-b) + c/(-c) )= -3 X -3 = 9

therefore, (-c/c + -a/a + -b/b)(a/(-a) + b/(-b) + c/(-c) )= -3 X -3 = 9

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