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Algebra Tricks – 3 [Approximation]

This is another useful article for you all and requires no prior knowledge of any sort.

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-

Q. 1

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. 

Q. 2
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. 3
 

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)

Q. 4


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. 5
x = 2 + √3 = 2 + 1.73 = 3.73
Now you have to find the value of √x + 1/√x

√x = √3.73
You 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. 6

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)

Q. 7

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. 8)


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)

Below are some important formulas for Algebra, see if you can memorize them 🙂

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