Square Root & Cube Root
Step 1: First of all group the number in pairs of 2 starting from the right.
Step 2: To get the ten’s place digit, Find the nearest square (equivalent or greater than or less than) to the first grouped pair from left and put the square root of the square.
Step 3: To get the unit’s place digit of the square root
Remember the following
If number ends in  Unit’s place digit of the square root 
1  1 or 9(101) 
4  2 or 8(102) 
9  3 or 7(103) 
6  4or 6(104) 
5  5 
0  0 
Lets see the logic behind this for a better understanding
We know,
1^{2}=1
2^{2}=4
3^{2}=9
4^{2}=16
5^{2}=25
6^{2}=36
7^{2}=49
8^{2}=64
9^{2}=81
10^{2}=100
Now, observe the unit’s place digit of all the squares.
Do you find anything common?
We notice that,
Unit’s place digit of both 1^{2} and 9^{2 }is 1.
Unit’s place digit of both 2^{2} and 8^{2} is 4
Unit’s place digit of both 3^{2} and 7^{2} is 9
Unit’s place digit of both 4^{2}
Step 4: Multiply the ten’s place digit (found in step 1) with its consecutive number and compare the result obtained with the first pair of the original number from left.
Remember,
If first pair of the original number > Result obtained on multiplication then select the greater number out of the two numbers as the unit’s place digit of the square root.
If firstpair of the original number < the result obtained on multiplication,then select the lesser number out of the two numbers as the unit’s place digit of the square root.
Let us consider an example to get a better understanding of the method
Example 1: √784=?
Step 1: We start by grouping the numbers in pairs of two from right as follows
7 84
Step 2: To get the ten’s place digit,
We find that nearest square to first group (7) is 4 and √4=2
Therefore ten’s place digit=2
Step 3: To get the unit’s place digit,
We notice that the number ends with 4, So the unit’s place digit of the square root should be either 2 or 8(Refer table).
Step 4: Multiplying the ten’s place digit of the square root that we arrived at in step 1(2) and its consecutive number(3) we get,
2×3=6
ten’s place digit of original number > Multiplication result
7>6
So we need to select the greater number (8) as the unit’s place digit of the square root.
Unit’s place digit =8
Ans:√784=28
Cube roots of perfect cubes
It may take twothree minutes to find out cube root of a perfect cube by using conventional method. However we can find out cube roots of perfect cubes very fast, say in onetwo seconds using Vedic Mathematics.
We need to remember some interesting properties of numbers to do these quick mental calculations which are given below.
Points to remember for speedy calculation of cube roots
 To calculate cube root of any perfect cube quickly, we need to remember the cubes of 1 to 10 which is given below.
1^{3}  =  1 
2^{3}  =  8 
3^{3}  =  27 
4^{3}  =  64 
5^{3}  =  125 
6^{3}  =  216 
7^{3}  =  343 
8^{3}  =  512 
9^{3}  =  729 
10^{3}  =  1000 
 From the above cubes of 1 to 10, we need to remember an interesting property.
1^{3} = 1  =>  If last digit of the perfect cube = 1, last digit of the cube root = 1 
2^{3} = 8  =>  If last digit of the perfect cube = 8, last digit of the cube root = 2 
3^{3} = 27  =>  If last digit of the perfect cube = 7, last digit of the cube root = 3 
4^{3} = 64  =>  If last digit of the perfect cube = 4, last digit of the cube root = 4 
5^{3} = 125  =>  If last digit of the perfect cube =5, last digit of the cube root = 5 
6^{3} = 216  =>  If last digit of the perfect cube = 6, last digit of the cube root = 6 
7^{3} = 343  =>  If last digit of the perfect cube = 3, last digit of the cube root = 7 
8^{3} = 512  =>  If last digit of the perfect cube = 2, last digit of the cube root = 8 
9^{3} = 729  =>  If last digit of the perfect cube = 9, last digit of the cube root = 9 
10^{3} = 1000  =>  If last digit of the perfect cube = 0, last digit of the cube root = 0 
It’s very easy to remember the relations given above because
1  >  1  (Same numbers) 
8  >  2  (10’s complement of 8 is 2 and 8+2 = 10) 
7  >  3  (10’s complement of 7 is 3 and 7+3 = 10) 
4  >  4  (Same numbers) 
5  >  5  (Same numbers) 
6  >  6  (Same numbers) 
3  >  7  (10’s complement of 3 is 7 and 3+7 = 10) 
2  >  8  (10’s complement of 2 is 8 and 2+8 = 10) 
9  >  9  (Same numbers) 
0  >  0  (Same numbers) 
Also see
8 > 2 and 2 > 8
7 > 3 and 3> 7
Questions
LevelI
1.  The cube root of .000216 is:  

2. 



3.  The least perfect square, which is divisible by each of 21, 36 and 66 is:  

4.  1.5625 = ?  

5.  If 35 + 125 = 17.88, then what will be the value of 80 + 65 ?  


6. 
If a = 0.1039, then the value of 4a^{2} – 4a + 1 + 3a is: 


7. 



8.  A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 59.29, the number of the member is the group is:  

9.  The square root of (7 + 35) (7 – 35) is  

10. 


LevelII


11. 



12.  0.0169 x ? = 1.3  

13. 



14.  How many twodigit numbers satisfy this property.: The last digit (unit’s digit) of the square of the twodigit number is 8 ?  

15.  The square root of 64009 is:  

16. √29929 = ?  

17. √106.09 = ?  
A.  10.6  
B.  10.5  
C.  10.3  
D.  10.2  
18. ?/√196 = 5 

A.  76  
B.  72  
C.  70  
D.  75  
Answers
LevelI
Answer:1 Option B
Explanation:
(.000216)^{1/3}  =  216  1/3  
10^{6} 
=  6 x 6 x 6  1/3  
10^{2} x 10^{2} x 10^{2} 
=  6 
10^{2} 
=  6 
100 
= 0.06
Answer:2 Option A
Explanation:
Let  x  =  162 
128  x 
Then x^{2} = 128 x 162
= 64 x 2 x 18 x 9
= 8^{2} x 6^{2} x 3^{2}
= 8 x 6 x 3
= 144.
x = 144 = 12.
Answer:3 Option A
Explanation:
L.C.M. of 21, 36, 66 = 2772.
Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11
To make it a perfect square, it must be multiplied by 7 x 11.
So, required number = 2^{2} x 3^{2} x 7^{2} x 11^{2} = 213444
Answer:4 Option B
Explanation:
11.5625( 1.25
1
——
22 56
 44
——
245 1225
 1225
——
 X
——
1.5625 = 1.25.
Answer:5 Option D
Explanation:
35 + 125 = 17.88
35 + 25 x 5 = 17.88
35 + 55 = 17.88
85 = 17.88
5 = 2.235
80 + 65 = 16 x 5 + 65
= 45 + 65
= 105 = (10 x 2.235) = 22.35
Answer:6 Option C
Explanation:
4a^{2} – 4a + 1 + 3a = (1)^{2} + (2a)^{2} – 2 x 1 x 2a + 3a
= (1 – 2a)^{2} + 3a
= (1 – 2a) + 3a
= (1 + a)
= (1 + 0.1039)
= 1.1039
Answer:7 Option C
Explanation:
x =  (3 + 1)  x  (3 + 1)  =  (3 + 1)^{2}  =  3 + 1 + 23  = 2 + 3. 
(3 – 1)  (3 + 1)  (3 – 1)  2 
y =  (3 – 1)  x  (3 – 1)  =  (3 – 1)^{2}  =  3 + 1 – 23  = 2 – 3. 
(3 + 1)  (3 – 1)  (3 – 1)  2 
x^{2} + y^{2} = (2 + 3)^{2} + (2 – 3)^{2}
= 2(4 + 3)
= 14
Answer:8 Option C
Explanation:
Money collected = (59.29 x 100) paise = 5929 paise.
Number of members = 5929 = 77
Answer:9 Option B
Explanation:
(7 + 35)(7 – 35)  =  (7)^{2} – (35)^{2}  = 49 – 45 = 4 = 2 
Answer:10 Option B
Explanation:
5  –  10  + 125  =  (5)^{2} – 20 + 25 x 55 
2  5  25 
=  5 – 20 + 50 
25 
=  35  x  5 
25  5 
=  355 
10 
=  7 x 2.236 
2 
= 7 x 1.118  
= 7.826  
LevelII
Answer:11 Option A
Explanation:
Given Expression =  25  x  14  x  11  = 5. 
11  5  14 
Answer:12 Option B
Explanation:
Let 0.0169 x x = 1.3.
Then, 0.0169x = (1.3)^{2} = 1.69
x =  1.69  = 100 
0.0169 
Answer:13 Option C
Explanation:
3 –  1  2  = (3)^{2} +  1  2  – 2 x 3 x  1  
3  3  3 
= 3 +  1  – 2 
3 
= 1 +  1 
3 
=  4 
3 
Answer:14 Option D
Explanation:
A number ending in 8 can never be a perfect square.
Answer:15 Option A
Explanation:
2 64009( 253 4 ———45 240 225 ———503 1509  1509 ———  X ———
64009 = 253.
Answer:16 Option A
Explanation:
√29929 = So, √29929 = 173
Answer:17 Option C
Answer:18 Option C
CGPCS Notes brings Prelims and Mains programs for CGPCS Prelims and CGPCS Mains Exam preparation. Various Programs initiated by CGPCS Notes are as follows: CGPCS Mains Tests and Notes Program
 CGPCS Prelims Exam  Test Series and Notes Program
 CGPCS Prelims and Mains Tests Series and Notes Program
 CGPCS Detailed Complete Prelims Notes