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

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.

(a

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

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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.

^{2}- bc)/(a^{2}+ bc) = 1/3 ... (1)
(b

^{2}- ca)/(b^{2}+ ca) = 1/3 ... (2)
(b

^{2}- ca)/(b^{2}- ca) = 1/3 ... (3)
From (1), we get a

^{2}= 2bc ... (4)
. Similarly from (2) and (3), we get b

^{2}= 2ca ... (5)
and c

^{2}= 2ab ... (6)
Put the values of a

^{2}, b^{2}and c^{2}from (4), (5) and (6) in the expression whose value we have to find...
You will get 2 as the answer

**Answer : (B)**

Therefore, (m – a

^{2})/(b^{2}+ c^{2}) = 1
or m = a

^{2}+ b^{2}+ c^{2}**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.

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.

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

Put x = 0

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 x

Put A = 1 in all the 4 options to check which option will give '1' as the output.^{3}and 1/ x^{3}= 1**Answer : (D)**

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

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

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

To buy the super-hit SSC Hack Book, follow the below link-

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Pls tell 27 kese aea qtn 122

ReplyDeletePut x = 2 in x^4 + 2x^3 -2x - 1

Delete= 16 + 2*8 - 2*2 - 1

= 32 - 5

= 27

Okok ... m mistk thnkyou:)

DeleteSir one doubt for me plz reply me sir I applied for ssc cgl sir but I done a mistake in giving my dob can I reapply once again sir a fresh application some of them are saying ssc will reject both applications plz tel me sir whether I can apply second tym a new one

ReplyDeleteSSC won't cancel your applications. You can fill the form 10 times if you want. Just do a new registration with correct DOB.

DeleteSir I have by mistaken filled from two regions will my candidature get rejected?

Delete@vishal...no problem...bus dono regions se exam dene mat chale jana :P

DeleteWhere is question no 122?

ReplyDeleteSir How do we identify that equation is symmetrical or not??

ReplyDeleteis there any way to identify these types of equations?

Hi Prashant, I applied for SSC. In educational qualification, I have given only the graduation details. Is it necessary to specify school? I forgot to give details about my schooling(10th and 12th). What should I do now?

ReplyDeleteIf you have received your registration no., then no issue...

DeleteYeah I got my registration number... Thanks a lot :)

DeleteWell thanks for making me feel like a dumbass 👍

ReplyDeleteIf you liked the blog, please don't forget to share it with others. After all sharing is caring :)

DeleteAwesomse hacks sir upload more plsssssssssssss

DeleteSir in question 11 option b also gives 3 as output

ReplyDeleteThanks for pointing it out. Although x^2 -1 is the correct answer, but I shud just delete that question :)

DeleteSir in question 11 option b also gives 3 as output

ReplyDeleteSir ques 12 me why you a=4 directly after trying with 1

ReplyDeleteBecause we have to deal with square roots. Hence taking perfect squares like 4, 9, etc. will make the calculations simpler.

DeleteThanks sir pls keep doing good work. Your blog is awesome :-)

ReplyDeleteWaiting for your next blog

ReplyDeleteSir u forgot to post the hacks for solving asymmetrical algebric ques....

ReplyDeleteplease update those as well....

and please some hacks related to geometry as well, as these days it is the most weighted section in SSC CGL

and it will b realy helpful for us if u can create and manage a whatsapp grp

I am asking a bit more frm you just becoz ur way of solving these questions have outclassed evry other blog out there in the jungles of internet.....

as i m personally following so many blog, i nvr found any blog so superb as ur blog is....

waiting for ur response regarding more tricks for Algebra, Trigo n geometry and also for whatsapp grp

thanks in advance

Thanks for the words...I will take up Geometry after Indo-Pak :)

DeleteI see only Q Nos. 1 , 10 ,12 and 13.bu6 in comments section i see guys having doubts on Q 11 ,i can't see it,neither i can see Q 2 to 8..

ReplyDeleteI messed up the numbering...You didn't miss anything :)

DeleteSIR INDO -PAK OVER.WE ARE WAITING FOR RAJ VS SSC.

DeleteYes sir i too cant see all questions

ReplyDeleteI messed up the numbering...You didn't miss anything :)

DeleteThank you for making this blog and sharing your knowledge. It is going to help us to crack cgl for sure. Personally, I am bit slow in QA section. I'll need these tricks to increase my speed. Your updates are highly awaited.

ReplyDeleteGood day

x^4-1/x^4 = 119 find x^3 -1/x^3

ReplyDeletea) 36 b) -36 c) +- 36 d) +-33

is there any short-cut to solve this type of sum.

I think it's x^4 + 1/x^4 = 119...right ?

Deletethankyou very much sir...

ReplyDeleteAwesome sir !

ReplyDeleteWaiting for ur nxt blog

In q.72 how can we assume x=0 directly ?? thank you in advance

ReplyDeletex is just a variable and that's the beauty of variables, you can assume anything as their value(in this context). Take x=1 and you will still get the value of the expression as zero...

DeleteCan i asked quetions without ssc cgl papers

ReplyDeletesir all of these tricks are soo gud pls share some more......

ReplyDeleteThis comment has been removed by the author.

ReplyDeletePut x=1, still you will get the same answer. 'x' is a variable, you can put anything as its value.

Deletethank you SIR .........if possible make some awesome tricks Videos & upload youtube. It will take less time than writing Blog.BTW your tricks/HACKS really awesome. Thank you.

ReplyDeleteSir...is there any short trick for chapters like ratios, average, partnership..if yes kindly update on these chapters.

ReplyDeletex+1/x=5, thn value of x^6+1/x^6 is ?

ReplyDeleteApply the formula which I have mentioned in this article-

Deletehttp://sschacks.blogspot.in/2016/03/algebra-tricks-3.html

This comment has been removed by the author.

ReplyDeletesir explain q.no 5

ReplyDeletethanks in advance.

updated the explanation...please read again...

DeleteSir Q.4 how did you solve taking x=1.25. I'm not getting the approach. Can you explain it lil bit more.

ReplyDeleteThank u sir

ReplyDeleteoh God!, how simple you made algebra... I wish I would have visited your blog earlier...a thousand of thanks for sharing this...... God Bless you.

ReplyDeleteSir any shortcuts for a/b+b/a=3 then a^3+b^3=?

ReplyDelete