#dailyquizadda #Algebra #tricks #33Question #AdvanceMaths
Algebra is the easiest topic for SSC because you don't have to memorize any formula for it and all the questions can be solved within 10 seconds with jugaad.
First let me share with you the concept of symmetrical expressions(as I call it). A symmetrical expression is the one in which the weight of all the variables (a, b, c, etc.) is equal. Examples will make things clear.
Examples of symmetrical expressions -
In this question we have to find the value of x.
Here the equation is completely symmetrical, hence we assume a = b = c
Put b=a, c=a (so that the whole equation is in terms of 'a')
Now LHS becomes 3(x - a2)/2a
RHS = 12a
Solving this, you will get, x = 9a2
From here we get that the value of x is 9a2
Now put a = b = c in all the 4 options and check which option gives you the value 9a2
A) 9a2
B) 3a2
C) 3a2
D) 0
Answer : (A)
bc + ab + ca = abc is symmetrical and hence we can assume a=b=c
Now put b=a and c=a in this equation. We will get -
3a2 = a3
So a = 3
Now put a=b=c=3 in the expression whose value we have to find. You will get the answer as 1.
Answer : (B)
Examples :
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.
(a2
- bc)/(a2 + bc) = 1/3 ... (1)
Please note that this hack is also applicable only for symmetrical equations.
x = √5 + 2 = 4.23
(x^4 - 1)x^2 = x^2 - 1/x^2
Neglect 1/x^2
x^2 = 4.23^2 = 17 (approx)
Answer : (A)
Again if you dont know how to solve the above question, then observe the above equation
If a=1, then LHS = 5.33 (which is little more than RHS, i.e., 5). We need to decrease the value of 'a'.
Hence let's take a = 0.9
5a + 1/3a = 4.8
Now LHS is more than RHS. We need to increase the value of 'a' slightly
So lets lock the final value a = 0.95 (Now no need to check the value of LHS for a=0.95)
9a^2 + 1/25a^2
Neglect 1/25a^2
9a^2 = 8 (approx)
Answer : (D)
You know that 19^2 = 361
Hence √3.73 = 1.9 (approx)
3x - 1/4y = 6
Put x=1 and solve the equation for y
y = -1/12
Put x = 1 and y=-1/12 in the expression (4x - 1/3y)
You will get 8
Answer : (D)
Here again we can say that the approx value of x is 8, because 8^2 = 64
Put x = 8 in the expression
= (64 - 1 + 16)/8
= 10 (approx)
Answer : (A)
Q. 22
Put x=2, the LHS becomes 6 and it is little more than RHS. So we need to decrease its value slightly
Let us take x = 1.8
LHS = (1.8)^2 + 1.8 = 5(approx)
LHS is almost equal to RHS, hence x=1.8 is a perfect value
Neglect 1/(x + 3)^3.
Now we only need to find the value of (x + 3)^3
(x + 3)^3 = (1.8 + 3)^3 = 110 (approx)
Answer : (A)
Algebra is the easiest topic for SSC because you don't have to memorize any formula for it and all the questions can be solved within 10 seconds with jugaad.
First let me share with you the concept of symmetrical expressions(as I call it). A symmetrical expression is the one in which the weight of all the variables (a, b, c, etc.) is equal. Examples will make things clear.
Examples of symmetrical expressions -
- a3 + b3 + c3
- 3a + 3b + 3c
- a2 + b2 + c2
- a + b + c
- ab + bc + ca
Examples of non - symmetrical expressions -
- a - b + c
- 2a + 3b + 3c
- a3 + b2 + c3
- a + b + c2
Hack - 1 : "Whenever you encounter a symmetrical equation in any question, you can safely assume : a = b = c (even if it is not given in the question)"
Let's solve previous year questions -
Q . 1.
Here you can see that the LHS as well as the RHS of the equation is symmetrical, hence a = b = c
Answer : (A)
Q .2. 
We put a = b = c, hence (a+c)/b becomes (a+a)/a, or 2
Answer : (B)
Q . 3.
Here the equation is completely symmetrical, hence we assume a = b = c
Put b=a, c=a (so that the whole equation is in terms of 'a')
Now LHS becomes 3(x - a2)/2a
RHS = 12a
Solving this, you will get, x = 9a2
From here we get that the value of x is 9a2
Now put a = b = c in all the 4 options and check which option gives you the value 9a2
A) 9a2
B) 3a2
C) 3a2
D) 0
Answer : (A)
Q . 4. 
bc + ab + ca = abc is symmetrical and hence we can assume a=b=c
Now put b=a and c=a in this equation. We will get -
3a2 = a3
So a = 3
Now put a=b=c=3 in the expression whose value we have to find. You will get the answer as 1.
Answer : (B)
Hack - 2 : "When only a single equation is given and based on that you have to find the value of an expression, you can assume the value of variables yourself. But make sure to assume only such values that will not make the denominator zero"
Examples :
Q . 5
In this question, only a single equation is given, i.e., x + y + z = 0,
and based on this equation we have to find the value of an expression
We can assume x = -1, y = 1 and z = 0 (such that x + y + z = 0)
Now on putting these values in the expression, we get the answer as 2
Answer : (D)
Q . 6. 
a + b = 1
Let's assume a = 1 and b = 0
Put the values in the expression, and you will get 0.
Answer : (A)
Q . 7.
In this question the values of x and y both depend on a constant 'a'. We
can assume any value for 'a' and this will give the values of x and y.
Let us assume a=1
This will give x = 2 and y = 0
Put these values in the expression and you will get the answer as 16.
Answer : (A)
Q . 8
Pick values for a, b and c, such that their sum is 2s. Let us assume a =
2s, b = s and c = -s (here you should not assume a,b or c to be zero
because that will make the elimination of options difficult)
Put these values in the expression and you will get 2s2
Now check all the four options to see which of them will give the value 2s2 on putting a=2s, b=s and c=-s
Answer : (C)
Sometimes the equations are complex and you will find it difficult to assume values for the variables. Example -
Q . 9
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.
(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. 10 
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)
Now let us see some other questions where you can assume the values.
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
Q. 11
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 . 12
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. 13.
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)
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
Q. 14. (a2 + 2a)2 + 12(a2 + 2a) – 45 can be expressed as:
(a) (a - 1)(a - 3)(a2 + 2a + 15)
(b) (a - 1)(a + 3)(a2 + 2a + 15)
(c) (a + 1)(a + 3)(a2 + 2a + 15)
(d) (a + 1)(a - 3)(a2 + 2a + 15)
(a) (a - 1)(a - 3)(a2 + 2a + 15)
(b) (a - 1)(a + 3)(a2 + 2a + 15)
(c) (a + 1)(a + 3)(a2 + 2a + 15)
(d) (a + 1)(a - 3)(a2 + 2a + 15)
Put a = 0
Hence, (a2 + 2a)2 + 12(a2 + 2a) – 45 = -45
Now put a = 0 in all the four options and check which one is giving -45 as output
(a) 45
(b) -45
(c) 45
(d) -45
Hence (b) and (d), two options are possible
Now put a = 1
(a2 + 2a)2 + 12(a2 + 2a) – 45 = 0
Put a = 1 in options (b) and (d) to see which one will give zero as the output
Answer: (b)
Hence, (a2 + 2a)2 + 12(a2 + 2a) – 45 = -45
Now put a = 0 in all the four options and check which one is giving -45 as output
(a) 45
(b) -45
(c) 45
(d) -45
Hence (b) and (d), two options are possible
Now put a = 1
(a2 + 2a)2 + 12(a2 + 2a) – 45 = 0
Put a = 1 in options (b) and (d) to see which one will give zero as the output
Answer: (b)
Approximation in Algebra:
Approximation is a very important tool that can help you solve some
complex and time taking questions. I will solve the below questions from
SSC CGL with approximation technique to give you an idea of how it
works. But before that, some basic rules of approximation:
1. Establish a limit within which the variable is falling.
2. Neglect the smaller terms of the expression (fractions with Denominator>Numerator)
3. Please use this technique only when the options have a significant difference between them. E.g. If in a question the 4 options are A. 4, B. 5, C. 6, D. 7, you can't use the approximation technique because the options are fairly close.
These rules will make sense once you go through the below CGL questions-
Now how will you approach this question if you dont know how to solve it?
Given, x^4 + 1/x^4 = 119
We can safely assume that 3<x<4 because 3^4 = 81 and 4^4 = 256 (119 lies between 81 and 256). Moreover x will be closer to 3 as 119 is more close to 81 than 256
We have established the limit of the variable.
Let us take our first value. Go with x = 3.2
3.2^4 = 104 (approx), which is still a little away from 119
Hence let us take x = 3.3 as our second value
3.3^4 = 118 (approx) [PERFECT]
Now we have to find x^3 - 1/x^3
Note that 1/x^3 is negligible and hence we can neglect it
So just find the value of 3.3^3
Answer : (C)
Note : You won't take much time in calculating 3.2^4 or 3.3^4 if you know a fast method to calculate squares. I have written an article about it. You can check it here.
Answer : (C
1. Establish a limit within which the variable is falling.
2. Neglect the smaller terms of the expression (fractions with Denominator>Numerator)
3. Please use this technique only when the options have a significant difference between them. E.g. If in a question the 4 options are A. 4, B. 5, C. 6, D. 7, you can't use the approximation technique because the options are fairly close.
These rules will make sense once you go through the below CGL questions-
Q. 15
Now how will you approach this question if you dont know how to solve it?
Given, x^4 + 1/x^4 = 119
We can safely assume that 3<x<4 because 3^4 = 81 and 4^4 = 256 (119 lies between 81 and 256). Moreover x will be closer to 3 as 119 is more close to 81 than 256
We have established the limit of the variable.
Let us take our first value. Go with x = 3.2
3.2^4 = 104 (approx), which is still a little away from 119
Hence let us take x = 3.3 as our second value
3.3^4 = 118 (approx) [PERFECT]
Now we have to find x^3 - 1/x^3
Note that 1/x^3 is negligible and hence we can neglect it
So just find the value of 3.3^3
Answer : (C)
Note : You won't take much time in calculating 3.2^4 or 3.3^4 if you know a fast method to calculate squares. I have written an article about it. You can check it here.
Q. 16
Here again no need to figure out how to solve the question
√3 = 1.73, √5 = 2.23
Hence √x = 1.73 - 2.23 or x = 0.25
Put the value of x
(0.25)^2 - 16*0.25 + 6
= 0.0625 - 4 + 6
= 2 (approx)Answer : (C
Q. 17
(x^4 - 1)x^2 = x^2 - 1/x^2
Neglect 1/x^2
x^2 = 4.23^2 = 17 (approx)
Answer : (A)
Q. 18
Again if you dont know how to solve the above question, then observe the above equation
If a=1, then LHS = 5.33 (which is little more than RHS, i.e., 5). We need to decrease the value of 'a'.
Hence let's take a = 0.9
5a + 1/3a = 4.8
Now LHS is more than RHS. We need to increase the value of 'a' slightly
So lets lock the final value a = 0.95 (Now no need to check the value of LHS for a=0.95)
9a^2 + 1/25a^2
Neglect 1/25a^2
9a^2 = 8 (approx)
Answer : (D)
Q. 19
x = 2 + √3 = 2 + 1.73 = 3.73
Now you have to find the value of √x + 1/√x
√x = √3.73You know that 19^2 = 361
Hence √3.73 = 1.9 (approx)
1/√x = 1/1.9 = 0.5
√x + 1/√x = 1.9 + 0.5 = 2.4 (which is close to √6)
Answer : (B)
Q. 20
Put x=1 and solve the equation for y
y = -1/12
Put x = 1 and y=-1/12 in the expression (4x - 1/3y)
You will get 8
Answer : (D)
Q. 21
Put x = 8 in the expression
= (64 - 1 + 16)/8
= 10 (approx)
Answer : (A)
Q. 22
Put x=2, the LHS becomes 6 and it is little more than RHS. So we need to decrease its value slightly
Let us take x = 1.8
LHS = (1.8)^2 + 1.8 = 5(approx)
LHS is almost equal to RHS, hence x=1.8 is a perfect value
Neglect 1/(x + 3)^3.
Now we only need to find the value of (x + 3)^3
(x + 3)^3 = (1.8 + 3)^3 = 110 (approx)
Answer : (A)
Q. (23)
= (2 + 2 + 2)(1/2 + 1/2 + 1/2)
(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. (24) 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. (25)
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. (26)
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. (27)
(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. (28) 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
Q.29.
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. 30)
(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. (31)
(A) -2 (B) -1/2 (C) 0 (D) 1/2
Same logic
Value of the expression = 0
Answer: (C)
Q. (32)
(A) 9 (B) 0 (C) 8 (D)
Although a + b + c = 0, is a symmetrical equation, and the expression
((a+b)/c+ (b+c)/a+ (c+a)/b)(a/(b+c)+ b/(c+a)+c/(a+b)) is also
symmetrical. But the numerator of the terms is not 1. Hence we can't say
that the answer is zero.
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)
Q. (33)
ab + bc + ca = 0 is symmetrical
Hence value of the expression = 0
Answer: (B)
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