Monday, 5 October 2015

Filled Under:

How to solve questions of Algebra

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.Firstly, 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.{(x + y)^2} = {x^2} + {y^2} + 2xy

        
2.{(x - y)^2} = {x^2} + {y^2} - 2xy

3.{x^2} - {y^2} = (x + y)(x - y)


4.{(x + y)^3} = {x^3} + {y^3} + 3xy(x + y)


5.{(x - y)^3} = {x^3} - {y^3} - 3xy(x - y)


6.{x^3} + {y^3} = (x + y)({x^2} + {y^2} - xy)


7.{x^3} - {y^3} = (x - y)({x^2} + {y^2} + xy)


8.{x^3} + {y^3} + {z^3} - 3xyz = (x + y + z)({x^2} + {y^2} + {z^2} - xy - yz - zx)


when x + y + z = 0,


{x^3} + {y^3} + {z^3} - 3xyz = 0 ( because RHS = 0)


{x^3} + {y^3} + {z^3} = 3xyz



NOTE THIS:


 when , {x^2} + {y^2} + {z^2} = 0


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





1. If a + \frac{1}{{a - 2}} = 4, then the value of {(a - 2)^2} + \frac{1}{{{{(a - 2)}^2}}}


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 {(a - 2)^{}} + \frac{1}{{{{(a - 2)}^{}}}} to get the desire result,



a + \frac{1}{{a - 2}} = 4


so, {(a - 2)^{}} + \frac{1}{{{{(a - 2)}^{}}}} = 2


and now square both sides,

{(a - 2)^2} + \frac{1}{{{{(a - 2)}^2}}} + 2 = 4

or, {(a - 2)^2} + \frac{1}{{{{(a - 2)}^2}}} = 2





2. xy(x + y) = 1 then the value of \frac{1}{{{x^3}{y^3}}} - {x^3} - {y^3}


Approach :


Since,

\begin{array}{l}  xy(x + y) = 1\\  xy = \frac{1}{{(x + y)}}  \end{array}

Since we have to get \frac{1}{{{x^3}{y^3}}} - {x^3} - {y^3} which is in term of x^3 and Y^3
so, we have to take cube on both the sides

 \Rightarrow {(x + y)^3} = \frac{1}{{{x^3}{y^3}}}
or, {x^3} + {y^3} + 3xy(x + y) = \frac{1}{{{x^3}{y^3}}}

or, 3xy(x + y) = \frac{1}{{{x^3}{y^3}}} - {x^3} - {y^3}

or,  3 \times 1 = \frac{1}{{{x^3}{y^3}}} - {x^3} - {y^3} (since xy(x + y) = 1)

so, \frac{1}{{{x^3}{y^3}}} - {x^3} - {y^3} = 3 (answer)



3. If {a^3} - {b^3} - {c^3} = 0 then the value of {a^9} - {b^9} - {c^9} - 3{a^3}{b^3}{c^3} is,


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


{x^3} + {y^3} + {z^3} - 3xyz = (x + y + z)({x^2} + {y^2} + {z^2} - xy - yz - zx)


And,



when x + y + z = 0,


then, {x^3} + {y^3} + {z^3} - 3xyz = 0




here x = {a^3},y =  - {b^3},z =  - {c^3}


therefore, {a^9} - {b^9} - {c^9} - 3{a^3}{b^3}{c^3} = 0





4. If  (x + 7954 \times 7956) be a square number, then the value of 'x' is


Approach:


we know that,



1.{(x + y)^2} = {x^2} + {y^2} + 2xy

      
2.{(x - y)^2} = {x^2} + {y^2} - 2xy

The problem is,


(x + 7954 \times 7956)


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 {x^2} + {y^2} + 2xy or  ,{x^2} + {y^2} - 2xy

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

lets take 7956 = 7954 + 2


then, (x + 7954 \times 7956) = x + 7954 \times (7954 + 2)


 = x + {7954^2} + 2.1.7954


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


\begin{array}{l}   = 1 + {7954^2} + 2.1.7954\\   = {(1 + 7954)^2}  \end{array}

 therefore , x = 1




5. If {a^2} + {b^2} + {c^2} + 3 = 2(a - b - c), then the value of 2a - b + c


Approach


Here, no relation is apparent between {a^2} + {b^2} + {c^2} + 3 = 2(a - b - c) and 2a - b + c


start the working, why it is given 2(a - b - c) or 2a - 2b - 2c ?


{a^2} + {b^2} + {c^2} + 3 = 2(a - b - c)


or, {a^2} + {b^2} + {c^2} + 3 - 2a + 2b + 2c = 0


same variable put together,


{a^2} - 2a + {b^2} + 2b + {c^2} + 2c + 3 = 0


{a^2} - 2a or {b^2} + 2bor {c^2} + 2c can make square if we add 1,


write 3 = 1 + 1 + 1


then, {a^2} + 1 - 2a + {b^2} + 1 + 2b + {c^2} + 1 + 2c = 0


or, {(a - 1)^2} + {(b + 1)^2} + {(c + 1)^2} = 0


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


therefore, a - 1 = 0b + 1 = 0 and c + 1 = 0


therefore,

a = 1,b =  - 1,c =  - 1

and so, 2a - b + c = 2.1-(-1)+(-1) = 2




6. If \frac{x}{a} = \frac{1}{a} - \frac{1}{x} , then the value of x - {x^2}


 adjustment approach:


you have to find  x - {x^2}


and given,


\frac{x}{a} = \frac{1}{a} - \frac{1}{x}


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


therefore,



\frac{x}{a} - \frac{1}{a} =  - \frac{1}{x}

or, \frac{{x - 1}}{a} =  - \frac{1}{x}

cross multiplication,


{x^2} - x =  - a

therefore, x - {x^2} = a



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 33 - 4\sqrt {35}  is :

Sol: To solve this type of problems, we have to find the number whose square is equal to  33 - 4\sqrt {35}



Approach:

we know that,



1.{(x + y)^2} = {x^2} + {y^2} + 2xy

      
2.{(x - y)^2} = {x^2} + {y^2} - 2xy

The problem is,


33 - 4\sqrt {35}


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 
\sqrt {35}   , 35 is the product of two prime numbers , 35 = 7 X 5

\sqrt {35}  = \sqrt 7 \sqrt 5


or, 4\sqrt {35}  = 2 \times 2\sqrt 7 \sqrt 5


 , they can become a square formula  ,{x^2} + {y^2} - 2xy

if, you write, 


\begin{array}{l}  {x^2} = 33 - 4\sqrt {35} \\  {x^2} = 28 + 5 - 4\sqrt {35}  = {(2\sqrt 7 )^2} + {(\sqrt 5 )^2} - 2.2\sqrt 7 \sqrt 5   \end{array}


{x^2} = {(2\sqrt 7  - \sqrt 5 )^2}

x =  \pm \sqrt {2\sqrt 7  - \sqrt 5 }



9. If a + b + c = 0, then the value of\left( {\frac{{a + b}}{c} + \frac{{b + c}}{a} + \frac{{c + a}}{b}} \right)\left( {\frac{a}{{b + c}} + \frac{b}{{c + a}} + \frac{c}{{a + b}}} \right)


Sol: 


Approach:


it is like an adjustment problem,

simple approach,

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






10. {a^2} + 1 = a, then the value of {a^{12}} + {a^6} + 1 is:


Sol: 

Approach,

its a tricky one,


if you take square of this, you cant find desire result


{a^2} + 1 = a


then what to do,


see the question, you have to find the value of  {a^{12}} + {a^6} + 1


we know that, {x^3} - {y^3} = (x - y)({x^2} + {y^2} + xy)


Pay attention, {a^2} + 1 - a = 0 can be converted to {a^3} + 1 ,


\begin{array}{l}  {a^2} + 1 = a\\   \Rightarrow {a^2} + 1 - a = 0\\  (a + 1)({a^2} + 1 - a) = 0\\  {a^3} + 1 = 0\\  {a^3} =  - 1  \end{array}


{a^{12}} + {a^6} + 1 = {({a^3})^4} + {({a^3})^2} + 1 = {( - 1)^4} + {( - 1)^2} + 1 = 3





11. If p - 2q = 4 then the value of {p^3} - 8{q^3} - 24pq - 64 is :


since we have to find the value of {p^3} - 8{q^3} - 24pq - 64 and it does not contain the multiple of p^2q and pq^2 , so we cant get the value of {p^3} - 8{q^3} - 24pq - 64 by taking cube of p - 2q = 4


Approach: Adjustment problem i.e. to use the given value and find the value of other



So we take,{p^3} - 8{q^3} - 24pq - 64 and use p - 2q = 4 to find the value,

we know that, {x^3} - {y^3} = (x - y)({x^2} + {y^2} + xy)

therefore,
{p^3} - 8{q^3} - 24pq - 64 = (p - 2q)({p^2} + 4{q^2} + 2pq) - 24pq - 64


since we know the value of p - 2q = 4, so we can know the value of {p^2} + 4{q^2} 
taking square of the both the side,
\begin{array}{l}  {p^2} + 4{q^2} - 4pq = 16\\  {p^2} + 4{q^2} = 16 + 4pq  \end{array}

put the values of p - 2q and {p^2} + 4{q^2} on above equation,

{p^3} - 8{q^3} - 24pq - 64 = 4(16 + 4pq + 2pq) - 24pq - 64

\begin{array}{l}   = 64 + 24pq - 24pq - 64\\   = 0  \end{array}




12.If the value of \frac{x}{{{x^2} - 2x + 1}} = \frac{1}{3} then what is the value of {x^3} + \frac{1}{{{x^3}}}

Approach: Adjustment problem i.e. to use the given value and find the value of other


since,\frac{x}{{{x^2} - 2x + 1}} = \frac{1}{3}

or, 3x = {x^2} - 2x + 1

or, {x^2} + 1 = 5x  ..........................(i)

we know that
{x^3} + {y^3} = (x + y)({x^2} + {y^2} - xy),

consider, 
{x^3} + \frac{1}{{{x^3}}} = \frac{{{x^6} + 1}}{{{x^3}}} = \frac{{{{({x^2})}^3} + 1}}{{{x^3}}}

or, {x^3} + \frac{1}{{{x^3}}} = \frac{{({x^2} + 1)({x^4} + 1 - {x^2})}}{{{x^3}}}


since, {x^2} + 1 = 5x

squaring both sides,

\begin{array}{l}  {x^4} + 1 + 2{x^2} = 25{x^2}\\  {x^4} + 1 = 23{x^2}  \end{array}

putting the value of x^2 + 1 and x^4 + 1

{x^3} + \frac{1}{{{x^3}}} = \frac{{({x^2} + 1)({x^4} + 1 - {x^2})}}{{{x^3}}} = \frac{{(5x)(23{x^2} - {x^2})}}{{{x^3}}}

 = \frac{{5x.22{x^2}}}{{{x^3}}}

= 110



13. If x + \frac{1}{x} = 4, then find the value of {x^4} + \frac{1}{{{x^4}}} ?

Approach:

Remember, square of x + \frac{1}{x}  has no coefficient of x

Thus, take square of  both side of x + \frac{1}{x} = 4

 {(x + \frac{1}{x})^2} = {4^2}

{x^2} + \frac{1}{{{x^2}}} + 2 = 16

or, {x^2} + \frac{1}{{{x^2}}} = 14

Again take square of both the sides,

{x^4} + \frac{1}{{{x^4}}} + 2 = 196

thus, {x^4} + \frac{1}{{{x^4}}} = 194



14. If value of a = xb and b  = ay then find the value of  \frac{1}{{x + 1}} + \frac{1}{{y + 1}}

Approach:

Its a simple question put x = a/b and y = b/a

\frac{1}{{x + 1}} + \frac{1}{{y + 1}} = \frac{1}{{a/b + 1}} + \frac{1}{{b/a + 1}}


 = \frac{b}{{a + b}} + \frac{a}{{b + a}}
 = \frac{{a + b}}{{a + b}}

= 1




15. If x , y are real and {x^2} + {y^2} + 2 = 2(y - x) then find the value of x

Approach: 

here is only one equation so we have to solve the equation to get the value of x

{x^2} + {y^2} + 2 = 2(y - x)

{x^2} + {y^2} + 2 - 2y + 2x = 0

same variable put together,

{x^2} + 2x + {y^2} - 2y + 2 = 0

{x^2} + 2x and  {y^2} - 2y can make square if we add 1,


{x^2} + 2x + {y^2} - 2y + 1 + 1 = 0

{x^2} + 2x + 1 + {y^2} - 2y + 1 = 0

or, {(x + 1)^2} + {(y - 1)^2} = 0

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

therefore, 
x + 1 = 0 and y - 1 = 0

or, x = -1




16. If x \ne 0,y \ne 0,z \ne 0 and \frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} + \frac{1}{{{z^2}}} = \frac{1}{{xy}} + \frac{1}{{yz}} + \frac{1}{{zx}} then find the relation among x, y and z


Approach:

\frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} + \frac{1}{{{z^2}}} = \frac{1}{{xy}} + \frac{1}{{yz}} + \frac{1}{{zx}}


options are a) 1/x +1/y +1/z = 0  b) x = y = z  c) x + y = z

clearly if you put x = y = z = a 
both sides will become equal 
therefore option b) is correct 


How to solve ?


make the square formula by multiplication by 2 on both sides,

\frac{2}{{{x^2}}} + \frac{2}{{{y^2}}} + \frac{2}{{{z^2}}} = \frac{2}{{xy}} + \frac{2}{{yz}} + \frac{2}{{zx}}

or, \frac{1}{{{x^2}}} + \frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} + \frac{1}{{{y^2}}} + \frac{1}{{{z^2}}} + \frac{1}{{{z^2}}} = \frac{2}{{xy}} + \frac{2}{{yz}} + \frac{2}{{zx}}

or, \frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} - \frac{2}{{xy}} + \frac{1}{{{y^2}}} + \frac{1}{{{z^2}}} - \frac{2}{{yz}} + \frac{1}{{{x^2}}} + \frac{1}{{{z^2}}} - \frac{2}{{zx}} = 0

or, {\left( {\frac{1}{x} - \frac{1}{y}} \right)^2} + {\left( {\frac{1}{y} - \frac{1}{z}} \right)^2} + {\left( {\frac{1}{z} - \frac{1}{x}} \right)^2} = 0

since sum of the square of the numbers never be zero unless all  brackets are individually zero,

therefore, 
{\frac{1}{x} - \frac{1}{y} = 0} => y = x, {\frac{1}{y} - \frac{1}{z} = 0} => y = z , {\frac{1}{z} - \frac{1}{x} = 0} => z = x

therefore x = y = z

0 comments:

DOWNLOAD EBOOKS