**Please read Algebra Part - 1 and Algebra Part - 2, before reading this post.**

**Q. (1)**

**(A) 3 (B) 4 (C) 6 (D) 9**

Although this equation is symmetrical, and hence we can assume x = y = z to solve it. But you should know one more thing about such equations.

**If the sum of certain number of terms is zero, you can assume each term to be zero.**That means,
(4x - 3)/x = 0 or x = 3/4

(4y - 3)/y = 0 or y = 3/4

(4z - 3)/z = 0 or z = 3/4

So, 1/x + 1/y + 1/z = 4

**Answer: (B)**

**Q. (2) If x^2 = y + z, y^2 = z + x, z^2 = x + y, then find the value of**

**(A) 1 (B) 2 (C) 0 (D) -1**

Symmetrical equation, hence x = y = z

x^2 = x + x

x^2 = 2x

x = 2

Hence x = y = z = 2

Put in the expression

= 1/3 + 1/3 + 1/3

= 1

**Answer: (A)**

**Q. (3)**

This is a very famous question-type. In such questions we take everything on RHS to LHS and then try to make squares. You will get,

(x - 1)^2 + (y + 1)^2 + (z + 1)^2 = 0

And like I said before, if the sum of certain number of terms is zero, we can assume each term to be zero.

(x - 1)^2 = 0, (y + 1)^2 = 0, (z + 1)^2 = 0

Hence, x = 1, y = -1 and z = -1

Put these values in (2x - 3y + 4z)

= 2(1) -3(-1) + 4(-1)

= 1

**Answer: (D)**

**Q. (4)**

Take everything to LHS,

(x - 1)^2 + y^2 = 0

Hence, x = 1 and y = 0

Put these values in the expression

= (1)^3 + 0^5

= 1

**Answer: (D)**

**Q. (5)**

**(A) 5/12 (B) 12/5 (C) 5/7 (D) 7/5**

**By Componendo and Dividendo, whenever you see any equation written in the form**

**(m + n)/(m - n) = p**

**You can directly write m/n = (p + 1)/(p - 1)**

In this question

m

**=**√(3 + x), n = √(3 - x), p = 2
Hence, by Componendo-Dividendo

Squaring both sides

(3 + x)/(3 - x) = 9

Again apply componendo-dividendo

3/x = (9 + 1)/(9 - 1)

3/x = 5/4

x = 12/5

**Answer: (B)**

**Q. (6) If x = 332, y = 333, z = 335, then the value of x^3 + y^3 + z^3 - 3xyz is**

**(A) 10000 (B) 7000 (C) 9000 (D) 8000**

There is one more formula for a^3 + b^3 + c^3 - 3abc, apart from the one which you know

Hence,

x^3 + y^3 + z^3 - 3xyz = 1/2(332 + 333 + 335)[(332 - 333)^2 + (333 - 335)^2 + (335 - 332)^2]

= 1/2 (1000)[1 + 4 + 9]

= 7000

**Answer: (B)**

**Q. 7)**

**(A) -1 (B) 3abc (C) 1 (D) 0**

a + b + c = 0 is symmetrical

**Whenever any symmetrical equation is equal to zero, and the expression whose value is asked, is also symmetrical( and the numerator of the terms is also 1), then the value of that expression will also be zero.**

**Answer: (D)**

**Q. (8)**

**(A) -2 (B) -1/2 (C) 0 (D) 1/2**

Same logic

Value of the expression = 0

**Answer: (C)**

**Q. (9)**

**(A) 9 (B) 0 (C) 8 (D)**

We will solve it by assuming a=b=c. Hence

= (2 + 2 + 2)(1/2 + 1/2 + 1/2)

= 6 * 3/2

= 9

**Answer: (A)**

ab + bc + ca = 0 is symmetrical

Hence value of the expression = 0

**Answer: (B)**

**If you have any doubts in this article, please drop a comment...**

**As the exam date has been extended, I have updated the book-list for SSC, because now you have more time to practise :)**

**Keep reading :)**

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

This comment has been removed by the author.

ReplyDeletesir, ur blog is soooooo good.doing a wonderful job. thankxxxxx a lott....

ReplyDeletesir in question no 7 how we find the numerator of term is 1 can you please explain ...

ReplyDeletenumerator = 1 to dikh raha hai, isme kuch find thodi karna hai...

DeleteGreat article prashant !!

ReplyDeleteAlways enjoy reading your tricks !!

Keep up the good work !!

Eagerly waiting for more articles !!

This comment has been removed by a blog administrator.

ReplyDeleteSir, please thoda help kijiye. Maine sbi po, sbi it officer, lic aao exam diye hai par mai kisi mai bhi interview stage tak nahi pahuch paaya hoon. Mai cgl key through css key liye padhai kar raha hoon, mock tier 1 mei 115 tak score kar leta hoon. Kya mera css mei hone ka chance hai? 20L candidates appear honge..kaise manage karu pls bataiye.

ReplyDeleteStill 4 months are left. You can easily improve your score. 115 is a little low because competition has increased in the recent years. Try increasing it to 130...you can do it :)

DeleteSir... your blog is excellent.... not only the tips you give are extremely helpful... it also keeps me wondering these are so simple, why did i not think this way... It would be great help if you could post some hacks for english too.... Eagerly waiting for youe next post....

ReplyDeleteSir, why haven't we applied the concept used in Q7 in Q1. In Q1, all the three conditions are met. Though, we don't have 0 in option. But, suppose if we had then. Plz , explain.

ReplyDeleteI too have the same doubt.Please explain sir......

DeleteThat is the difference between tricks and theorems. Tricks are not universal. But don't worry, that doesn't mean what I have written in this article is wrong (at least for SSC, it has 100% applicability). It simply means that you will have to be a little observant. In the first question, we can easily see x=y=z=3/4 (they have designed the equation that way), so we don't need to apply any trick. But in rest of the questions, where a + b + c = 0 or ab + bc + ca = 0, you can't figure out the exact values of a,b,c and hence the trick will work. In other words, when you are given a+b+c=0, you can assume any values for a,b,c like a=-1, b=-1 and c=2 or a=-1, b=1 and c=0, and you will get the same answer(irrespective of the values) but when any such equation (like the equation in Q.1) is given, you can't assume the values on your own, because they have given you the values (a=b=c=3/4). Solve the previous year papers and you will be able to identity the questions easily.

Deleteye hui na baat.You truly are a genius sir.Sir,aapko MEA me kahaa posting mila?

DeleteMEA = Delhi :)

DeleteThanx sir, vey well explained.

DeleteI thought you were posted abroad:-)

Deletesir geometry ke part 2 me kuch question post kiye h mene... un 3 questions ko explain kr do sir...please

ReplyDeleteI have posted the solutions

Deletehello sir, pl. help me out with dz ques....

ReplyDeleteif a+b-c=14 then find value of 2b^2c^2+2c^2a^2+2a^2b^2-a^4-b^4-c^4???....

ny shrt trick fo dz????

pl. rpl.

this question appeared in CGL and it is wrong...

DeleteSo don't bother :)

okay...

ReplyDeletepractise paper vallo ko mention krna chaiye buk mey nd ans. b diya h. unhone toh anyways thank you......

...:)

Sir ..is there any news about cgl paper's pattern that's why we proceed preparation accordingly ....pls sir share as as soon as possible

ReplyDeleteNic sir ..kp uploading thnkew :)

ReplyDelete_/\_ _/\_

ReplyDeletemy Doubt...Thanks.:)

ReplyDeleteThank you so much sir. Your tricks are so useful and giving me so much confidence.

ReplyDeleteTime and again we are curiously opening your blog for new concepts.....and updates.that much curiosity u inculcated in us as like facebook..WhatsApp! A big thanks to u sir.....

ReplyDeleteDaily I check yr blog for new update/post.. but from last 5 day's no new post

ReplyDeleteHey Prashant. I wanna know that the short tricks you've posted here, are these present in Kiran's quickest mathematics?

ReplyDeleteI don't think so...

Deletebhai,ye sub tricks Prashant raj sir ke tricks hai.Koi book me nahi milega.Sir apni mehnat se tricks nilkati hai.We should also try to learn these types of tricks.It can be learnt only if we constantly practise questions with different approaches.

Deletesir ssc ki sbhi post pe govt quarter milta h kya?

ReplyDeletegood morning sir.sir, can u please provide video tutorial for making of sample power point presentation,and excel.

ReplyDeleteSir please write about pipes and cistern, facing difficulties in this.

ReplyDeleteSir please write about pipes and cistern, facing difficulties in this.

ReplyDeleteCheck my article on Time and Work

Delete(b-c)/a +(a+c)/b +(a-b)/c =1

ReplyDeleteIf a-b+c # 0

1) 1/a=1/b+1/c 2) 1/b=1/a-1/c

3) 1/b= 1/a+1/c 4) 1/c= 1/a+1/b

Please solve it sir i could not get its ans

Sir...in q.9 assuming a=b=c wouldn't actually fetch the given condn a+b+c=0...den how can we assume dat??if we put a=b=c in the given condn it will make a=b=c=0 and so everything ll be incongruent...kindly make it clear

ReplyDeleteI have same doubt.Sir, Plz reply.

Deletesame doubt.

Deletethough the solution coming is correct, so this must be some trick, will this work in all such qustions ?

Hello ..Can you please solve this : (a-b)^2/(b-c)(c-a) + (b-c)^2/(c-a)(a-b) + (c-a)^2/(a-b)(b-c)

ReplyDeleteGood evening sir. Amazing blogs you share :). I wanted to know where is algebra hack-3 ?? Its 4 directly. And how can i get your updates and complete access?

ReplyDeleteDear prashant,

ReplyDeleteAll these maths are are created by you only or they are taken from a book. if it is so, kindly provide the book name pls. thnks in advance

Sir...if X+1/X=1

ReplyDeleteThen. (X+1)^5 + 1/(X+1)^5=???

sir algebra-4 ka 3rd question clear nhi hua..agr RHS ki terms LHS p le jayge to -3 bhi to h..to (x-1)^2=0 kese hua???

ReplyDeleteif (a^3)b=abc=180,a,b,c are positive integers,then find value of c??

ReplyDeletea)110

b)1

c)4

d)25

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ReplyDeleteSir in Question 2 of algebra tricks -4

ReplyDeleteIf x^2 = y + z, y^2 = z + x, z^2 = x + y, then find the value of

1/(1+x) + 1/(1+y) + 1/(1+z)= ?

even the expression is symmetrical and numerator is 1

Answer in not 0 ....Please sir Help me out

And How to recognize this correctly...???

Sir in Question 2 of algebra tricks -4

ReplyDeleteIf x^2 = y + z, y^2 = z + x, z^2 = x + y, then find the value of

1/(1+x) + 1/(1+y) + 1/(1+z)= ?

even the expression is symmetrical and numerator is 1

Answer in not 0 ....Please sir Help me out

same query how it is different from Q7

IF a^2+b^2+c^2=20 and a+b+c=0 then ab+bc+ac=?

ReplyDeleteSir how to solve this

This comment has been removed by the author.

ReplyDelete